Prasanna

Refer to our Texas Go Math Grade 5 Answer Key Pdf to score good marks in the exams. Test yourself by practicing the problems from Texas Go Math Grade 5 Module 2 Assessment Answer Key.

Texas Go Math Grade 5 Module 2 Assessment Answer Key

Vocabulary

Choose the best term from the box.

Vocabulary
compatible numbers
partial quotients
place value

Question 1.
You can use _____________ to estimate quotients because they are easy to compute mentally. (p. 53)
Answer:
We know that,
You can use “Compatible numbers” to estimate quotients because they are easy to compute mentally.
Hence, from the above,
We can conclude that
The best term that is suitable for the given statement is: Compatible numbers

Grade 5 Module 2 Answer Key Go Math Question 2.
To decide where to place the first digit in the quotient you can estimate or use __________. (p. 37)
Answer:
We know that,
To decide where to place the first digit in the quotient you can estimate or use “Place Value”
Hence, from the above,
We can conclude that
The best term that is suitable for the given statement is: Place Value

Concept and Skills

Estimate. Then find the product. (TEKS 5.3.A, 5.3.B)

Question 3.
Estimate: ____________

Answer:
The given expression is: 576 × 3
Now,
Estimate:
576 × 3 = 580 × 3
= 1,740
Now,
By using the Long Multiplication,

Hence, from the above,
We can conclude that
The value of the given expression is: 1,728

Question 4.
Estimate: ____________

Answer:
The given expression is: 219 × 6
Now,
Estimate:
219 × 6 = 220 × 6
= 1,320
Now,
By using the Long Multiplication,

Hence, from the above,
We can conclude that
The value of the given expression is: 1,314

Question 5.
Estimate: _____________

Answer:
The given expression is: 72 × 37
Now,
Estimate:
72 × 37 = 70 × 40
= 2,800
Now,
By using the Long Multiplication,

Hence, from the above,
We can conclude that
The value of the given expression is: 2,664

Module 2 Test Answer Key Go Math Grade 5 Question 6.
Estimate: _____________

Answer:
The given expression is: 359 × 18
Now,
Estimate:
359 × 18 = 360 × 18
= 6,480
Now,
By using the Long Multiplication,

Hence, from the above,
We can conclude that
The value of the given expression is: 6,462

Question 7.
Estimate: _____________

Answer:
The given expression is: 804 × 37
Now,
Estimate:
804 × 37 = 800 × 40
= 32,000
Now,
By using the Long Multiplication,

Hence, from the above,
We can conclude that
The value of the given expression is: 29,748

Use compatible numbers to estimate the quotient. (TEKS 5.3. A)

Question 8.
522 ÷ 60
Answer:
The given division expression is: 522 ÷60
Now,
By using the Long Division,

Hence, from the above,
We can conclude that
522 ÷ 60 = 8 R 42

Question 9.
1,285 ÷ 32
Answer:
The given division expression is: 1,285 ÷ 32
Now,
By using the Long Division,

Hence, from the above,
We can conclude that
1,285 ÷ 32 = 40 R 5

Go Math Grade 5 Module 2 Answer Key Question 10.
6,285 ÷ 89
Answer:
The given division expression is: 6,285 ÷ 89
Now,
By using the Long Division,

Hence, from the above,
We can conclude that
6,285 ÷ 89 = 70 R 55

Divide. (TEKS 5.3.C)

Question 11.
156 ÷ 13
Answer:
The given division expression is: 156 ÷13
Now,
By using the Long Division,

Hence, from the above,
We can conclude that
156 ÷ 13 = 12

Question 12.
318 ÷ 53
Answer:
The given division expression is: 318 ÷ 53
Now,
By using the Long Division,

Hence, from the above,
We can conclude that
318 ÷ 53 = 6

Go Math Grade 5 Module 2 Answer Key Pdf Question 13.
1,562 ÷ 34
Answer:
The given division expression is: 1,562 ÷ 34
Now,
By using the Long Division,

Hence, from the above,
We can conclude that
1,562 ÷ 34 = 45 R 32

Question 14.
4,024 ÷ 68
Answer:
The given division expression is: 4,024 ÷ 68
Now,
By using the Long Division,

Hence, from the above,
We can conclude that
4,024 ÷ 68 = 59 R 12

Fill in the bubble completely to show your answer.

Question 15.
Dylan’s dog weighs 12 times as much as his pet rabbit. The dog and rabbit weigh 104 pounds altogether. How much does Dylan’s dog weigh? (TEKS 5.3.C)
(A) 88 pounds
(B) 96 pounds
(C) 104 pounds
(D) 8 pounds
Answer:
It is given that
Dylan’s dog weighs 12 times as much as his pet rabbit. The dog and rabbit weigh 104 pounds altogether
Now,
Let the weight of Dylan’s pet rabbit be: x Pounds
So,
The weight of Dylan’s dog is: 12x Pounds
Now,
According to the given information,
12x + x = 104 Pounds
13x = 104 Pounds
x = \(\frac{104}{13}\)
x = 8 Pounds
So,
The weight of Dylan’s dog is: 96 Pounds
Hence, from the above,
We can conclude that
The weight of Dylan’s dog is:

Module 2 Assessment Answers Grade 5 Go Math Question 16.
Tickets for the basketball game cost $14 each. If the sale of the tickets brought in $2,212, how many tickets were sold? (TEKS 5.3.C)
(A) 172
(B) 158
(C) 150
(D) 168
Answer:
It is given that
The tickets for the basketball game cost $14 each and the sale of the tickets brought in $2,212
So,
According to the given information,
The number of tickets that were sold = \(\frac{$2,212}{$14}\)
Now,

Hence, from the above,
We can conclude that
The tickets that were sold are:

Question 17.
Adele earns $194 every two weeks delivering newspapers. She spends $30 of each paycheck and saves the rest. If she is paid 26 times this year, how much will Adele save? (TEKS 5.3.B)
(A) $3,244
(B) $4,264
(C) $772
(D) $5,044
Answer:
It is given that
Adele earns $194 every two weeks delivering newspapers. She spends $30 of each paycheck and saves the rest. If she is paid 26 times this year
Now,
The amount of money saved by Adele = $194 – $30
= $164
Now,
According to the given information,
The total amount of money saved by Adele in this year = $164 × 26
Now,

Hence, from the above,
We can conclude that
The total amount of money saved by Adele this year is:

Go Math Answer Key Grade 5 Module 2 Assessment Question 18.
A local orange grower processes 2,330 oranges from his grove this year. The oranges are packaged in crates that each holds 96 oranges. All but one crate is full. How many oranges are in this last crate? Record your answer and fill in the bubbles on the grid. Be sure to use the correct place value. (TEKS 5.3.C)

Answer:
It is given that
A local orange grower processes 2,330 oranges from his grove this year. The oranges are packaged in crates that each holds 96 oranges. All but one crate is full
So,
According to the given information,
The number of crates = \(\frac{2,330}{96}\)
Now,

So,
The number of oranges that are present in the last crate is: 26 oranges
Hence, from the above,
We can conclude that
The number of oranges that are present in the last crate is:

Refer to our Texas Go Math Grade 5 Answer Key Pdf to score good marks in the exams. Test yourself by practicing the problems from Texas Go Math Grade 5 Lesson 1.7 Answer Key Add and Subtract Decimals Through Thousandths.

Texas Go Math Grade 5 Lesson 1.7 Answer Key Add and Subtract Decimals Through Thousandths

Unlock the Problem

At the 2010 Winter Olympics, Armin Zoeggeler r won the bronze medal in the men’s luge event, r In his fastest run, he completed the first interval in 21.261 seconds. It took him another 27.198 b seconds to reach the finish line. What was Zoeggeler’s finish time?

• Underline the sentence that tells you what you are trying to find.
• Circle the numbers you need to use.

Add decimals through thousandths. 21.261 + 27.198
THINK

• Line up the numbers in each place.
  
• First, add the thousandths.
  
• Then, add the hundredths, tenths, ones, and tens. Regroup as needed.
  
• Place the decimal point in the sum.
  

So,
Zoeggeler’s finish time was 48.459 seconds.

Math Talk
Mathematical Processes

Why would you use paper and pencil instead of mental math to solve this problem?
Answer:
We can use mental math to solve simple equations. To do this, we put the equation into words and then form a question with those words. We then answer the question in our heads. We can also make use of inverse operations to reword questions if the answer isn’t obvious to us.
But,
In the above situation,
It is not possible because the given numbers are decimal numbers with three digits after the decimal point

Try This!

Use subtraction to check your work. Subtract one of the addends from the sum.

Addition and subtraction are opposite, or Inverse operations. You can use subtraction to check your answer to an addition problem.
Answer:
The given numbers from the above problem are: 21.261 and 27.198
Now,
To find out the answer to the above problem is correct or not,
We have to subtract the answer from one of the given numbers to get the second number,
If we didn’t get the second number, then the answer is not correct
So,
48.459 – 27.198 = 21.261
Hence, from the above,
We can conclude that
The answer to the above problem is correct

Example
Zoeggeler finished the first half of the run in 31.116 seconds. The second half took only 17.343 seconds. How many seconds faster was the second half of the run?
Estimate. 31 – 17 = ___________

STEP 1
Line up the place values. Subtract the thousandths.

STEP 2
Subtract the hundredths. Subtract the tenths. Regroup as needed.

STEP 3
Subtract the ones and tens. Place the decimal point in the difference.

So,
Zoeggeler was 13.773 seconds faster in the second half of the run.

To check subtraction using the inverse operation, add the number you subtracted to the difference. The sum should equal the number you subtracted from.

Math talk
Mathematical Processes

Explain how you know without adding that the sum of 2.475 and 6.43 will result in a 5 in the thousandths place.
Answer:
The given numbers are: 2.475 and 6.43
Now,
When we observe the given numbers,
2.475 has the thousandths place value but 6.43 does not have the thousandths place value
So,
The representation of 6.43 with the thousandth place will be: 6.430
Now,
When we add the thousandths place of both the numbers, i.e., 5 + 0,
We will get the result as 5
Hence, from the above,
We can conclude that
To make the numbers have the same place values when there are unequal place values, add zeros to the right side of the number that has the number of place values less than the other number

Try This! Subtract. Then check your work.

Lesson 1.7 Add and Subtract Decimals Answer Key Question 1.

Answer:
The given numbers are: 42.784 and 6.980
So,

Share and Show

Find the sum or difference.

Question 1.

Answer:
The given numbers are: 1.845 and 0.357
So,

Hence, from the above,
We can conclude that
The result for the given expression is: 1.488

Question 2.
3.46 + 6.834
Answer:
The given numbers are: 3.46 and 6.834
So,

Hence, from the above,
We can conclude that
The result for the given expression is: 10.294

Question 3.
13 – 0.943
Answer:
The given numbers are: 13 and 0.943
So,

Hence, from the above,
We can conclude that
The result for the given expression is: 12.057

Question 4.

Answer:
The given numbers are: 0.254, 1.3, and 0.79
So,

Hence, from the above,
We can conclude that
The result for the given expression is: 2.344

Problem Solving

Find the sum or difference.

Question 5.

Answer:
The given numbers are: 3.704 and 1.325
So,

Hence, from the above,
We can conclude that
The result for the given expression is: 2.379

Go Math Lesson 1.7 5th Grade Subtraction of Decimals Question 6.

Answer:
The given numbers are: 14.467 and 12.33
So,

Hence, from the above,
We can conclude that
The result for the given expression is: 26.797

Question 7.

Answer:
The given numbers are: 23.002 and 1.74
So,

Hence, from the above,
We can conclude that
The result for the given expression is: 21.262

Question 8.

Answer:
The given numbers are: 9.94, 0.318, and 1.283
So,

Hence, from the above,
We can conclude that
The result for the given expression is: 11.541

Practice: Copy and Solve Find the sum or difference.

Question 9.
21.54 + 4.758
Answer:
The given numbers are: 21.54 and 4.758
So,

Hence, from the above,
We can conclude that
The result for the given expression is: 26.298

Question 10.
6.328 – 3.62
Answer:
The given numbers are: 6.328 and 3.62
So,

Hence, from the above,
We can conclude that
The result for the given expression is: 2.708

Question 11.
15.87 – 3.274
Answer:
The given numbers are: 15.87 and 3.274
So,

Hence, from the above,
We can conclude that
The result for the given expression is: 12.596

Question 12.
2.45 + 3.247 + 1.8
Answer:
The given numbers are: 2.45, 3.247, and 1.8
So,

Hence, from the above,
We can conclude that
The result for the given expression is: 7.497

H.O.T. Algebra Find the unknown numbers in the pattern. Then write a description for the pattern.

Question 13.
2.1, 3.3, 4.5, 5.7, ______ , 8.1, ________
Description: _____________
Answer:
The given pattern is:
2.1, 3.3, 4.5, 5.7, ______ , 8.1, ________
Now,
When we observe the above pattern,
We can say that
2.1 + 1.2 = 3.3
3.3. + 1.2 = 4.5
Hence, from the above,
We can conclude that
The completed pattern is:
2.1, 3.3, 4.5, 5.7, 6.9, 8.1, 9.3
We have to add 1.2 to the previous number to complete the given pattern

Adding and Subtracting Decimals Answer Key Lesson 1.7 Question 14.
4.05, 4.00, 3.95, ________, 3.85, ________
Description: ______________
Answer:
The given pattern is:
4.05, 4.00, 3.95, ________, 3.85, ________
Now,
When we observe the above pattern,
We can say that
4.05 – 0.05 = 4.00
4.00 – 0.05 = 3.95
Hence, from the above,
We can conclude that
The completed pattern is:
4.05, 4.00, 3.95, 3.90, 3.85, 3.80
We have to subtract 0.05 from the previous number to complete the given pattern

Problem-Solving

Use the table to solve 15-16.

Question 15.
Evaluate Apolo Ohno won the men’s 500-meter speed skating final at the 2006 Winter Olympics. His time for the race was 4 1.935 seconds. Francois-Louis Tremblay came in second, finishing 0.067 second behiñd Ohno. What was Tremblay’s time?
Answer:
It is given that
Apolo Ohno won the men’s 500-meter speed skating final at the 2006 Winter Olympics. His time for the race was 4 1.935 seconds. Francois-Louis Tremblay came in second, finishing 0.067 second behiñd Ohno
Now,
The given table is:

So,
The time is taken by Tremblay to complete the final = 41.935 – 0.067
Now,
By using the Long Subtraction,

Hence, from the above,
We can conclude that
The time took by Tremblay to complete the final is: 41.868 seconds

Question 16.
Apply Jon Eley came in fifth in the men’s speed skating final. How many seconds after Apolo Ohno did Eley finish?
Answer:
It is given that
Jon Eley came in fifth in the men’s speed skating final
Now,
The given table is:

So,
The time took Eley to finish the final after Apolo Ohno = 42.497 – 41.935
Now,
By using the Long subtraction,

Hence, from the above,
We can conclude that
The time took Eley to finish the final 0.0562 seconds after Apolo Ohno

Question 17.
H.O.T. Multi-Step The sum of two numbers is 4.004. One number has a 4 in the tenths place and a 3 in the thousandths place. The other number has a 1 in the one’s place and an 8 in the hundredths place. What are the two numbers?

Answer:
It is given that
The sum of two numbers is 4.004. One number has a 4 in the tenths place and a 3 in the thousandths place. The other number has a 1 in the one’s place and an 8 in the hundredths place.
Now,

Hence, from the above,
We can concldue that
The two numbers are: 2.423 and 1.581

Daily Assessment Task

Fill in the bubble completely to show your answer.

Question 18.
Multi-Step Team A spent 61.242 seconds and 59.438 seconds running a relay race. Team B spent 60.561 seconds and 60.289 seconds running the relay race. Which statement is true?
A. Team A won by 0.17 seconds.
B. Team B won by 0.17 seconds.
C. Team A won by 1.17 seconds.
D. Team B won by 1.17 seconds.
Answer:
It is given that
Team A spent 61.242 seconds and 59.438 seconds running a relay race. Team B spent 60.561 seconds and 60.289 seconds running the relay race
Now,
According to the given information,
The  time taken by Team A running the relay race = 61.242 – 59.438
= 1.804 seconds
The  time taken by Team B running the relay race = 60.561 – 60.289
= 0.272 seconds
So,
The difference between the times to finish the relay race = 1.804 – 0.272
= 1.532 seconds
Hence, from the above,
We can conclude that
The statements that are true is:

Question 19.
Manuel is walking from his home to his school. He has walked 0.142 miles so far and has 0.088 miles left to walk. How far is Manuel’s home from school?
A. 0.054 mile
B. 0.12 mile
C. 0.23 mile
D. 1.022 miles
Answer:
It is given that
Manuel is walking from his home to his school. He has walked 0.142 miles so far and has 0.088 miles left to walk
So,
According to the given information,
The distance of Manuel’s home from school = 0.142 + 0.088
= 0.230 miles
Hence, from the above,
We can conclude that
The distance of Manuel’s home from school is:

Question 20.
Multi-Step Music & Movie Mart is selling used CDs for $6.99 each and used DVDs for $8.49 each. You have $30 to spend. Which of these combinations could you buy?
A. TWO CDS and two DVDs
B. Three CDs and one DVD
C. Three CDs and three DVDs
D. One CD and three DVDs
Answer:
It is given that
Music & Movie Mart is selling used CDs for $6.99 each and used DVDs for $8.49 each. You have $30 to spend.
Now,
By using the options,
A.
2 ($6.99) + 2 ($8.49) = $30.96
B.
3 ($6.99) + $8.49 = $29.46
C.
3 ($6.99) + 3 ($8.49) = $46.44
D.
$6.99 + 3 ($8.49) = $32.46
Hence, from the above,
We can conclude that
The combinations that you could buy is:

Texas Test Prep

Question 21.
Toni has a ribbon that is 2.75 meters long. She cuts off 0.345 meters. How much of the ribbon does Toni have left?
A. 3.095 meters
B. 2.715 meters
C. 2.785 meters
D. 2.405 meters
Answer:
It is given that
Toni has a ribbon that is 2.75 meters long. She cuts off 0.345 meters
So,
The amount of ribbon does Toni have left = 2.75 – 0.345
= 2.405 meters
Hence, from the above,
We can conclude that
The amount of ribbon does Toni have left is:

Texas Go Math Grade 5 Lesson 1.7 Homework and Practice Answer Key

Find the sum or difference.

Question 1.

Answer:
The given numbers are: 13.87 and 6.06
So,

Hence, from the above,
We can conclude that
The result for the given addition expression is: 19.93

Question 2.

Answer:
The given numbers are: 26.25 and 5.73
So,

Hence, from the above,
We can conclude that
The result for the given subtraction expression is: 20.52

Question 3.

Answer:
The given numbers are: 2.50 and 0.926
So,

Hence, from the above,
We can conclude that
The result for the given addition expression is: 3.426

Question 4.

Answer:
The given numbers are: 43.66 and 9.08
So,

Hence, from the above,
We can conclude that
The result for the given subtraction expression is: 34.58

Lesson 1.7 Add and Subtract Decimals Answer Key Question 5.

Answer:
The given numbers are: 6.27, 0.133, and 4.31
So,

Hence, from the above,
We can conclude that
The result for the given addition expression is: 10.713

Question 6.

Answer:
The given numbers are: 25.75 and 8.2
So,

Hence, from the above,
We can conclude that
The result for the given subtraction expression is: 17.55

Find the sum or difference.

Question 7.
6.389 + 17.39
Answer:
The given numbers are: 6.389 and 17.39
So,

Hence, from the above,
We can conclude that
The result for the given addition expression is: 23.779

Question 8.
8.747 – 4.8
Answer:
The given numbers are: 8.747 and 4.8
So,

Hence, from the above,
We can conclude that
The result for the given subtraction expression is: 3.947

Question 9.
2.09 + 12.639
Answer:
The given numbers are: 2.09 and 12.639
So,

Hence, from the above,
We can conclude that
The result for the given addition expression is: 14.729

Find the unknown numbers in the pattern. Then write a description for the pattern

Question 10.
1.0, 2.1, 3.2, 4.3. ________ 6.5, 7.6
Description: _____________
Answer:
The given pattern is:
1.0, 2.1, 3.2, 4.3. ________ 6.5, 7.6
Now,
From the given pattern,
We can observe that
1.0 + 1.1 = 2.1
2.1 + 1.1 = 3.2
Hence, from the above,
We can conclude that
The completed pattern is:
1.0, 2.1, 3.2, 4.3. 5.4, 6.5, 7.6
The description of the given pattern is:
Add 1.1 to the first number and continue the same pattern

Question 11.
5.03,5.00, 4.97, 4.94 ________ 4.88
Description: _____________
Answer:
The given pattern is:
5.03,5.00, 4.97, 4.94 ________ 4.88
Now,
From the above pattern,
We can observe that
5.03 – 0.03 = 5.00
5.00 – 0.03 = 4.97
Hence, from the above,
We can conclude that
The completed pattern is:
5.03,5.00, 4.97, 4.94, 4.91, 4.88
The description of the given pattern is:
Subtract 0.03 from the first number and continue the same pattern

Problem Solving

Question 12.
Abby finished a biking race in 21.39 minutes. Juanita took 19.59 minutes to finish. How much longer did Abby take to finish the race than Juanita?
Answer:
It is given that
Abby finished a biking race in 21.39 minutes. Juanita took 19.59 minutes to finish.
So,
The difference between the time took by Abby and Junaita = 21.39 – 19.59
Now,
By using the Long Subtraction,

Hence, from the above,
We can conclude that
Abby took 1.80 seconds more than Junaita to finish the race

Question 13.
The sum of the two numbers is 5.036. One number has a 4 in the tenths place and a 7 in the thousandths place. The other number has a 1 in the one’s place and a 2 in the hundredths place. What are the two numbers?
Answer:
It is given that
The sum of the two numbers is 5.036. One number has a 4 in the tenths place and a 7 in the thousandths place. The other number has a 1 in the one’s place and a 2 in the hundredths place.
Now,

Hence, from the above,
We can conclude that
The two numbers are: 1.629 and 3.407

Lesson Check

Fill in the bubble completely to show your answer.

Question 14.
Dara spends $13.02 on sandals, $9.85 on shorts, and $15.20 on a hat. How much change does she receive back from $50?
A. $10.93
B. $11.93
C. $11.34
D. $12.67
Answer:
It is given that
Dara spends $13.02 on sandals, $9.85 on shorts, and $15.20 on a hat.
So,
According to the given information,
The amount of money Dara received back from $50 = $50 – ($13.02 + $9.85 + $15.20)
= $50 – $38.07
= $11.93
Hence, from the above,
We can conclude that
The amount of money Dara received back from $50 is:

Question 15.
Ling’s relay team ran the first part of the race in 28.134 seconds, the second part in 17.922 seconds, and the third part in 34.023 seconds. By how much did the relay team beat the champions, who ran the race in 83.736 seconds?
A. 3.412 seconds
B. 3.657 seconds
C. 3.327 seconds
D. 4.073 seconds
Answer:
It is given that
Ling’s relay team ran the first part of the race in 28.134 seconds, the second part in 17.922 seconds, and the third part in 34.023 seconds
So,
According to the given information,
The difference of time between Ling;s team and champions = 83.736 – (28.134 + 17.922 + 34.023)
= 3.657 seconds
Hence, from the above,
We can conclude that
The difference of time between the relay team and the Champions is:

Question 16.
Juan has a batting average of 0.334. Gary has a batting average of 0.284. What is the difference between the batting averages of the two baseball players?
A. 0.50
B. 0.05
C. 0.028
D. 0.042
Answer:
It is given that
Juan has a batting average of 0.334. Gary has a batting average of 0.284
Now,
The difference between the batting averages of the two baseball players = 0.334 – 0.284
= 0.050 seconds
Hence, from the above,
We can conclude that
The difference between the batting averages of the two baseball players is:

Question 17.
Jonas is jogging from the park to the school. He has jogged 0.424 miles so far. He has 0.384 miles left to jog. How far is the park located away from the school?
A. 0.730 mile
B. 0.808 mile
C. 1.032 miles
D. 0.40 mile
Answer:
It is given that
Jonas is jogging from the park to the school. He has jogged 0.424 miles so far. He has 0.384 miles left to jog.
So,
According to the given information,
The distance from the park to the school = 0.424 + 0.384
= 0.808 miles
Hence, from the above,
We can conclude that
The distance from the park to the school is:

Question 18.
Multi-Step Jimmie adds 3.24, 4.294, and 2.073. How much more than 5.31 is the sum?
A. 4.297
B. 9.607
C. 4.607
D. 4.477
Answer:
It is given that
Jimmie adds 3.24, 4.294, and 2.073
So,
According to the given information,
The difference between 5.31 and the numbers that will be added by Jimmie = (3.24 + 4.294 + 2.073) – 5.31
= 4.297
Hence, from the above,
We can conclude that
The difference between 5.31 and the numbers that will be added by Jimmie is:

Question 19.
Multi-Step A gallon of milk costs $3.32 and a container of orange juice costs $2.95. Max buys two gallons of milk and two containers of orange juice. How much does Max spend?
A. $123.69
B. $6.64
C. $12.54
D. $6.27
Answer:
It is given that
A gallon of milk costs $3.32 and a container of orange juice costs $2.95. Max buys two gallons of milk and two containers of orange juice
So,
The total amount of money spent by Max = 2 ($3.32) + 2 ($2.95)
= $6.64 + $5.9
= $12.54
Hence, from the above,
We can conclude that
The amount of money spent by Max is:

Refer to our Texas Go Math Grade 5 Answer Key Pdf to score good marks in the exams. Test yourself by practicing the problems from Texas Go Math Grade 5 Lesson 2.8 Answer Key Division.

Texas Go Math Grade 5 Lesson 2.8 Answer Key Division

Unlock the Problem

Sean and his crew operate a fishing charter company. They caught a blue marlin and amberjack. The weight of the blue marlin was 12 times as great as the weight of the amberjack. The combined weight of both fish was 1,014 pounds. How much did each fish weigh?

Read

What do I need to find?
I need to find the weight of each fish.

What information am I given?

I know that they caught a total of 1,014 pounds of fish and the weight of the blue marlin was 12 times as great as the weight of the amberjack.

Plan

I can use the strategy of Addition and Substitution and then divide. I can draw and use a strip diagram to write the division problem that helps me find the weight of each fish.

Solve

I will draw one box to show the weight of the amberjack. Then I will draw a strip of 12 boxes of the same size to show the weight of the blue marlin. I can divide the total weight of the two fish by the total number of boxes.

Write the quotient in each box. Multiply it by 12 to find the weight of the blue marlin.
So,
The amberjack weighed 78 pounds and the blue marlin weighed 936 pounds.

Try Another Problem

Jason, Murray, and Dana went fishing. Dana caught red snapper. Jason caught a tuna with a weight 3 times as great as the weight of the red snapper. Murray caught a sailfish with a weight 12 times as great as the weight of the red snapper. If the combined weight of the three fish was 208 pounds, how much did the tuna weigh?

Read

What do I need to find?
Answer:
I need to find the weight of tuna

What information am I given?
Answer:
I know that Jason, Murray, and Dana went fishing. Dana caught red snapper. Jason caught a tuna with a weight 3 times as great as the weight of the red snapper. Murray caught a sailfish with a weight 12 times as great as the weight of the red snapper and the combined weight of the three fish was 208 pounds

Plan

What is my plan or strategy?
Answer:
I can use the strategy of Addition and Substitution and then divide

Solve
It is given that
Jason, Murray, and Dana went fishing. Dana caught red snapper. Jason caught a tuna with a weight 3 times as great as the weight of the red snapper. Murray caught a sailfish with a weight 12 times as great as the weight of the red snapper and the combined weight of the three fish was 208 pounds
Now,
Let the weight of the red snapper be: x Pounds
So,
The weight of the fish caught by Dana = x Pounds
The weight of the tuna caught by Jason = 3x Pounds
The weight of the saltfish caught by Murray = 12x Pounds
Now,
According to the combined information,
x + 3x + 12x = 208 Pounds
16x = 208
x = \(\frac{208}{16}\)
= 13 Pounds
So,
The weight of the tuna caught by Jason = 3x
= 3 × 13
= 39 Pounds
So,
The tuna weighed 39 pounds.

How can you check if your answer is correct?
Answer:
Substitute the weight of the tuna in the total weight of the three fishes and check whether the answer is equal to the combined weight of the fishes caught by Dana and Murray

Math Talk
Mathematical Processes

Explain how you could use another strategy to solve this problem.
Answer:
The other strategy to solve the above problem is:
Step 1:
Read the given information
Step 2:
Plan the strategy how to find the required quantity from the given information
Step 3:
Solve the problem and find the required quantity

Share and Show

Go Math Lesson 2.8 5th Grade Answer Key Question 1.
Paula caught a tarpon with a weight that was 10 times as great as the weight of a permit fish she caught. The total weight of the two fish was 132 pounds. How much did each fish weigh?
First, draw one box to represent the weight of the permit fish and ten boxes to represent the weight of the tarpon.
Next, divide the total weight of the two fish by the total number of boxes you drew. Place the quotient in each box.
Last, find the weight of each fish.
The permit fish weighed __________ pounds. The tarpon weighed ___________ pounds.
Answer:
It is given that
Paula caught a tarpon with a weight that was 10 times as great as the weight of a permit fish she caught. The total weight of the two fish was 132 pounds
Now,
Let the weight of the permit fish be: x pounds
So,
The weight of the tarpon is: 10x Pounds
Now,
According to the given information,
x + 10x = 132 Pounds
11x = 132 pounds
x = \(\frac{132}{11}\)
Now,

So,
The weight of the permit fish is: 12 pounds
The weight of the tarpon is: 120 pounds
Hence, from the above,
We can conclude that
The weight of the permit fish is: 12 pounds
The weight of the tarpon is: 120 pounds

Question 2.
What if The weight of the tarpon was 11 times the weight of the permit fish, and the total weight of the two fish was 132 pounds? How much would each fish weigh?
permit fish: __________ pounds
tarpon: __________pounds
Answer:
It is given that
The weight of the tarpon was 11 times the weight of the permit fish, and the total weight of the two fish was 132 pounds
Now,
Let the weight of the permit fish be: x pounds
So,
The weight of the tarpon is: 11x pounds
So,
According to the given information,
11x + x = 132 pounds
12x = 132 pounds
x = \(\frac{132}{12}\)
Now,

So,
The weight of the permit fish is: 11 pounds
The weight of the tarpon is: 121 pounds
Hence, from the above,
We can conclude that
The weight of the permit fish is: 11 pounds
The weight of the tarpon is: 121 pounds

Problem Solving

Question 3.
H.O.T. Multi-Step The crew on a fishing boat caught four fish that weighed a total of 1,092 pounds. The tarpon weighed twice as much as the amberjack and the white marlin weighed twice as much as the tarpon. The weight of the tuna was 5 times the weight of the amberjack. How much did each fish weigh?

Answer:
It is given that
The crew on a fishing boat caught four fish that weighed a total of 1,092 pounds. The tarpon weighed twice as much as the amberjack and the white marlin weighed twice as much as the tarpon. The weight of the tuna was 5 times the weight of the amberjack.
Now,
Let the weight of the amberjack be: x Pounds
So,
The weight of the tarpon = 2x Pounds
The weight of the white marlin = 2 × 2x = 4x Pounds
The weight of the tuna = 5x Pounds
Now,
According to the given information,
x + 2x + 4x + 5x = 1,092 Pounds
12x = 1,092 Pounds
So,
x = \(\frac{1,092}{12}\)
Now,

So,
The weight of the amberjack is: 91 Pounds
The weight of the tarpon is: 182 Pounds
The weight of the white marlin is: 364 Pounds
The weight of the tuna is: 455 Pounds
Hence, from the above,
We can conclude that
The weight of the amberjack is: 91 Pounds
The weight of the tarpon is: 182 Pounds
The weight of the white marlin is: 364 Pounds
The weight of the tuna is: 455 Pounds

Go Math Grade 5 Lesson 2.8 Answer Key Question 4.
H.O.T. Multi-Step A fish market bought two swordfish at a rate of $13 per pound. The cost of the larger fish was 3 times as great as the cost of the smaller fish. The total cost of the two fish was $3,952. How much did each fish weigh?
Answer:
It is given that
A fish market bought two swordfish at a rate of $13 per pound. The cost of the larger fish was 3 times as great as the cost of the smaller fish. The total cost of the two fish was $3,952
Now,
Let the cost of the smaller fish be: $x
So,
The cost of the larger fish is: $3x
Now,
The total cost of the two fishes per pound = ($3x + $x) × $13
= $52
Now,
According to the given information,
The weight of each fish = $3,952 ÷ $52
Now,

So,
The weight of the smaller fish is: 76 Pounds
The weight of the larger fish is: 228 Pounds
Hence, from the above,
We can conclude that
The weight of the smaller fish is: 76 Pounds
The weight of the larger fish is: 228 Pounds

Daily Assessment Task

Fill in the bubble completely to show your answer.

Question 5.
An ostrich at the zoo weighs 3 times as much as Mark’s pet dog. Together, the ostrich and the dog weigh 448 pounds. How much does the ostrich weigh?
(A) 149 pounds
(B) 112 pounds
(C) 224 pounds
(D) 336 pounds
Answer:
It is given that
An ostrich at the zoo weighs 3 times as much as Mark’s pet dog. Together, the ostrich and the dog weigh 448 pounds
Now,
Let the weight of Mark’s pet dog be: x Pounds
So,
The weight of the ostrich is: 3x Pounds
Now,
According to the given information,
3x + x = 448 Pounds
4x = 448 Pounds
x = \(\frac{448}{4}\)
x = 112 Pounds
So,
The weight of the ostrich is: 336 Pounds
Hence, from the above,
We can conclude that
The weight of the ostrich at the zoo is:

Lesson 2.8 Go Math 5th Grade Question 6.
Use Diagrams A student drew this diagram to solve a word problem. Which statement matches what she drew?

(A) A man weighs 12 times as much as a cat.
(B) A cat weighs 12 times as much as a man.
(C) A man weighs 13 times as much as a cat.
(D) A cat weighs 13 times as much as a man.
Answer:
It is given that
A student drew this diagram to solve a word problem
Now,
The given diagram is:

Hence, from the above,
We can conclude that
The statement that matches with what the student drew is:

Question 7.
Multi-Step A juice factory uses 4,650 pounds of pears and apples a day. The factory uses 30 times as many apples as pears. How many pounds of pears and how many pounds of apples does the factory use each day?
(A) 93 lb pears and 4,557 lb apples
(B) 150 lb pears and 4,500 lb apples
(C) 155 lb pears and 4,495 lb apples
(D) 310 lb pears and 4,340 lb apples
Answer:
It is given that
A juice factory uses 4,650 pounds of pears and apples a day. The factory uses 30 times as many apples as pears
Now,
Let the weight of the pears be: x Pounds
So,
The weight of the apples is: 30x Pounds
Now,
According to the given information,
30x + x = 4,650
31x = 4,650
x = \(\frac{4,650}{31}\)
Now,

So,
The weight of pears is: 150 Pounds
The weight of apples is: 4,500 Pounds
Hence, from the above,
We can conclude that
The number of pounds of pears and the number of pounds of apples does the factory use each day is:

Texas Test Prep

Go Math Lesson 2.8 5th Grade Answer Key Question 8.
Apply Captain James offers a deep-sea fishing tour. He charges $2,940 for a 14-hour trip. How much does each hour of the tour cost?
(A) $201
(B) $138
(C) $210
(D) $294
Answer:
It is given that
Captain James offers a deep-sea fishing tour. He charges $2,940 for a 14-hour trip
Now,
According to the given information,
The cost of each hour of the tour = $2,940 ÷ 14
Now,

Hence, from the above,
We can conclude that
The cost of each hour of the tour is:

Texas Go Math Grade 5 Lesson 2.8 Homework and Practice Answer Key

Eddie’s Moving Company moved 3 boxes out of Mrs. Diaz’s house. The box labeled “Kitchen” was 4 times heavier than the box labeled “Bedroom.” The box labeled “Library” was 14 times heavier than the box marked “Bedroom.” If the combined weight of the boxes was 817 pounds, how much did the box labeled “Kitchen” weigh?

Question 1.
What do I need to find?
Answer:
I need to find the weight of the box labeled “Kitchen”

Question 2.
What information am I given?
Answer:
I know that Eddie’s Moving Company moved 3 boxes out of Mrs. Diaz’s house. The box labeled “Kitchen” was 4 times heavier than the box labeled “Bedroom.” The box labeled “Library” was 14 times heavier than the box marked “Bedroom” and the combined weight of the boxes was 817 pounds

Go Math Grade 5 Practice and Homework Lesson 2.8 Answer Key Question 3.
Draw a diagram.

Answer:
The representation of the number of boxes in the form of a strip diagram is:

Question 4.
Solve.
Let the weight of the box labeled “Bedroom” be: x Pounds
So,
x + 4x + 14x = 817 Pounds
19x = 817
So,
817 ÷ 19 = 43
So,
I can multiply 4 by 43 to find the weight of the “Kitchen” box.
So,
The box labeled “Kitchen” weighed 172 pounds.

Problem Solving

Question 5.
An athlete ran 1,200 laps around a track over a two-month period. He ran 4 times the number of laps in the second month than he ran in the first month. How many laps did the athlete run the first month?
Answer:
It is given that
An athlete ran 1,200 laps around a track over a two-month period. He ran 4 times the number of laps in the second month than he ran in the first month
Now,
Let the number of laps run by an athlete in the first month be: x laps
So,
The number of laps run by an athlete in the first month be: 4x laps
Now,
According to the given information,
x + 4x = 1,200
5x = 1,200
x = \(\frac{1,200}{5}\)
x = 240 laps
Hence, from the above,
We can conclude that
The number of laps did the athlete run in the first month is: 240 laps

Question 6.
The diagram below compares the weights of a boy and a fish. Tell me what the diagram means.

Answer:
It is given that
The diagram below compares the weights of a boy and a fish
Now,
The given diagram is:

Now,
From the given diagram,
We can observe that
The weight of the boy is 11 times as many as the weight of the fish
Hence, from the above,
We can conclude that
The given diagram represents:
The weight of the boy is 11 times as many as the weight of the fish

Lesson Check

Fill in the bubble completely to show your answer.

Question 7.
Yani drew this diagram to solve a word problem. Which statement matches what she drew?

(A) A hamster weighs 7 times as much as a duck.
(B) A duck weighs 7 times as much as a hamster.
(C) A hamster weighs 6 times as much as a duck.
(D) A duck weighs 6 times as much as a hamster.
Answer:
It is given that
Yani drew this diagram to solve a word problem
Now,
The given diagram is:

Hence, from the above,
We can conclude that
The statement that matches with what she drew is:

Practice and Homework Lesson 2.8 Go Math Grade 5 Question 8.
An oak tree in Jake’s yard is 13 times taller than a model tree that he is making for a performance set. Together, the oak tree and model tree are 112 feet tall. How tall is the model tree?
(A) 8 feet
(B) 14 feet
(C) 26 feet
(D) 96 feet
Answer:
It is given that
An oak tree in Jake’s yard is 13 times taller than a model tree that he is making for a performance set. Together, the oak tree and model tree are 112 feet tall
Now,
Let the height of the model oak tree be: x feet
So,
The height of the oak tree is: 13x feet
Now,
According to the given information,
13x + x = 112 feet
14x = 112 feet
x = \(\frac{112}{14}\)
x = 8 feet
Hence, from the above,
We can conclude that
The height of the model tree is:

Question 9.
Lori’s book has 6 times more pages than Sal’s book. The combined number of pages for the two books is 532. How many pages does Lori’s book have?
(A) 76
(B) 88
(C) 456
(D) 532
Answer:
It is given that
Lori’s book has 6 times more pages than Sal’s book. The combined number of pages for the two books is 532
Now,
Let the number of pages in Sai’s book be: x pages
So,
The number of pages in Lori’s book is: 6x pages
Now,
According to the given information,
x + 6x = 532
7x = 532
x = \(\frac{532}{7}\)
x = 76 pages
So,
The number of pages in Lori’s book = 76 × 6
= 456 pages
Hence, from the above,
We can conclude that
The number of pages in Lori’s book is:

Question 10.
Raj drives his car 5 times farther than Gary drives his car. Together they drive 624 miles. How many miles does Rai drive?
(A) 520 miles
(B) 416 miles
(C) 104 miles
(D) 312miles
Answer:
It is given that
Raj drives his car 5 times farther than Gary drives his car. Together they drive 624 miles
Now,
Let the distance driven by Gary be: x miles
So,
The distance driven by Raj is: 5x miles
Now,
According to the given information,
x + 5x = 624
6x = 624
x = \(\frac{624}{6}\)
x = 104 miles
So,
The distance driven by Raj is: 3,120 miles
Hence, from the above,
We can conclude that
The distance driven by Raj is:

Go Math 5th Grade Lesson 2.8 Homework Answers Question 11.
Multi-Step The total number of books in the community’s mobile library is 1,155. There are 20 times more nonfiction books than fiction books. How many fiction books and how many nonfiction books are in the mobile library?
(A) 110 fiction and 1,045 nonfiction
(B) 55 fiction and 1,100 nonfiction
(C) 55 fiction and 1,045 nonfiction
(D) 1,100 fiction and 55 nonfiction
Answer:
It is given that
The total number of books in the community’s mobile library is 1,155. There are 20 times more nonfiction books than fiction books
Now,
Let the number of fiction books be: x books
So,
The number of non-fiction books is: 20x books
Now,
According to the given information,
x + 20x = 1,155
21x = 1,155
x = \(\frac{1,155}{21}\)
x = 55 books
So,
The number of fiction books is: 55 books
The number of non-fiction books is: 1,100
Hence, from the above,
We can conclude that
The number of fiction books and nonfiction books that are in the mobile library is:

Question 12.
Multi-Step Charlie’s weight is 5 times as great as the weight of his little sister, Tali. If the total weight of the two children is 132 pounds, how much does each child weigh?
(A) Tali: 22 pounds; Charlie: 110 pounds
(B) Tali: 110 pounds; Charlie: 22 pounds
(C) Tali: 44 pounds; Charlie: 88 pounds
(D) Tali: 88 pounds; Charlie: 44 pounds
Answer:
It is given that
Charlie’s weight is 5 times as great as the weight of his little sister, Tali and the total weight of the two children is 132 pounds
Now,
Let the weight of Charlie’s little sister be: x Pounds
So,
The weight of Charlie is: 5x Pounds
Now,
According to the given information,
x + 5x = 132
6x = 132 Pounds
x = \(\frac{132}{6}\)
x = 22 Pounds
So,
The weight of Charlie’s little sister is: 22 Pounds
The weight of Charlie is: 110 Pounds
Hence, from the above,
We can conclude that
The weight of each child is:

Refer to our Texas Go Math Grade 5 Answer Key Pdf to score good marks in the exams. Test yourself by practicing the problems from Texas Go Math Grade 5 Lesson 1.5 Answer Key Round Decimals.

Texas Go Math Grade 5 Lesson 1.5 Answer Key Round Decimals

Unlock the Problem

The Gold Frog of South America is one of the smallest frogs in the world. It is 0.386 of an inch long. What is this length rounded to the nearest hundredth of an inch?

• Underline the length of the Gold Frog.
• Is the frog’s length about the same as the length or the width of a large paper clip?

One Way
Use a place-value chart.

• Write the number in a place-value chart and circle the digit in the place value to which you want to round.
• In the place-value chart, underline the digit to the right of the place to which you are rounding.
• If the digit to the right is less than 5, the digit in the place value to which you are rounding stays the same.
• If the digit to the right is 5 or greater, the digit in the rounding place increases by 1.
• Drop the digits after the place to which you are rounding.
  
  Think: Does the digit in the rounding place stay the same or increase by 1?

So, to the nearest hundredth of an inch, a Gold Frog is about 0.39 of an inch long.

Another Way

Use place value.
The Little Grass Frog is the smallest frog in North America. It is 0.437 of an inch long

A. What is the length of the frog to the nearest hundredth of an inch?

So, to the nearest hundredth of an inch,
The frog is about 0.44 of an inch long.

B. What is the length of the frog to the nearest tenth of an inch?

So, to the nearest tenth of an inch,
The frog is about 0.4 of an inch long.

Share and Show

Write the place value of the underlined digit. Round each number to the place of the underlined digit.

Question 1.
0.673
Answer:
The given decimal number is: 0.673
Now,
The representation of the given number in the place value chart is:

So,
From the given place value chart,
The place value of 7 is: Hundredths
So,
The rounded number that is to the place of the underlined digit is: 0.67
Hence, from the above,
We can conclude that
The place value of 7 is: Hundredths
The rounded number that is to the place of the underlined digit is: 0.67

Question 2.
4.282
Answer:
The given decimal number is: 4.282
Now,
The representation of the given number in the place value chart is:

So,
From the given place value chart,
The place value of 2 is: Tenths
So,
The rounded number that is to the place of the underlined digit is: 4.2
Hence, from the above,
We can conclude that
The place value of 2 is: Tenths
The rounded number that is to the place of the underlined digit is: 4.2

Go Math Grade 5 Lesson 1.5 Answer Key Question 3.
12.917
Answer:
The given decimal number is: 12.917
Now,
The representation of the given number in the place value chart is:

So,
From the given place value chart,
The place value of 2 is: Ones
So,
The rounded number that is to the place of the underlined digit is: 13
Hence, from the above,
We can conclude that
The place value of 2 is: Ones
The rounded number that is to the place of the underlined digit is: 13

Name the place value to which each number was rounded.

Question 4.
0.982 to 0.98
Answer:
The given decimal number is: 0.982
Now,
The representation of the given number in the place value chart is:

Now,
When we observe that above place value chart,
We can observe that
The thousandths place is less than 5
Hence, from the above,
We can conclude that
The place value that the given number was rounded is: Thousandths

Question 5.
3.695 to 4
Answer:
The given decimal number is: 3.695
Now,
The representation of the given number in the place value chart is:

Now,
When we observe that above place value chart,
We can observe that
The tenths place is greater than 5
Hence, from the above,
We can conclude that
The place value that the given number was rounded is: Tenths

Question 6.
7.486 to 7.5
Answer:
The given decimal number is: 7.486
Now,
The representation of the given number in the place value chart is:

Now,
When we observe that above place value chart,
We can observe that
The hundredths place is greater than 5
Hence, from the above,
We can conclude that
The place value that the given number was rounded is: Hundredths

Problem Solving

Write the place value of the underlined digit. Round each number to the place of the underlined digit.

Question 7.
0.592
Answer:
The given decimal number is: 0.592
Now,
The representation of the given number in the place value chart is:

Now,
From the above place value chart,
The place value of 5 is: 5 tenths
So,
The rounded number that is to the place of the underlined digit for the given number is: 0.6
Hence, from the above,
We can conclude that
The place value of 5 is: 5 tenths
The rounded number that is to the place of the underlined digit for the given number is: 0.6

Go Math 5th Grade Lesson 1.5 Answers Question 8.
6.518
Answer:
The given decimal number is: 6.518
Now,
The representation of the given number in the place value chart is:

Now,
From the above place value chart,
The place value of 6 is: 6 ones
So,
The rounded number that is to the place of the underlined digit for the given number is: 7
Hence, from the above,
We can conclude that
The place value of 6 is: 6 ones
The rounded number that is to the place of the underlined digit for the given number is: 7

Question 9.
0.809
Answer:
The given decimal number is: 0.809
Now,
The representation of the given number in the place value chart is:

Now,
From the above place value chart,
The place value of 0 is: 0 hundredths
So,
The rounded number that is to the place of the underlined digit for the given number is: 0.8
Hence, from the above,
We can conclude that
The place value of 0 is: 0 hundredths
The rounded number that is to the place of the underlined digit for the given number is: 0.8

Round 16.748 to the place named.

Question 10.
tenths __________
Answer:
The given number is: 16.748
Now,
The representation of the given number in the place value chart is:

Now,
From the given place value chart,
We can observe that
The tenths place value is greater than 5
Hence, from the above,
We can conclude that
The rounded number that is to the nearest tenth for the given number is: 16.8

Question 11.
hundredths __________
Answer:
The given number is: 16.748
Now,
The representation of the given number in the place value chart is:

Now,
From the given place value chart,
We can observe that
The hundredths place value is less than 5
Hence, from the above,
We can conclude that
The rounded number that is to the nearest hundredth for the given number is: 16.74

Question 12.
ones _____________
Answer:
The given number is: 16.748
Now,
The representation of the given number in the place value chart is:

Now,
From the given place value chart,
We can observe that
The ones place value is greater than 5
Hence, from the above,
We can conclude that
The rounded number that is to the nearest one for the given number is: 20

Question 13.
Explain why any number less than 12.5 and greater than or equal to 11.5 would round to 12 when rounded to the nearest whole number.
Answer:
Let the number be x
Now,
We know that,
When the last digit of a number is greater than 5, then it will be rounded to the next whole number
When the last digit of a number is less than 5, then it will be rounded to the nearest previous whole number
So,
If x = 11.6, then
x = 12 when x > 11.5 and x < 12.5
Hence, from the above,
We can conclude that
When the last digit of a number is greater than 5, then it will be rounded to the next whole number
When the last digit of a number is less than 5, then it will be rounded to the nearest previous whole number

Texas Go Math Grade 5 Lesson 1.5 Reteach Answer Key Question 14.
Write Math Explain what happens when you round 4.999 to the nearest tenth.
Answer:
The given decimal number is: 4.999
Now,
The representation of 4.999 in the place value chart is:

Now,
When we observe the tenth’s place value in the above lace value chart
We can observe that
The number is greater than 5
Now,
We know that,
When the last digit of a number is greater than 5, then it will be rounded to the next whole number
When the last digit of a number is less than 5, then it will be rounded to the nearest previous whole number
Hence, from the above,
We can conclude that
The rounded number for 4.999 when we rounded off to the nearest tenth to the whole number is: 5

Problem Solving

Use the table for 15-18.

Question 15.
The speeds of two insects, when rounded to the nearest whole number, are the same. Which two insects are they?
Answer:
The given table is:

Now,
From the given table,
We can say that
The rounded speeds nearest to the whole number of the different insects are:
Dragonfly      –         7 m/s
Horsefly         –         4 m/s
Bumblebee    –         3 m/s
Honeybee      –         3 m/s
Housefly        –          2 m/s
Hence, from the above,
We can conclude that
The  names of the insects that have the speeds of the two insects when rounded to the nearest to the whole number that are the same are: Bumblebee and Honeybee

Question 16.
What is the speed of the housefly rounded to the nearest hundredth?
Answer:
The given table is:

Now,
From the given table,
We can observe that
The speed of Housefly is: 1.967 m/s
Now,
The representation of the speed of housefly in the place value chart is:

Now,
From the above place value chart,
We can say that
The speed of the housefly when rounded to the nearest hundredth is: 1.97 m/s
Hence, from the above,
We can conclude that
The speed of the housefly when rounded to the nearest hundredth is: 1.97 m/s

Question 17.
The speed of an insect is about 3.9 meters per second. Which insect could it be?
Answer:
It is given that
The speed of an insect is about 3.9 meters per second
Now,
The given table is:

Now,
From the given table,
We can observe that
The speed of the horsefly is: 3.934 m/s
Now,
When we rounded off the speed of horsefly to the nearest hundredth,
The speed of horsefly will be: 3.9 m/s
Hence, from the above,
We can conclude that
The insect that has the speed of 3.9 m/s is: Horsefly

Question 18.
H.O.T. What’s the Error? Mark said that the speed of a dragonfly rounded to the nearest tenth was 6.9 meters per second. Is he correct? If not, what is his error?
Answer:
It is given that
Mark said that the speed of a dragonfly rounded to the nearest tenth was 6.9 meters per second.
Now,
The given table is:

Now,
From the given table,
We can observe that
The speed of dragonfly is: 6.974 m/s
Now,
When we round off the speed of dragonfly to the nearest tenth,
The speed of dragonfly will be: 6.9 m/s
Hence, from the above,
We can conclude that
Mark is correct

Question 19.
H.O.T. Multi-Step A rounded number for the speed of an insect is 5.67 meters per second. What are the fastest and slowest speeds to the thousandths that could round to 5.67? Explain.
Answer:
It is given that
A rounded number for the speed of an insect is 5.67 meters per second.
Now,
When we round off the given speed of insect to the nearest whole number,
The speed of insect will become: 6 m/s
When we round off the given speed of insect to the hundredth,
The speed of insect will become: 5.7 m/s
Hence, from the above,
We can conclude that
The fastest speed is: 6 m/s
The slowest speed is: 5.7 m/s

Daily Assessment Task

Fill in the bubble completely to show your answer.

Question 20.
Chef Round uses 1.257 kilograms of meat for his famous meatballs. There are four packages at the market. Which amount is closest to 1.257 kilograms?
A. 1.26 kg
B. 1.2 kg
C. 1.3 kg
D. 1.35 kg
Answer:
It is given that
Chef Round uses 1.257 kilograms of meat for his famous meatballs. There are four packages at the market.
Now,
The given number is: 1.257
When we round off the given number to the nearest thousandth,
The number will be: 1.26
Hence, from the above,
We can conclude that
The amount that is closest to 1.257 kilograms is:

Question 21.
Representations The table shows the neck lengths for four giraffes in the Long-Necked Friends Sanctuary. Which two giraffes have necks that are the same length when rounded to the nearest tenth?

A. Spot and Stretch
B. Mittens and Spot
C. Mittens and Stretch
D. Spot and Zippy
Answer:
It is given that
The table shows the neck lengths for four giraffes in the Long-Necked Friends Sanctuary.
Now,
The given table is:

Now,
From the given table,
We can say that
The length of the necks when rounded off to the nearest tenth is:
Mittens            –               5.6 feet
Spot                 –               5.7 feet
Stretch             –               5.4 feet
Zippy               –               5.7 feet
So,
The two giraffes that have the same neck length are: Spot and Zippy
Hence, from the above,
We can conclude that
The two giraffes that have the same neck lengths when rounded to the nearest tenth are:

Question 22.
Multi-Step Janis sells friendship bracelets at a fair. She has three bracelets that are 16.53 cm, 16.755 cm, and 16.55 cm long. She wants to label the lengths to the nearest tenth of a cm. In order, how should she label the bracelets?
A. 16.5 cm, 16.76 cm, 16.6 cm
B. 16.6 cm, 16.7 cm, 16.5 cm
C. 16.5 cm, 16.8 cm, 16.6 cm
D. 16.3 cm, 16.5 cm, 16.5 cm
Answer:
It is given that
Janis sells friendship bracelets at a fair. She has three bracelets that are 16.53 cm, 16.755 cm, and 16.55 cm long
Now,
The given numbers are: 16.53, 16.755, and 16.55
So,
The given numbers when rounded to the nearest tenth will become: 16.5, 16.76, and 16.6
So,
The order of the numbers from the greatest to the least are: 16.76, 16.6, and 16.5
The order of the numbers from the least to the greatest are: 16.5, 16.6, and 16.76
Hence, from the above,
We can conclude that
The order that Janis should label the bracelets are:

Texas Test Prep

Question 23.
To which place value is the number rounded? 6.706 to 6.71
A. ones
B. tenths
C. hundredths
D. thousandths
Answer:
The given decimal number is: 6.706
Now,
The representation of 6.706 in the place value chart is:

Now,
From the above place value chart,
We can say that
The given rounded can be rounded to thousandths place to become 6.71
Hence, from the above,
We can conclude that
The place value that the given number is rounded will be:

Texas Go Math Grade 5 Lesson 1.5 Homework and Practice Answer Key

Write the place value of the underlined digit. Round each number to the place of the underlined digit.

Question 1.
0.782
Answer:
The given decimal number is: 0.782
Now,
The representation of the given number in the place value chart is:

Now,
From the above place value chart,
The place value of 7 is: 7 tenths
So,
The rounded number that is to the place of the underlined digit for the given number is: 0.8
Hence, from the above,
We can conclude that
The place value of 7 is: 7 tenths
The rounded number that is to the place of the underlined digit for the given number is: 0.8

Question 2.
4.071
Answer:
The given decimal number is: 4.071
Now,
The representation of the given number in the place value chart is:

Now,
From the above place value chart,
The place value of 7 is: 7 hundredths
So,
The rounded number that is to the place of the underlined digit for the given number is: 4.07
Hence, from the above,
We can conclude that
The place value of 7 is: 7 hundredths
The rounded number that is to the place of the underlined digit for the given number is: 4.07

Question 3.
21.939
Answer:
The given decimal number is: 21.939
Now,
The representation of the given number in the place value chart is:

Now,
From the above place value chart,
The place value of 1 is: 1 ones
So,
The rounded number that is to the place of the underlined digit for the given number is: 22
Hence, from the above,
We can conclude that
The place value of 1 is: 1 ones
The rounded number that is to the place of the underlined digit for the given number is: 22

Go Math Lesson 1.5 Answer Key 5th Grade Question 4.
8.155
Answer:
The given decimal number is: 8.155
Now,
The representation of the given number in the place value chart is:

Now,
From the above place value chart,
The place value of 1 is: 1 tenth
So,
The rounded number that is to the place of the underlined digit for the given number is: 8.16
Hence, from the above,
We can conclude that
The place value of 1 is: 1 tenth
The rounded number that is to the place of the underlined digit for the given number is: 8.16

Question 5.
34.09
Answer:
The given decimal number is: 34.09
Now,
The representation of the given number in the place value chart is:

Now,
From the above place value chart,
The place value of 3 is: 3 tens
So,
The rounded number that is to the place of the underlined digit for the given number is: 34
Hence, from the above,
We can conclude that
The place value of 3 is: 3 tens
The rounded number that is to the place of the underlined digit for the given number is: 34

Question 6.
4.276
Answer:
The given decimal number is: 4.276
Now,
The representation of the given number in the place value chart is:

Now,
From the above place value chart,
The place value of 7 is: 7 hundredths
So,
The rounded number that is to the place of the underlined digit for the given number is: 4.28
Hence, from the above,
We can conclude that
The place value of 7 is: 7 hundredths
The rounded number that is to the place of the underlined digit for the given number is: 4.28

Round 8.293 to the place named.

Question 7.
hundredths ___________
Answer:
The given number is: 8.293
Now,
The representation of 8.293 in the place value chart is:

Now,
From the above place value chart,
We can say that
The rounded value of 8.293 nearest to the place value of hundredths is: 8.29
Hence, from the above,
We can conclude that
The rounded value of 8.293 nearest to the place value of hundredths is: 8.29

Question 8.
tenths ______________
Answer:
The given number is: 8.293
Now,
The representation of 8.293 in the place value chart is:

Now,
From the above place value chart,
We can say that
The rounded value of 8.293 nearest to the place value of tenths is: 8.3
Hence, from the above,
We can conclude that
The rounded value of 8.293 nearest to the place value of tenths is: 8.3

Question 9.
ones ____________
Answer:
The given number is: 8.293
Now,
The representation of 8.293 in the place value chart is:

Now,
From the above place value chart,
We can say that
The rounded value of 8.293 nearest to the place value of ones is: 8
Hence, from the above,
We can conclude that
The rounded value of 8.293 nearest to the place value of ones is: 8

Round 12.462 to the place named.

Question 10.
tenths __________
Answer:
The given number is: 12.462
Now,
The representation of 8.293 in the place value chart is:

Now,
From the above place value chart,
We can say that
The rounded value of 12.462 nearest to the place value of tenths is: 12.5
Hence, from the above,
We can conclude that
The rounded value of 12.462 nearest to the place value of tenths is: 12.5

Question 11.
hundredths ___________
Answer:
The given number is: 12.462
Now,
The representation of 8.293 in the place value chart is:

Now,
From the above place value chart,
We can say that
The rounded value of 12.462 nearest to the place value of hundredths is: 12.46
Hence, from the above,
We can conclude that
The rounded value of 12.462 nearest to the place value of hundredths is: 12.46

Question 12.
ones _____________
Answer:
The given number is: 12.462
Now,
The representation of 8.293 in the place value chart is:

Now,
From the above place value chart,
We can say that
The rounded value of 12.462 nearest to the place value of ones is: 12
Hence, from the above,
We can conclude that
The rounded value of 12.462 nearest to the place value of ones is: 12

Problem Solving

Question 13.
What is Anita’s speed rounded to the nearest hundredth?
Answer:
The given table is:

Now,
From the given table,
We can observe that
Anita’s speed is: 0.277 m/s
Now,
The value of Anita’s speed when rounded to the nearest hundredth will become: 0.28 m/s
Hence, from the above,
We can conclude that
Anita’s speed that is rounded to the nearest hundredth is: 0.28 m/s

Question 14.
The speeds of two runners when rounded to the nearest tenth are the same. Which two runners are they?
Answer:
The given table is:

Now,
From the given table,
We can observe that
The values of runners speeds when rounded to the nearest tenth will be:
Luke             –            0.2 m/s
Margola       –            0.2 m/s
Anita            –            0.28 m/s
Shateel         –            0.30 m/s
So,
From the above,
We can observe that
The speed of the two runners that have the same speed when rounded to the nearest tenth are: Luke and Margola
Hence, from the above,
We can conclude that
The speed of the two runners that have the same speed when rounded to the nearest tenth are: Luke and Margola

Lesson Check

Fill in the bubble completely to show your answer.

Question 15.
To which place value is the number rounded? 8.293 to 8.29
A. ones
B. tenths
C. hundredths
D. thousandths
Answer:
The given decimal number is: 8.293
Now,
The representation of 8.293 in the place value chart is:

So,
From the above place value chart
We can observe that
The number is rounded at the thousandths place
Hence, from the above,
We can conclude that
The place value that the given number is rounded is:

Question 16.
To which place value is the number rounded? 0.799 to 0.8
A. ones
B. tenths
C. hundredths
D. thousandths
Answer:
The given decimal number is: 0.799
Now,
The representation of 0.799 in the place value chart is:

So,
From the above place value chart
We can observe that
The number is rounded at the hundredths place
Hence, from the above,
We can conclude that
The place value that the given number is rounded is:

Question 17.
Which number is 5.389 rounded to the nearest hundredth?
A. 5.4
B. 5.38
C. 5.39
D. 5.399
Answer:
The given decimal number is: 5.389
Now,
The representation of 5.389 in the place value chart is:

Now,
From the above place value chart,
We can say that
The value of 5.389 when rounded to the nearest hundredth is: 5.39
Hence, from the above,
We can conclude that
The number that is to the nearest hundredth for 5.389 is:

Question 18.
Which number is 7.323 rounded to the nearest tenth?
A. 7.3
B. 7.4
C. 7.32
D. 7.33
Answer:
The given decimal number is: 7.323
Now,
The representation of 7.323 in the place value chart is:

Now,
From the above place value chart,
We can say that
The value of 7.323 when rounded to the nearest tenth is: 7.3
Hence, from the above,
We can conclude that
The number that is to the nearest tenth for 7.323 is:

Question 19.
A baker uses 2.327 kilograms of blueberries for muffins. There are four packages at the fruit market. Which amount is closest to 2.327 kilograms?
A. 2.3 kg
B. 2.42 kg
C. 2.33 kg
D. 2.4 kg
Answer:
It is given that
A baker uses 2.327 kilograms of blueberries for muffins. There are four packages at the fruit market
Now,
The representation of 2.327 in the place value chart is:

Now,
From the above place value chart,
We can say that
The number that is nearest to 2.327 is: 2.33
Hence, from the above,
We can conclude that
The amount that is closest to 2.327 kilograms is:

Question 20.
Multi-Step The chart shows the length of each student’s pencil. Which person has a pencil length that can be rounded up to the nearest tenth of an inch?

A. Lori
B. Ping
C. Jason
D. Annie
Answer:
It is given that
The chart shows the length of each student’s pencil.
Now,
The given table is:

Now,
From the given table,
We can observe that
The values of  pencil lengths of Lori, Ping, and Jason are represented in hundredths whereas the value of pencil length for Annie is represented in tenths only
Hence, from the above,
We can conclude that
The person that has a pencil length that can be rounded up to the nearest tenth of an inch is:

Refer to our Texas Go Math Grade 5 Answer Key Pdf to score good marks in the exams. Test yourself by practicing the problems from Texas Go Math Grade 5 Lesson 1.4 Answer Key Compare and Order Decimals.

Texas Go Math Grade 5 Lesson 1.4 Answer Key Compare and Order Decimals

Unlock the Problem

The table lists some of the mountains in the United States that are over two miles high. How does the height of Cloud Peak in Wyoming compare to the height of Boundary Peak in Nevada?

One Way
Use place value.
Line up the decimal points. Start at the left. Compare the digits in each place-value position until the digits are different.

Since 9 > 8, then 2.495 > 2.488, and 2.488 < 2.495.
So, the height of Cloud Peak is greater than the height of Boundary Peak.

Another Way
Use a place-value chart to compare
Compare the height of Cloud Peak to Wheeler Peak.

Since 5 > 3, then 2.495 > 2.493, and 2.493 < 2.495.
So, the height of Cloud Peak is greater than the height of Wheeler Peak.

Math Talk
Mathematical Processes

Explain why it is important to line up the decimal points when comparing decimals.
Answer:
Decimals extend the number system beyond the simple ‘hundreds, tens, units’ into ‘tenths of units’, ‘hundredths of units’ and so on. The most important rule to remember is to line up the decimal points in your calculation, ensuring that the decimal point in the answer also lines up with the decimal points above it

Example

Mount Whitney in California is 2.745 miles high, Mount Rainier in Washington * is 2.729 miles high, and Mount Harvard in Colorado is 2.731 miles high. Order the heights of these mountains from least to greatest. Which mountain has the least height? Which mountain has the greatest height?

STEP 1
Line up the decimal points. There are the same number of ones. Circle the tenths and compare.
2.745 Whitney
2.729 Rainier
2.731 Harvard
There are the same number of tenths.

STEP 2
Underline the hundredths and compare. Order from least to greatest.
2.745 Whitney
2.729 Rainier
2.731 Harvard
Since 2.729 < 2.731 < 2.745, the heights in order from least to greatest are 2.729, 2.731, 2.745.
So,
Rainier has the least height and Whitney has the greatest height.

Math Talk
Mathematical Processes

Explain why you do not have to compare the digits in the thousandths place to order the heights of the 3 mountains.
Answer:
We know that,
The place-value of “Hundredths” is greater than the place-value of “Thousandths”
Hence, from the above,
We can conclude that
Since the place-value of hundredths is greater than the thousandths, you do not have to compare the digits in the thousandths place to order the heights of the 3 mountains

Share and Show

Question 1.
Use the place-value chart to compare the two numbers. What is the greatest place-value position where the digits differ?

Answer:
The given place-value chart is:

Now,
From the given place-value chart,
We can observe that
The given decimal numbers are: 3.472 and 3.445
Now,
We know that,
The place-value of “Hundredths” is greater than the place-value of “Thousandths”
Now,
When we compare the place-values,
We can observe that
7 hundredths > 4 hundredths
Hence, from the above,
We can conclude that
The greatest place-value position where the digits differ is: Hundredths

Compare. Write <, >, or =.

Question 2.
4.563 4.536
Answer:
The given decimal numbers are: 4.563 and 4.536
Now,
We know that,
The place-value of “Hundredths” is greater than the place-value of “Thousandths”
Hence,
4.563 > 4.536

Question 3.
5.640 5.64
Answer:
The given decimal numbers are: 5.640 and 5.64
Now,
From the given numbers,
We can observe that
All the place-values are the same
Hence,
5.640 = 5.64

Question 4.
8.673 8.637
Answer:
The given decimal numbers are: 8.673 and 8.637
Now,
We know that,
The place-value of “Hundredths” is greater than the place-value of “Thousandths”
Hence,
8.673 > 8.637

Name the greatest place-value position where the digits differ. Name the greater number.

Question 5.
3.579; 3.564
Answer:
The given decimal numbers are: 3.579 and 3.564
Now,
We know that,
The place-value of “Hundredths” is greater than the place-value of “Thousandths”
Hence,
3.579 > 3.564

Question 6.
9.572; 9.637
Answer:
The given decimal numbers are: 9.572 and 9.637
Now,
We know that,
The place-value of “Tenths” is greater than the place-value of “Hundredths”
Hence,
9.572 < 9.637

Go Math Grade 5 Lesson 1.4 Answer Key Question 7.
4.159; 4.152
Answer:
The given decimal numbers are: 4.159 and 4.152
Now,
We know that,
The place-value of “Thousandths” is greater than the place-value of “Ten thousandths”
Hence,
4.159 > 4.152

Problem Solving

Order from greatest to least.

Question 8.
2.007; 2.714; 2.09; 2.97
Answer:
The given decimal numbers are: 2.007, 2.714, 2.09, and 2.97
Now,
We know that,
The place-value of “Tenths” is greater than the place-value of “Hundredths” is greater than the “Thousandths”
Hence, from the above,
We can conclude that
The order of numbers from the greatest to the least is: 2.97, 2.714, 2.09, and 2.007

Question 9.
0.386; 0.3; 0.683; 0.836
Answer:
The given decimal numbers are: 0.386, 0.3, 0.683, and 0.836
Now,
We know that,
The place-value of “Tenths” is greater than the place-value of “Hundredths” is greater than the “Thousandths”
Hence, from the above,
We can conclude that
The order of numbers from the greatest to the least is: 0.836, 0.683, 0.386, and 0.3

H.O.T. Algebra Find the unknown digit to make each statement true.

Question 10.
3.59 > 3.5 1 > 3.572
Answer:
Let the unknown digit be x
Hence, from the above,
We can conclude that
The unknown digit that make the given statement true is:
3.59 > 3.581 > 3.572
Where,
x = 0, 1, 2, ……. 8

Question 11.
6.837 > 6.83 > 6.835
Answer:
Let the unknown digit be x
Hence, from the above,
We can conclude that
The unknown digit that make the given statement true is:
6.837 > 6.836 > 6.835
Where,
x = 0, 1, 2, ……. 6

Question 12.
2.45 < 2. 6 < 2.461
Answer:
Let the unknown digit be x
Hence, from the above,
We can conclude that
The unknown digit that make the given statement true is:
2.45 < 2.46 < 2.461
Where,
x = 0, 1, 2, ……. 4

Question 13.
Dawn keeps track of her softball batting average each year. In the first year, her batting average was .783. In the second year, her batting average is .81. In which year did she have the greater batting average?
Answer:
It is given that
Dawn keeps track of her softball batting average each year. In the first year, her batting average was .783. In the second year, her batting average is .81
Now,
The given decimal numbers are: 0.783 and 0.81
Now,
We know that,
The place-value of “Tenths” is greater than the place-value of “Hundredths” is greater than the “Thousandths”
So,
0.81 > 0.783
Hence, from the above,
We can conclude that
In the second year, Dawn has the greatest batting average

Problem Solving

Use the table for 14-16.

Question 14.
Use Math Language How does the height of Steele Mountain compare to the height of Blackburn Mountain? Compare the heights using words.
Answer:
The given table is:

Now,
From the given table,
We can observe that
The height of Steele Mountain is: 3.152 miles
The height of Blackburn Mountain is: 3.104 miles
Now,
We know that,
The place-value of “Tenths” is greater than the place-value of “Hundredths” is greater than the “Thousandths”
So,
3.152 > 3.104
Hence, from the above,
We can conclude that
The height of Steele Mountain is greater than the height of Blackburn Mountain

Question 15.
Write Math Explain how to order the height of the mountains from greatest to least.
Answer:
The given table is:

Now,
We know that,
The place-value of “Tenths” is greater than the place-value of “Hundredths” is greater than the “Thousandths”
So,
The order of the heights from the greatest to the least is:
3.152 > 3.134 > 3.104
Hence, from the above,
We can conclude that
The order of the height of mountains from the greatest to the least is:
3.152 miles > 3.134 miles > 3.104 miles

Question 16.
Multi-Step What if the height of Blackburn Mountain were 0.05 mile greater. Would it then be the mountain with the greatest height? Explain.

Answer:
The given table is:

Now,
From the given table,
We can observe that
The height of the Blackburn Mountain is: 3.104 miles
The greatest height is: 3.152 miles
So,
The increased height of  Blackburn Mountain = 0.05 + 3.104
= 3.154 miles
So,
3.154 miles > 3.152 miles
Hence, from the above,
We can conclude that
With the height of Blackburn Mountain 0.05 miles greater, Blackburn Mountain will have the greatest height

Daily Assessment Task

Fill in the bubble completely to show your answer.

Question 17.
The length of a piece of plastic for a science kit needs to be greater than 22.4 inches and less than 22.5 inches. Which length of plastic can be used?
A. 22.35 in
B. 22.51 in
C. 22.47 in.
D. 22.40 in.
Answer:
It is given that
The length of a piece of plastic for a science kit needs to be greater than 22.4 inches and less than 22.5 inches.
So,
The numbers ca also be written as: 22.40 inches and 22.50 inches
Hence, from the above,
We can conclude that
The length of the plastic that can be used is:

Question 18.
Louis is comparing the numbers 8.402 and 8.451. What is the least place value he needs to compare to decide which number is greater?
A. ones
B. tenths
C. hundredths
D. thousandths
Answer:
It is given that
Louis is comparing the numbers 8.402 and 8.451
Now,
The given numbers are: 8.402 and 8.451
Now,
We know that,
The place-value of “Tenths” is greater than the place-value of “Hundredths” is greater than the “Thousandths”
So,
We have to compare the hundredths place value to decide which number is greater
Hence, from the above,
We can conclude that
The least place value that he needs to compare to decide which number is greater is:

Question 19.
Multi-Step Melinda compares four numbers. Which shows the numbers from least to greatest?
A. 3.04; 3.10; 3.529; 3.685
B. 3.10; 3.04; 3.529; 3.685
C. 3.10; 3.04; 3.685; 3.529
D. 3.04; 3.685; 3.529; 3.10
Answer:
It is given that
Melinda compares four numbers
Now,
We know that,
The place-value of “Tenths” is greater than the place-value of “Hundredths” is greater than the “Thousandths”
Hence, from the above,
We can conclude that
The numbers that have the order from the least to the greatest is:

Texas Test Prep

Question 20.
Mount Logan in the Yukon is 3.702 miles high. Mount McKinley in Alaska is 3.848 miles high and Pico de Orizaba in Mexico is 3.571 miles high. Order these mountains by height from greatest to least.
A. Logan, McKinley, Pico de Orizaba
B. McKinley, Logan, Pico de Orizaba
C. Pico de Orizaba, Logan, McKinley
D. Logan, Pico de Orizaba, McKinley
Answer:
It is given that
Mount Logan in the Yukon is 3.702 miles high. Mount McKinley in Alaska is 3.848 miles high and Pico de Orizaba in Mexico is 3.571 miles high
Now,
The given decimal numbers are: 3.702, 3.848, and 3.571
Now,
We know that,
The place-value of “Tenths” is greater than the place-value of “Hundredths” is greater than the “Thousandths”
So,
The order of the numbers from the greatest to the least is:
3.848 > 3.702 > 3.571
Hence, from the above,
We can conclude that
The height of the mountains fro the least to the greatest is:

Texas Go Math Grade 5 Lesson 1.4 Homework and Practice Answer Key

Compare. Write <, >, or =.

Question 1.
9.23 9.32
Answer:
The given decimal numbers are: 9.23 and 9.32
Now,
We know that,
The place-value of “Tenths” is greater than the place-value of “Hundredths”
Hence,
9.23 < 9.32

Question 2.
7.20 7.2
Answer:
The given decimal numbers are: 7.20 and 7.2
Now,
We know that,
The number that has the last digit zero and the same number that does not have the last digit zero are the same
Hence,
7.20 = 7.2

Practice and Homework Lesson 1.4 Compare Decimals Question 3.
1.994 1.493
Answer:
The given decimal numbers are: 1.994 and 1.493
Now,
We know that,
The place-value of “Tenths” is greater than the place-value of “Hundredths”
Hence,
1.994 > 1.493

Order from greatest to least.

Question 7.
7.081; 7.81; 7.002; 7.14
Answer:
The given decimal numbers are: 7.081, 7.81, 7.002, and 7.14
Now,
We know that,
The place-value of “Tenths” is greater than the place-value of “Hundredths” is greater than the “Thousandths”
Hence, from the above,
We can conclude that
The order of numbers from the greatest to the least is: 7.81, 7.14, 7.081, and 7.002

Question 8.
1.001; 1.1; 1.403; 1.078
Answer:
The given decimal numbers are: 1.001, 1.1, 1.403, and 1.078
Now,
We know that,
The place-value of “Tenths” is greater than the place-value of “Hundredths” is greater than the “Thousandths”
Hence, from the above,
We can conclude that
The order of numbers from the greatest to the least is: 1.403, 1.1, 1.078, and 1.001

Question 9.
0.04; 0.5; 0.021; 0.133
Answer:
The given decimal numbers are: 0.04, 0.5, 0.021, 0.133
Now,
We know that,
The place-value of “Tenths” is greater than the place-value of “Hundredths” is greater than the “Thousandths”
Hence, from the above,
We can conclude that
The order of numbers from the greatest to the least is: 0.5, 0.133, 0.040, and 0.021

Question 10.
4; 4.022; 4.002; 4.221
Answer:
The given decimal numbers are: 4, 4.022, 4.002, and 4.221
Now,
We know that,
The place-value of “Tenths” is greater than the place-value of “Hundredths” is greater than the “Thousandths”
Hence, from the above,
We can conclude that
The order of numbers from the greatest to the least is: 4.221, 4.022, 4.002, and 4

Problem Solving

Find the unknown digit to make each statement true,

Question 11.
5.345 > 5.34  > 5.343
Answer:
Let the unknown digit be x
Hence, from the above,
We can conclude that
The unknown digit that make the given statement true is:
5.345 > 5.344 > 5.343
Where,
x = 0, 1, 2, 3, and 4

Question 12.
8.25 < 8. 6 < 8.361
Answer:
Let the unknown digit be x
Hence, from the above,
We can conclude that
The unknown digit that make the given statement true is:
8.25 < 8.26 < 8.361
Where,
x = 0, 1, 2

Question 13.
6.48 > 6.4 1 > 6.470
Answer:
Let the unknown digit be x
Hence, from the above,
We can conclude that
The unknown digit that make the given statement true is:
6.48 > 6.471 > 6.470
Where,
x = 0, 1, 2, ……. 7

Question 14.
Jack travels 2.45 miles to school. Wanda travels 2.31 miles to school. Compare the distances using words.
Answer:
It is given that
Jack travels 2.45 miles to school. Wanda travels 2.31 miles to school
Now,
We know that,
The place-value of “Tenths” is greater than the place-value of “Hundredths” is greater than the “Thousandths”
So,
2.45 > 2.31
Hence, from the above,
We can conclude that
Jack travels more miles than Wanda to school

Practice and Homework Lesson 1.4 Answer Key 5th Grade Question 15.
In a jumping contest, Marcus jumped 1.02 meters, Gustavo jumped 1.29 meters, and Loreena jumped 1.202 meters. Who won the jumping contest?
Answer:
It is given that
In a jumping contest, Marcus jumped 1.02 meters, Gustavo jumped 1.29 meters, and Loreena jumped 1.202 meters
Now,
The given decimal numbers are: 1.02, 1.29, and 1.202
Now,
We know that,
The place-value of “Tenths” is greater than the place-value of “Hundredths” is greater than the “Thousandths”
So,
1.29 > 1.202 > 1.02
Hence, from the above,
We can conclude that
Gustavo won the jumping contest

Lesson Check

Fill in the bubble completely to show your answer.

Question 16.
Which number has the greatest value?
A. 9.382
B. 9.47
C. 9.09
D. 9.7
Answer:
We know that,
The place-value of “Tenths” is greater than the place-value of “Hundredths” is greater than the “Thousandths”
Hence, from the above,
We can conclude that
The number that has the greatest value is:

Question 17.
Kerry is comparing the numbers 9.207 and 9.210. What is the least place value she, needs to compare to decide which number is greater?
A. ones
B. thousandths
C. hundredths
D. tenths
Answer:
It is given that
Kerry is comparing the numbers 9.207 and 9.210
Now,
We know that,
The place-value of “Tenths” is greater than the place-value of “Hundredths” is greater than the “Thousandths”
So,
We have to compare the “Hundredths” place value to find the greatest number in the given decimal numbers
Hence, from the above,
We can conclude that
The least place value Kerry needs to compare to decide which number is greater is:

Question 18.
Which number has the least value?
A. 7.034
B. 7.304
C. 7.403
D. 7.003
Answer:
We know that,
The place-value of “Tenths” is greater than the place-value of “Hundredths” is greater than the “Thousandths”
Hence, from the above,
We can conclude that
The number that has the least value is:

Question 19.
In 5.820, what is the place-value position of the 0?
A. thousandths
B. tenths
C. hundredths
D. ones
Answer:
The given decimal number is: 5.820
Now,
The representation of 5.820 in the place value chart is:

Hence, from the above,
We can conclude that
The place value position of 0 in 5.820 is:

Question 20.
Multi-Step Jenna compares the lengths of her pet fish. The table shows the lengths of the fish. Which shows the lengths in order from greatest to least?

A. Spots, Oscar, Gus, Gem
B. Oscar, Spots, Gem, Gus
C. Gus, Oscar, Gem, Spots
D. Spots, Oscar, Gem, Gus
Answer:
It is given that
Jenna compares the lengths of her pet fish. The table shows the lengths of the fish
Now,
The given table is:

Now,
We know that,
The place-value of “Tenths” is greater than the place-value of “Hundredths” is greater than the “Thousandths”
So,
From the given table,
We can observe that
The order of the numbers that are from the greatest to the least is:
4.7 > 4.32 > 4.21 > 4.055
So,
The order of the lengths of Jenna’s pet fish is:
Sports > Oscar > Gem > Gus
Hence, from the above,
We can conclude that
The lengths of Jenna’s pet fishes in order from the least to the greatest is:

Question 21.
Multi-Step Look at the table in Question 20. What if Gem grew 0.5 inches. Which would then show the lengths in order from greatest to least?
A. Gem, Spots, Gus, Oscar
B. Gus, Oscar, Gem, Spots
C. Gem, Spots, Oscar, Gus
D. Spots, Oscar, Gem, Gus
Answer:
The given table is:

Now,
The present length of Gem = (The old length of gem) + 0.5
= 0.5 + 4.21
= 4.71 inches
So,
The order of the lengths of Jenna’s pet fish according to the new size of Gem is:
Gem > Sports > Oscar > Gus
Hence, from the above,
We can conclude that
The order of the lengths of pet fish from the greatest to the least after the increase of Gem’s length by 0.5 inches is:

Refer to our Texas Go Math Grade 5 Answer Key Pdf to score good marks in the exams. Test yourself by practicing the problems from Texas Go Math Grade 5 Lesson 1.6 Answer Key Estimate Decimals Sums and Differences.

Texas Go Math Grade 5 Lesson 1.6 Answer Key Estimate Decimals Sums and Differences

Unlock the Problem

A singer is recording a CD. The lengths of the three songs are 3.4 minutes, 2.78 minutes, and 4.19 minutes. About how much recording time will be on the CD?

Use rounding to estimate.

Round to the nearest whole number. Then add.

Remember
To round a number, determine the place to which you want to round.
. If the digit to the right is less than 5, the digit in the rounding place stays the same.
. If the digit to the right is 5 or greater, the digit in the rounding place increases by 1.
So,

So,
There will be about 10 minutes of recording time on the CD.

Try This!

Use rounding to estimate.

A.
Round to the nearest whole dollar. Then subtract.

To the nearest dollar, $27.95 – $11.72 is about _________.
Answer:
The given numbers are: 27.95 and 11.72
Now,
When we round off the given numbers to the nearest whole numbers,
The numbers will become: 28 and 12
Now,

Hence, from the above,
We can conclude that
To the nearest dollar, the subtraction of the given values is about $16

B.
Round to the nearest ten dollars. Then subtract.

To the nearest ten dollar, $27.95 – $11.72 is about _________.
Answer:
The given numbers are: 27.95 and 11.72
Now,
When we round off the given numbers to the nearest tenths,
The numbers will become: 28.0 and 11.7
So,

Hence, from the above,
We can conclude that
The value of the given numbers when rounded off to the nearest tenth is about $16.2

Do you want an overestimate or an underestimate when you estimate the total cost of items you want to buy? Explain.
Answer:
Easy all you have to do it estimate.
Example:
$11.62 + $7.86 = $19.48
$11.62 rounds to 12 and $7.80 rounds to $8.00
So,
The round to the total would be $20 so this would be an over estimation

Use Benchmarks Benchmarks are familiar numbers used as points of reference. You can use the benchmarks 0, 0.25, 0.50, 0.75, and 1 to estimate decimal sums and differences.

Example

Use benchmarks to estimate. 0.76 – 0.22
Locate and graph a point on the number line for each decimal. Identify which benchmark each decimal is closer to.

So, 0.76 – 0.22 is about ____________.
Answer:
We know that,
“Benchmarks” are the numbers that act as a point of reference and the benchmark numbers are: 0, 0.25, 0.50, and 0.75
Now,


So,
The difference between 0.76 and 0.22 with the reference to the benchmarks will be:

Hence, from the above,
We can conclude that

Math Talk
Mathematical Processes

Use the Example to explain how using rounding or benchmarks to estimate a decimal difference can give you different answers.
Answer:
To estimate means to make a rough guess or calculation. To round means to simplify a known number by scaling it slightly up or down. Rounding is a type of estimation. Both methods can help you make educated approximations and can be used in everyday life for tasks related to money, time, or distance.
A benchmark helps a student estimate the general number a fraction or decimal number is. For example, a student can quickly learn that the fraction \(\frac{1}{2}\) means a half, 0.50, or 50 percent because of intuition.

Share and Show

Use rounding to estimate.

Question 1.

Answer:
The given numbers are: 2.34, 1.9, and 5.23
So,

Hence, from the above,
We can conclude that
The rounded value for the given addition of given numbers is: 9.5

Go Math Lesson 1.6 5th Grade Answer Key Question 2.

Answer:
The given numbers are: 10.39, 4.28
So,

Hence, from the above,
We can conclude that
The rounded value for the subtraction of given numbers is: 6.1

Question 3.

Answer:
The given numbers are: 19.75, 3.98
So,

Hence, from the above,
We can conclude that
The rounded value for the addition of given numbers is: $23.7

Use benchmarks to estimate.

Question 4.

Answer:
The given numbers are: 0.34, 0.1, and 0.25
Now,
We know that,
The benchmarks are: 0, 0.25, 0.75, and 1
So,

Hence, from the above,
We can conclude that
The benchmark result for the given addition of numbers is: 0.75

Question 5.

Answer:
The given numbers are: 10.39, 4.28
Now,
We know that,
The benchmarks are: 0, 0.25, 0.75, and 1
So,

Hence, from the above,
We can conclude that
The benchmark result for the given subtraction of numbers is: 6

Math Talk
Mathematical Processes

Describe the difference between an estimate and an exact answer.
Answer:
An estimate is an answer to a problem that is close to the exact number, but not necessarily exact
An exact answer is an answer that is precise and need not be in a closed condition

Problem Solving

Use rounding or benchmarks to estimate.

Question 6.

Answer:
The given numbers are: 0.93, 0.18
So,

Hence, from the above,
We can conclude that
The additional result of the given numbers by using the benchmark method is: 1

Estimate Decimal Sums and Differences Lesson 1.6 Answer Key Question 7.

Answer:
The given numbers are: 8.12, 5.52
So,

Hence, from the above,
We can conclude that
The addition result of the given numbers by using the rounding method is: 13.6

Question 8.

Answer:
The given numbers are: 9.75, 3.47
So,

Hence, from the above,
We can conclude that
The subtraction result of the given numbers by using the rounding method is: 6.3

Practice: Copy and Solve Use rounding or benchmarks to estimate.

Question 9.
12.83 + 16.24
Answer:
The given numbers are: 12.83 and 16.24
So,

Hence, from the above,
We can conclude that
The addition result of the given numbers by using the rounding method is: 29.1

Question 10.
$26.92 – $11.13
Answer:
The given numbers are: 26,92 and 11.13
So,

Hence, from the above,
We can conclude that
The subtraction result of the given numbers by using the rounding method is: 15.8

Question 11.
9.41 + 3.82
Answer:
The given numbers are: 9.41 and 3.82
So,

Hence, from the above,
We can conclude that
The addition result of the given numbers by using the rounding method is: 13.23

H.O.T. Estimate to compare. Write < or >.

Question 12.
2.74 + 4.22 3.13 + 1.87

Answer:
The given addition expressions are: 2.74 + 4.22 and 3.13 + 1.87
So,

So,
The estimated values of the given addition expressions respectively are: 7 and 5
Hence, from the above,
We can conclude that

Practice and Homework Lesson 1.6 Answers 5th Grade Question 13.
6.25 – 2.39 9.79 – 3.84

Answer:
The given subtraction expressions are: 6.25 – 2.39 and 9.79 – 2.84
So,

So,
The estimated values of the given subtraction expressions respectively are: 4 and 7
Hence, from the above,
We can conclude that

Problem Solving

Use the table to solve 14. Show your work.

Question 14.
Representations For the week of April 4, 1964, the Beatles had the top four songs. About how long would it take to listen to these four songs?
Answer:
It is given that
For the week of April 4, 1964, the Beatles had the top four songs
Now,
The given table is:

So,
The total duration that took to listen these four songs = 2.30 + 2.50 + 2.75 + 2.00
Now,
By using the Long Addition,

Hence, from the above,
We can conclude that
The time that it took to listen to the given four songs is about 10 minutes

Use the table and estimation to solve 15-16.

Question 15.
Representations Gina had a scrambled egg, an oat bran muffin, and a cup of low-fat milk for breakfast. About how many grams of protein did Gina have at breakfast?
Answer:
It is given that
Gina had a scrambled egg, an oat bran muffin, and a cup of low-fat milk for breakfast
Now,
The given table is:

Now,
From the given table,
We can observe that
The total amount of grams of protein did Gina have at breakfast = 6.75 + 3.99 + 8.22
Now,
By using the Long Addition,

Hence, from the above,
We can conclude that
The total amount of grams of protein Gina has at breakfast is about 19 grams

Estimate Decimals Sums and Differences Lesson 1.6 Answer Key Question 16.
H.O.T. Multi-Step Pablo had a cup of shredded wheat cereal, a cup of low-fat milk, and one other item for breakfast. He had about 21 grams of protein. What was the third item Pablo had for breakfast?

Answer:
It is given that
Pablo had a cup of shredded wheat cereal, a cup of low-fat milk, and one other item for breakfast. He had about 21 grams of protein
Now,
The given table is:

Now,
Let the amount of proteins present in the third item Pable had for breakfast be x grams
So,
According to the given information,
5.56 + 8.22 + x = 21
x = 21 – (5.56 + 8.22)
x = 21 – 13.78
Now,
By using the Long Subtraction,

Hence, from the above,
We can conclude that
The third item Padro had for breakfast is: 1 Scrambled egg

Daily Assessment Task

Fill in the bubble completely to show your answer.

Question 17.
Multi-Step Marco has room for 15 more minutes of songs on his mp3 player. He has three songs to upload with the following lengths: 5.15 minutes, 3.8 minutes, and 4.4 minutes. After uploading all three songs, how many minutes will he have left?
A. 3 minutes
B. 1 minute
C. 2 minutes
D. 0 minutes
Answer:
It is given that
Marco has room for 15 more minutes of songs on his mp3 player. He has three songs to upload with the following lengths: 5.15 minutes, 3.8 minutes, and 4.4 minutes.
So,
The number of minutes Marco has left = 15 – (5.15 + 3.8 + 4.4)
= 15 – 13.35
= 1.65 minutes
≅ 2 minutes
Hence, from the above,
We can conclude that
The number of minutes Marco has left is:

Question 18.
Connect Geoff and Andrea want to combine their money to buy pizza. Geoff has $5.95. Andrea has $8.30. Which benchmarks should they use to estimate the money they have?
A. $6.00 + $8.25
B. $5.75 + $8.00
C. $6.00 + $8.50
D. $5.00 + $8.00
Answer:
It is given that
Geoff and Andrea want to combine their money to buy pizza. Geoff has $5.95. Andrea has $8.30
So,
The total amount of money they should have = $5.95 + $8.30
Now,
By using the Benchmarks,
The estimated total amount of money they should have = $6 + $8.25
Hence, from the above,
We can conclude that
The benchmarks should they use to estimate the money they have is:

Question 19.
Multi-Step Maiko wrote two songs. The first song was 2.9 minutes long. Then he added 0.4 minutes. The second song was 5.6 minutes long. Then he subtracted 1.1 minutes from it. Use benchmarks to estimate the length of each song.
A. first song: 2.5 minutes, second song: 4 minutes
B. first song: 3.5 minutes, second song: 4.5 minutes
C. first song: 3.5 minutes, second song: 7 minutes
D. first song: 2.5 minutes, second song: 5.5 minutes
Answer:
It is given that
Maiko wrote two songs. The first song was 2.9 minutes long. Then he added 0.4 minutes. The second song was 5.6 minutes long. Then he subtracted 1.1 minutes from it.
Now,
According to the given information,
The duration of the first song = 2.9 + 0.4
= 3.3 minutes
The duration of the second song = 5.6 – 1.1
= 4.5 minutes
So,
By using the Benchmarks,
The duration of the first song is: 3.5 minutes
The duration of the second song is: 4.5 minutes
Hence, from the above,
We can conclude that
The estimation of the lengths of each song by using the benchmarks is:

Texas Test Prep

Question 20.
Fran bought sneakers for $54.26 and a shirt for $34.34. If Fran started with $100, how much money does she have left?
A. $35
B. $20
C. $5
D. $80
Answer:
It is given that
Fran bought sneakers for $54.26 and a shirt for $34.34 and Fran started with $100
So,
The total amount of Fran has left = $100 – ($54.26 + 34.34)
= $100 – $88.60
= $11.40
Hence, from the above,
We can conclude that
The amount of money Fran has left is about:

Texas Go Math Grade 5 Lesson 1.6 Homework and Practice Answer Key

Use rounding or benchmarks to estimate.

Question 1.

Answer:
The given numbers are: 3.39 and 4.58
So,

Hence, from the above,
We can conclude that
The result of the given addition expression by rounded method is: 8

Question 2.

Answer:
The given numbers are: 6.77 and 3.23
So,

Hence, from the above,
We can conclude that
The result of the given addition expression by benchmarks method is:10

Question 3.

Answer:
The given numbers are: 2.09 and 5.92
So,

Hence, from the above,
We can conclude that
The result of the given addition expression by benchmarks method is: 8

Question 4.

Answer:
The given numbers are: 9.76 and 3.28
So,

Hence, from the above,
We can conclude that
The result of the given subtraction expression by rounding method is: 6.5

Additional Practice 1.6 Round Decimals Answer Key 5th Grade Question 5.

Answer:
The given numbers are: 7.36 and 0.83
So,

Hence, from the above,
We can conclude that
The result of the given subtraction expression by rounding method is: 6.5

Question 6.

Answer:
The given numbers are: 5.55 and 0.25
So,

Hence, from the above,
We can conclude that
The result of the given subtraction expression by rounding method is: 5.3

Estimate to compare. Write < or >.

Question 7.
8.23 + 2.22 3.21 + 5.89

Answer:
The given addition expressions are:
8.23 + 2.22 and 3.21 + 5.89
Now,
By using the Long Addition,

So,
The estimations of the given addition expressions are: 10.5 and 9.1
Hence, from the above,
We can conclude that

Question 8.
7.32 – 3.76 9.23 – 4.28

Answer:
The given addition expressions are:
7.32 – 3.76 and 9.23 – 4.28
Now,
By using the Long Subtraction,

So,
The estimations of the given addition expressions are: 3.6 and 5
Hence, from the above,
We can conclude that

Question 9.
$18.25 + $5.50 $10.75 + $15.00

Answer:
The given addition expressions are:
$18.25 + $5.50 and $10.75 + $15.00
So,

So,
The estimated values of the given addition expressions respectively are: 24 and 26
Hence, from the above,
We can conclude that

Question 10.
6.2 – 4.8 7.23 – 5.08

Answer:
The given addition expressions are:
6.2 – 4.8 and 7.23 – 5.08
Now,
By using the Long Subtraction,

So,
The estimations of the given addition expressions are: 1.4 and 2.1
Hence, from the above,
We can conclude that

Problem Solving

Question 11.
Li is running a race around the school track. He runs 0.327 kilometers during the first half of the race. What is the distance that he travels in the first half of the race, rounded to the nearest hundredth?
Answer:
It is given that
Li is running a race around the school track. He runs 0.327 kilometers during the first half of the race.
Now,
When we rounded to the nearest hundredth,
The distance that Li traveled around the school track will become: 0.33 kilometers
Hence, from the above,
We can conclude that
The distance that Li travels in the first half of the race, rounded to the nearest hundredth is: 0.33 kilometers

Question 12.
A video lasts 6.44 minutes. About how long does the video last, rounded to the nearest minute?
Answer:
It is given that
A video lasts 6.44 minutes
Now,
The duration of video when rounded to the nearest tenth is: 6 minutes
Hence, from the above,
We can conclude that
The video lasts  about 6 minutes when rounded to the nearest minute

Lesson Check

Fill in the bubble completely to show your answer.

Question 13.
Dylan wants to buy a book for $16.95 and a magazine for $4.50. About how much will he need to buy the items without having to round down to the nearest benchmark?
A. $15
B. $30
C. $25
D. $20
Answer:
It is given that
Dylan wants to buy a book for $16.95 and a magazine for $4.50
So,
The amount of money Dylan will need = $16.95 + $4.50
Now,
By using the Long Addition,

SO,
The amount of money Dylan will need, when rounded to the nearest benchmark is: $20
Hence, from the above,
We can conclude that
The amount of money will Dylan need to buy the items without having to round down to the nearest benchmark is:

Question 14.
Sasha downloads a song that is 4.32 minutes long and one that is 8.28 minutes long. About how many minutes will it take all of her songs to play if she already has 13.8 minutes of songs?
A. 15 minutes
B. 30 minutes
C. 20 minutes
D. 26 minutes
Answer:
It is given that
Sasha downloads a song that is 4.32 minutes long and one that is 8.28 minutes long
Now,
Let x be the total time
So,
According to the given information,
x = 13.8 + (4.32 + 8.28)
x  = 13.8 + 12.6
x = 13.8 + 12.6
x = 26.4 minutes
x ≅ 26 minutes
Hence, from the above,
We can conclude that
The number of minutes will it take all of sasha’s songs to play is:

Question 15.
Mike spends $23.40 on a present for his mom and $18.95 on a present for his dad. If Mike started with $75, how much money does he have left?
A. $30
B. $50
C. $20
D. $40
Answer:
It is given that
Mike spends $23.40 on a present for his mom and a $18.95 on a present for his dad. If Mike started with $75
So,
The amount of money does Mike spend = $75 – ($23.40 + $18.95)
= $75 – $42.35
= $32.65
≅ $30
Hence, from the above,
We can conclude that
The amount of money does Mike had left is:

Question 16.
Damelle and Raj want to combine their money to buy a soccer ball. Danielle has $7.95. Raj has $6.35. Which benchmarks should they use to estimate the money they have?
A. $7.75 + $6.50
B. $8.00 + $6.25
C. $7.50 + $6.50
D. $6.00 + $6.50
Answer:
It is given that
Damelle and Raj want to combine their money to buy a soccer ball. Danielle has $7.95. Raj has $6.35.
So,
The amount of money they have = $7.95 + $6.35
So,
The amount of money they have when rounded to the nearest benchmark = $7.75 + $6.50
Hence, from the above,
We can conclude that
The benchmarks should they use to estimate the money they have is:

Question 17.
Multi-Step Jesse, Max, and Kalel each ate 0.32 of a pizza pie. Which number shows the best estimate of how much of the pie they ate together?
A. 1.0
B. 0.5
C. 0.25
D. 0.10
Answer:
It is given that
Jesse, Max, and Kalel each ate 0.32 of a pizza pie
So,
The amount of pizza pie they ate = 0.32 + 0.32 + 0.32
= 0.64 + 0.32
= 0.96
≅ 1
Hence, from the above,
We can conclude that
The number that shows the best estimate of the amount of pie they ate together is:

Question 18.
Multi-Step George is giving two speeches at school. The first was 3.4 minutes long. Then he added 0.9 minutes to the speech. The second speech was 4.1 minutes long. Then he subtracted 1.4 minutes from it. Use benchmarks to estimate the length of each speech.
A. first speech: 4.0 minutes; second speech: 3.5 minutes
B. first speech: 4.0 minutes; second speech: 3 minutes
C. first speech: 4.5 minutes; second speech: 2.5 minutes
D. first speech: 4.5 minutes; second speech: 3.5 minutes
Answer:
It is given that
George is giving two speeches at school. The first was 3.4 minutes long. Then he added 0.9 minutes to the speech. The second speech was 4.1 minutes long. Then he subtracted 1.4 minutes from it
So,
The duration of the first speech = 3.4 + 0.9
= 4.3 minutes
The duration of the second speech = 4.1 – 1.4
= 2.7 seconds
Now,
By using the Benchmarks,
The duration of the first speech is: 4 minutes
The duration of the second speech is: 3 minutes
Hence, from the above,
We can conclude that
By uing Benchmarks, the length of each speech is:

Refer to our Texas Go Math Grade 5 Answer Key Pdf to score good marks in the exams. Test yourself by practicing the problems from Texas Go Math Grade 5 Lesson 1.3 Answer Key Place Value of Decimals.

Texas Go Math Grade 5 Lesson 1.3 Answer Key Place Value of Decimals

Unlock the Problem

The Brooklyn Battery Tunnel in New York City is 1.726 miles long. It is the longest underwater tunnel for vehicles in the United States. To understand this distance, you need to understand the place value of each digit in 1.726.

You can use a place-value chart to understand decimals. Whole numbers are to the left of the decimal point. Decimals are to the right of the decimal point. The thousandths place is to the right of the hundredths place.

The place value of the digit 6 in 1.726 is thousandths. The value of 6 in 1.726 is 6 \(\frac{1}{1,000}\) or 0.006.
Standard Form: 1.726
Word Form: one and seven hundred twenty-six thousandths
Expanded Form: 1 + 0.7 + 0.02 + 0.006

Math Talk
Mathematical Processes

Explain how the value of the last digit in a decimal can help you read a decimal.
Answer:
We know that,
Values to the left of the decimal point are greater than one.
Now,
38 means 3 tens and 8 ones.
The word name of the decimal is determined by the place value of the digit in the last place.
Ex:
In 1.3215,
The last digit (5) is in the ten-thousandth place

Try This!

Use place value to read and write decimals.

A. Standard Form: 2.35
Word Form: two and ____________
Expanded Form: 2 + ___________
Answer:
The given standard form is: 2.35
Now,
We know that,
Expanded form means writing the number so you show the value of each digit. Word form means writing the number with words instead of numerals
So,
The word from of 2.35 is: Two and Thirty-five hundredths
The Expanded form of 2.35 is: 2 + 0.3 + 0.05
Hence, from the above,
We can conclude that
The word from of 2.35 is: Two and Thirty-five hundredths
The Expanded form of 2.35 is: 2 + 0.3 + 0.05

B. Standard Form: ____________
Word Form: three and six hundred fourteen thousandths
Expanded Form: ____________ + 0.6 + ____________ + ____________
Answer:
The given word form is: Three and six hundred fourteen thousandths
Now,
We know that,
The standard form is the combination of digits written together as a single number. In the expanded form the digits of the number are split into each of the individual digits with their place value and written in expanded form
So,
The standard form of the given word form is: 3.614
The expanded form of the given word form is: 3 + 0.6 + 0.01 + 0.004
Hence, from the above,
We can conclude that
The standard form of the given word form is: 3.614
The expanded form of the given word form is: 3 + 0.6 + 0.01 + 0.004

Example

Use place value to compare.
The silk spun by a common garden spider is about 0.003 mm thick. A commonly used sewing thread is about 0.3 mm thick. How does the thickness of the spider silk and the thread compare?
Count the number of decimal place-value positions to the digit 3 in 0.3 and 0.003.
0.3 has 2 fewer decimal places than 0.003
2 fewer decimal places: 10 × 10 = 100
So,
0.3 is 100 times as much as 0.003
0.003 is 0.01 of 0.3
So, the thread is 0.01 times as thick as the garden spider’s silk.
So,
The thickness of the garden spider’s silk is 0.01 times that of the thread.

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Go Math Grade 5 Lesson 1.3 Answer Key Question 1.
Complete the place-value chart to find the value of each digit.

Answer:
The given number is: 3.524
Hence,
The completed place-value chart for the given number is:

Write the value of the underlined digit.

Question 2.
0.543
Answer:
The given number is: 0.543
So,
The representation of the given number in the place-value table is:

Hence, from the above,
We can conclude that
The value of the underlined digit in 0.543 is: 4 Hundredths

Question 3.
6.234
Answer:
The given number is: 6.234
So,
The representation of the given number in the place-value table is:

Hence, from the above,
We can conclude that
The value of the underlined digit in 6.234 is: 2 Tenths

Question 4.
3.954
Answer:
The given number is: 3.954
So,
The representation of the given number in the place-value table is:

Hence, from the above,
We can conclude that
The value of the underlined digit in 3.954 is: 4 Thousandths

Write the number in two other forms.

Go Math 5th Grade Lesson 1.3 Express as Decimal Numerals Question 5.
0.253
Answer:
The given number is: 0.253
Now,
From the given number,
We can observe that
The given number is in standard form
So,
The Word form of 0.253 is: Two hundred and fifty-three thousandths
The Expanded form of 0.253 is: 0 + 0.2 + 0.05 + 0.003
Hence, from the above,
We can conclude that
The representation of the given number in other two forms is:
The Word form of 0.253 is: Two hundred and fifty-three thousandths
The Expanded form of 0.253 is: 0 + 0.2 + 0.05 + 0.003

Question 6.
7.632
Answer:
The given number is: 7.632
Now,
From the given number,
We can observe that
The given number is in standard form
So,
The Word form of 7.632 is: Seven and six hundred and thirty-two thousandths
The Expanded form of 7.632 is: 7 + 0.6 + 0.03 + 0.002
Hence, from the above,
We can conclude that
The representation of the given number in other two forms is:
The Word form of 7.632 is: Seven and six hundred and thirty-two thousandths
The Expanded form of 7.632 is: 7 + 0.6 + 0.03 + 0.002

Problem Solving

Question 7.
Connect You can use place-value patterns to rename a decimal. Complete the chart to rename 0.3 using other place values.

Answer:
The given number is: 0.3
So,
From the given number,
We can observe that
There is 30 hundredths and 300 thousandths place value
Hence,
The complete place-value chart for 0.3 using other place values is:

Go Math Lesson 1.3 Grade 5 Answer Key Question 8.
Write Math
Explain how you know that the digit 6 in the numbers 3.675 and 3.756 does not have the same value.
Answer:
The given decimal numbers are: 3.675 and 3.756
So,
The representation of 3.675 in the place-value table is:

So,
The representation of 3.756 in the place-value table is:

Hence, from the above,
We can conclude that
The value of the digit 6 from 3.675 and 3.756 does not have the same value by using the place-value tables

Problem Solving

Use the table for 9-11.

Question 9.
What is the value of the digit 7 in New Mexico’s average annual rainfall?
Answer:
The given table is:

Now,
From the given table,
We can observe that
The average rainfall in New Mexico is: 0.372 meters
Now,
The representation of New Mexico’s rainfall in the place-value table is:

Hence, from the above,
We can conclude that
The value of the digit 7 in New Mexico’s average rainfall is: 7 hundredths

Lesson 1.3 5th Grade Go Math Understand Decimal Place Value Question 10.
The average annual rainfall in Maine is one and seventy-four thousandths meters per year. Complete the table by writing that amount in standard form.
Answer:
It is given that
The average annual rainfall in Maine is one and seventy-four thousandths meters per year
Now,
The representation of the given word form in the standard form is: 1.074 meters
Hence, from the above,
We can conclude that
The complete table with Maine’s average rainfall is:

Question 11.
Which of the states has an average annual rainfall with the least number in the thousandth place?
Answer:
From Exercise 10,
We know that,
The complete table with Maine’s average rainfall is:

Now,
When we compare the thousandths place in all the numbers,
We can observe that
0.820 has the least value in thousandths place
Hence, from the above,
We can conclude that
Wisconsin has the average rainfall with the least number in the thousandths place

Question 12.
H.O.T. Multi-Step Dan used a meter stick to measure some seedlings in his garden. One day, a corn stalk was 0.85 m tall. A tomato plant was 0.850 m. A carrot top was 0.085 m. Which plant was shortest?
Answer:
It is given that
Dan used a meter stick to measure some seedlings in his garden. One day, a corn stalk was 0.85 m tall. A tomato plant was 0.850 m. A carrot top was 0.085 m
So,
The representation of 0.85 in the place value table is:

So,
The representation of 0.850 in the place value table is:

So,
The representation of 0.085 in the place value table is:

Now,
When we compare the tenths position of the three numbers,
We can observe that
0.8 = 0.850 > 0.085
Hence, from the above,
We can conclude that
A carrot top has the shortest length

Go Math 5th Grade Lesson 1.3 Understand Decimal Place Value Question 13.
H.O.T. What’s the Error? Damian wrote the number four and twenty-three thousandths as 4.23. Describe and correct his error.
Answer:
It is given that
Damian wrote the number four and twenty-three thousandths as 4.23
Now,
The representation of the given word form in the standard form is: 4.023
Now,
The representation of Damian’s number in the place value chart is:

Now,
The actual representation of Damian’s number in the place value chart is:

Now,
When we compare the place value charts,
We can observe that
Damian’s number does not have thousandths place but he mentioned thousandths in his number’s word form
Hence, from the above,
We can conclude that
Damian’s error is: The misrepresentation of the word form into the correct standard form
The correct representation of Damian’s number is: 4.023

Daily Assessment Task

Fill in the bubble completely to show your answer.

Question 14.
Although the average turkey sandwich in Turkeyton costs 3.456 dollhorts, the cost of a gourmet turkey sandwich at Sandwich Palace is 5.032 dollhorts. What does the 3 in 5.032 dollhorts represent?
A. 3 dollhorts
B. 3 tenths of a dollhort
C. 3 hundredths of a dollhort
D. 4 hundredths of a dollhort
Answer:
It is given that
Although the average turkey sandwich in Turkeyton costs 3.456 dollhorts, the cost of a gourmet turkey sandwich at Sandwich Palace is 5.032 dollhorts.
Now,
The representation of 5.032 in the place value chart is:

Hence, from the above,
We can conclude that
The digit 3 in 5.032 represents:

Question 15.
Use Math Language One meter is equal to about 3.281 feet. What is the word form of this decimal?
A. three and twenty-eight hundredths and one thousandth
B. three and two hundred eighty-one thousandths
C. three and two tenths and eighty-one thousandths
D. three point two eighty-one
Answer:
It is given that
One meter is equal to about 3.281 feet
Hence, from the above,
We can conclude that
The representation of 3.281 feet in the word form is:

Go Math Grade 5 Lesson 1.3 Understand Decimal Place Value Question 16.
Multi-Step Sam had a piece of ribbon 8.469 cm long. He cut off exactly one-tenth of a cm from the ribbon. Then he cut off two-hundredths of a cm. How long is the ribbon now?
A. 8.169 cm
B. 8.269 cm
C. 8.349 cm
D. 8.466 cm
Answer:
It is given that
Sam had a piece of ribbon 8.469 cm long. He cut off exactly one-tenth of a cm from the ribbon. Then he cut off two-hundredths of a cm
Now,
The present length of the ribbon = 8.469 – 0.1 – 0.02
= 8.349 cm
Hence, from the above,
We can conclude that
The length of the ribbon now is:

Texas Test Prep

Question 17.
In 24.736, which digit is in the thousandths place?
A. 4
B. 7
C. 3
D. 6
Answer:
The given number is: 24.736
Now,
The representation of 24.736 in the place value chart is:

Hence, from the above,
We can conclude that
The digit that is present in thousandths place in 24.736 is:

Texas Go Math Grade 5 Lesson 1.3 Homework and Practice Answer Key

Rename 0.6 using other place values.

Question 1.

Answer:
The given number is: 0.6
Now,
When we observe the given number,
We can say that there is 60 hundredths and 600 thousandths in the given number
Hence,
The complete table of 0.6 using other place values is:

Rename 0.2 using other place values.

Question 2.

Answer:
The given number is: 0.2
Now,
When we observe the given number,
We can say that there is 20 hundredths and 200 thousandths in the given number
Hence,
The complete table of 0.6 using other place values is:

Write the number in two other forms.

Question 3.
0.632
Answer:
The given number is: 0.632
Now,
From the given number,
We can observe that
The given number is in standard form
So,
The Word form of 0.632 is: Six hundred and thirty-two thousandths
The Expanded form of 0.253 is: 0 + 0.6 + 0.03 + 0.002
Hence, from the above,
We can conclude that
The representation of the given number in other two forms is:
The Word form of 0.632 is: Six hundred and thirty-two thousandths
The Expanded form of 0.253 is: 0 + 0.6 + 0.03 + 0.002

Question 4.
4.293
Answer:
The given number is: 4.293
Now,
From the given number,
We can observe that
The given number is in standard form
So,
The Word form of 4.293 is: Four and two hundred and ninety-three thousandths
The Expanded form of 4.293 is: 4 + 0.2 + 0.09 + 0.003
Hence, from the above,
We can conclude that
The representation of the given number in other two forms is:
The Word form of 4.293 is: Four and two hundred and ninety-three thousandths
The Expanded form of 4.293 is: 4 + 0.2 + 0.09 + 0.003

Problem Solving

Question 5.
A weather reporter reads aloud the number of meters of snowfall. The amount is 0.103 meter. What words should the reporter use on the air to explain the amount of snowfall?
Answer:
It is given that
A weather reporter reads aloud the number of meters of snowfall. The amount is 0.103 meter
So,
The representation of 0.103 in the Word form is: One hundred and three thousandths
Hence, from the above,
We can conclude that
The words the reporter should use on the air to explain the amount of snowfall is: One hundred and three-thousandths meters

Homework and Practice 1.3 Decimals to Thousandths Answer Key Question 6.
Amil says that the snow on the ground is 5 × \(\frac{1}{100}\) meter high. How can this number be expressed as a decimal?
Answer:
It is given that
Amil says that the snow on the ground is 5 × \(\frac{1}{100}\) meter high
Now,
5 × \(\frac{1}{100}\) = 5 × 0.01
= 0.05 meters
Hence, from the above,
We can conclude that
The representation of 5 × \(\frac{1}{100}\) as a decimal is: 0.05

Lesson Check

Fill in the bubble completely to show your answer.

Question 7.
In the number 5.923, which number is in the thousandths place?
A. 5
B. 9
C. 2
D. 3
Answer:
The given number is: 5.923
Now,
The representation of 5.923 in the place-value chart is:

Hence, from the above,
We can conclude that
The digit that is present in the thousandths place in the given number is:

Question 8.
Which number has the same value as 7 × \(\frac{1}{1,000}\) ?
A. 0.7
B. 0.007
C. 0.07
D. 7.0
Answer:
The given expression is: 7 × \(\frac{1}{1,000}\)
Now,
7 × \(\frac{1}{1,000}\) = 7 × 0.001
= 0.007
Hence, from the above,
We can conclude that
The number that has the same value as 7 × \(\frac{1}{1,000}\) is:

Question 9.
Multi-Step Which value correctly represents the number that is 10 times as great as five hundred eighty-one thousandths?
A. 5.81
B. 581
C. 0.581
D. 500.81
Answer:
The given Word form is: five hundred eighty-one thousandths
So,
The representation of the given Word form in the standard form is: 0.581
Now,
According to the given information,
10 × 0.581 = 5.81
Hence, from the above,
We can conclude that
The correct value that represents the number that is 10 times as great as five hundred eighty-one thousandths is:

Question 10.
Which number has the same value as 3 × \(\frac{1}{10}\)?
A. 0.003
B. 3.10
C. 0.3
D. 0.03
Answer:
The given expression is: 3 × \(\frac{1}{10}\)
So,
3 × \(\frac{1}{10}\) = 3 × 0.1
= 0.3
Hence, from the above,
We can conclude that
The number that has the same value as 3 × \(\frac{1}{10}\) is:

Question 11.
Jack’s baby sister crawled three and five tenths meters today. Which number represents that distance?
A. 3.5
B. 3.05
C. 3.510
D. 3.005
Answer:
It is given that
Jack’s baby sister crawled three and five tenths meters today
So,
The representation of the given Word form in the Standard form is: 3.5 meters
Hence, from the above,
We can conclude that
The number that represents the Jack’s baby sister’s crawling distance is:

Question 12.
Multi-Step Leon measured the distance each of his four toy cars rolled across the floor. Which car measurement that has a 4 in the thousandths place represents the shortest distance rolled?
A. 0.741 meter
B. 0.714 meter
C. 0.224 meter
D. 0.213 meter
Answer:
It is given that
Leon measured the distance each of his four toy cars rolled across the floor.
So,
From the given options,
We can observe that
There are two options that have 4 in the thousandths place.
They are: 0.714 meters and 0.224 meters
Now,
From the above numbers,
We can observe that
The shortest number is: 0.224 meters
Hence, from the above,
We can conclude that
The car measurement that has a 4 in the thousandths place and represents the shortest distance rolled is:

Refer to our Texas Go Math Grade 5 Answer Key Pdf to score good marks in the exams. Test yourself by practicing the problems from Texas Go Math Grade 5 Lesson 1.1 Answer Key Properties.

Texas Go Math Grade 5 Lesson 1.1 Answer Key Properties

Unlock the Problem

The table shows the number of bones in several parts of the human body. What is the total number of bones in the ribs, the skull, and the spine?

To find the sum of addends using mental math, you can use the Commutative and Associative Properties.

Use properties to find 24 + 28 + 26.
24 + 28 + 26 = 28 + ________ + 26 Use the ______________ Property to reorder the addends.
= 28 + (24 + ________) Use the _____________ Property to group the addends.
= 28 + ________ Use mental math to add.
= ________
So, there are ________ bones in the ribs, the skull, and the spine.
Answer:
It is given that
The table shows the number of bones in several parts of the human body
Now,
The given table is:

Now,
From the given table,
We can observe that
The total number of bones in the ribs, the skull, and the spine = 24 + 28 + 26
Now,


Hence, from the above,
We can conclude that
The total number of bones in the ribs, the skull and the spine are: 78 bones

Math Talk
Mathematical Processes

Explain why grouping 24 and 26 makes the problem easier to solve.
Answer:
The given table is:

Now,
From the given table,
We can observe that
The number of bones in the ribs is: 24 bones
The number of bones in the skull is: 28 bones
The number of bones in the Spine is: 26 bones
Now,
If we add 24 and 28, then we will get 52
If we add 28 and 26, then we will get 54
If we add 24 and 26, then we will get 50
Now,
When we observe the addition values of 24 and 28, and 28 and 26, it will be difficult to do mental calculation
But,
When we observe the addition value of 24 and 26, the result is 50 which is the multiple of 10. So, it will be easy to do mental calculation
Hence, from the above,
We can conclude that
The grouping of 24 and 26 makes the problem easier because the result of grouping will be the multiple of 10

Example

Use the Distributive Property to find the product.
One Way
Use addition.
8 × 59 = 8 × (50 + 9) Use a multiple of 10 to write 59 as a sum.
= (8 × 50) + (8 × 9)    Use the Distributive Property.
= 400 + 72                 Use mental math to multiply.
= 472                         Use mental math to add.

Another Way
Use subtraction.
8 × 59 = 8 × (60 – 1) Use a multiple of 10 to write 59 as a difference.
= (8 × 60) – (8 × 1)    Use the Distributive Property.
= 480 – 8                   Use mental math to multiply.
= 472                         Use mental math to subtract.

Math Talk
Mathematical Processes

Explain how you could find the product 3 × 299 by using mental math.
Answer:
The given multiplication expression is: 3 × 299
Now,
One way:
3 × 299 = 3 × (300 – 1)
= (3 × 300) – (3 × 1)
= 900 – 3
= 897
Other way:
3 × 299 = 3 × (200 + 99)
= (3 × 200) + (3 × 99)
= 600 + 297
= 897
Hence, from the above,
We can conclude that
The value for the given multiplication expression is: 897

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Go Math 5th Grade Lesson 1.1 Answer Key Question 1.
Use properties to find 4 × 23 × 25.
23 × ______ × 25 __________ Property of Multiplication
23 × (_______ × ________) __________ Property of Multiplication
23 × _______
________
Answer:
The given multiplication expression is: 4 × 23 × 25
Now,
The given multiplication can also be written as: 23 × 4 × 25
Now,

Use properties to find the sum or product.

Question 2.
89 + 27 + 11
Answer:
The given addition expression is: 89 + 27 + 11
Now,
89 + 27 + 11
= 27 + 89 + 11                         Commutative Property of Addition
= 27 + (89 + 11)                       Associative Property of Addition
= 27 + 100                                Mental Math
= 127
Hence, from the above,
We can conclude that
The value of the given addition expression is: 127

Question 3.
9 × 52
Answer:
The given multiplication expression is: 9 × 52
Now,
9 × 52
= 9 × (50 + 2)                           Use a multiple of 10 to write 52 as a sum
= (9 × 50) + (9 × 2)                   Use the Distributive Property
= 450 + 18                                 Use Mental Math to calculate
= 468
Hence, from the above,
We can conclude that
The value of the given multiplication expression is: 468

Question 4.
107 + 0 + 39 + 13
Answer:
The given addition expression is: 100 + 0 + 39 + 13
Now,
100 + 0 + 39 + 13
= 100 + 13 + 39 + 0                 Commutative Property of addition
= 100 + 13 + 39                        Identity Property of Addition
= 100 + (39 + 13)                     Associative Property of Addition
= 52 + 100                                Mental Math
= 152
Hence, from the above,
We can conclude that
The value of the given addition expression is: 152

Problem Solving

Practice: Copy and Solve Use properties to find the sum or product.

5th Grade Go Math Lesson 1.1 Answer Key Question 5.
3 × 78
Answer:
The given multiplication expression is: 3 × 78
Now,
3 × 78
= 3 × (70 + 8)                           Use a multiple of 10 to write 52 as a sum
= (3 × 70) + (3 × 8)                   Use the Distributive Property
= 210 + 24                                 Use Mental Math to calculate
= 234
Hence, from the above,
We can conclude that
The value of the given multiplication expression is: 234

Question 6.
4 × 60 × 5
Answer:
The given multiplication expression is: 4 × 60 × 5
Now,
4 × 60 × 5
= 4 × 5 × 60                       Commutative Property of Multiplication
= (4 × 5) × 60                     Associative Property of Multiplication
= 20 × 60                            Use Mental Math
= 1,200
Hence, from the above,
We can conclude that
The value of the given multiplication expression is: 1,200

Question 7.
21 + 25 + 39 + 5
Answer:
The given addition expression is: 21 + 25 + 39  + 5
Now,
21 + 25 + 39 + 5
= 21 + 25 + 5 + 39                Commutative Property of Addition
= (21 + 39) + 30                    Associative Property of Addition
= 60 + 30                               Mental Math
= 90
Hence, from the above,
We can conclude that
The value of the given addition expression is: 90

Complete the equation, and tell which property you used.

Question 8.
11+ (19 + 6) = (11 + ______) + 6
Answer:
The given addition expression is:
11+ (19 + 6) = (11 + ______) + 6
Now,
We know that,
According to the Associative Property of Addition,
If the grouping of addends changes, the sum stays the same
So,
The completed equation is:
11+ (19 + 6) = (11 + 19) + 6
Hence, from the above,
We can conclude that
The completed equation is:
11+ (19 + 6) = (11 + 19) + 6
The property used is: Associative Property of Addition

Question 9.
25 + 14 = _______ + 25
Answer:
The given addition expression is:
25 + 14 = _______ + 25
Now,
We know that,
According to the Commutative Property of Addition,
If the order of addends changes, the sum stays the same
So,
The completed equation is:
25 + 14 = 14 + 25
Hence, from the above,
We can conclude that
The completed equation is:
25 + 14 = 14 + 25
The property used is: Commutative Property of Addition

Lesson 1.1 Reteach Answer Key Go Math Grade 5 Question 10.
H.O.T. Show how you can use the Distributive Property to rewrite and find (32 × 6) + (32 × 4).
Answer:
The given expression is:
(32 × 6) + (32 × 4)
Now,
According to the Distributive Property,
Multiplying a sum by a number is the same as multiplying each addend by the number and then adding the products
So,
(32 × 6) + (32 × 34)
= 32 × (6 + 34)
= 32 × 40
= 1,280
Hence, from the above,
We can conclude that
The value of the given expression is: 1,280

Problem-Solving

Question 11.
Multi-Step Three friends at a restaurant cost $13, $14, and $11. Use parentheses to write two different expressions and find how much the friends spent in all. Which property does your pair of expressions demonstrate?
Answer:
It is given that
Three friends meals at a restaurant cost $13, $14, and $11
Now,
The total amount of money the three friends spent in all = $13 + $14 + $11
= ($13 + $14) + $11                            Associative Property of Addition
= $13 + ($11 + $14)                            Associative Property of Addition
= $13 + $25
= $38
Hence, from the above,
We can conclude that
The above pair of expressions demonstrate “Associative Property of Addition”
The total amount of money the three friends spent in all is: $38

Question 12.
Multi-Step Jacob is designing an aquarium for a doctor’s office. He plans to buy 6 red blond guppies, 1 blue neon guppy, and 1 yellow guppy. The table shows the price list for the guppies. How much will the guppies for the aquarium cost?

Answer:
It is given that
Jacob is designing an aquarium for a doctor’s office. He plans to buy 6 red blond guppies, 1 blue neon guppy, and 1 yellow guppy. The table shows the price list for the guppies
Now,
The given table is:

Now,
From the given table,
We can observe that
The total cost of guppies for the aquarium = 6 × (The cost of each red-blond guppy) + 1 × (The cost of each blue neon guppy) + 1 × (The cost of each yellow guppy)
= (6 × $22) + (1 × $11) + (1 × $19)
= $132 + $11 + $19
= $162
Hence, from the above,
We can conclude that
The total cost of guppies for the aquarium is: $162

Go Math Grade 5 Lesson 1.1 Answer Key Question 13.
H.O.T. Sense or Nonsense? Julie wrote (15 – 6) – 3 = 15 – (6 – 3). Is Julie’s equation sense or nonsense? Do you think the Associative Property works for subtraction? Explain.

Answer:
It is given that
Julie wrote (15 – 6) – 3 = 15 – (6 – 3)
Now,
The given expression is:
(15 – 6) – 3 = 15 – (6 – 3)
Now,
We know that,
According to the Associative Property,
If the groupings of addends changes, the sum stays the same
Now,
(15 – 6) – 3 = 9 – 3 = 6
15 – (6 – 3) = 15 – 3 = 12
So,
(15 – 6) – 3 ≠ 15 – (6 – 3)
Hence, from the above,
We can conclude that
Julie’s equation is nonsense
“Associative Property” does not work for subtraction

Daily Assessment Task

Fill in the bubble completely to show your answer.

Question 14.
Connect In a flea circus, 12 fleas pull carts, 23 fleas ride on tiny bicycles, and 18 fleas sit on seesaws. You can add 12 + 23 + 18 to find the total number of fleas. Which equation shows the Commutative Property of Addition?
A. 12 + 23 + 18 = 23 + 12 + 18
B. 12 + 23 + 18 = (12 + 23) + 18
C. 12 + 23 + 18 = 12 + (23 + 18)
D. 12 + 23 + 18 = 35 + 18
Answer:
It is given that
In a flea circus, 12 fleas pull carts, 23 fleas ride on tiny bicycles, and 18 fleas sit on seesaws. You can add 12 + 23 + 18 to find the total number of fleas
Now,
We know that,
According to the Commutative Property of Addition,
If the order of addends changes, the sum stays the same
Hence, from the above,
We can conclude that
The equation that shows the Commutative Property of Addition for the given equation is:

Question 15.
Use Symbols Complete the equation.
15 + (25 + 27) = (15 + ______) + 27
A. 40
B. 25
C. 10
D. 52
Answer:
The given equation is:
15 + (25 + 27) = (15 + ______) + 27
Now,
We know that,
According to the Associative Property of Addition,
If the groupings of addends changes, the sum stays the same
So,
15 + (25 + 27) = (15 + 25) + 27
Hence, from the above,
We can conclude that
The symbol that will be used to complete the given equation is:

Go Math Lesson 1.1 Grade 5 Answer Key Question 16.
Multi-Step Nick read 24 pages in his book one day, 29 pages the next day, and 16 pages on the third day. Megan read 26 pages for three days in a row. Who read more pages? How many more?
A. Megan read 9 more pages.
B.Megan read 11 more pages.
C. Nick read 9 more pages.
D. Megan read 11 more pages.
Answer:
It is given that
Nick read 24 pages in his book one day, 29 pages the next day, and 16 pages on the third day. Megan read 26 pages for three days in a row
Now,
The total number of pages read by Nick = (The number of pages read by Nick on the first day) + (The number of pages read by Nick on the second day) + (The number of pages read by Nick on the third day)
= 24 + 29 + 16
= 40 + 29
= 69 pages
Now,
The total number of pages read by Megan = 26 + 26 + 26
= 78 pages
Now,
The difference between the number of pages read by Megan and Nick respectively = 78 – 69
= 9 pages
Hence, from the above,
We can conclude that

Texas Test Prep

Question 17.
Canoes rent for $29 per day. Which expression can be used to find the cost in dollars of renting 6 canoes for a day?
A. (6 + 20) + (6 + 9)
B. (6 × 20) + (6 × 9)
C. (6 + 20) × (6 + 9)
D. (6 × 20) × (6 × 9)
Answer:
It is given that
Canoes rent for $29 per day
Now,
The amount in dollars for renting 6 canoes for a day = (The number of canoes) × (The amount of rent in dollars for canoes per day)
= 6 × $29
= 6 × ($20 + $9)
= (6 × $20) + (6 × $9)
Hence, from the above,
We can conclude that
The expression that can be used to find the cost in dollars of renting 6 canoes for a day is:

Texas Go Math Grade 5 Lesson 1.1 Homework and Practice Answer Key

Use properties to find the sum or product.

Question 1.
4 × 47
Answer:
The given multiplication expression is: 4 × 47
Now,
4 × 47
= 4 × (40 + 7)                  Write 47 as a sum of a multiple of 10 and a number
= (4 × 40) + (4 × 7)          Use the Distributive Property
= 160 + 28                        Use Mental Math
= 188
Hence, from the above,
We can conclude that
The value for the given multiplication expression is: 188

Go Math Lesson 1.1 5th Grade Answer Key Question 2.
3 × 40 × 6
Answer:
The given multiplication expression is: 3 × 40 × 6
Now,
3 × 40 × 6
= (3 × 6) × 40                              Use the Associative Property of Multiplication
= 18 × 40
= 720
Hence, from the above,
We can conclude that
The value for the given multiplication expression is: 720

Question 3.
34 + 21 + 84 + 7
Answer:
The given addition expression is: 34 + 21 + 84 + 7
Now,
34 + 21 + 84 + 7
= 34 + 7 + 21 + 84                           Commutative Property of Addition
= 41 + 105
= 146
Hence, from the above,
We can conclude that
The value for the given addition expression is: 146

Complete the equation, and tell which property you used.

Question 4.
103 + 21 = _______ + 103
Answer:
The given equation is:
103 + 21 = _______ + 103
Now,
We know that,
According to the Commutative Property of Addition,
If the order of addends changes, the sum stays the same
So,
The completed equation is:
103 + 21 = 21 + 103
Hence, from the above,
We can conclude that
The completed equation is:
103 + 21 = 21 + 103
The property used to complete the given equation is: Commutative Property of Addition

Question 5.
16 + (70 + 8) = (16 + ______) + 8
Answer:
The given equation is:
16 + (70 + 8) = (16 + ______) + 8
Now,
We know that,
According to the Associative Property of Addition,
If the grouping of addends changes, the sum stays the same
So,
The completed equation is:
16 + (70 + 8) = (16 + 70) + 8
Hence, from the above,
We can conclude that
The completed equation is:
16 + (70 + 8) = (16 + 70) + 8
The property used to complete the given equation is the associative Property of Addition

Problem-Solving

Lesson 1.1 5th Grade Math Answer Key Question 6.
Maggie’s class bought three books at the book fair. The books cost $15, $9, and $12. Use parentheses to write two different expressions and find how much the class spent in all. Which property does the pair of expressions demonstrate?
Answer:
It is given that
Maggie’s class bought three books at the book fair. The books cost $15, $9, and $12
Now,
The total amount of money the class spent in all = $15 + $9 + $12
= ($15 + $9) + $12                            Associative Property of Addition
= $15 + ($9 + $12)                            Associative Property of Addition
= $15 + $21
= $36
Hence, from the above,
We can conclude that
The above pair of expressions demonstrate “Associative Property of Addition”
The total amount of money the class spent in all is: $36

Question 7.
Leo’s basketball team makes 16 points in the first game, 22 points in the second game, and 18 points in the third game. Use the Associative Property of Addition to show two different ways Leo can group the addends to show the total number of points.
Answer:
It is given that
Leo’s basketball team makes 16 points in the first game, 22 points in the second game, and 18 points in the third game
Now,
The total number of points earned = (The number of points the team makes in the first game) + (The number of points the team makes in the second game) + (The number of points the team makes in the third game)
= $16 + $22 + $18
= ($16 + $22) + $18                          Associative Property of Addition
= $16 + ($22 + $18)                          Associative Property of Addition
= $16 + $40
= $56
Hence, from the above,
We can conclude that
The above pair of expressions demonstrate “Associative Property of Addition”
The total number of points earned is: $56

Lesson Check

Fill in the bubble completely to show your answer.

Question 8.
Which property is shown by the following equation?
44 + (13 + 11) = (44 + 13) + 11
A. Commutative Property of Addition
B. Associative Property of Addition
C. Identity Property of Addition
D. Distributive Property
Answer:
The given equation is:
44 + (13 + 11) = (44 + 13) + 11
Hence, from the above,
We can conclude that
The property that is showed by the given equation is:

Question 9.
Janel rakes leaves for $8 per lawn. Which expression can be used to find the amount she earns if she rakes 14 lawns?
A. (8 + 10) × (8 + 4)
B. (8 + 10) + (8 + 4)
C. (8 × 10) × (8 × 4)
D. (8 × 10) + (8 × 4)
Answer:
It is given that
Janel rakes leaves for $8 per lawn
So,
The total amount of money she earned if she rakes 14 lawns = 14 × $8
= $8 × (10 + 4)
= (10 × $8) + (4 × $8)
Hence, from the above,
We can conclude that
The expression that can be used to find the amount she earns if she rakes 14 lawns is:

Go Math 5th Grade Answers Lesson 1.1 Question 10.
Sara bought 7 tickets to the school play. Each ticket costs $11. To find the total cost in dollars, she added the product 7 × 1 to the product 7 × 10, for a total of $77. Which property did Sara use?
A. Commutative Property of Multiplication
B. Associative Property of Multiplication
C. Identity Property of Multiplication
D. Distributive Property
Answer:
It is given that
Sara bought 7 tickets to the school play. Each ticket costs $11. To find the total cost in dollars, she added the product 7 × 1 to the product 7 × 10, for a total of $77
Now,
The total cost of tickets in dollars = 7 × $11
= 7 × (10 + 1)
= (7 × 1) + (7 × 10)
Hence, from the above,
We can conclude that
The property did Sara used is:

Question 11.
There are 6 granola bars in each box. Which expression can be used to find the number of granola bars in a dozen boxes?
A. (6 + 10) × (6 + 2)
B. (6 × 10) + (6 × 2)
C. (6 + 10) + (6 + 2)
D. (6 × 10) × (6 × 2)
Answer:
It is given that
There are 6 granola bars in each box
So,
The number of granola bars in a dozen boxes = 6 × 12
= 6 × (10 + 2)
= (6 × 10) + (6 × 2)
Hence, from the above,
We can conclude that
The expression that can be used to find the number of granola bars in a dozen boxes is:

Question 12.
Multi-Step Kara’s garden has 4 sections. Each section has 7 rows of carrots and 9 rows of celery. How many rows are in the garden?
A. 64
B. 56
C. 20
D. 36
Answer:
It is given that
Kara’s garden has 4 sections. Each section has 7 rows of carrots and 9 rows of celery
Now,
The total number of rows in each section = 7 + 9
= 16 rows
So,
The total number of rows in 4 sections = 4 × 16
= 64 rows
Hence, from the above,
We can conclude that
The total number of rows that are present in the garden is:

Question 13.
Multi-Step Max’s school bought boxes of pencils. Each box holds 12 pencils. The fourth-grade classes need 25 boxes. The fifth-grade classes need 22 boxes. How many pencils are needed by both grades?
A. 300
B. 264
C. 564
D. 600
Answer:
It is given that
Max’s school bought boxes of pencils. Each box holds 12 pencils. The fourth-grade classes need 25 boxes. The fifth-grade classes need 22 boxes
Now,
The total number of boxes = 25 + 22
= 47 boxes
So,
The number of pencils present in 47 boxes = 47 × 12
= (40 + 7) × (10 + 2)
= (40 × 10) + (40 × 2) + (7 × 10) + (7 × 2)
= 400 + 80 + 70 + 14
= 564 pencils
Hence, from the above,
We can conclude that
The number of pencils that are needed by both grades is:

Refer to our Texas Go Math Grade 5 Answer Key Pdf to score good marks in the exams. Test yourself by practicing the problems from Texas Go Math Grade 5 Lesson 1.2 Answer Key Thousandths.

Texas Go Math Grade 5 Lesson 1.2 Answer Key Thousandths

Investigate

Materials: color pencils: straightedge

Thousandths are smaller parts than hundredths. If one hundredth is divided into ten equal parts, each part is one thousandth.
Use the model at the right to show tenths, hundredths, and thousandths.

A. Divide the larger square into 10 equal columns or rectangles. Shade one rectangle. What part of the whole is the shaded rectangle? Write that part as a decimal and a fraction.
Answer:
It is given that
Divide the larger square into 10 equal columns or rectangles. Shade one rectangle
Now,
The representation of the larger square with one shaded rectangle is:

Now,
From the given figure,
We can observe that
The shaded part is one-hundredth of the whole part
So,
The representation of one-hundredth as a fraction is: \(\frac{1}{100}\)
The representation of one-hundredth as a decimal is: 0.01
Hence, from the above,
We can conclude that
The shaded part is one-hundredth of the whole part
The representation of one-hundredth as a fraction is: \(\frac{1}{100}\)
The representation of one-hundredth as a decimal is: 0.01

B. Divide each rectangle into 10 equal squares. Use a second color to shade in one of the squares. What part of the whole is the shaded square? Write that part as a decimal and a fraction.
Answer:
It is given that
Divide each rectangle into 10 equal squares. Use a second color to shade in one of the squares
Now,
The representation of the square that is divided into 10 squares with one shaded part is:

Now,
From the given figure,
We can observe that
The shaded part of square is one-tenth of the whole square
So,
The representation of one-tenth as a fraction is: \(\frac{1}{10}\)
The representation of one-tenth as a decimal is: 0.1
Hence, from the above,
We can conclude that
The shaded part is one-hundredth of the whole part
The representation of one-tenth as a fraction is: \(\frac{1}{10}\)
The representation of one-tenth as a decimal is: 0.1

C. Divide the enlarged hundredths square into 10 equal columns or rectangles. If each hundredths square is divided into ten equal rectangles, how many parts will the model have?
Answer:
It is given that
Divide the enlarged hundredths square into 10 equal columns or rectangles and each hundredths square is divided into ten equal rectangles
Now,
From the given information,
We can observe that
The total number of parts the model will have = \(\frac{100}{10}\)
= 10
Hence, from the above,
We can conclude that
The total number of parts the given model is: 10 parts

Use a third color to shade one rectangle of the enlarged hundredths square. What part of the whole is the shaded rectangle? Write that part as a decimal and a fraction.
Answer:
It is given that
Use a third color to shade one rectangle of the enlarged hundredths square
Now,
The representation of the figure with the given information is:

Now,
From the given figure,
We can observe that
The whole part that is the shaded rectangle = \(\frac{1}{100}\) × \(\frac{1}{10}\)
= \(\frac{1 × 1}{100 × 10}\)
= \(\frac{1}{1,000}\)
So,
The representation of the whole part that is the shaded rectangle in the form of a fraction is: \(\frac{1}{1,000}\)
The representation of the whole part that is the shaded rectangle in the form of a decimal number is: 0.001
Hence, from the above,
We can conclude that
The representation of the whole part that is the shaded rectangle in the form of a fraction is: \(\frac{1}{1,000}\)
The representation of the whole part that is the shaded rectangle in the form of a decimal number is: 0.001

Math Talk
Mathematical Processes
There are 10 times as many hundredths as there are tenths. Explain how the model shows this.
Answer:
It is given that
There are 10 times as many hundredths as there are tenths
Hence, from the above,
Hence,
The representation of the model according to the given information is:

Make Connections

The relationship of a digit in different place-value positions is the same with decimals as it is with whole numbers. You can use your understanding of place-value patterns and a place-value chart to write decimals that are 10 times as much as or \(\frac{1}{10}\) of any given decimal.

Use the steps below to complete the table.

STEP 1 Write the given decimal in a place-value chart.
STEP 2 Use the place-value chart to write a decimal that is 10 times as much as the given decimal.
STEP 3 Use the place-value chart to write a decimal that is \(\frac{1}{10}\) 0f the given decimal.

Answer:
Let the decimal number be 0.0x (or) 0.x
Now,
According to the given steps,
10 times of 0.0x is: 0.x
\(\frac{1}{10}\) of 0.0x is: 0.00x
Now,
According to the given steps,
10 times of 0.x is: x
\(\frac{1}{10}\) of 0.x is: 0.0x
So,
The completed table by using the above steps is:

Hence, from the above,
We can conclude that
The completed table is:

Math Talk
Mathematical Processes

Describe the pattern you see when you move one decimal place value to the right and one decimal place value to the left.
Answer:
Let the decimal number be 0.0x (or) 0.x
Now,
When you move one decimal place to the right,
The value of 0.0x will be: 0.00x
The value of 0.x will be: 0.0x
Now,
When you move one decimal place to the left,
The value of 0.0x will be: 0.x
The value of 0.x will be: x
Hence, from the above,
We can conclude that
a.
When you move one decimal place to the right,
The value of 0.0x will be: 0.00x
The value of 0.x will be: 0.0x
b.
When you move one decimal place to the left,
The value of 0.0x will be: 0.x
The value of 0.x will be: x

Share and Show

Place Value of Whole Numbers Lesson 1.2 Go Math Grade 5 Answer Key Question 1.
Write the decimal shown by the shaded part of the model.

Answer:
The given figure is:

Now,
From the given figure,
We can observe that
The total number of parts are: 100
The number of shaded parts is: 1
So,
The representation of the shaded part from the whole part is: \(\frac{1}{100}\)
Now,
The representation of the shaded part from the whole part in the form of a decimal number is: 0.01
Hence, from the above,
We can conclude that
The representation of the shaded part from the whole part in the form of a decimal number is: 0.01

Use place-value patterns to complete the table.

Answer:
Let the decimal number be 0.0x (or) 0.x
Now,
10 times of 0.0x is: 0.x
\(\frac{1}{10}\) of 0.0x is: 0.00x
Now,
10 times of 0.x is: x
\(\frac{1}{10}\) of 0.x is: 0.0x
Hence,
The completed table by using the place-value patterns is:

Problem Solving

Use the table for 6-9.

Go Math Grade 5 Answer Key Lesson 1.2 Place Value of Whole Numbers Question 6.
What is the value of the digit 2 in the carpenter bee’s length?
Answer:
The given table is:

Now,
From the given table,
We can observe that
The length of the carpenter Bee is: 0.025 m
Now,
The representation of the place-value table for the length of the carpenter Bee is:

So,
From the given table,
We can observe that
The value of the digit 2 is: 0.02
Hence, from the above,
We can conclude that
The value of the digit 2 in the Carpenter Bee’s length is: 0.02

Question 7.
If you made a model of a bumblebee that was 10 times as large as the actual bee, how long would the model be in meters? Write your answer as a decimal.
Answer:
It is given that
you made a model of a bumblebee that was 10 times as large as the actual bee
Now,
The given table is:

Now,
From the given table,
We can observe that
The length of the Bumblebee is: 0.019 meters
So,
According to the given information,
(The length of the bumble bee) × 10 = The length of the model of bumblebee
= 0.019 × 10
= 0.19 meters
Hence, from the above,
We can conclude that
The length of the model of Bumble bee is: 0.19 meters

Question 8.
Record The sweat bee’s length is 6 thousandths of a meter. Complete the table by recording the sweat bee’s length.
Answer:
It is given that
The sweat bee’s length is 6 thousandths of a meter
Now,
The given table is:

Now,
According to the given information,
The length of the sweat bee is: 0.006 meters
Now,
The representation of the length of sweat bee in the place value table is:

Hence, from the above,
We can conclude that
The completed table with sweat bee’s length is:

Go Math Grade 5 Lesson 1.2 Answer Key Question 9.
H.O.T. An atlas beetle is about 0.14 of a meter long. How does the length of the atlas beetle compare to the length of a leaf cutting bee?

Answer:
It is given that
An atlas beetle is about 0.14 of a meter long
Now,
The given table is:

Now,
From the given table,
We can observe that
The length of the Leaf cutting Bee is: 0.014 meters
Now,
According to the given information,
The length of an Atlas Bee is 10 times as long as the length of the Leaf cutting bee
Hence, from the above,
We can conclude that
The length of an Atlas Bee is 10 times as long as the length of the Leaf cutting bee

Question 10.
H.O.T. Multi-Step Terry, Sasha, and Harry each chose a number. Sasha’s number is 10 times as much as Harry’s number. Terry’s number is 10 times as much as Sasha’s number. Harry’s number is 0.075. What number did each person choose?
Answer:
It is given that
Terry, Sasha, and Harry each chose a number. Sasha’s number is 10 times as much as Harry’s number. Terry’s number is 10 times as much as Sasha’s number. Harry’s number is 0.075
Now,
According to the given information,
Sasha’s number = 10 × Harry’s number
Terry’s number = 10 × Sasha’s number
So,
Sasha’s number is: 0.75
Terry’s number is: 7.5
Hence, from the above,
We can conclude that
Sasha’s number is: 0.75
Terry’s number is: 7.5

Question 11.
Write Math Explain how you can use place value to describe how 0.05 and 0.005 compare.
Answer:
The given decimal numbers are: 0.05 and 0.005
Now,
The representation of the given decimal numbers in the place-value table is:


Now,
From the given place-values tables,
We can observe that
The hundredth’s place in 0.05 is greater than the hundredth’s place in 0.005
Hence, from the above,
We can conclude that
0.05 is greater than 0.005

Daily Assessment Task

Fill in the bubble completely to show your answer.

Question 12.
A squirrel is 1.1 feet tall. The squirrel shrinks to \(\frac{1}{10}\) of its height. How tall is the squirrel now?
A. 11.0 feet
B. 0.11 feet
C. 1.10 feet
D. 0.011 feet
Answer:
It is given that
A squirrel is 1.1 feet tall. The squirrel shrinks to \(\frac{1}{10}\) of its height
So,
The present height of squirrel = 1.1 × \(\frac{1}{10}\)
= \(\frac{11}{10}\) × \(\frac{1}{10}\)
= \(\frac{11 × 1}{10 ×10}\)
= 0.11 feet
Hence, from the above,
We can conclude that
The height of the squirrel now is:

Go Math Lesson 1.2 5th Grade Answer Key Question 13.
Use Diagrams Write the decimal shown by the model.


Answer:
The given figure is:

Now,
From the given figure,
We can observe that
From the bigger rectangle,
The part of the shaded from the whole part = (The number of shaded parts) ÷ (The total number of parts)
= \(\frac{18}{100}\)
= 0.18
Now,
From the square,
The part of the shaded from the whole part = (The 1 part of the shaded part in the rectangle) × (The part of te shaded part of the square)
= \(\frac{1}{100}\) × \(\frac{1}{10}\)
= \(\frac{1}{1,000}\)
= 0.001
So,
The total shaded part fro the given figure = 0.18 + 0.001
= 0.181
Hence, from the above,
We can conclude that
The decimal number that is shown by the given model is:

Question 14.
Multi-Step A chef uses 0.03 liter olive oil to make soup. He uses 10 times as much chicken broth as olive oil. He uses 10 times as much vegetable broth as chicken broth. How much vegetable broth does he use?
A. 0.003 liter
B. 0.03 liter
C. 0.3 liter
D. 3 liters
Answer:
It is given that
A chef uses 0.03 liter olive oil to make soup. He uses 10 times as much chicken broth as olive oil. He uses 10 times as much vegetable broth as chicken broth
So,
The amount of chicken broth a chef used = 10 × (The amount of olive oil used by a chef)
The amount of vegetable broth a chef used = 10 × (The amount of chicken broth a chef used)
So,
The amount of chicken broth a chef used = 10 ×0.03
= 0.3 liters
The amount of vegetable broth a chef used = 10 × 0.3
= 3 liters
Hence, from the above,
We can conclude that
The amount of vegetable broth the chef used is:

Texas Test Prep

Texas Go Math Grade 5 Answer Key Pdf Lesson 1.2 Question 15.
What is the relationship between 1.0 and 0.1?
A. 0.1 is 10 times as much as 1.0
B. 1.0 is \(\frac{1}{10}\) of 0.1
C. 0.1 is \(\frac{1}{10}\) of 1.0
D. 1.0 is equal to 0.1
Answer:
The given decimal numbers are: 1.0 and 0.1
Now,
1.0 is 10 times as 0.1
0.1 is \(\frac{1}{10}\) of 1
Hence, from the above,
We can conclude that
The relationship between 1.0 and 0.1 is:

Texas Go Math Grade 5 Lesson 1.2 Homework and Practice Answer Key

Write the decimal shown by the shaded part of the model.

Question 1.

Answer:
The given figure is:

Now,
From the given figure,
We can observe that
The total number of parts are: 100
The number of shaded parts are: 33
So,
The part of the shaded from the whole part = (The number of shaded parts) ÷ (The total number of parts)
= \(\frac{33}{100}\)
= 0.33
Hence, from the above,
We can conclude that
The decimal shown by the shaded part is: 0.33

Question 2.

Answer:
The given figure is:

Now,
From the given figure,
We can observe that
The total number of parts are: 100
The number of shaded parts are: 75
So,
The part of the shaded from the whole part = (The number of shaded parts) ÷ (The total number of parts)
= \(\frac{75}{100}\)
= 0.75
Hence, from the above,
We can conclude that
The decimal shown by the shaded part is: 0.75

Use place-value patterns to complete the table.

Question 3.

Answer:
Let the decimal number be 0.0x (or) 0.x
Now,
10 times of 0.0x is: 0.x
\(\frac{1}{10}\) of 0.0x is: 0.00x
Now,
10 times of 0.x is: x
\(\frac{1}{10}\) of 0.x is: 0.0x
Hence,
The completed table by using the place-value patterns is:

Practice and Homework Lesson 1.2 Answer Key 5th Grade Question 4.

Answer:
Let the decimal number be 0.0x (or) 0.x
Now,
10 times of 0.0x is: 0.x
\(\frac{1}{10}\) of 0.0x is: 0.00x
Now,
10 times of 0.x is: x
\(\frac{1}{10}\) of 0.x is: 0.0x
Hence,
The completed table by using the place-value patterns is:

Problem Solving

Question 5.
Andy makes a model of a new tool that is 0.24 meter long. His model is 10 times as large as the actual tool. How big is the model?
Answer:
It is given that
Andy makes a model of a new tool that is 0.24 meter long. His model is 10 times as large as the actual tool
So,
The length of the model = 10 × (The length of the actual model)
= 10 × 0.24
= 2.4 meters
Hence, from the above,
We can conclude that
The length of the model is: 2.4 meters

Question 6.
Jamie makes a necklace with beads that are 0.08 meter long. Her next necklace will have beads that are \(\frac{1}{10}\) the size of the beads of the first necklace. What will be the length of the beads for the second necklace?
Answer:
It is given that
Jamie makes a necklace with beads that are 0.08 meter long. Her next necklace will have beads that are \(\frac{1}{10}\) the size of the beads of the first necklace
So,
The length of the beads for the second necklace = \(\frac{1}{10}\) × (The length of the beads for the first necklace)
= \(\frac{1}{10}\) × 0.08
= 0.008 meters
Hence, from the above,
We can conclude that
The length of the beads for the second necklace is: 0.008 meters

Lesson Check

Fill in the bubble completely to show your answer.

Question 7.
What is the value of the digit 1 in the number 0.413?
A. 1 tenth
B. 1 hundredth
C. 1 hundreds
D. 1 tens
Answer:
The given decimal number is: 0.413
So,
The representation of the given number in the place-value table is:

Hence, from the above,
We can conclude that
The value of the digit 1 in the given number is:

Question 8.
How does 0.03 compare to 0.003?
A. 0.03 is 10 times as much as 0.003
B. 0.03 is 100 times as much as 0.003
C. 0.03 is \(\frac{1}{10}\) of 0.003
D. 0.03 is \(\frac{1}{100}\) of 0.003
Answer:
The given decimal numbers are: 0.03 and 0.003
Now,
From the given numbers,
We can observe that
0.03 is \(\frac{1}{10}\) of 0.003
Hence, from the above,
We can conclude that
The relationship between 0.03 and 0.003 is:

Go Math 5th Grade Lesson 1.2 Answer Key Question 9.
The ice on a pond is 0.03 meters thick. If the ice were 10 times as thick, how many meters thick would the ice be?
A. 0.003 meter
B. 0.3 meter
C. 3 meters
D. 3.03 meters
Answer:
It is given that
The ice on a pond is 0.03 meters thick
So,
The new thickness of the ice = 10 × (The thickness of the old ice on the pond)
= 10 × 0.03
= 0.3 meters
Hence, from the above,
We can conclude that
The thickness of the new ice is:

Question 10.
How does 0.008 compare to 0.08?
A. 0.008 is 10 times as much as 0.08
B. 0.008 is 100 times as much as 0.08
C. 0.008 is \(\frac{1}{10}\) of 0.08
D. 0.008 is \(\frac{1}{100}\) of 0.08
Answer:
The given decimal numbers are: 0.008 and 0.08
Now,
From the given numbers,
We can observe that
0.008 is \(\frac{1}{10}\) of 0.08
Hence, from the above,
We can conclude that
The relationship between 0.008 and 0.08 is:

Question 11.
Multi-Step A green ribbon is 9 inches long. A red ribbon is \(\frac{1}{10}\) the length of the green ribbon. A purple ribbon is 10 times as long as the red ribbon. What is the length of the purple ribbon?
A. 0.09 inch
B. 9 inches
C. 0.9 inch
D. 90 inches
Answer:
It is given that
A green ribbon is 9 inches long. A red ribbon is \(\frac{1}{10}\) the length of the green ribbon. A purple ribbon is 10 times as long as the red ribbon
So,
The length of the red ribbon = \(\frac{1}{10}\) × (The length of the green ribbon)
= \( × 9
= 0.9 meters
So,
The length of the purple ribbon = 10 × (The length of the red ribbon)
= 10 ×0.9
= 9 meters
Hence, from the above,
We can conclude that
The length of the purple ribbon is:

Question 12.
Multi-Step Alan lives 10 times as far from school as Linda. Linda lives 0.7 mile from school. Jenny lives [latex]\frac{1}{10}\) as far from school as Alan. Pedro lives \(\frac{1}{10}\) as far from school as Jenny. How far from school does Pedro live?
A. 7 miles
B. 0.07 mile
C. 70 miles
D. 0.7 mile
Answer:
It is given that
Alan lives 10 times as far from school as Linda. Linda lives 0.7 mile from school. Jenny lives \(\frac{1}{10}\) as far from school as Alan. Pedro lives \(\frac{1}{10}\) as far from school as Jenny
So,
The distance from school to Alan’s house = 10 × (The distance from the school to Linda’s house)
= 10 × 0.7
= 7 miles
So,
The distance from school to Jenny’s house = \(\frac{1}{10}\) × (The distance from the school to Alan’s house)
= \(\frac{1}{10}\) × 7
= 0.7 miles
The distance from school to Pedro’s house = \(\frac{1}{10}\) × (The distance from the school to Jenny’s house)
= \(\frac{1}{10}\) × 0.7
= 0.07 miles
Hence, from the above,
We can conclude that
The distance from school to Pedro’s house is:

Refer to our Texas Go Math Grade 6 Answer Key Pdf to score good marks in the exams. Test yourself by practicing the problems from Texas Go Math Grade 6 Module 18 Quiz Answer Key.

Texas Go Math Grade 6 Module 18 Quiz Answer Key

Texas Go Math Grade 6 Module 18 Ready to Go On? Answer Key

18.1 Choosing a bank

Question 1.
Ruby is trying to choose a bank. She does not write checks. She makes a withdrawal from the ATM once a week and buys coffee with her debit card 5 times a week.

What is Ruby’s monthly cost for each bank?
Answer:
Determine the monthly cost.
FN Bank
$2 × 4 = $8 ATM fee for withdrawal in a month
$8 + $1.50 = $9.50 monthly cost for FN Bank
Community Bank
5 × 4 = 20 use of debit card in a month
$0.50 × 20 = $10 monthly cost for Community Bank
The monthly cost for FN Bank is $9.50 while for Community Bank is $10.

18.2 Protecting Your Credit

Go Math Grade 6 Answer Key Module 18 Quiz Answer Question 2.
Would one late credit card payment 2 years ago make you more likely or less likely to get a car loan?
Answer:
Having late payment on a credit card bill may affect the credit score of a customer. If a customer has a low credit score, he would be less likely to be approved of a car loan. If a customer has a high credit score, he would be more likely to get a car loan.

18.3 Paying for College

Question 3.
The cost for Roberta’s college next year is $11,000. She has a grant for $1,500 and a scholarship of $5,000. She plans to earn 40% of the remaining amount. How much will still be left to pay?
Answer:
Determine the amount left to be paid.
= $11,000 – ($1,500 + $5,000) add the grant and scholarship
= $11,000 – $6,500 subtract the total from the college cost
= $4,500
40% of $4, 500 = $1,800
$4500 $1,800 = $2,700 amount left to pay
The amount left to pay is $2,700.

18.4 Wages, Salaries, and Careers

Question 4.
The annual median income for an entry-level auto mechanic is $33,934. The annual median income for a senior auto mechanic is $50,856. How much more can a senior auto mechanic earn than an entry-level auto mechanic over 10 years?
Answer:
Compare the income of a senior auto mechanic and an entry-level auto mechanic over 10 years.
$33,934 × 10 = $339,340 income in 10 years as an entry-level auto mechanic
$50,856 × 10 = $508,560 income in 10 years as a senior auto mechanic
$508,560 – $339,340 = $169,220 difference on the income
A senior auto mechanic earns $169,220 more than the entry-level auto mechanic.

Essential Question

Question 5.
What are some things you can do to be a knowledgeable consumer?
Answer:
Being a knowledgeable consumer entails a lot of responsibilities. If you use a credit card for your purchases, make sure that you know the details of the charges given by the bank on the credit card. Avoid paying late the credit card bill because it will affect your credit history and the credit score. And if your credit score is low, it might result to nonapproval of loan.

Texas Go Math Grade 6 Module 18 Mixed Review Texas Test Prep Answer Key

Selected Response

Texas Go Math Module 18 Grade 6 Quiz Key Answer Question 1.
Which situation describes the use of a credit card?
(A) Ryan entered his PIN to complete a transaction.
(B) Ryan overdrew his checking account.
(C) Ryan bought groceries and paid for them at the end of the month.
(D) After making a purchase, Ryan recorded the information in his check register.
Answer:
(C) Ryan bought groceries and paid for them at the end of the month.

Explanation:
Credit card is used when you purchase an item and pay for it at a later date.

Question 2.
Which of these can have a positive effect on your credit score?
(A) paying credit card balance monthly
(B) late payments
(C) a new job
(D) high monthly debt repayment
Answer:
(A) paying credit card balance monthly

Explanation:
Factors that have an effect on attaining a high credit score are payment history, amount owed to the credit company, duration of the credit, new credit, and type of credit.

Question 3.
Joel has an annual salary of $34,840. He works 40 hours a week, 52 weeks a year. How much does Joel earn per hour?
(A) $13
(B) $15
(C) $16.75
(D) $18
Answer:
(C) $16.75

Explanation:
Determine how much Joel earn in an hour.
$34,840 ÷ 52 weeks = $670 earnings in a week
$670 ÷ 40 hours = $16.75 earnings in an hour

Grade 6 Module 18 Answer Key Texas Go Math Question 4.
Traci’s bank charges $2 for withdrawals at nonbank ATMs and a fee of $2.50 per month for a debit card. Last month she paid $10.50 in ATM and debit card fees. How many nonbank ATM withdrawals did she make?
(A) 2
(B) 4
(C) 5
(D) 8
Answer:
(B) 4

Explanation:
Determine the number of nonbank ATM withdrawals she made.
$10.50 – $2.50 = $8 subtract the debit card fee from the total amount paid
$8 ÷ $2 = 4 number of nonbank ATM withdrawals

Question 5.
Lacey is attending a university where the cost for one year is $10,500. She has a scholarship worth $6,000 and a grant worth $900. She earns $45 a day at her job. How many days does she need to work to pay for 50% of the remaining amount?
(A) 40
(B) 50
(C) 60
(D) 80
Answer:
(A) 40

Explanation:
Determine the number of days needed to pay for 50% of the remaining amount.
= $10,500 – ($6,000 + $900) add the scholarship and grant
= $10,500 – $6,900 subtract the total from the university cost
= $3,600 remaining amount
50% of 83,600 = $1,800 half of the remaining amount
81,800 ÷ 45 = 40 days number of days needed to work

Question 6.
The annual median income for a senior accounting clerk is $42,294. The annual median income for a senior accountant is $64,473. How much more can a senior accountant earn than a senior accounting clerk over 30 years?
(A) $22,179
(B) $66,537
(C) $16.75
(D) $665,370
Answer:
(D) $665,370

Explanation:
Compare the income of a senior accountant and senior accounting clerk over 30 years.
$64,473 × 30 = $1,934,190 income in 30 years of a senior accountant
$42,294 × 30 = $1,268,820 income in 30 years of a senior accounting clerk
$1,934,190 $1,268,820 = $665,370 difference in the income
The senior accountant earns D. $665,370 more than the senior accounting clerk over 30 years.

Gridded Response

Texas Go Math Grade 6 Answer Key Pdf Module 18 Test Question 7.
Sally recorded the distances she ran for 5 days: 5 miles, 3.5 miles, 6 miles, 4.5 miles, and 5.5 miles. What is the mean number of miles Sally ran per day?

Answer:
Determine the mean number of miles

The mean number of miles run by Sally is 49 miles.