Texas Go Math

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 7 Assessment Answer Key.

Texas Go Math Grade 5 Module 7 Assessment Answer Key

Vocabulary

Choose the best term from the box.

Vocabulary
composite number
numerical expression
prime number

Question 1.
A ___________ is a whole number greater than 1 that has exactly two factors, 1 and itself. (p. 297)
Answer:
A Prime number is a whole number greater than 1 that has exactly two factors, 1 and itself.

Question 2.
A ____________ is a mathematical phrase that has numbers and operation signs but does not have an equal sign. (p. 303)
Answer:
A numerical expression is a mathematical phrase that has numbers and operation signs but does not have an equal sign.

Concepts and Skills

Decide if the number Is prime or composite. If it is composite, list the factor pairs. (TEKS 5.4.A)

Question 3.
54
Answer:
The number 54 is a composite number.
The factor pairs are as below.
1 × 54 = 54
2 × 27 = 54
3 × 18 = 54
6 × 9 = 54
9 × 6 = 54
Explanation:
In Mathematics, composite numbers are numbers that have more than two factors. The number 54 is a composite number. The factors of 54 are 1, 2, 3, 6, 9, 18, 27, 54. The factor pairs of 54 are (1, 54), (2, 27), (3, 18), (6, 9), (9,6).

Go Math Answer Key Grade 5 Module 7 Question 4.
28
Answer:
The number 28 is a composite number.
The factor pairs areas below.
1 x 28 = 28
2 x 14 = 28
4 x 7 = 28
7 x 4 = 28
Explanation:
In Mathematics, composite numbers are numbers that have more than two factors. The number 28 is a composite number. The factors of 28 are 1, 2, 4, 7, 14, 28. The factor pairs of 28 are (1, 28), (2, 14), (4, 7), (7,4).

Tell whether the number Is prime or composite. (TEKS 5.4.A)

Question 5.
33
Answer:
The factors of 33 are 1, 3, 11, 33.
So, the number 33 is a composite number.
Explanation:
A composite number is a number that can be divided evenly by more numbers than 1 and itself. The number 33 can be evenly divided by 1, 3, 11 and 33, with no remainder. Since 33 cannot be divided by just 1 and 33. So, 33 is a composite number.

Question 6.
47
Answer:
The factors of 47 are 1, 47. So, 47 is a prime number.
Explanation:
47 is a prime number. The number 47 is divisible only by 1 and the number itself. The number 47 is classified as a prime number, because it have exactly two factors.

Question 7.
91
Answer:
The factors of 91 are 1, 7, 13, 91. So, 91 is a composite number.
Explanation:
A composite number is a number that can be divided evenly by more numbers than 1 and itself. It is the opposite of a prime number. The number 91 can be evenly divided by 1, 7, 13, 91 with no remainder. Since 91 cannot be divided by just 1 and 91. So, it is a composite number.

Grade 5 Module 7 Test Answer Key Texas Go Math Question 8.
81
Answer:
The factors of 81 are 1, 3, 9, 27, 81
So, the number 81 is a composite number.
Explanation:
A composite number is a number that can be divided evenly by more numbers than 1 and itself. It is the opposite of a prime number. The number 81 can be evenly divided by 1, 3, 9, 27, and 81, with no remainder. Since 81 cannot be divided by just 1 and 81. So, it is a composite number.

Simplify the numerical expression. (TEKS 5.4.F)

Question 9.
18 – (8 × 3) ÷ 4
Answer:
18 – (8 × 3) ÷ 4
18 – 24 ÷ 4
18 – 6
12
Explanation:
The numerical expression is 18 – (8 × 3) ÷ 4. Perform the operations in the parentheses first 18 – 24 ÷ 4. Next perform the division operation 18 – 6. Then perform subtraction operation the difference is 12. The simplified form of given numerical expression is 12.

Question 10.
35 – [(4 × 5) + (2 × 5)]
Answer:
35 – [(4 × 5) + (2 × 5)]
35 – [20 + 10]
35 – 30
5
Explanation:
The numerical expression using parentheses and brackets is 35 – [(4 × 5) + (2 × 5)]. Perform the operations in the parentheses first 35 – [20 + 10]. Next perform the operations in the brackets 35 – 30. Then perform subtraction operation the difference is 5. The simplified form of given numerical expression is 5.

Fill in the bubble completely to show your answer.

Question 11.
Students in a math contest are asked to simplify a numerical expression. The correct answer is 34. Which could be the expression? (TEKS 5.4.F)
(A) 6 + 3 × 4 – 2
(B) 6 + 3 × (4 – 2)
(C) (6 + 3) × 4 – 2
(D) 6 + (3 × 4) – 2
Answer:

(6 + 3) × 4 – 2
9 x 4 – 2
36 – 2
34
So, option C is correct.
Explanation:
Students in a math contest are asked to simplify a numerical expression. The correct answer is 34. The numerical expression is (6 + 3) × 4 – 2. The simplified form of numerical expression is 34.

Texas Go Math Grade 5 Module 7 Assessment Question 12.
Ana writes the expression (8 × 4) + (6 × 3) to represent the number of cards in her sports card collection. Which could be Ana’s sports card collection? (TEKS 5.4.E)
(A) 8 soccer cards and 4 baseball cards in one box, 6 soccer cards and 3 baseball cards in another box
(B) 8 soccer cards were separated into 4 boxes, and 6 baseball cards were separated into 3 boxes
(C) 8 boxes with 4 soccer cards in each box, 6 boxes with 3 baseball cards in each box
(D) 12 soccer cards, 9 baseball cards
Answer:

Option C is correct.
Explanation:
Ana writes the expression (8 × 4) + (6 × 3) to represent the number of cards in her sports card collection. Ana’s sports card collection is 8 boxes with 4 soccer cards in each box, 6 boxes with 3 baseball cards in each box.

Question 13.
A florist has 9 daffodils and twice as many tulips. He donates the flowers equally to 3 parks. Which expression represents the number of flowers each park receives? (TEKS 5.4.E)
(A) (9 × 2) ÷ 3
(B) [9 + (2 × 9)] ÷ 3
(C) [9 × (2 × 9)] ÷ 3 .
(D) [9 × (2 × 9)] ÷ 3
Answer:

Option B is correct.
Explanation:
A florist has 9 daffodils and twice as many tulips. He donates the flowers equally to 3 parks. The expression represents the number of flowers each park receives is [9 + (2 × 9)] ÷ 3.

Texas Go Math Grade 5 Module 7 Answer Key Question 14.
David washes 10 cars and waxes 4 cars every Saturday. He earns $5 for each car he washes and $12 for each car he waxes. How much money does he earn on 3 Saturdays in dollars? Simplify the expression 3 × [(10 × 5) + (4 × 12)] to find the answer. (TEKS 5.4.F)

Record your answer and fill in the bubbles on the grid. Be sure to use the correct place value.
Answer:

3 × [(10 × 5) + (4 × 12)]
3 x [50 + 48]
3 x 98
$294.00
On 3 Saturdays he earns $294.00.
Explanation:
David washes 10 cars and waxes 4 cars every Saturday. He earns $5 for each car he washes and $12 for each car he waxes. The expression is 3 × [(10 × 5) + (4 × 12)]. The simplified form of the expression is $294.00. On 3 Saturdays he earns $294.00.

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 7.5 Answer Key Grouping Symbols.

Texas Go Math Grade 5 Lesson 7.5 Answer Key Grouping Symbols

Unlock the Problem

Mary’s weekly allowance is $8 and David’s weekly allowance is $5. Every week they each spend $2 on lunch. Write a numerical expression to show how many weeks it will take them together to save enough money to buy a video game for $45.

• Underline Mary’s weekly allowance and how much she spends.
• Circle David’s weekly allowance and how much he spends.

Use parentheses and brackets to write an expression.

You can use parentheses and brackets to group operations that go together. Operations in parentheses and brackets are performed first.
STEP 1: Write an expression to represent how much Mary and David save each week.

• How much money does Mary save each week?
  Think: Each week Mary gets $8 and spends $2.
  (__________)
• How much money does David save each week?
  Think: Each week David gets $5 and spends $2.
  (__________)
• How much money do Mary and David save together each week? ___________

STEP 2: Write an expression to represent how many weeks it will take Mary and David to save enough money for the video game.

• How many weeks will it take Mary and David to save enough for a video game?
  Think: I can use brackets to group operations a second time. $45 is divided by the total amount of money saved each week.
  _____________ ÷ [_______________]

Answer:
STEP 1: Each week Mary gets $8 and spends $2.
$8 – $2
$6
Each week Mary saves $6.
Each week David gets $5 and spends $2.
$5 – $2 
$3
Each week David saves $3.
The expression to represent the how much Mary and David save each week is $6 + $3.
Mary and David saves together each week is $9.
STEP 2:
In this step we are calculating how many weeks will it take Mary and David to save enough money for the video game.
Here we are using brackets to group operations a second time. $45 is divided by the total amount of money saved each week.
The expression is $45 ÷ [$6 + $3]
$45 ÷ $9
5
They took 5 weeks to save enough money to buy a video game.

Math Talk
Mathematical Processes

Explain why brackets are placed around the part of the expression that represents the amount of money Mary and David save each week.
Answer:

Example

John gets $6 for his weekly allowance and spends $4 of it. His sister Tina gets $7 for her weekly allowance and spends $3 of it. Their mother’s birthday is in 4 weeks. If they save the same amount each week, how much money can they save together in that time to buy her a present?

• Write the expression using parentheses and brackets. 4 × [($6 – $4) + ($7 – $3)]
• Perform the operations in the parentheses first. 4 × [_______ + _______]
• Next perform the operations in the brackets. 4 × _______
• Then multiply. _______

So, John and Tina will be able to save _______ for their mother’s birthday present.
Answer:
John gets $6 for his weekly allowance and spends $4 of it.
$6 – $4
His sister Tina gets $7 for her weekly allowance and spends $3 of it.
$7 – $3
Their mother’s birthday is in 4 weeks.
The expression using parentheses and brackets.
4 × [($6 – $4) + ($7 – $3)]
Perform the operations in the parentheses first.
4 × [$2 + $4]
Next perform the operations in the brackets.
4 × $6
Then multiply.
$24
So, John and Tina will be able to save $24 for their mother’s birthday present.
H.O.T. What if only Tina saves money? Will this change the numerical expression? Explain.
Answer:
If Tina only saves money then the numerical expression changes to 4 × ($7 – $3).
Explanation:
Tina gets $7 for her weekly allowance and spends $3 of it.
$7 – $3
Their mother’s birthday is in 4 weeks.
The expression using parentheses.
4 × ($7 – $3)
Perform the operations in the parentheses first.
4 × ($4)
Then multiply.
$16
So, Tina will be able to save $16 for her mother’s birthday present.

Share and Show

Simplify the numerical expression.

Question 1.
12 + [(15 – 5) + (9 – 3)]
12 + [10 + ________]
12 + _______
Answer:
12 + [(15 – 5) + (9 – 3)]
12 + [10 + 6]
12 + 16
28
Explanation:
The numerical expression using parentheses and brackets is 12 + [(15 – 5) + (9 – 3)]. Perform the operations in the parentheses first 12 + [10 + 6]. Next, perform the operations in the brackets 12 + 16. Then perform the addition operation the sum is 28. The simplified form of the given numerical expression is 28.

Grouping Symbols Lesson 7.5 Answer Key 5th Grade Question 2.
5 × [(26 – 4) – (4 + 6)]
Answer:
5 × [(26 – 4) – (4 + 6)]
5 x [22 – 10]
5 x 12
60
Explanation:
The numerical expression using parentheses and brackets is 5 × [(26 – 4) – (4 + 6)]. Perform the operations in the parentheses first 5 x [22 -10]. Next perform the operations in brackets 5 x 12. Then perform multiplication operation the product is 60. The simplified form of the given numerical expression is 60.

Question 3.
36 ÷ [(18 – 10) – (8 – 6)]
Answer:
36 ÷ [(18 – 10) – (8 – 6)]
36 ÷ [8 – 2]
36 ÷ 6
6
Explanation:
The numerical expression using parentheses and brackets is 36 ÷ [(18 – 10) – (8 – 6)]. Perform the operations in the parentheses first 36 ÷ [8 – 2]. Next, perform the operations in the brackets 36 ÷ 6. Then perform the division operation the result is 6. The simplified form of the given numerical expression is 6.

Problem Solving

Question 4.
H.O.T. Use Symbols Write the expression 2 × 8 + 20 – 12 + 6 with parentheses and brackets two different ways so its value is less than 10 and greater than 50.
Answer:

Unlock the Problem

Question 5.
Reasoning Dan has a flower shop. Each day he displays 24 roses. He gives away 10 and sells the rest. Each day he displays 36 carnations. He gives away 12 and sells the rest. What expression can you use to find out how many roses and carnations Dan sells in a week?
a. What information are you given?
Answer:
Dan has a flower shop. Each day he displays 24 roses. He gives away 10 and sells the rest. Each day he displays 36 carnations. He gives away 12 and sells the rest.

b. What are you being asked to do?
Answer:
I was asked to find out how many roses and carnations Dan sells in a week.

c. What expression shows how many roses Dan sells in one day?
Answer:
The expression is 24 – 10.
24 – 10
14
Dan sells 14 roses in one day.
d. What expression shows how many carnations Dan sells in one day?
Answer:
The expression is 36 – 12.
36 – 12
24
Dan sells 24 carnations in one day.

e. Write an expression to represent the total number of roses and carnations Dan sells in one day.
Answer:
(24 – 10) + (36 – 12)
14 + 24
38
Explanation:
The expression (24 – 10) + (36 – 12) represents the total number of roses and carnations Dan sells in one day is 38.

f. Write the expression that shows how many roses and carnations Dan sells in a week.
Answer:
7 x [(24 – 10) + (36 – 12)]
Explanation:
The expression 7 x [(24 – 10) + (36 – 12)] represents the roses and carnations Dan sells in a week.

Go Math Expressions Grade 5 Answer Key Lesson 7.5 Question 6.
Multi-Step Simplify the expression to find out how many roses and carnations Dan sells in a week.
Answer:
7 x [(24 – 10) + (36 – 12)]
7 x [14 + 24]
7 x 38
266
Dan sells 266 roses and carnations in a week.

Question 7.
H.O.T. How could you change the story in Problem 5 so there is only one expression inside parentheses?

Answer:

Daily Assessment Task

Fill in the bubble completely to show your answer.

Question 8.
A gift shop had 500 colored pencils. The shop sold 3 sets of 20 colored pencils, 6 sets of 12 colored pencils, and 10 sets of 18 colored pencils. Which expression shows how many colored pencils are left?
(A) 3 × 20 + 6 × 12 + 10 × 18 – 500
(B) 500 – [3 × (20 + 6) × (12 + 10) × 18)]
(C) 500 + [(3 × 20) + (6 × 12) + (10 × 18)]
(D) 500 – [(3 × 20) + (6 × 12) + (10 × 18)]
Answer:

The expression is 500 – [(3 × 20) + (6 × 12) + (10 × 18)].
So, option D is correct.
Explanation:
A gift shop had 500 colored pencils. The shop sold 3 sets of 20 colored pencils, 6 sets of 12 colored pencils, and 10 sets of 18 colored pencils. The expression that shows the colored pencils left is 500 – [(3 × 20) + (6 × 12) + (10 × 18)]. So, option D is correct.

Question 9.
Anya buys 8 oranges every Monday morning at the farmer’s market. She gives 6 away and eats the rest. Every Friday she buys 6 oranges, gives 5 away, and eats the rest. Simplify the expression 52 × [(8 – 6) + (6 – 5)] to find the number of oranges Anya eats in a year.
(A) 104
(B) 208
(C) 156
(D) 260
Answer:

52 × [(8 – 6) + (6 – 5)]
52 x [2 + 1]
52 x 3
156
Anya eats 156 oranges in a year.
So, option C is correct.
Explanation:
Anya buys 8 oranges every Monday morning at the farmer’s market. She gives 6 away and eats the rest. Every Friday she buys 6 oranges, gives 5 away, and eats the rest. The expression is 52 × [(8 – 6) + (6 – 5)]. The simplified form of the expression value is 156. Anya eats 156 oranges in a year. So, option C is correct.

Question 10.
Multi-Step A company can produce 300 ballpoint pens or 550 gel pens each hour. Each weekday, the company produces ballpoint pens for 5 hours and gel pens for 8 hours. How many pens does the company produce in 5 weekdays?
(A) 7,500
(B) 24,500
(C) 29,500
(D) 5,900
Answer:

The expression is 5 x [(5 x 300) + (8 x 550)].
5 x [1,500 + 4,400]
5 x 5,900
29,500
The company produce 29,500 pens in 5 weekdays.
So, option C is correct.
Explanation:
A company can produce 300 ballpoint pens or 550 gel pens each hour. Each weekday, the company produces ballpoint pens for 5 hours and gel pens for 8 hours. The expression is 5 x [(5 x 300) + (8 x 550)]. The simplified form of the expression value is 29,500. The company produces 29,500 pens in 5 weekdays.
So, option C is correct.
Texas Test Prep

Grouping Symbols 5th Grade Lesson 7.5 Answer Key Question 11.
Which expression has a value of 4?
(A) [(4 × 5) + (9 + 7)] + 9
(B) [(4 × 5) + (9 + 7)] ÷ 9
(C) [(4 × 5) – (9 + 7)] × 9
(D) [(4 + 5) + (9 + 7)] – 9
Answer:

[(4 × 5) + (9 + 7)] ÷ 9
[20 + 16] ÷ 9
36 ÷ 9
4
Explanation:
The numerical expression using parentheses and brackets is [(4 × 5) + (9 + 7)] ÷ 9. Perform the operations in the parentheses first [20 + 16] ÷ 9. Next perform the operations in the brackets 36 ÷ 9. Then perform division operation the result is 4. The expression that has a value of 4 is [(4 × 5) + (9 + 7)] ÷ 9.

Texas Go Math Grade 5 Lesson 7.5 Homework and Practice Answer Key

Simplify the numerical expression.

Question 1.
14 + [(2 × 5) + (3 × 8)]
Answer:
14 + [(2 × 5) + (3 × 8)]
14 + [10 + 24]
14 + 34
48
Explanation:
The numerical expression using parentheses and brackets is 14 + [(2 × 5) + (3 × 8)]. Perform the operations in the parentheses first 14 + [10 + 24]. Next perform the operations in the brackets 14 + 34. Then perform addition operation the result is 48. The simplified form of given numerical expression is 48.

Question 2.
5 × [(8 + 2) – (16 – 9)]
Answer:
5 × [(8 + 2) – (16 – 9)]
5 x [10 – 7]
5 x 3
15
Explanation:
The numerical expression using parentheses and brackets is 5 × [(8 + 2) – (16 – 9)]. Perform the operations in the parentheses first 5 × [10 – 7]. Next perform the operations in the brackets 5 x 3. Then perform multiplication operation the result is 15. The simplified form of given numerical expression is 15.

Question 3.
40 ÷ [(18 – 9) – (13 – 12)]
Answer:
40 ÷ [(18 – 9) – (13 – 12)]
40 ÷ [9 – 1]
40 ÷ 8
5
Explanation:
The numerical expression using parentheses and brackets is 40 ÷ [(18 – 9) – (13 – 12)]. Perform the operations in the parentheses first 40 ÷ [9 – 1]. Next, perform the operations in the brackets 40 ÷ 8. Then perform the division operation the result is 5. The simplified form of the given numerical expression is 5.

Go Math Practice and Homework Lesson 7.5 Answer Key Question 4.
[(15 + 5) + (5 × 2)] ÷ 3
Answer:
[(15 + 5) + (5 × 2)] ÷ 3
[20 + 10] ÷ 3
30 ÷ 3
10
Explanation:
The numerical expression using parentheses and brackets is [(15 + 5) + (5 × 2)] ÷ 3. Perform the operations in the parentheses first [20 + 10] ÷ 3. Next perform the operations in the brackets 30 ÷ 3. Then perform division operation the result is 10. The simplified form of given numerical expression is 10.

Question 5.
[(21 – 13) + (32 – 24)] × 4
Answer:
[(21 – 13) + (32 – 24)] × 4
[8 + 8] × 4
16 x 4
64
Explanation:
The numerical expression using parentheses and brackets is [(21 – 13) + (32 – 24)] × 4. Perform the operations in the parentheses first [8 + 8] x 4. Next perform the operations in the brackets 16 x 4. Then perform multiplication operation the result is 64. The simplified form of given numerical expression is 64.

Question 6.
49 – [(3 × 4) + (9 × 2)]
Answer:
49 – [(3 × 4) + (9 × 2)]
49 – [12 + 18]
49 – 30
19
Explanation:
The numerical expression using parentheses and brackets is 49 – [(3 × 4) + (9 × 2)]. Perform the operations in the parentheses first 49 – [12 + 18]. Next perform the operations in the brackets 49 – 30. Then perform subtraction operation the result is 19. The simplified form of given numerical expression is 19.

Question 7.
32 + [(11 – 7) + (5 × 3)]
Answer:
32 + [(11 – 7) + (5 × 3)]
32 + [4 + 15]
32 + 19
51
Explanation:
The numerical expression using parentheses and brackets is 32 + [(11 – 7) + (5 × 3)]. Perform the operations in the parentheses first 32 + [4 + 15]. Next perform the operations in the brackets 32+ 19. Then perform addition operation the result is 51. The simplified form of given numerical expression is 51.

Question 8.
[(6 × 9) – (7 × 4)] – 17
Answer:
[(6 × 9) – (7 × 4)] – 17
[54 – 28] – 17
26 – 17
9
Explanation:
The numerical expression using parentheses and brackets is [(6 × 9) – (7 × 4)] – 17. Perform the operations in the parentheses first [54 – 28] – 17. Next perform the operations in the brackets 26 – 17. Then perform subtraction operation the result is 9. The simplified form of given numerical expression is 9.

Question 9.
[(13 – 9) × 3] + [(14 – 8) × 2]
Answer:
[(13 – 9) × 3] + [(14 – 8) × 2]
[4 × 3] + [6 × 2]
12 + 12
24
Explanation:
The numerical expression using parentheses and brackets is [(13 – 9) × 3] + [(14 – 8) × 2]. Perform the operations in the parentheses first [4 × 3] + [6 × 2]. Next perform the operations in the brackets 12 + 12. Then perform addition operation the result is 24. The simplified form of given numerical expression is 24.

Question 10.
[(2 × 6) + 3] + [35 – (7 × 3)]
Answer:
[(2 × 6) + 3] + [35 – (7 × 3)]
[12 + 3] + [35 – 21]
15 + 14
29
Explanation:
The numerical expression using parentheses and brackets is [(2 × 6) + 3] + [35 – (7 × 3)]. Perform the operations in the parentheses first [12 + 3] + [35 – 21]. Next perform the operations in the brackets 15 + 14. Then perform addition operation the result is 29. The simplified form of given numerical expression is 29.

Problem Solving

Question 11.
Fred’s Car Dealership has a three-floor parking garage with cars for sale. Each floor has 3 rows of 5 compact cars and 4 rows of 8 sedans. Write an expression you can use to find the number of cars in Fred’s garage. Simplify the expression.
Answer:
3 x [(3 x 5) + (4 x 8)]
3 x [15 + 32]
3 x 47
141
The number of cars in Fred’s garage is 141.
Explanation:
Fred’s Car Dealership has a three-floor parking garage with cars for sale. Each floor has 3 rows of 5 compact cars and 4 rows of 8 sedans. The expression is 3 x [(3 x 5) + (4 x 8)]. The number of cars in Fred’s garage is 141.

Question 12.
A carpenter has a supply of 84 large and small boards for the cabinets he builds. One week, he uses 3 sets of 8 small boards. Then he buys 3 more sets of 6 large boards. Write an expression you can use to find the number of boards the carpenter has now. Simplify the expression.
Answer:

Lesson Check

Fill in the bubble completely to show your answer.

Question 13.
Which expression has a value of 1?
(A) 30 + [(9 × 2) + (4 × 3)]
(B) 30 ÷ [(9 × 2) + (4 × 3)]
(C) 30 – [(9 × 2) + (4 × 3)]
(D) 30 × [(9 + 2) – (4 + 3)]
Answer:

30 ÷ [(9 × 2) + (4 × 3)]
30 ÷ [18 + 12]
30 ÷ 30
1
So, option B is correct.
Explanation:
The expression that has the value of 1 is 30 ÷ [(9 × 2) + (4 × 3)]. Perform the operations in the parentheses first 30 ÷ [18 + 12]. Next, perform the operations in the brackets 30 ÷ 30. Then perform the division operation the value is 1.

Go Math Expressions 5th Grade Answer Key Grouping Symbols Question 14.
Which expression has a value equal to the value of the expression 4 × [(12 + 4) – (12 ÷ 3)]?
(A) (4 × 12) + 4
(B) 4 × (16 – 4)
(C) 4 + (12 × 4)
(D) 4 × 12 + 4
Answer:

4 × [(12 + 4) – (12 ÷ 3)]
4 x [16 – 4]
4 x 12
48
So, option B is correct.
Explanation:
The value of this expression 4 x (16 – 4) is 48. The value of this expression 4 × [(12 + 4) – (12 ÷ 3)] is 48. So, the expression 4 x (16 – 4) has a value  is equal to the value of the expression 4 × [(12 + 4) – (12 ÷ 3)]. So, option B is correct.

Question 15.
The school math coach takes an inventory of math materials. He counts the materials for 5 different classes. Each class has 7 boxes of 10 pattern blocks, 6 boxes of 9 rulers, and 3 boxes of 2 calculators. Which expression shows the number of math materials?
(A) 5 × [(7 × 10) + (6 × 9) + (3 × 2)]
(B) 5 + [(7 × 10) + (6 × 9) + (3 × 2)]
(C) 5 × 7 × 10 + 6 × 9 + 3 × 2
(D) [(7 × 10) + (6 × 9) + (3 × 2)] – 5
Answer:

option A is correct.
Explanation:
The school math coach takes an inventory of math materials. He counts the materials for 5 different classes. Each class has 7 boxes of 10 pattern blocks, 6 boxes of 9 rulers, and 3 boxes of 2 calculators. The expression that shows the number of math materials is 5 × [(7 × 10) + (6 × 9) + (3 × 2)]. So, option A is correct.

Question 16.
At the beginning of the year, the teacher’s supply closet contained 250 markers. In September, 5 sets of 6 black markers, 4 sets of 5 red markers, and 6 sets of 3 yellow markers are used. Which expression shows the number of markers left?
(A) 250 + [(5 × 6) + (4 × 5) + (6 × 3)]
(B) (250 – 5) × 6 + [(4 × 5) + (6 × 3)]
(C) 250 + [(5 + 6) × (4 + 5) × (6 + 3)]
(D) 250 – [(5 × 6) + (4 × 5) + (6 × 3)]
Answer:

Option D is correct.
Explanation:
At the beginning of the year, the teacher’s supply closet contained 250 markers. In September, 5 sets of 6 black markers, 4 sets of 5 red markers, and 6 sets of 3 yellow markers are used. The expression that shows the number of markers left is 250 – [(5 × 6) + (4 × 5) + (6 × 3)].

Question 17.
Multi-Step The principal conducted a school assembly every school day for a week. On Monday, 78 students attended. Then 6 classes with 25 students in each class attended each day for the next three days. On Friday, 8 classes with 32 students in each class attended the assembly. How many students attended the assembly?
(A) 606
(B) 484
(C) 706
(D) 784
Answer:

78 + 3 x (6 x 25) + (8 x 32)
78 + 3 x 150 + 256
78 + 450 + 256
784
So, option D is correct.
Explanation:
The principal conducted a school assembly every school day for a week. On Monday, 78 students attended. Then 6 classes with 25 students in each class attended each day for the next three days. On Friday, 8 classes with 32 students in each class attended the assembly. 784 students attended the assembly. So, option D is correct.

Question 18.
Multi-Step Employees from a local store donated picnic supplies for the end-of-the-school-year picnic. They donated 20 packs of 12 forks, 10 packs of 12 spoons, 5 packs of 10 knives, and 175 paper plates. How many picnic items were donated?
(A) 244
(B) 410
(C) 585
(D) 695
Answer:

(20 x 12) + (10 x 12) + (5 x 10) + 175
240 + 120 + 50 + 175
585
So, option C is correct.
Explanation:
Employees from a local store donated picnic supplies for the end-of-the-school-year picnic. They donated 20 packs of 12 forks, 10 packs of 12 spoons, 5 packs of 10 knives, and 175 paper plates. The expression is (20 x 12) + (10 x 12) + (5 x 10) + 175. Employees donated 585 picnic items.

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 5 Assessment Answer Key.

Texas Go Math Grade 5 Module 5 Assessment Answer Key

Vocabulary

Choose the best term from the box.

Vocabulary
common denominator
common multiple
equivalent fraction

Question 1.
A ________________ is a common multiple of two or more denominators. (p. 213)
Answer:
A common denominator is a common multiple of two or more denominators.

Concepts and Skills

Estimate the sum or difference. (TEKS 5.3.A)

Question 2.
\(\frac{8}{9}\) + \(\frac{4}{7}\)
Answer:

a. Round \(\frac{8}{9}\) to its closest benchmark.
Answer:  \(\frac{9}{9}\)

b. Round \(\frac{4}{7}\) to its closest benchmark.
Answer: \(\frac{4}{7}\)

c. Add to find the estimate.   \(\frac{9}{9}\) +\(\frac{4}{7}\)  = \(\frac1{1}{2}\)
Answer: \(\frac1{1}{2}\)
Explanation:
used benchmarks to find reasonable estimates
by rounding fractions to 0, \(\frac{1}{2}\), or 1.

Go Math Answer Key Grade 5 Module 5 Assessment Question 3.
3\(\frac{2}{5}\) – \(\frac{5}{8}\)
Answer:

a. Round \(\frac{17}{5}\) to its closest benchmark.
Answer:  \(\frac{20}{5}\)

b. Round \(\frac{5}{8}\) to its closest benchmark.
Answer: \(\frac{4}{8}\)

c. Add to find the estimate.   \(\frac{20}{4}\) – \(\frac{4}{8}\)  = 3\(\frac{1}{2}\)
Answer: 3\(\frac{1}{2}\)
Explanation:
used benchmarks to find reasonable estimates
by rounding fractions to 0, \(\frac{1}{2}\), or 1.

Question 4.
1\(\frac{5}{6}\) + 2\(\frac{2}{11}\)
Answer:

a. Round \(\frac{11}{6}\) to its closest benchmark.
Answer:  \(\frac{12}{6}\)

b. Round \(\frac{24}{11}\) to its closest benchmark.
Answer: \(\frac{22}{11}\)

c. Add to find the estimate.   \(\frac{22}{11}\) – \(\frac{12}{6}\)  = 4
Answer: 4
Explanation:
used benchmarks to find reasonable estimates
by rounding fractions to 0, \(\frac{1}{2}\), or 1.

Use the least common denominator to write ah equivalent fraction for each fraction. (TEKS 5.3)

Question 5.
\(\frac{2}{5}\), \(\frac{1}{10}\)
least common denominator: ___________
Answer: 10
Explanation:
least common denominator: 5 and 10 is 10

Question 6.
\(\frac{5}{6}\), \(\frac{3}{8}\)
least common denominator: ____________
Answer: 48
Explanation:
least common denominator: 6 and 8 is 48

Go Math Grade 5 Module 5 Answer Key Pdf Question 7.
\(\frac{1}{3}\), \(\frac{2}{7}\)
least common denominator: _____________
Answer: 21
Explanation:
least common denominator: 3 and 7 is 21

Use models or strategies to find the sum or difference. Write your answer in simplest form. (TEKS 5.3.H, 5.3.K)

Question 8.
\(\frac{11}{8}\) – \(\frac{1}{6}\)
Answer:
\(\frac{11}{8}\) – \(\frac{1}{6}\)
\(\frac{13-4}{24}\)
\(\frac{29}{24}\)
Explanation:
Step 1: The least common denominator is found
Step 2: written equivalent fractions with equal denominators
Step 3: write the answer in simplest form.

Question 9.
\(\frac{2}{7}\) + \(\frac{2}{5}\)
Answer:
\(\frac{2}{7}\) + \(\frac{2}{5}\)
\(\frac{10}{35}\) + \(\frac{14}{35}\)
\(\frac{24}{35}\)
Explanation:
Step 1: The least common denominator is found
Step 2: written equivalent fractions with equal denominators
Step 3: write the answer in simplest form.

Question 10.
\(\frac{3}{4}\) – \(\frac{3}{10}\)
Answer:
\(\frac{3}{4}\) – \(\frac{3}{10}\)
\(\frac{15-6}{20}\)
\(\frac{9}{20}\)
Explanation:
Step 1: The least common denominator is found
Step 2: written equivalent fractions with equal denominators
Step 3: write the answer in simplest form.

Use the properties and mental math to solve. Write your answer in simplest form. (TEKS 5.3.H)

Question 11.
(\(\frac{3}{8}\) + \(\frac{2}{3}\)) + \(\frac{1}{3}\)
Answer:
Answer:
(\(\frac{3}{8}\) + \(\frac{2}{3}\)) + \(\frac{1}{3}\)
(\(\frac{9}{24}\) + \(\frac{16}{24}\)) + \(\frac{8}{24}\)
\(\frac{9+16+28}{24}\) =
\(\frac{33}{24}\)
Explanation:
Written the number sentence to represent the problem.
Used the Associative Property to group fractions with equal denominators together.
Used mental math to add the fractions with
equal denominators.
Written equivalent fractions with equal denominators and then added

Go Math Grade 5 Module 5 Answer Key Question 12.
1\(\frac{4}{5}\) + (2\(\frac{3}{20}\) + \(\frac{3}{5}\))
Answer:
1\(\frac{4}{5}\) + (2\(\frac{3}{20}\) + \(\frac{3}{5}\))
\(\frac{36+43+12}{20}\)
\(\frac{91}{20}\)
Explanation:
Written the number sentence to represent the problem.
Used the Associative Property to group fractions with equal denominators together.
Used mental math to add the fractions with
equal denominators.
Written equivalent fractions with equal denominators and then added

Question 13.
3\(\frac{5}{9}\) + (1\(\frac{7}{9}\) + 2\(\frac{5}{12}\))
Answer:
3\(\frac{5}{9}\) + (1\(\frac{7}{9}\) + 2\(\frac{5}{12}\))
\(\frac{32}{9}\) + \(\frac{16}{9}\) + \(\frac{29}{12}\)
\(\frac{128+64+87}{36}\)
Explanation:
Written the number sentence to represent the problem.
Used the Associative Property to group fractions with equal denominators together.
Used mental math to add the fractions with
equal denominators.
Written equivalent fractions with equal denominators and then added

Fill in the bubble completely to show your answer.

Question 14.
Samuel walks in the Labor Day parade. He walks 3\(\frac{1}{4}\) miles along the parade route and 2\(\frac{5}{6}\) miles home. How many miles does Samuel walk? (TEKS 5.3.K)
(A) \(\frac{5}{10}\) mile
(B) 6\(\frac{1}{12}\) miles
(C) 5\(\frac{1}{2}\) miles
(D) 5\(\frac{11}{12}\) miles
Answer: B
Explanation:
Samuel walks in the Labor Day parade.
He walks 3\(\frac{1}{4}\) miles along the parade route and 2\(\frac{5}{6}\) miles home.
3\(\frac{1}{4}\) + 2\(\frac{5}{6}\)
6\(\frac{1}{12}\) miles

Go Math Answer Key 5th Grade Module 5 Test Answers Question 15.
Mrs. Michaels bakes a pie for her book club meeting. The shaded part of the diagram shows the amount of pie left after the meeting. That evening, Mr. Michaels eats \(\frac{1}{4}\) of the whole pie. Which fraction represents the amount of pie remaining? (TEKS 5.3.H, 5.3.K)

(A) \(\frac{1}{4}\)
(B) \(\frac{3}{8}\)
(C) \(\frac{5}{8}\)
(D) \(\frac{3}{4}\)
Answer: A
Explanation:
Mrs. Michaels bakes a pie for her book club meeting.
The shaded part of the diagram shows the amount of pie left after the meeting.
That evening, Mr. Michaels eats \(\frac{1}{4}\) of the whole pie.
\(\frac{1}{4}\) fraction represents the amount of pie remaining

Question 16.
Aaron is practicing for a triathlon. On Sunday, he bikes 12\(\frac{5}{8}\) miles and swims 5\(\frac{2}{3}\) miles. On Monday, he runs 6\(\frac{3}{8}\) miles. How many total miles does Aaron cover on the two days? (TEKS 5.3.K)
(A) 23\(\frac{1}{6}\) miles
(B) 25\(\frac{7}{12}\) miles
(C) 24\(\frac{7}{12}\) miles
(D) 24\(\frac{2}{3}\) miles
Answer: D
Explanation:
Aaron is practicing for a triathlon. On Sunday,
he bikes 12\(\frac{5}{8}\) miles and swims 5\(\frac{2}{3}\) miles.
On Monday, he runs 6\(\frac{3}{8}\) miles.
24\(\frac{2}{3}\) miles total miles does Aaron cover on the two days
12\(\frac{5}{8}\) + 6\(\frac{3}{8}\) + 6\(\frac{3}{8}\)
\(\frac{303+136+153}{24}\) = \(\frac{592}{24}\)
24\(\frac{2}{3}\) miles

Grade 5 Go Math Module 5 Assessment Answer Key Question 17.
Mario is painting his walls. He needs a total of 5\(\frac{2}{3}\) gallons of paint for the job. He has 3\(\frac{3}{4}\) gallons of paint. How much more paint does he need? (TEKS 5.3.K)
(A) 2\(\frac{5}{6}\) gallons
(B) 9\(\frac{1}{12}\) gallons
(C) 2\(\frac{1}{12}\) gallons
(D) 1\(\frac{11}{12}\) gallons
Answer: D
Explanation:
Mario is painting his walls.
He needs a total of 5\(\frac{2}{3}\) gallons of paint for the job.
He has 3\(\frac{3}{4}\) gallons of paint.
5\(\frac{2}{3}\) – 3\(\frac{3}{4}\)
1\(\frac{11}{12}\) gallons

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 5.5 Answer Key Add and Subtract Fractions.

Texas Go Math Grade 5 Lesson 5.5 Answer Key Add and Subtract Fractions

Unlock the Problem

Malia bought shell beads and glass beads to weave into designs in her baskets. She bought \(\frac{1}{4}\) pound of shell beads and \(\frac{3}{8}\) pound of glass beads. How many pounds of beads did she buy?

• Underline the question you need to answer.
• Draw a circle around the information you will use.

Add. \(\frac{1}{4}\) + \(\frac{3}{8}\). Write your answer in simplest form.

One Way
Find a common denominator by multiplying the denominators.
4 × 8 = ________ ← common denominator
Use the common denominator to write equivalent fractions with equal denominators. Then add, and write your answer in simplest form.

Another Way

Find the least common denominator.
The least common denominator of \(\frac{1}{4}\) and \(\frac{3}{8}\) is ___________.

So, Malia bought _________ pound of beads.
Answer:

One Way
Find a common denominator by multiplying the denominators.
4 × 8 = 24 ← common denominator
Use the common denominator to write equivalent fractions with equal denominators. Then add, and write your answer in simplest form.

Another Way
The least common denominator of 14 and 38 is 8

So, Malia bought \(\frac{5}{8}\)  pound of beads.

Lesson 5.5 Go Math Grade 5 Answer Key Question 1.
Explain how you know whether your answer is reasonable.
Answer: Both methods are the same
they both give the same answer
The least common denominator is the simplest method

Example

When subtracting two fractions with unequal denominators, follow the same steps you follow when adding two fractions. However, instead of adding the fractions, subtract.

Subtract. \(\frac{9}{10}\) – \(\frac{2}{5}\) Write your answer in simplest form.

Describe the steps you took to solve the problem.
Answer:

Explanation:
Step 1: The least common denominator is found
Step 2: written equivalent fractions with equal denominators
Step 3: write the answer in simplest form.

Question 2.
Explain how you know whether your answer is reasonable.
Answer:
The fraction solved into simplest form is reasonable
which found by least common denominator

Share and Show

Find the sum or difference. Write your answer in simplest form.

Question 1.
\(\frac{5}{12}\) + \(\frac{1}{3}\)
Answer:
\(\frac{5}{12}\) + \(\frac{1}{3}\) = \(\frac{5}{12}\) + \(\frac{4}{12}\) = \(\frac{9}{12}\)
Explanation:
Step 1: The least common denominator is found
Step 2: written equivalent fractions with equal denominators
Step 3: write the answer in simplest form.

Question 2.
\(\frac{2}{5}\) + \(\frac{3}{7}\)
Answer:
\(\frac{2}{5}\) + \(\frac{3}{7}\) = \(\frac{14}{35}\) + \(\frac{15}{35}\) =\(\frac{29}{35}\)
Explanation:
Step 1: The least common denominator is found
Step 2: written equivalent fractions with equal denominators
Step 3: write the answer in simplest form.

Question 3.
\(\frac{1}{6}\) + \(\frac{3}{4}\)
Answer:
\(\frac{1}{6}\) + \(\frac{3}{4}\) = \(\frac{2}{12}\) + \(\frac{9}{12}\) = \(\frac{11}{12}\)
Explanation:
Step 1: The least common denominator is found
Step 2: write equivalent fractions with equal denominators
Step 3: write the answer in simplest form.

Lesson 5.5 Answer Key Go Math Grade 5 Question 4.
\(\frac{3}{4}\) – \(\frac{1}{8}\)
Answer:
\(\frac{3}{4}\) – \(\frac{1}{8}\) = \(\frac{6}{8}\) – \(\frac{1}{8}\) = \(\frac{5}{8}\)
Explanation:
Step 1: The least common denominator is found
Step 2: write equivalent fractions with equal denominators
Step 3: write the answer in simplest form.

Question 5.
\(\frac{1}{4}\) – \(\frac{1}{7}\)
Answer:
\(\frac{1}{4}\) – \(\frac{1}{7}\) = \(\frac{7}{28}\) – \(\frac{4}{28}\)= \(\frac{3}{28}\)
Explanation:
Step 1: The least common denominator is found
Step 2: written equivalent fractions with equal denominators
Step 3: write the answer in simplest form.

Question 6.
\(\frac{9}{10}\) – \(\frac{1}{4}\)
Answer:
\(\frac{9}{10}\) – \(\frac{1}{4}\) = \(\frac{18}{20}\) – \(\frac{5}{20}\)= \(\frac{13}{20}\)
Explanation:
Step 1: The least common denominator is found
Step 2: written equivalent fractions with equal denominators
Step 3: write the answer in simplest form.

Math Talk
Mathematical Processes

Explain why it is important to check your answer for reasonableness.
Answer:

Problem Solving

Practice: Copy and Solve Find the sum or difference. Write your answer in simplest form.

Question 7.
\(\frac{1}{3}\) + \(\frac{4}{18}\)
Answer:
\(\frac{1}{3}\) + \(\frac{4}{18}\) = \(\frac{6}{18}\) + \(\frac{4}{18}\) =\(\frac{10}{18}\)
Explanation:
Step 1: The least common denominator is found
Step 2: written equivalent fractions with equal denominators
Step 3: write the answer in simplest form.

Question 8.
\(\frac{3}{5}\) + \(\frac{1}{3}\)
Answer:
\(\frac{3}{5}\) + \(\frac{1}{3}\) = \(\frac{9}{15}\) + \(\frac{5}{15}\) = \(\frac{14}{15}\)
Explanation:
Step 1: The least common denominator is found
Step 2: written equivalent fractions with equal denominators
Step 3: write the answer in simplest form.

Question 9.
\(\frac{3}{10}\) + \(\frac{1}{6}\)
Answer:
\(\frac{3}{10}\) + \(\frac{1}{6}\) = \(\frac{9}{30}\) + \(\frac{5}{30}\) = \(\frac{14}{30}\)
Explanation:
Step 1: The least common denominator is found
Step 2: written equivalent fractions with equal denominators
Step 3: write the answer in simplest form.

Question 10.
\(\frac{1}{2}\) + \(\frac{4}{9}\)
Answer:
\(\frac{1}{2}\) + \(\frac{4}{9}\) = \(\frac{9}{18}\) + \(\frac{8}{18}\) = \(\frac{17}{18}\)
Explanation:
Step 1: The least common denominator is found
Step 2: written equivalent fractions with equal denominators
Step 3: write the answer in simplest form.

Go Math Grade 5 Lesson 5.5 Answer Key Question 11.
\(\frac{1}{2}\) – \(\frac{3}{8}\)
Answer:
\(\frac{1}{2}\) – \(\frac{3}{8}\) = \(\frac{4}{8}\) – \(\frac{3}{8}\) = \(\frac{1}{8}\)
Explanation:
Step 1: The least common denominator is found
Step 2: written equivalent fractions with equal denominators
Step 3: write the answer in simplest form.

Question 12.
\(\frac{5}{7}\) – \(\frac{2}{3}\)
Answer:
\(\frac{5}{7}\) – \(\frac{2}{3}\) = \(\frac{15}{21}\) – \(\frac{14}{21}\) = \(\frac{1}{21}\)
Explanation:
Step 1: The least common denominator is found
Step 2: written equivalent fractions with equal denominators
Step 3: write the answer in simplest form.

Question 13.
\(\frac{4}{9}\) – \(\frac{1}{6}\)
Answer:
\(\frac{4}{9}\) – \(\frac{1}{6}\) = \(\frac{8}{18}\) – \(\frac{3}{18}\) = \(\frac{5}{18}\)
Explanation:
Step 1: The least common denominator is found
Step 2: written equivalent fractions with equal denominators
Step 3: write the answer in simplest form.

Question 14.
\(\frac{11}{12}\) – \(\frac{7}{15}\)
Answer:
\(\frac{11}{12}\) – \(\frac{7}{15}\) = \(\frac{55}{60}\) – \(\frac{28}{60}\) = \(\frac{27}{60}\)
Explanation:
Step 1: The least common denominator is found
Step 2: written equivalent fractions with equal denominators
Step 3: write the answer in simplest form.

H.O.T. Algebra Find the unknown number.

Question 15.
\(\frac{9}{10}\) – ☐ = \(\frac{1}{5\)
☐ = ___________
Answer:
\(\frac{9}{10}\) – \(\frac{7}{10}\)= \(\frac{1}{5\)
Explanation:
Step 1: The least common denominator is found
Step 2: written equivalent fractions with equal denominators
Step 3: write the answer in simplest form.

Question 16.
\(\frac{5}{12}\) + ☐ = \(\frac{1}{2}\)
☐ = ____________
Answer:
\(\frac{5}{12}\) + \(\frac{1}{12}\) =\(\frac{6}{12}\) =\(\frac{1}{2}\)
Explanation:
Step 1: The least common denominator is found
Step 2: written equivalent fractions with equal denominators
Step 3: write the answer in simplest form.

Problem Solving

Use the picture for 17-18

Go Math Grade 5 Lesson 5.5 Practice Answer Key Question 17.
Sara is making a key chain using the bead design shown. What fraction of the beads in her design are either blue or red?
Answer:
Explanation:
Let us consider dark black as red
and light black-as-blue
The number of beads is 15
Number of red beads are \(\frac{5}{15}\)
Number of black beads are \(\frac{6}{15}\)

Question 18.
H.O.T. Multi-Step In making the key chain, Sara uses the pattern of beads 3 times. After the key chain is complete, what fraction of the beads in the key chain are either white or blue?

Answer:

Question 19.
Write Math Jamie had \(\frac{4}{5}\) of a spool of twine. He then used \(\frac{1}{2}\) of a spool of twine to make friendship knots. He claims to have \(\frac{3}{10}\) of the original spool of twine leftover. How you know whether Jamie’s claim is reasonable.
Answer: Yes. Jamie’s claim is reasonable.
Explanation:
Jamie had \(\frac{4}{5}\) of a spool of twine. He then used \(\frac{1}{2}\) of a spool of twine to make friendship knots.  So \(\frac{3}{10}\) of the original spool of twine leftover. Since
\(\frac{4}{5}\) –\(\frac{1}{2}\) = \(\frac{8}{10}\) – \(\frac{5}{10}\) = \(\frac{3}{10}\)
He claims to have \(\frac{3}{10}\) of the original spool of twine leftover. So it is equla to what he leftover. So his claim is reasonabale.

Daily Assessment Task

Fill in the bubble completely to show your answer.

Question 20.
Apply Students are voting for a new school mascot. So far, the results show that \(\frac{3}{10}\) of the students voted for “Fightin’ Titan,” \(\frac{1}{2}\) of the students voted for “Nifty Knight,” and the rest of the students have not voted yet. What fraction of the student population has not voted yet?
(A) \(\frac{3}{10}\)
(B) \(\frac{2}{5}\)
(C) \(\frac{1}{5}\)
(D) \(\frac{4}{5}\)
Answer: (C) \(\frac{1}{5}\)
Explanation:
So far, the results show that \(\frac{3}{10}\) of the students voted for “Fightin’ Titan,” \(\frac{1}{2}\) of the students voted for “Nifty Knight,” Then \(\frac{8}{10}\) voted. Since
\(\frac{3}{10}\) +\(\frac{1}{2}\) = \(\frac{8}{10}\)
So \(\frac{1}{5}\) of the students have not voted yet. Since
1- \(\frac{8}{10}\) = \(\frac{1}{5}\)

Question 21.
Tina spent \(\frac{3}{5}\) of her paycheck on a trip to the beach. She spent \(\frac{3}{8}\) of her paycheck on new clothes for the trip. What fraction of her paycheck did Tina spend on the trip and clothes together?
(A) \(\frac{9}{40}\)
(B) \(\frac{3}{4}\)
(C) \(\frac{7}{8}\)
(D) \(\frac{39}{40}\)
Answer: (D) \(\frac{39}{40}\)
Explanation:
Tina spent \(\frac{3}{5}\) of her paycheck on a trip to the beach. She spent \(\frac{3}{8}\) of her paycheck on new clothes for the trip. So Tortal Spent is \(\frac{39}{40}\)
\(\frac{3}{5}\) + \(\frac{3}{8}\)  = \(\frac{39}{40}\)

Question 22.
Multi-Step On Friday, \(\frac{1}{6}\) of band practice was spent trying on uniforms. The band spent \(\frac{1}{4}\) of practice on marching. What fraction of practice time was left for playing music?
(A) \(\frac{5}{12}\)
(B) \(\frac{1}{2}\)
(C) \(\frac{7}{12}\)
(D) \(\frac{1}{4}\)
Answer: (C) \(\frac{7}{12}\)
Explanation:
\(\frac{1}{6}\) of band practice was spent trying on uniforms. The band spent \(\frac{1}{4}\) of practice on marching. So Total time spent is \(\frac{5}{12}\). So Time left is \(\frac{7}{12}\)
\(\frac{1}{6}\) + \(\frac{1}{4}\) = \(\frac{5}{12}\)
1-\(\frac{5}{12}\) =\(\frac{7}{12}\)

Texas Test Prep

Question 23.
Which equation represents the fraction of beads that are green or yellow?

Answer:

Texas Go Math Grade 5 Lesson 5.5 Homework and Practice Answer Key

Find the sum or difference. Write your answer in simplest form.

Question 1.
\(\frac{1}{5}\) + \(\frac{1}{2}\) ____________
Answer:
\(\frac{1}{5}\) + \(\frac{1}{2}\) = \(\frac{2}{10}\) + \(\frac{5}{10}\) = \(\frac{7}{10}\)
Explanation:
Step 1: The least common denominator is found
Step 2: write equivalent fractions with equal denominators
Step 3: write the answer in simplest form.

Go Math Grade 5 Lesson 5.5 Answer Key Question 2.
\(\frac{2}{3}\) + \(\frac{1}{6}\) ____________
Answer:
\(\frac{2}{3}\) + \(\frac{1}{6}\) = \(\frac{4}{6}\) + \(\frac{1}{6}\) = \(\frac{5}{6}\)
Explanation:
Step 1: The least common denominator is found
Step 2: written equivalent fractions with equal denominators
Step 3: write the answer in simplest form.

Question 3.
\(\frac{1}{4}\) + \(\frac{2}{3}\) ____________
Answer:
\(\frac{1}{4}\) + \(\frac{2}{3}\) = \(\frac{3}{12}\) + \(\frac{8}{12}\) = \(\frac{11}{12}\)
Explanation:
Step 1: The least common denominator is found
Step 2: written equivalent fractions with equal denominators
Step 3: write the answer in simplest form.

Question 4.
\(\frac{3}{4}\) + \(\frac{1}{8}\) ____________
Answer:
\(\frac{3}{4}\) + \(\frac{1}{8}\) = \(\frac{6}{8}\) + \(\frac{1}{8}\) = \(\frac{7}{8}\)
Explanation:
Step 1: The least common denominator is found
Step 2: written equivalent fractions with equal denominators
Step 3: write the answer in simplest form.

Question 5.
\(\frac{2}{9}\) + \(\frac{1}{3}\) ____________
Answer:
\(\frac{2}{9}\) + \(\frac{1}{3}\) = \(\frac{2}{9}\) + \(\frac{3}{9}\) = \(\frac{5}{9}\)
Explanation:
Step 1: The least common denominator is found
Step 2: written equivalent fractions with equal denominators
Step 3: write the answer in simplest form.

Question 6.
\(\frac{1}{2}\) + \(\frac{2}{6}\) ____________
Answer:
\(\frac{1}{2}\) + \(\frac{2}{6}\) = \(\frac{3}{6}\) + \(\frac{2}{6}\) = \(\frac{5}{6}\)
Explanation:
Step 1: The least common denominator is found
Step 2: written equivalent fractions with equal denominators
Step 3: write the answer in simplest form.

Question 7.
\(\frac{3}{10}\) + \(\frac{1}{3}\) ____________
Answer:
\(\frac{3}{10}\) + \(\frac{1}{3}\) = \(\frac{9}{30}\) + \(\frac{10}{30}\) = \(\frac{19}{30}\)
Explanation:
Step 1: The least common denominator is found
Step 2: written equivalent fractions with equal denominators
Step 3: write the answer in simplest form.

Question 8.
\(\frac{4}{18}\) + \(\frac{2}{6}\) ____________
Answer:
\(\frac{4}{18}\) + \(\frac{2}{6}\) = \(\frac{4}{18}\) + \(\frac{6}{18}\) = \(\frac{10}{18}\)
Explanation:
Step 1: The least common denominator is found
Step 2: written equivalent fractions with equal denominators
Step 3: write the answer in simplest form.

Question 9.
\(\frac{6}{12}\) – \(\frac{1}{3}\) ____________
Answer:
\(\frac{6}{12}\) – \(\frac{1}{3}\) = \(\frac{6}{12}\) – \(\frac{4}{12}\) = \(\frac{2}{12}\)
Explanation:
Step 1: The least common denominator is found
Step 2: written equivalent fractions with equal denominators
Step 3: write the answer in simplest form.

Question 10.
\(\frac{3}{4}\) – \(\frac{1}{6}\) ____________
Answer:
\(\frac{3}{4}\) – \(\frac{1}{6}\) = \(\frac{9}{12}\) – \(\frac{2}{12}\) = \(\frac{7}{12}\)
Explanation:
Step 1: The least common denominator is found
Step 2: written equivalent fractions with equal denominators
Step 3: write the answer in simplest form.

Question 11.
\(\frac{5}{7}\) – \(\frac{1}{2}\) ____________
Answer:
\(\frac{5}{7}\) – \(\frac{1}{2}\) = \(\frac{10}{14}\) – \(\frac{7}{14}\) = \(\frac{3}{14}\)
Explanation:
Step 1: The least common denominator is found
Step 2: written equivalent fractions with equal denominators
Step 3: write the answer in simplest form.

Question 12.
\(\frac{8}{9}\) – \(\frac{2}{3}\) ____________
Answer:
\(\frac{8}{9}\) – \(\frac{2}{3}\) = \(\frac{8}{9}\) – \(\frac{6}{9}\) = \(\frac{2}{9}\)
Explanation:
Step 1: The least common denominator is found
Step 2: written equivalent fractions with equal denominators
Step 3: write the answer in simplest form.

Question 13.
\(\frac{5}{9}\) – \(\frac{1}{6}\) ____________
Answer:
\(\frac{5}{9}\) – \(\frac{1}{6}\) = \(\frac{10}{18}\) – \(\frac{3}{18}\) = \(\frac{7}{18}\)
Explanation:
Step 1: The least common denominator is found
Step 2: written equivalent fractions with equal denominators
Step 3: write the answer in simplest form.

Question 14.
\(\frac{2}{3}\) – \(\frac{1}{4}\) ____________
Answer:
\(\frac{2}{3}\) – \(\frac{1}{4}\) = \(\frac{8}{12}\) – \(\frac{3}{12}\) = \(\frac{5}{12}\)
Explanation:
Step 1: The least common denominator is found
Step 2: written equivalent fractions with equal denominators
Step 3: write the answer in simplest form.

Question 15.
\(\frac{7}{14}\) – \(\frac{2}{7}\) ____________
Answer:
\(\frac{7}{14}\) – \(\frac{2}{7}\) = \(\frac{7}{14}\) – \(\frac{4}{14}\) = \(\frac{3}{14}\)
Explanation:
Step 1: The least common denominator is found
Step 2: written equivalent fractions with equal denominators
Step 3: write the answer in simplest form.

Question 16.
\(\frac{5}{6}\) – \(\frac{3}{4}\) ____________
Answer:
\(\frac{5}{6}\) – \(\frac{3}{4}\) = \(\frac{10}{12}\) – \(\frac{9}{12}\) = \(\frac{1}{12}\)
Explanation:
Step 1: The least common denominator is found
Step 2: written equivalent fractions with equal denominators
Step 3: write the answer in simplest form.

Find the unknown number.

Question 17.
\(\frac{7}{12}\) – ☐ = \(\frac{1}{6}\)
☐ = _____________
Answer:
\(\frac{7}{12}\) – \(\frac{5}{12}\) = \(\frac{1}{6}\)
Explanation:
Step 1: The least common denominator is found
Step 2: written equivalent fractions with equal denominators
Step 3: write the answer in simplest form.

Question 18.
\(\frac{5}{18}\) + ☐ = \(\frac{1}{2}\)
☐ = _____________
Answer:
\(\frac{5}{18}\) + \(\frac{4}{18}\) = \(\frac{1}{2}\)
Explanation:
Step 1: The least common denominator is found
Step 2: written equivalent fractions with equal denominators
Step 3: write the answer in simplest form.

Question 19.
\(\frac{7}{10}\) – ☐ = \(\frac{2}{5}\)
☐ = ______________
Answer:
\(\frac{7}{10}\) – \(\frac{3}{10}\) = \(\frac{2}{5}\)
Explanation:
Step 1: The least common denominator is found
Step 2: written equivalent fractions with equal denominators
Step 3: write the answer in simplest form.

Question 20.
☐ + \(\frac{1}{9}\) = \(\frac{1}{3}\)
☐ = _______________
Answer:
\(\frac{2}{9}\) + \(\frac{1}{9}\) = \(\frac{1}{3}\)
Explanation:
Step 1: The least common denominator is found
Step 2: written equivalent fractions with equal denominators
Step 3: write the answer in simplest form.

Problem Solving

Question 21.
There are 12 students in the pep squad. Three students are wearing white shirts. Six students are wearing blue shirts. What fraction of the students in the pep squad are wearing either white or blue shirts?
Answer: \(\frac{1}{4}\)  wearing the white shirts and \(\frac{1}{2}\) wearing the blue shirts.
Explanation:
There are 12 students in the pep squad. Three students are wearing white shirts.
\(\frac{3}{12}\)  = \(\frac{1}{4}\)
Six students are wearing blue shirts.
\(\frac{6}{12}\)  = \(\frac{1}{2}\)

 

Question 22.
Tiffany ran \(\frac{5}{6}\) mile. Shayne ran \(\frac{3}{4}\) mile. Who ran farther? How much farther?
Answer:

Lesson Check

Fill in the bubble completely to show your answer.

Question 23.
Mr. Benson spent \(\frac{2}{5}\) of the monthly budget on rent and \(\frac{3}{10}\) of the budget on food. What fraction of Mr. Benson’s budget was spent on rent and food?
(A) \(\frac{1}{3}\)
(B) \(\frac{3}{10}\)
(C) \(\frac{7}{10}\)
(D) \(\frac{1}{2}\)
Answer: (C) \(\frac{7}{10}\)
Explanation:
Mr. Benson spent \(\frac{2}{5}\) of the monthly budget on rent and \(\frac{3}{10}\) of the budget on food.
Sum of \(\frac{2}{5}\) and \(\frac{3}{10}\)  is \(\frac{7}{10}\) .
Since
\(\frac{3}{10}\)+ \(\frac{2}{5}\)= \(\frac{7}{10}\) .

Question 24.
The Ortega family made \(\frac{15}{16}\) pound of confetti for the annual Fiesta celebration in San Antonio. They used \(\frac{1}{4}\) pound to make confetti filled eggs. How much confetti is left to use next year?
(A) \(\frac{11}{16}\) pound
(B) \(\frac{9}{16}\) pound
(C) \(\frac{4}{5}\) pound
(D) \(\frac{3}{4}\) pound
Answer: (A) \(\frac{11}{16}\) pound
Explanation:
The Ortega family made \(\frac{15}{16}\) pound of confetti for the annual Fiesta celebration in San Antonio. They used \(\frac{1}{4}\) pound to make confetti filled eggs. confetti is left to use next year is \(\frac{11}{16}\) pound. Since

\(\frac{15}{16}\) – \(\frac{1}{4}\) = \(\frac{11}{16}\) pound

Use the recipe for 25-26.

Question 25.
If Rory measures the lemon juice and the vanilla extract into one spoon before adding them to the blender, how much liquid will be in the spoon?
(A) \(\frac{5}{8}\) teaspoon
(B) \(\frac{1}{5}\) teaspoon
(C) \(\frac{1}{4}\) teaspoon
(D) \(\frac{3}{8}\) teaspoon
Answer: (A) \(\frac{5}{8}\) teaspoon
Explanation:
Sum of lemon juice and the vanilla extract is
\(\frac{1}{2}\) teaspoon + \(\frac{1}{8}\) teaspoon = \(\frac{5}{8}\) teaspoon

Question 26.
Multi-Step Rory has \(\frac{5}{8}\) cup of milk. How much milk does she have left after she doubles the recipe for the smoothie?
(A) \(\frac{3}{8}\) cup
(B) \(\frac{1}{8}\) cup
(C) \(\frac{3}{4}\) cup
(D) \(\frac{1}{2}\) cup
Answer: (B) \(\frac{1}{8}\) cup
Explanation:
she doubles the recipe for the smoothie. So it is \(\frac{1}{2}\) cup. Since
\(\frac{1}{4}\) cup + \(\frac{1}{4}\) cup  = \(\frac{1}{2}\) cup.
Rory has \(\frac{5}{8}\) cup of milk. She left \(\frac{1}{8}\) cup of milk.
Since
\(\frac{5}{8}\)  –\(\frac{1}{2}\) cup. = \(\frac{1}{8}\)

Question 27.
Multi-Step Torn has \(\frac{7}{8}\) cup of olive oil. He uses \(\frac{1}{2}\) cup to make salad dressing and \(\frac{1}{4}\) cup to make tomato sauce. How much olive oil does Torn have left?
(A) \(\frac{5}{4}\) cups
(B) \(\frac{5}{8}\) cup
(C) \(\frac{3}{8}\) cup
(D) \(\frac{1}{8}\) cup
Answer: (D) \(\frac{1}{8}\) cup
Explanation:
\(\frac{1}{2}\) + \(\frac{1}{4}\)  = \(\frac{3}{4}\)
and
\(\frac{7}{8}\) cup – \(\frac{3}{4}\) cup = \(\frac{1}{8}\) cup

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 5.3 Answer Key Estimate Fraction Sums and Differences.

Texas Go Math Grade 5 Lesson 5.3 Answer Key Estimate Fraction Sums and Differences

Unlock the Problem

Kimberly will be riding her bike to school this year. The distance from her house to the end of the Street is \(\frac{1}{62}\)mile. The distance from the end of the Street to the school is \(\frac{3}{8}\) mile. About how far is Kimberly’s house from school?

You can use benchmarks to find reasonable estimates by rounding fractions to 0, \(\frac{1}{2}\), or 1.

One Way:

Use a number line.
Estimate. \(\frac{1}{6}\) + \(\frac{3}{8}\)
STEP 1:
Place a point at \(\frac{1}{6}\) on the number line.
The fraction is between ________ and __________.
The fraction \(\frac{1}{6}\) is closer to the benchmark _________.
Round to ________.

STEP 2:
Place a point at \(\frac{3}{8}\) on the number line.
The fraction is between __________ and _________.
The fraction \(\frac{3}{8}\) is closer to the benchmark ___________.
Round to _________.

STEP 3:
Add the rounded fractions.

So, Kimberly’s house is about ________ mile from the school.
Answer:\(\frac{1}{6}\)
Use a number line.
Estimate. \(\frac{1}{6}\) + \(\frac{3}{8}\)
STEP 1:
Place a point at \(\frac{1}{6}\) on the number line.
The fraction is between \(\frac{0}{6}\) and \(\frac{6}{6}\)
The fraction \(\frac{1}{6}\) is closer to the benchmark \(\frac{0}{6}\)
Round to \(\frac{0}{6}\)

STEP 2:
Place a point at \(\frac{3}{8}\) on the number line.
The fraction is between \(\frac{0}{8}\) and \(\frac{3}{8}\)
The fraction \(\frac{3}{8}\) is closer to the benchmark \(\frac{4}{8}\)
Round to \(\frac{4}{8}\)

STEP 3:
Add the rounded fractions.

So, Kimberly’s house is about \(\frac{1}{2}\)mile from the school.

Another Way

Use mental math.
You can compare the numerator and the denominator to round a fraction and find a reasonable estimate.

Estimate. \(\frac{9}{10}\) – \(\frac{5}{8}\)
STEP 1:
Round \(\frac{9}{10}\).
Think: The numerator is about the same as the denominator.
Round the fraction \(\frac{9}{10}\) to __________.

Remember
A fraction with the same numerator and denominator, such as \(\frac{2}{2}, \frac{5}{5}, \frac{12}{12}\) or \(\frac{96}{96}\), is equal to 1.

STEP 2:
Round \(\frac{5}{8}\)
Think: The numerator is about half the denominator.
Round the fraction \(\frac{5}{8}\) to ___________.

STEP 3:
Subtract

So, \(\frac{9}{10}\) – \(\frac{5}{8}\) is about __________.
Answer:

STEP 1:
Round \(\frac{9}{10}\).
Think: The numerator is about the same as the denominator.
Round the fraction \(\frac{9}{10}\) to \(\frac{10}{10}\)

Remember
A fraction with the same numerator and denominator, such as \(\frac{2}{2}, \frac{5}{5}, \frac{12}{12}\) or \(\frac{96}{96}\), is equal to 1.

STEP 2:
Round \(\frac{5}{8}\)
Think: The numerator is about half the denominator.
Round the fraction \(\frac{5}{8}\) to \(\frac{4}{8}\)

STEP 3:
Subtract

So, \(\frac{9}{10}\) – \(\frac{5}{8}\) is about \(\frac{1}{2}\)

Math Talk
Mathematical Processes

Explain another way you could use benchmarks to estimate \(\frac{9}{10}\) – \(\frac{5}{8}\).
Answer:
\(\frac{9}{10}\) – \(\frac{5}{8}\) = \(\frac{1}{6}\)
\(\frac{1}{6}\) is very near to \(\frac{1}{5}\)
Explanation:
Used bench marks to find the sum

Share and Show

Estimate the sum or difference.

Question 1.
\(\frac{5}{6}\) + \(\frac{3}{8}\)
a. Round \(\frac{5}{6}\) to its closest benchmark.
Answer:  \(\frac{6}{6}\)

b. Round \(\frac{3}{8}\) to its closest benchmark.
Answer: \(\frac{4}{8}\)

c. Add to find the estimate.   \(\frac{6}{6}\) +\(\frac{4}{8}\)  = 1\(\frac{1}{2}\)
Answer: 1\(\frac{1}{2}\)
Explanation:
used benchmarks to find reasonable estimates
by rounding fractions to 0, \(\frac{1}{2}\), or 1.

Go Math Lesson 5.3 5th Grade Answer Key Question 2.
\(\frac{5}{9}\) – \(\frac{3}{8}\)
Answer:
a. Round \(\frac{5}{9}\) to its closest benchmark.
Answer:  \(\frac{5}{9}\)

b. Round \(\frac{3}{8}\) to its closest benchmark.
Answer: \(\frac{4}{8}\)

c. Add to find the estimate.   \(\frac{5}{9}\) – \(\frac{4}{8}\)  = 1\(\frac{1}{18}\)
Answer: 1\(\frac{1}{18}\)
Explanation:
used benchmarks to find reasonable estimates
by rounding fractions to 0, \(\frac{1}{2}\), or 1.

Question 3.
\(\frac{5}{6}\) + \(\frac{2}{5}\)
Answer:
a. Round \(\frac{5}{6}\) to its closest benchmark.
Answer:  \(\frac{6}{6}\)

b. Round \(\frac{2}{5}\) to its closest benchmark.
Answer: \(\frac{2}{5}\)

c. Add to find the estimate.   \(\frac{6}{6}\) +\(\frac{2}{5}\)  = 1\(\frac{1}{2}\)
Answer: 1\(\frac{1}{2}\)
Explanation:
used benchmarks to find reasonable estimates
by rounding fractions to 0, \(\frac{1}{2}\), or 1.

Question 4.
\(\frac{9}{10}\) – \(\frac{1}{9}\)
Answer:
a. Round \(\frac{9}{10}\) to its closest benchmark.
Answer:  \(\frac{10}{10}\)

b. Round \(\frac{1}{9}\) to its closest benchmark.
Answer: \(\frac{0}{9}\)

c. Add to find the estimate.   \(\frac{10}{10}\) – \(\frac{0}{9}\)  = 1
Answer: 1
Explanation:
used benchmarks to find reasonable estimates
by rounding fractions to 0, \(\frac{1}{2}\), or 1.

Problem Solving

Lesson 5.3 Answer Key 5th Grade Go Math Question 5.
How do you know whether your estimate for \(\frac{9}{10}\) + 3\(\frac{6}{7}\) would be greater than or less than the actual sum? Explain.
Answer: Greater than the actual sum
\(\frac{9}{10}\) + 3\(\frac{6}{7}\) =
close to bench marks \(\frac{10}{10}\) + 3\(\frac{7}{7}\) =  4
Explanation:
Is greater than the actual sum
used benchmarks to find reasonable estimates
by rounding fractions to 0, \(\frac{1}{2}\), or 1.

Question 6.
Write Math Nick estimated that \(\frac{5}{8}\) + \(\frac{4}{7}\) is about 2. Explain how you know his estimate is not reasonable.
Answer: \(\frac{5}{8}\) + \(\frac{4}{7}\)
close to benchmarks \(\frac{4}{8}\) + \(\frac{4}{7}\) = 1
Explanation:
Nick estimated that \(\frac{5}{8}\) + \(\frac{4}{7}\) is about 2.
used benchmarks to find reasonable estimates
by rounding fractions to 0, \(\frac{1}{2}\), or 1.
so, his estimation is wrong

Problem Solving

Question 7.
Lisa and Valerie are picnicking in Trough Creek State Park in Pennsylvania. Lisa has brought a salad that she made with \(\frac{3}{4}\) cup of strawberries, \(\frac{7}{8}\) cup of peaches, and \(\frac{1}{6}\) cup of blueberries. About how many total cups of fruit are in the salad?
Answer:
\(\frac{3}{4}\) + \(\frac{7}{8}\) + \(\frac{1}{6}\) very close to bench marks
\(\frac{4}{4}\) + \(\frac{8}{8}\) + \(\frac{0}{6}\) =2 \(\frac{1}{2}\)
Explanation:
Lisa and Valerie are picnicking in Trough Creek State Park in Pennsylvania.
Lisa has brought a salad that she made with \(\frac{3}{4}\) cup of strawberries,
\(\frac{7}{8}\) cup of peaches, and \(\frac{1}{6}\) cup of blueberries.
2\(\frac{1}{2}\)   total cups of fruit are in the salad

Question 8.
Multi-Step At Trace State Park in Mississippi, there is a 40-mile mountain bike trail. Tommy rode A of the trail on Saturday and \(\frac{1}{5}\) of the trail on Sunday. He estimates that he rode more than 22 miles over the two days. Is Tommy’s estimate reasonable?

Answer: yes
Explanation:
\(\frac{1}{5}\) + \(\frac{1}{5}\) = 1
20 + 20 = 40
one represents the whole
so, his estimation is reasonable

Go Math 5th Grade Lesson 5.3 How to Estimate Fractions Question 9.
H.O.T Explain how you know that \(\frac{5}{8}\) + \(\frac{6}{10}\) is greater than 1.
Answer: No
Explanation:
Close to the bench marks
\(\frac{8}{8}\) + \(\frac{5}{10}\) = 1
actual sum is greater than 1

Daily Assessment Task

Fill in the bubble completely to show your answer.

Question 10.
Mia uses \(\frac{1}{5}\) of a bag of gravel in the morning and \(\frac{11}{12}\) of a bag in the afternoon. About how much gravel does she use in one day?
(A) 0 bags
(B) \(\frac{1}{2}\) bag
(C) 1 bag
(D) 2\(\frac{1}{2}\) bags
Answer:  C
\(\frac{1}{5}\) + \(\frac{11}{12}\)
nearest benchmarks are
\(\frac{0}{5}\) + \(\frac{12}{12}\)  = 1
Explanation:
Mia uses \(\frac{1}{5}\) of a bag of gravel in the morning and \(\frac{11}{12}\) of a bag in the afternoon.
she use 1 bag of gravel

Question 11.
Evaluate Reasonableness Hector and Veronica are going hiking. They made a trail mix that has \(\frac{2}{3}\) cup of almonds, \(\frac{7}{8}\) cup of peanuts, and \(\frac{4}{5}\) cup of raisins in it. Hector estimates that they made about 3 cups of trail mix. Is the estimate greater than or less than the actual sum? How do you know?
(A) The estimate is greater because each fraction is rounded up to a benchmark.
(B) The estimate is less because each fraction is rounded down to a benchmark.
(C) The estimate is greater because they really made more than 3 cups.
(D) The estimate is less because each fraction is rounded up to a benchmark.
Answer: A
Explanation:
\(\frac{2}{3}\) + \(\frac{7}{8}\) + \(\frac{4}{5}\)
rounded to the nearest benchmarks
\(\frac{3}{3}\) + \(\frac{8}{8}\) + \(\frac{5}{5}\) = 3
Evaluated Reasonableness Hector and Veronica are going hiking.
They made a trail mix that has \(\frac{2}{3}\) cup of almonds, ”
\(\frac{7}{8}\) cup of peanuts,
and \(\frac{4}{5}\) cup of raisins in it.
Hector estimates that they made about 3 cups of trail mix.

Lesson 5.3 Go Math 5th Grade Answer Key Question 12.
Multi-Step Amanda picked \(\frac{3}{5}\) pound of blueberries at her local farm yesterday. She used \(\frac{3}{8}\) pound of blueberries. Today she picked \(\frac{4}{5}\) pound of blueberries. About how many pounds of blueberries does Amanda have now?
(A) \(\frac{1}{5}\)lb
(B) 1 lb
(C) \(\frac{1}{2}\)lb
(D) 1\(\frac{1}{2}\)lbs
Answer: B
Explanation:
what she bought is that she used yesterday
in today marked to nearest benchmarks \(\frac{4}{5}\)  is \(\frac{5}{5}\)
that is 1

Texas Test Prep

Question 13.
Jake added \(\frac{1}{8}\) cup of sunflower seeds and \(\frac{4}{5}\) cup of banana chips to his sundae. Which is the best estimate of the total amount of toppings Jake added to his sundae?
(A) about 2 cups
(B) about 1 cup
(C) about 1\(\frac{1}{2}\) cups
(D) about \(\frac{1}{2}\) cup
Answer: B
Explanation:
Jake added \(\frac{1}{8}\) cup of sunflower seeds and
\(\frac{4}{5}\) cup of banana chips to his sundae.
The best estimate of the total amount of toppings Jake added to his sundae is 1 cup

Texas Go Math Grade 5 Lesson 5.3 Homework and Practice Answer Key

Estimate the sum or difference.

Question 1.
\(\frac{3}{8}\) + \(\frac{4}{5}\) = ___________
Answer:
\(\frac{3}{8}\) + \(\frac{4}{5}\) rounded to the nearest benchmarks
\(\frac{4}{8}\) + \(\frac{5}{5}\) = 1 \(\frac{1}{2}\)
Explanation:
used benchmarks to find reasonable estimates
by rounding fractions to 0, \(\frac{1}{2}\), or 1.

5th Grade Go Math Lesson 5.3 Answer Key Question 2.
\(\frac{9}{10}\) – \(\frac{3}{8}\) = ___________
Answer:
\(\frac{9}{10}\) – \(\frac{3}{8}\) rounded to the nearest benchmarks
\(\frac{10}{10}\) – \(\frac{4}{8}\) = \(\frac{1}{2}\)
Explanation:
used benchmarks to find reasonable estimates
by rounding fractions to 0, \(\frac{1}{2}\), or 1.

Question 3.
\(\frac{5}{8}\) + \(\frac{2}{5}\) = ___________
Answer:
\(\frac{5}{8}\) + \(\frac{2}{5}\) rounded to the nearest benchmarks
\(\frac{4}{8}\) + \(\frac{2}{5}\) = 1
Explanation:
used benchmarks to find reasonable estimates
by rounding fractions to 0, \(\frac{1}{2}\), or 1.

Question 4.
\(\frac{6}{7}\) + \(\frac{3}{5}\) = ___________
Answer:
\(\frac{6}{7}\) + \(\frac{3}{5}\) rounded to the nearest benchmarks
\(\frac{7}{7}\) + \(\frac{2}{5}\) = 1\(\frac{1}{2}\)
Explanation:
used benchmarks to find reasonable estimates
by rounding fractions to 0, \(\frac{1}{2}\), or 1.

Question 5.
\(\frac{3}{8}\) – \(\frac{1}{6}\) = ___________
Answer:
\(\frac{3}{8}\) – \(\frac{1}{6}\) rounded to the nearest benchmarks
\(\frac{4}{8}\) – \(\frac{0}{6}\) = \(\frac{1}{2}\)
Explanation:
used benchmarks to find reasonable estimates
by rounding fractions to 0, \(\frac{1}{2}\), or 1.

Question 6.
\(\frac{7}{12}\) + \(\frac{1}{7}\) = ___________
Answer:
\(\frac{7}{12}\) + \(\frac{1}{7}\) rounded to the nearest benchmarks
\(\frac{6}{12}\) + \(\frac{0}{7}\) = \(\frac{1}{2}\)
Explanation:
used benchmarks to find reasonable estimates
by rounding fractions to 0, \(\frac{1}{2}\), or 1.

Go Math Lesson 5.3 5th Grade Homework Answer Key Question 7.
\(\frac{4}{9}\) – \(\frac{5}{8}\) = ___________
Answer:
\(\frac{4}{9}\) – \(\frac{5}{8}\) rounded to the nearest benchmarks
\(\frac{5}{9}\) – \(\frac{4}{8}\) = 0
Explanation:
used benchmarks to find reasonable estimates
by rounding fractions to 0, \(\frac{1}{2}\), or 1.

Question 8.
\(\frac{1}{9}\) + \(\frac{5}{6}\) = ___________
Answer:
\(\frac{1}{9}\) + \(\frac{5}{6}\) rounded to the nearest benchmark
\(\frac{0}{9}\) + \(\frac{6}{6}\) = 1
Explanation:
used benchmarks to find reasonable estimates
by rounding fractions to 0, \(\frac{1}{2}\), or 1.

Question 9.
\(\frac{7}{8}\) + \(\frac{4}{7}\) = ___________
Answer:
\(\frac{7}{8}\) + \(\frac{4}{7}\) rounded to the nearest bench mark
\(\frac{8}{8}\) + \(\frac{4}{7}\) =1\(\frac{1}{2}\)
Explanation:
used benchmarks to find reasonable estimates
by rounding fractions to 0, \(\frac{1}{2}\), or 1.

Question 10.
\(\frac{1}{5}\) + \(\frac{3}{8}\) = ___________
Answer:
\(\frac{1}{5}\) + \(\frac{3}{8}\) rounded to the nearest benchmark
\(\frac{0}{5}\) + \(\frac{4}{8}\) = \(\frac{1}{2}\)
Explanation:
used benchmarks to find reasonable estimates
by rounding fractions to 0, \(\frac{1}{2}\), or 1.

Question 11.
\(\frac{7}{9}\) – \(\frac{2}{6}\) = ___________
Answer:
\(\frac{7}{9}\) – \(\frac{2}{6}\) rounded to the nearest benchmark
\(\frac{9}{9}\) – \(\frac{3}{6}\) = \(\frac{1}{2}\)
Explanation:
used benchmarks to find reasonable estimates
by rounding fractions to 0, \(\frac{1}{2}\), or 1.

Go Math Grade 5 Lesson 5.3 Homework Answer Key Question 12.
\(\frac{9}{10}\) – \(\frac{7}{8}\) = ___________
Answer:
\(\frac{9}{10}\) – \(\frac{7}{8}\) rounded to the benchmarks
\(\frac{10}{10}\) – \(\frac{8}{8}\) = 0
Explanation:
used benchmarks to find reasonable estimates
by rounding fractions to 0, \(\frac{1}{2}\), or 1.

Question 13.
Explain how you can estimate the sum of \(\frac{4}{5}\) and \(\frac{1}{6}\).
Answer:
\(\frac{4}{5}\) + \(\frac{1}{6}\) rounded to the nearest bench marks
\(\frac{5}{5}\) + \(\frac{0}{6}\) = 1
Explanation:
used benchmarks to find reasonable estimates
by rounding fractions to 0, \(\frac{1}{2}\), or 1.

Problem Solving

Question 14.
Jena uses \(\frac{7}{8}\) cup of raisins for muffins and \(\frac{5}{8}\) cup of raisins for a bowl of oatmeal. Does lena need more than or less than 1 cup of raisins to make muffins and oatmeal? Explain.
Answer: more than 1 cup of raisins
Explanation:
Jena uses \(\frac{7}{8}\) cup of raisins for muffins and
\(\frac{5}{8}\) cup of raisins for a bowl of oatmeal.
\(\frac{7}{8}\) + \(\frac{5}{8}\) rounded the benhmark
\(\frac{8}{8}\) + \(\frac{4}{8}\) = 1\(\frac{1}{2}\)

Question 15.
A group of students ate \(\frac{5}{12}\) of a cheese pizza, \(\frac{7}{8}\) of a pepperoni pizza, and \(\frac{5}{8}\) of a veggie pizza. About how many pizzas were eaten?
Answer:
\(\frac{5}{12}\) + \(\frac{7}{8}\) + \(\frac{5}{8}\) rounded to the nearest benchmark
\(\frac{6}{12}\) + \(\frac{8}{8}\) + \(\frac{4}{8}\) = 2
Explanation:
A group of students ate \(\frac{5}{12}\) of a cheese pizza,
\(\frac{7}{8}\) of a pepperoni pizza,
and \(\frac{5}{8}\) of a veggie pizza.
2 pizzas were eaten in whole.

Lesson Check

Fill in the bubble completely to show your answer.

Question 16.
On Saturday, the scouts hiked \(\frac{4}{5}\) mile up the mountain. On Sunday, they hiked \(\frac{1}{4}\) mile up the mountain. About how far did the scouts hike up the mountain in all?
(A) \(\frac{1}{2}\) mile
(B) 1 mile
(C) 1\(\frac{1}{2}\) miles
(D) 2 miles
Answer:
\(\frac{4}{5}\) + \(\frac{1}{4}\) rounded to nearest benchmark
\(\frac{5}{5}\) + \(\frac{0}{4}\)  is 1 mile
Explanation:
On Saturday, the scouts hiked \(\frac{4}{5}\) mile up the mountain.
On Sunday, they hiked \(\frac{1}{4}\) mile up the mountain.
1 mile far the scouts hike up the mountain in all

Question 17.
Which of the following best describes the difference for \(\frac{11}{12}\) – \(\frac{7}{10}\) ?
(A) less than \(\frac{1}{2}\)
(B) greater than \(\frac{1}{2}\)
(C) greater than 1
(D) greater than 1\(\frac{1}{2}\)
Answer: A
Explanation:
\(\frac{11}{12}\) – \(\frac{7}{10}\) is 0
that is less than \(\frac{1}{2}\)

Practice and Homework Lesson 5.3 Answer Key 5th Grade Question 18.
Which sum is greatest? Use estimation to decide.
(A) \(\frac{2}{7}\) + \(\frac{3}{8}\)
(B) \(\frac{1}{10}\) + \(\frac{3}{8}\)
(C) \(\frac{1}{6}\) + \(\frac{1}{8}\)
(D) \(\frac{2}{9}\) + \(\frac{1}{8}\)
Answer: A
Explanation:
\(\frac{2}{7}\) + \(\frac{3}{8}\) = 1

Question 19.
Which statement is not correct? Use estimation to decide.

Answer: B
Explanation:
used benchmarks to find reasonable estimates
by rounding fractions to 0, \(\frac{1}{2}\), or 1.

Question 20.
Multi-Step Michaela has \(\frac{11}{12}\) yard of orange fabric and \(\frac{7}{8}\) yard of green fabric. She uses \(\frac{1}{2}\) yard of each color for her sewing project. About how much fabric does Michaela have left if she combines the two colors?
(A) 1 yard
(B) \(\frac{1}{2}\) yard
(C) 1 \(\frac{1}{2}\) yards
(D) 2 yards
Answer:  D
\(\frac{11}{12}\) + \(\frac{7}{8}\) rounded to nearest bench marks
\(\frac{12}{12}\) + \(\frac{8}{8}\) = 2
Explanation:
2 yards fabric uses Michaela have left if she combines the two colors.

Question 21.
Multi-Step Dustin buys \(\frac{9}{10}\) yard of striped fabric. He uses \(\frac{3}{8}\) yard. He buys \(\frac{7}{8}\) yard more. About how much fabric does Dustin have now?
(A) 1 yard
(B) \(\frac{1}{2}\) yard
(C) 1\(\frac{1}{2}\) yards
(D) 2 yards
Answer: C
Explanation:
Dustin buys \(\frac{9}{10}\) yard of striped fabric.
He uses \(\frac{3}{8}\) yard.
He buys \(\frac{7}{8}\) yard more.
\(\frac{9}{10}\) + \(\frac{3}{8}\)  + \(\frac{7}{8}\)  rounded to nearest benchmarks
\(\frac{10}{10}\) – \(\frac{4}{8}\)  + \(\frac{8}{8}\)  = 1\(\frac{1}{2}\) yards

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 Unit 2 Answer Key Number and Operations: Fractions.

Texas Go Math Grade 5 Unit 2 Answer Key Number and Operations: Fractions

Show What You Know

Check your understanding of important skills.

Part of a Whole: Write a fraction to name the shaded part.

Question 1.

number of shaded parts ___________
number of total parts ___________
fraction ___________
Answer:
The above figure is a hexagon. It is divided into six parts.
number of shaded parts are 5.
number of total parts is 6
fraction 5/6

Texas Go Math Grade 5 Pdf Unit 2 Answer Key Question 2.

number of shaded parts ___________
number of total parts ___________
fraction ___________
Answer:
The shape of the above figure is a circle.
number of shaded parts are 1
number of total parts is 4
fraction 1/4

Add and Subtract Fractions
Write the sum or difference In simplest form.

Question 3.
\(\frac{3}{6}\) + \(\frac{1}{6}\) = ___________
Answer:
\(\frac{3}{6}\) + \(\frac{1}{6}\)
The denominator of both the fraction is same.
We can add the numerator directly.
\(\frac{3}{6}\) + \(\frac{1}{6}\) = \(\frac{4}{6}\)

Question 4.
\(\frac{4}{10}\) + \(\frac{1}{10}\) = ____________
Answer:
\(\frac{4}{10}\) + \(\frac{1}{10}\)
The denominator of both the fraction is same.
We can add the numerator directly.
\(\frac{4}{10}\) + \(\frac{1}{10}\) = (4 + 1)/10 = \(\frac{5}{10}\)

Question 5.
\(\frac{7}{8}\) – \(\frac{3}{8}\) = _____________
Answer:
\(\frac{7}{8}\) – \(\frac{3}{8}\)
The denominator of both the fraction is same.
We can subtract the numerator directly.
\(\frac{7}{8}\) – \(\frac{3}{8}\) = (7 – 3)/8 = \(\frac{4}{8}\)

Question 6.
\(\frac{9}{12}\) – \(\frac{2}{12}\) = _____________
Answer:
Given the fractions,
\(\frac{9}{12}\) – \(\frac{2}{12}\)
The denominator of both the fraction is same.
We can subtract the numerator directly.
\(\frac{9}{12}\) – \(\frac{2}{12}\) = (9-2)/12 = \(\frac{7}{12}\)

Equivalent Fractions
Write an equivalent fractions.

Question 7.
\(\frac{3}{4}\) _________
Answer:
Equivalent fractions are fractions with different numbers representing the same part of a whole.
\(\frac{3}{4}\) × \(\frac{2}{2}\) = \(\frac{6}{8}\)

Grade 5 Unit 2 Answer Key Go Math Question 8.
\(\frac{9}{15}\) __________
Answer:
Equivalent fractions are fractions with different numbers representing the same part of a whole.
\(\frac{9}{15}\) × \(\frac{2}{2}\) = \(\frac{18}{30}\)

Question 9.
\(\frac{24}{40}\) ____________
Answer:
Equivalent fractions are fractions with different numbers representing the same part of a whole.
\(\frac{24}{40}\) × \(\frac{2}{2}\) = \(\frac{48}{80}\)

Unit 2 Math Test Grade 5 Go Math Question 10.
\(\frac{5}{7}\) ____________
Answer:
Equivalent fractions are fractions with different numbers representing the same part of a whole.
\(\frac{5}{7}\) × \(\frac{2}{2}\) = \(\frac{10}{14}\)

Vocabulary Builder

Visualize It

Use the ✓ words to complete the H-diagram.

Understand Vocabulary

Draw a line to match the word with its definition.

Reading & Writing Math

Reading To get the right answer to a mathematics problem, you need to make sure you understand the question.

Problem 1.
Three friends ordered a pizza with 8 slices. Jeanette ate of the pizza. Marissa ate of the pizza. Ariel ate the rest. How many slices of Pizza did Ariel eat?
A. 1 slice
B. 2 slices
C. 4 slices
D. 6 slices

Thinking Through the Problem

Understand the question You want to know how many slices Ariel ate. Will your answer be a fraction or whole number?

Plan Find out what fraction of the pizza Ariel ate. Look at the numerator to tell how many pizza slices Ariel ate.

Solve Follow your plan. Write the answer to the problem.

Look Back Use fraction strips to check your answer.

The correct answer is B.

Writing Now it’s your turn. Answer Problem 2. Then write about how you solved the problem, step by step.:

Go Math Unit 2 5th Grade Answer Key Problem 2.
The Perez family ordered a large pizza for dinner, A large pizza is divided into 8 slices. Marco ate \(\frac{3}{8}\) of the pizza. Ramon ate 1 slice more than Marco. Emilio ate the rest. How much of the pizza did Emilio eat?

A. \(\frac{1}{8}\) of the pizza
B. \(\frac{2}{8}\) of the pizza
C. \(\frac{4}{8}\) of the pizza
D. \(\frac{5}{8}\) of the pizza
Answer:
Given,
The Perez family ordered a large pizza for dinner, A large pizza is divided into 8 slices.
Marco ate \(\frac{3}{8}\) of the pizza.
Ramon ate 1 slice more than Marco. Emilio ate the rest.
\(\frac{3}{8}\) + \(\frac{1}{8}\) = \(\frac{4}{8}\)
\(\frac{4}{8}\) + \(\frac{4}{8}\) = 1
Thus the correct answer is option C.

Get Ready Game

Action Fractions
Object of the Game Practice comparing fractions

Materials
Number/Symbol Cards: 2 sets labeled 1, 2, 3, 4, 6, 8
Number of Players 2

Set Up
Give each player 2 sets of number cards. Players shuffle their cards and place them face down in a stack.

How to Play

(1) One player shuffles arid deals all cards facedown. Players stack their cards.

(2) Players take 3 cards from the top of their stacks. Using 2 of the 3 cards, each player makes a fraction whose numerator is less than or equal to its denominator. The unused card is returned to the bottom of the player’s stack.

(3) Players compare the fractions. The player with the greater fraction earns 1 point. If the fractions are equivalent, each player earns 1 point.

(4) Repeat Steps 2 and 3. The player with more points after all the cards have been used is the winner.

Answer:

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 6.3 Answer Key Fraction and Whole-Number Multiplication.

Texas Go Math Grade 5 Lesson 6.3 Answer Key Fraction and Whole-Number Multiplication

Unlock the Problem

Charlene has five 1-pound bags of different color sands. For an art project, she will use \(\frac{3}{8}\) pound of each bag of sand to create a colorful sand-art jar. How much sand will be in Charlene’s sand-art jar?

• How much sand is in each bag?
  ____________________
• Will Charlene use all of the sand in each bag? Explain.
  ____________________

Multiply a fraction by a whole number.

So, there are _________ pounds of sand in Charlene’s sand -art jar.
Answer:

• How much sand is in each bag?
  1 pound
• Will Charlene use all of the sand in each bag? Explain.
  No, she will use 3/8 pound of the sand in each bag.

So, there are 1(7/8) pounds of sand in Charlene’s sand -art jar.
Math Talk
Mathematical Processes

Explain how you can find how much sand Charlene has left.
Answer:

Example

Multiply a whole number by a fraction.
Kirsten brought in 4 loaves of bread to make sandwiches for the class picnic. Her classmates used \(\frac{2}{3}\) of the bread. How many loaves of bread were used?

So, ________ loaves of bread were used.
Answer:

So, 2(2/3) loaves of bread were used.

Share and Show

Find the product. Write the product in simplest form.

Question 1.
3 × \(\frac{2}{5}\) = ___________

• Multiply the numerator by the whole number. Write the product over the denominator.
• Write the answer as a mixed number in simplest form.

Answer:

Explanation:
In the above image we can observe the expression 3 x (2/5). First multiply the two whole numbers in the numerator. Multiply 3 with 2 the product is 6. Write the product 6 over the denominator 5. Write the fraction 6/5 in simplest form as a mixed number. The mixed number is 1(1/5).

Go Math Grade 5 Lesson 6.3 Answer Key Question 2.
\(\frac{2}{3}\) × 5 = ___________
Answer:

Explanation:
In the above image we can observe the expression (2/3) x 5. First multiply the two whole numbers in the numerator. Multiply 2 with 5 the product is 10. Write the product 10 over the denominator 3. Write the fraction 10/3 in simplest form as a mixed number. The mixed number is 3(1/3).

Question 3.
6 × \(\frac{2}{3}\) = ___________
Answer:

Explanation:
In the above image we can observe the expression 6 x (2/3). First multiply the two whole numbers in the numerator. Multiply 6 with 2 the product is 12. Write the product 12 over the denominator 3. Write the fraction 12/3 in simplest form as 4.

Question 4.
\(\frac{5}{7}\) × 4 = ___________
Answer:

Explanation:
In the above image we can observe the expression (5/7) x 4. First multiply the two whole numbers in the numerator. Multiply 5 with 4 the product is 20. Write the product 20 over the denominator 7. Write the fraction 20/7 in simplest form as a mixed number. The mixed number is 2(6/7).

Unlock the Problem

Question 5.
The caterer wants to have enough turkey to feed 24 people. If he wants to provide \(\frac{3}{4}\) of a pound of turkey for each person, how much turkey does he need?
(A) 72 pounds
(B) 24 pounds
(C) 18 pounds
(D) 6 pounds

a. What do you need to find?
Answer:
He need to find that how much turkey he needs.

b. What operation will you use?
Answer:
He uses Multiplication operation.

c. What information are you given?
Answer:
The information given is 24 people. 3/4 of a pound person.

d. Solve the problem.
Answer:
24 x (3/4) = 72/4 = 18

e. Complete the sentences.
The caterer wants to serve 24 people __________ of a pound of turkey each.
He will need ________ × ________, or ________ pounds of turkey.
Answer:
The caterer wants to serve 24 people 3/4 of a pound of turkey each.
He will need 24 ×3/4, or 72/4 or 18 pounds of turkey.

f. Fill in the bubble for the correct answer choice.
Answer:

The correct option is C.
Explanation:
The caterer wants to have enough turkey to feed 24 people. He wants to provide 3/4 of a pound of turkey for each person. Multiply 3/4 with 24 the product is 18 pounds. He need 18 pounds of turkey. So draw a circle to option C.

H.O.T. Algebra Find the unknown digit.

Lesson 6.3 Answer Key 5th Grade Go Math Question 6.

Answer:

The unknown digit is 1.
Explanation:
In the above image we can observe that numerator digit is missing. If we place 1 in the numerator then the product is 4. Multiply 1/2 with 8 the product is 4. So, the unknown digit is 1.

Question 7.

Answer:

The unknown digit is 4.
Explanation:
In the above image we can observe that one digit is missing. If we place 4 then the product is 20/6. Multiply 4 with 5/6 the product is 20/6 or 3(1/3). So, the unknown digit is 4.

Question 8.

Answer:

The unknown digit is 6.
Explanation:
In the above image we can observe that denominator digit is missing. If we place 6 in the denominator then the product is 3. Multiply 1/6 with 18 the product is 3. So, the unknown digit is 6.

Question 9.
H.O.T. Multi-Step Patty wants to run \(\frac{5}{6}\) of a mile every day for 5 days. Keisha wants to run \(\frac{3}{4}\) of a mile every day for 6 days. Who will run the greater distance?

Answer:
Patty:
5 x (5/6) = 25/6 = 4(1/6)
Patty runs 4(1/6).
Keisha:
6 x (3/4) = 18/4 = 4(1/2)
Keisha runs 4(1/2).
Keisha runs the greater distance.
Explanation:
Patty wants to run 5/6 of a mile every day for 5 days. Multiply 5 with 5/6 the product is 25/6. The mixed fraction of 25/6 is 4(1/6). Patty runs 4(1/6). Keisha wants to run 3/4 a mile every day for 6 days. Multiply 6 with 3/4 the product is 18/4. The mixed fraction of 18/4 is 4(1/2). Keisha runs 4(1/2). Keisha runs the greater distance.

Daily Assessment Task

Fill in the bubble completely to show your answer.

Question 10.
A heavy-duty snowmaking machine makes \(\frac{3}{4}\) inch of snow each minute. How many inches of snow can the machine make in 8 minutes?
(A) 8 inches
(B) 6 inches
(C) 7\(\frac{1}{4}\) inches
(D) 4\(\frac{1}{2}\) inches
Answer:

(3/4) x 8 = 6 inches
The machine can make 6 inches of snow in 8 minutes.
So, option B is correct.
Explanation:
A heavy-duty snowmaking machine makes 3/4 inch of snow each minute. Multiply 8 minutes with 3/4 inch of snow the product is 6 inches. The machine can make 6 inches of snow in 8 minutes. So, option B is correct.

Lesson 6.3 Answer Key Go Math Grade 5 Question 11.
Connect Which has the same product as \(\frac{2}{3}\) × 8?
(A) \(\frac{5}{6}\) × 7
(B) \(\frac{1}{4}\) × 13
(C) \(\frac{3}{8}\) × 2
(D) \(\frac{1}{3}\) × 16
Answer:

(2/3) x 8 = 16/3
So, option D is correct.
Explanation:
The given expression (2/3) x 8. Multiply 2 with 8 the product is 16. The fraction is 16/3. So, option D is correct.

Question 12.
Multi-Step A baker made 5 pounds of icing. He used \(\frac{4}{9}\) of the icing to decorate cakes. How much of the icing is left over?
(A) 1 pound
(B) 1\(\frac{5}{9}\) pounds
(C) 1\(\frac{2}{3}\) pounds
(D) 2\(\frac{7}{9}\) pounds
Answer:

(4/9) x 5 = 20/9
He used 20/9 of the icing to decorate cakes.
5 – (20/9) = (45 – 20)/9 = 25/9 = 2(7/9)
2(7/9) of the icing is left over.
So, option D is correct.
Explanation:
A baker made 5 pounds of icing. He used 4/9 of the icing to decorate cakes. Multiply 4/9 with 5 the product is 20/9. He used 20/9 of the icing to decorate cakes. Subtract 20/9 from 5 the difference is 25/9. The mixed fraction of 25/9 is 2(7/9). The icing left over is 2(7/9). So, draw a circle for option D.

Texas Test Prep

Question 13.
Doug has 33 feet of rope. He wants to use \(\frac{2}{3}\) of it for his canoe. How many feet of rope will he use for his canoe?
(A) 66 feet
(B) 22 feet
(C) 33 feet
(D) 11 feet
Answer:

(2/3) x 33 = 22 feet
He used 22 feet of rope for his canoe.
So, option B is correct.
Explanation:
Doug has 33 feet of rope. He wants to use 2/3 of it for his canoe. Multiply (2/3) with 33 the product is 22 feet. He used 22 feet of rope for his canoe. So, option B is correct.

Texas Go Math Grade 5 Lesson 6.3 Homework and Practice Answer Key

Find the product. Write the product in the simplest form.

Question 1.
\(\frac{3}{7}\) × 4 = ____________
Answer:
(3/7) x 4 = (3 x 4)/7 = 12/7
The product is 12/7.
The simplest form of 12/7 is 1(5/7).
Explanation:
In the above image we can observe the expression (3/7) x 4. First multiply the two whole numbers in the numerator. Multiply 3 with 4 the product is 12. Write the product 12 over the denominator 7. Write the fraction 12/7 in simplest form as a mixed number. The mixed number is 1(5/7).

Go Math Lesson 6.3 5th Grade Answer Key Question 2.
\(\frac{3}{5}\) × 5 = ____________
Answer:
(3/5) x 5 = (3 x 5)/5 = 15/5
The product is 15/5.
The simplest form of 15/5 is 3.
Explanation:
In the above image we can observe the expression (3/5) x 5. First multiply the two whole numbers in the numerator. Multiply 3 with 5 the product is 15. Write the product 15 over the denominator 5. Write the fraction 15/5 in simplest form as 3.

Question 3.
\(\frac{2}{3}\) × 8 = ____________
Answer:
(2/3) x 8 = (2 x 8)/3 = 16/3
The product is 16/3.
The simplest form of 16/3 is 5(1/3).
Explanation:
In the above image we can observe the expression (2/3) x 8. First multiply the two whole numbers in the numerator. Multiply 2 with 8 the product is 16. Write the product 16 over the denominator 3. Write the fraction 16/3 in simplest form as a mixed number. The mixed number is 5(1/3).

Question 4.
16 × \(\frac{3}{4}\) = ____________
Answer:
16 x (3/4) = (16 x 3)/4 = 48/4
The product is 48/4.
The simplest form of 48/4 is 12.
Explanation:
In the above image we can observe the expression 16 x (3/4). First multiply the two whole numbers in the numerator. Multiply 16 with 3 the product is 48. Write the product 48 over the denominator 4. Write the fraction 48/4 in simplest form as a 12.

Question 5.
9 × \(\frac{5}{6}\) = ____________
Answer:
9 x (5/6) = (9 x 5)/6 = 45/6
The product is 45/6.
The simplest form of 45/6 is 7(3/6).
Explanation:
In the above image we can observe the expression 9 x (5/6). First multiply the two whole numbers in the numerator. Multiply 9 with 5 the product is 45. Write the product 45 over the denominator 6. Write the fraction 45/6 in simplest form as a mixed number. The mixed number is 7(3/6).

Question 6.
6 × \(\frac{3}{8}\) = ____________
Answer:
6 x (3/8) = (6 x 3)/8 = 18/8 = 9/4
The product is 9/4.
The simplest form of 9/4 is 2(1/4).
Explanation:
In the above image we can observe the expression 6 x (3/8). First multiply the two whole numbers in the numerator. Multiply 6 with 3 the product is 18. Write the product 18 over the denominator 8. Write the fraction 9/4 in simplest form as a mixed number. The mixed number is 2(1/4).

Question 7.
\(\frac{2}{9}\) × 5 = ____________
Answer:
(2/9) x 5 = (2 x 5)/9 = 10/9
The product is 10/9.
The simplest form of 10/9 is 1(1/9).
Explanation:
In the above image we can observe the expression (2/9) x 5. First multiply the two whole numbers in the numerator. Multiply 2 with 5 the product is 10. Write the product 10 over the denominator 9. Write the fraction 10/9 in simplest form as a mixed number. The mixed number is 1(1/9).

Go Math 5th Grade Lesson 6.3 Answer Key Question 8.
\(\frac{4}{7}\) × 3 = ____________
Answer:
(4/7) x 3 = (4 x 3)/7 = 12/7
The product is 12/7.
The simplest form of 12/7 is 1(5/7).
Explanation:
In the above image we can observe the expression (4/7) x 3. First multiply the two whole numbers in the numerator. Multiply 4 with 3 the product is 12. Write the product 12 over the denominator 7. Write the fraction 12/7 in simplest form as a mixed number. The mixed number is 1(5/7).

Question 9.
\(\frac{3}{10}\) × 7 = ____________
Answer:
(3/10) x 7 = (3 x 7)/10 = 21/10
The product is 21/10.
The simplest form of 21/10 is 2(1/10).
Explanation:
In the above image we can observe the expression (3/10) x 7. First multiply the two whole numbers in the numerator. Multiply 3 with 7 the product is 21. Write the product 21 over the denominator 10. Write the fraction 21/10 in simplest form as a mixed number. The mixed number is 2(1/10).

Find the unknown digit.

Question 10.

Answer:

The unknown digit is 1.
Explanation:
In the above image we can observe that numerator digit is missing. If we place 1 in the numerator then the product is 2. Multiply 1/4 with 8 the product is 2. So, the unknown digit is 1.

Question 11.

Answer:

The unknown digit is 6.
Explanation:
In the above image we can observe that one digit is missing. If we place 6 then the product is 30/7. Multiply 6 with 5/7 the product is 30/7 or 4(2/7). So, the unknown digit is 6.

Question 12.

Answer:

The unknown digit is 6.
Explanation:
In the above image we can observe that denominator digit is missing. If we place 6 in the denominator then the product is 4. Multiply 1/6 with 24 the product is 4. So, the unknown digit is 6.

Question 13.

Answer:

The unknown digit is 3.
Explanation:
In the above image we can observe that denominator digit is missing. If we place 3 in the denominator then the product is 3. Multiply 1/3 with 9 the product is 3. So, the unknown digit is 3.

Question 14.

Answer:

The unknown digit is 5.
Explanation:
In the above image we can observe that one digit is missing. If we place 5 then the product is 20/9. Multiply 5 with 4/9 the product is 20/9. So, the unknown digit is 5.

Question 15.

Answer:

The unknown digit is 3.
Explanation:
In the above image we can observe that numerator digit is missing. If we place 3 in the numerator then the product is 3. Multiply 3/4 with 4 the product is 3. So, the unknown digit is 3.

Problem Solving

Question 16.
Sandra exercised \(\frac{2}{3}\) hour every day for two weeks while she was on vacation. How many hours did Sandra exercise during her vacation?
Answer:
(2/3) x 14 = 28/3
Sandra exercised 28/3 hours during her vacation.
Explanation:
Sandra exercised 2/3 hours every day for two weeks while she was on vacation. In one week there are 7 days. Multiply 2/3 with 14 days the product is 28/3 hours. Sandra exercised 28/3 hours during her vacation.

Lesson 6.3 Go Math 5th Grade Independent Practice Answer Key Question 17.
Mike bought 15 baseball cards. Rookie players are featured on \(\frac{3}{5}\) of the cards. How many cards feature rookie players?
Answer:
15 x (3/5) = 9
The rookie players featured 9 cards.
Explanation:
Mike bought 15 baseball cards. Rookie players are featured on 3/5 of the cards. Multiply 15 with 3/5the product is 9. The rookie players featured 9 cards.

Lesson Check

Fill in the bubble completely to show your answer.

Question 18.
The florist arranges a bouquet with 12 flowers. He decides that \(\frac{3}{4}\) of the flowers in the bouquet will be carnations. How many carnations will the florist need to complete the bouquet?
(A) 10
(B) 0
(C) 6
(D) 3
Answer:

12 x (3/4) = 9
In the bouquet 9 flowers are the carnations.
The florist need 0 flowers to complete the bouquet.
So, option B is correct.
Explanation:
The florist arranges a bouquet with 12 flowers. He decides that 3/4 of the flowers in the bouquet will be carnations. Multiply 12 with 3/4 the product is 9. In the bouquet 9 flowers are the carnations. The florist need 0 flowers to complete the bouquet. So, draw a circle to option B.

Question 19.
The average rainfall for each week for the last 4 weeks was \(\frac{7}{12}\) inch. How much rain fell during the last 4 weeks?
(A) 2\(\frac{1}{3}\) inches
(B) 4\(\frac{1}{12}\) inches
(C) 2 inches
(D) \(\frac{11}{12}\) inch
Answer:

4 x (7/12) = 7/3 = 2(1/3)
The rain fell during the last 4 weeks 2(1/3) inches.
So, option A is correct.
Explanation:
The average rainfall for each week for the last 4 weeks was 7/12 inch. Multiply 4 with 7/3 the product is 7/3. The fraction form 7/3 in mixed fraction is 2(1/3). The rain fell during the last 4 weeks 2(1/3) inches. So, draw a circle to option A.

Question 20.
Eric practiced for his piano recital \(\frac{3}{4}\) hour every day last week. How many hours did Eric practice last week?
(A) 3\(\frac{3}{4}\) hours
(B) 7 hours
(C) 5\(\frac{1}{4}\) hours
(D) 2\(\frac{1}{2}\) h0urs
Answer:

7 x (3/4) = 21/4
Eric practiced 21/4 hours in last week.
So, option C is correct.
Explanation:
Eric practiced for his piano recital 3/4 hour every day last week. In one week there are 7 days. Multiply 7 with 3/4 the product is 21/4 hours. So, draw a circle to option C.

Question 21.
Which does not have the same product as 4 × \(\frac{5}{9}\)?
(A) 4 × \(\frac{9}{5}\)
(B) 5 × \(\frac{4}{9}\)
(C) 2 × \(\frac{10}{9}\)
(D) 10 × \(\frac{2}{9}\)
Answer:

4 x (9/5) = 36/5
So, option A is correct.
Explanation:
The product for the expression 4 x (5/9) is 20/9. The expression that does not have the product 20/9 is 4 x (9/5). The product for 4 x (9/5) is 36/5. So, draw a circle to option A.

Go Math 5th Grade Lesson 6.3 Homework Answer Key Question 22.
Multi-Step Rose bought a dozen eggs. She used \(\frac{2}{3}\) of the eggs to make custard and \(\frac{1}{4}\) of the eggs to make an omelet. How many eggs does Rose have left?
(A) 9
(B) 4
(C) 3
(D) 1
Answer:

(2/3) x 12 = 8
She used 8 of the eggs to make custard.
(1/4) x 12 = 3
She used 3 eggs to make an omelet.
8 + 3 = 11
12 – 11 = 1
So, option D is correct.
Explanation:
Rose bought a dozen eggs. She used 2/3 of the eggs to make custard. Multiply (2/3) with 12 the product is 8.  She used 8 of the eggs to make custard. She used 1/4 of the eggs to make an omelet. Multiply 1/4 with 12 the product is 3. Add 8 with 3 the sum is 11. Subtract 11 from 12 the difference is 1. So, draw a circle to option D.

Go Math Grade 5 Lesson 6.3 Homework Answer Key Question 23.
Multi-Step Meredith’s class has 21 students. Meredith rides the bus home with \(\frac{2}{3}\) of the students in her class. How many students in Meredith’s class do not ride the bus home with her?
(A) 14
(B) 16
(C) 7
(D) 6
Answer:

21 x (2/3) = 42/3
42/3 of students ride the bus.
21 – (42/3) = (63 – 42)/3 = 21/3 = 7 students
So, option C is correct.
Explanation:
Meredith’s class has 21 students. Meredith rides the bus home with 2/3 of the students in her class. Multiply 21 with 2/3 the product is 42/3. The students 42/3 ride the bus. Subtract 42/3 from 21 the difference is 21/3. The simplified form of 21/3 is 7. So, 7 students in Meredith’s class do not ride the bus home with her. So, draw a circle to Option C.

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 6.2 Answer Key Multiply Fractions and Whole Numbers.

Texas Go Math Grade 5 Lesson 6.2 Answer Key Multiply Fractions and Whole Numbers

Investigate

Martin is planting a vegetable garden. Each row is two meters long. He wants to plant carrots along \(\frac{3}{4}\) of each row. How many meters of each row will he plant with carrots?

Multiply. \(\frac{3}{4}\) × 2
Materials; fraction strips; MathBoard

A. Place two 1-whole fraction strips side-by-side to represent the length of the garden.

B. Find 4 fraction strips all with the same denominator that fit exactly under the two wholes.

C. Draw a picture of your model. ____________________

D. Circle \(\frac{3}{4}\) of 2 on the model you drew.

E. Complete the number sentence. \(\frac{3}{4}\) × 2 = _________
So, Martin will plant carrots along _________ meters of each row.
Answer:

Multiply (3/4) × 2
A. Place two 1-whole fraction strips side-by-side to represent the length of the garden.
B. Find 4 fraction strips all with the same denominator that fit exactly under the two wholes.
C. Draw a picture of our model.

D. Circle 3/4 of 2 on the model we drew.
E. Complete the number sentence.
(3/4) × 2 = 6/4 = 3/2 or 1(1/2)
So, Martin will plant carrots along 1(1/2) meters of each row.

Draw Conclusions

Question 1.
Explain why you placed four fraction strips with me the same denominator under the two 1-whole strips.
Answer:
I wanted to divide the entire length of the two wholes into four equal parts. The 4, 1/2 fractions strips did that.

Go Math Grade 5 Lesson 6.2 Answer Key Question 2.
Explain how you would model \(\frac{3}{10}\) of 2?
Answer:
I could divide the two wholes into ten equal parts using 10, 1/5 fraction strips. I would circle 3/10 of the 1/5 strips. This would equal 3/5.

Make Connections

You can also use a model to multiply a fraction by a whole number.

Margo was helping clean up after a class party. There were 3 boxes remaining with pizza in them. Each box had \(\frac{3}{8}\) of a pizza left. How much pizza was left in all?
Materials; fraction circles
STEP 1: Find 3 × \(\frac{3}{8}\). Model three 1-whole fraction circles to represent the number of boxes containing pizza.

STEP 2: Place \(\frac{1}{8}\) fraction circle pieces on each circle to represent the amount of pizza that was left in each box. Shade the fraction circles below to show your model.

Each circle shows ________ eighths of a whole.
The 3 circles show ________ eighths of a whole.

STEP 3: Complete the number sentences.
\(\frac{3}{8}\) + \(\frac{3}{8}\) + \(\frac{3}{8}\) = ____________
3 × \(\frac{3}{8}\) = ____________
So, Margo had __________ boxes of pizza left.
Answer:
STEP 1: Find 3 × 3/8. Model three 1-whole fraction circles to represent the number of boxes containing pizza.
STEP 2: Place 1/8 fraction circle pieces on each circle to represent the amount of pizza that was left in each box. Shade the fraction circles below to show your model.

Each circle shows 3 eighths of a whole.
The 3 circles show 9 eighths of a whole.
STEP 3: Complete the number sentences.
3/8 + 3/8 + 3/8 = 9/8
3 × 3/8 = 9/8
So, Margo had 9/8 or 1(1/8) boxes of pizza left.

Math Talk
Mathematical Processes

Explain how you would know there is more than one pizza left.
Answer:

Share and Show

Use the model to find the product.

Question 1.
\(\frac{5}{6}\) × 3 = __________

Answer:

5/6 x 3 = 15/6 = 5/2 or 2(1/2)
Explanation:
In the above image we can observe three 1-whole fraction strips side-by-side. The 6 fraction strips all with the same denominator that fit exactly under the three wholes. So, drawn a circle for 5/2 on the model given. The number sentence is (5/6) × 3 = 5/2. 

Lesson 6.2 Go Math 5th Grade Key to Fractions Answer Key Pdf Question 2.
2 × \(\frac{5}{6}\) = ____________

Answer:

2 x (5/6) = 10/6 = 5/3 or 1(2/3)
Explanation:
In the above image we can observe two circles. Each circle shows 5 six’s of a whole. The 2 circles show 10 six’s of a whole. The number sentence is 2 × 5/6 = 10/6 = 5/3 or 1(2/3).

Problem Solving

Pose a Problem

Question 3.
Tarique drew the model below for a problem. Write 2 problems that can be solved using this model. One of your problems should involve multiplying a whole number by a fraction and the other problem should involve multiplying a fraction by a whole number.

Pose a problem.

Solve your problems.
Answer:

Explanation:
Pose a problem:
1. A gardener is planting flowers in 6 rows of the garden. He will plant 2/5 of the 6 rows with 6 roses. How many rows will be filled with roses.
2. A gardener planted 2/5 of a row with roses. If he plants 5 more rows like first row, how many rows of roses will there be. When all the 6 rows are planted.
Solve your problems:
1. (2/5) of 6 = (2/5) x 6 = 12/5 or 2(2/5)
2(2/5) rows will be roses.
2. 6 x (2/5) = 12/5 or 2 (2/5)
2(2/5) rows will be roses.

Question 4.
H.O.T. Multi-Step How could you change the model to give you an answer of 4\(\frac{4}{5}\)? Explain and write a new equation.
Answer:

6 x (4/5) = 24/5 or 4(4/5)
Explanation:
In the above image we can observe 6 rectangles. I change the model to give an answer of 4(4/5). I would shade 2 more sections in each rectangle to get 24/5 or 4(4/5).

Daily Assessment Task

Fill In the bubble completely to show your answer.

Question 5.
Carly mixes vinegar and baking soda for a science project. She has a spoon that measures \(\frac{1}{4}\) teaspoon. If she fills the spoon 6 times, how much baking soda will she have?
(A) \(\frac{1}{10}\) teaspoon
(B) \(\frac{2}{3}\) teaspoon
(C) 1\(\frac{1}{2}\) teaspoons
(D) 1\(\frac{3}{4}\) teaspoons
Answer:

6 x (1/4) = 3/2 = 1(1/2)
Carly have 1(1/2) baking soda.
So, option C is correct.
Explanation:
Carly mixes vinegar and baking soda for a science project. She has a spoon that measures 1/4 teaspoon. She fills the spoon 6 times. Multiply 6 with 1/4 the product is 3/2. The fraction 3/2 in mixed fraction form is 1(1/2). Carly has 1(1/2) baking soda. So, draw a circle to option C.

Lesson 6.2 Answer Key 5th Grade Go Math Question 6.
Use Tools Which multiplication problem does the model represent?

Answer:

3/8 x 4 = 3/2
So, option B is correct.
Explanation:
In the above image we can observe four 1-whole fraction strips side-by-side. The 8 fraction strips all with the same denominator that fit exactly under the four wholes. So, drawn a circle for 3/8 of 4 on the model given. The number sentence is (3/8) × 4 = 3/2. So, the multiplication (3/8) x 4 represents the above model.

Question 7.
Multi-Step Josh brought 4 small spinach pies to his baseball team party. At the end of the party, \(\frac{3}{5}\) of each pie was left. If Josh gave 2 whole pies away, what part of a pie did he have left to take home?
(A) \(\frac{2}{5}\)
(B) \(\frac{5}{6}\)
(C) \(\frac{1}{3}\)
(D) \(\frac{1}{2}\)
Answer:

4 x (3/5) = 12/5 
(12/5) – 2 = 2/5
Josh have 2/5 part of a pie left to take home.
So, option A is correct.
Explanation:
Josh brought 4 small spinach pies to his baseball team party. At the end of the party, 3/5 of each pie was left. Multiply 4 with 3/5 the product is 12/5. Josh gave 2 whole pies away. Subtract 2 from 12/5 the difference is 2/5. Josh have 2/5 part of a pie left to take home. So, draw a circle to option A.

Texas Test Prep

Go Math Answer Key Grade 5 Lesson 6.2 Question 8.
Katana has a shelf that is 5 feet long. She wants to paint a design along \(\frac{7}{10}\) of the shelf. How many feet of the shelf will Katana paint a design?
(A) 1\(\frac{2}{5}\) feet
(B) 1\(\frac{1}{5}\) feet
(C) 3\(\frac{1}{2}\) feet
(D) 4\(\frac{3}{10}\) feet
Answer:

5 x (7/10) = 7/2 = 3(1/2)
Katana paint a design 3(1/2) feet of the shelf.
So, option C is correct.
Explanation:
Katana has a shelf that is 5 feet long. She wants to paint a design along 7/10 of the shelf. Multiply 5 with 7/10 the product is 7/2. The fraction 7/2 in mixed fraction form is 3(1/2). Katana paint a design 3(1/2) feet of the shelf.
So, draw a circle to option C.

Texas Go Math Grade 5 Lesson 6.1 Homework and Practice Answer Key

Use the model to find the product.

Question 1.

Answer:

Explanation:
In the above we can observe two 1-whole fraction strips side-by-side. The 10 fraction strips all with the same denominator that fit exactly under the two wholes. So, circle is already drawn for (9/10) x 2 on the model given. The number sentence is (9/10) × 2 = 9/5. 

Question 2.

Answer:

Explanation:
In the above image, we can observe three circles. Each circle is shaded in 3 parts out of 4. The 3 circles are shaded 9 parts out of 12. The number sentence is (3/4) x 3 = 9/4.

Go Math 5th Grade Practice and Homework Lesson 6.2 Answer Key Question 3.

Answer:

Explanation:
In the above we can observe three 1-whole fraction strips side-by-side. The 24 fraction strips all with the same denominator that fit exactly under the three wholes. So, circle is already drawn for (5/8) x 3 on the model given. The number sentence is (5/8) × 3 = 15/8. 

Question 4.

Answer:

Explanation:
In the above image we can observe four circles. Each circle is shaded 5 parts out of 6. The 4 circles are shaded 20 parts out of 24. The number sentence is (5/6) x 4 = 10/3.

Question 5.

Answer:

Explanation:
In the above we can observe two 1-whole fraction strips side-by-side. The 12 fraction strips all with the same denominator that fit exactly under the two wholes. So, circle is already drawn for (7/12) x 2 on the model given. The number sentence is (7/12) × 2 = 7/6. 

Question 6.

Answer:

Explanation:
In the above image we can observe two circles. Each circle is shaded 7 parts out of 10. The 2 circles are shaded 14 parts out of 20. The number sentence is (7/10) x 2 = 7/5.

Problem Solving

Question 7.
Chef Talbot is baking 6 blueberry pies. If he uses 3/4 pint of blueberries in each pie, how many pints of blueberries will he need?
Answer:
6 x 3/4 = 9/2 = 4(1/2)
He need 4(1/2) pints of blueberries.
Explanation:
Chef Talbot is baking 6 blueberry pies. He uses 3/4 pint of blueberries in each pie. Multiply 6 with 3/4 the product is 9/2. The fraction 9/2 in mixed fraction form is 4(1/2). He need 4(1/2) pints of blueberries.

Go Math Lesson 6.2 5th Grade Fractions Answer Key Question 8.
Mr. McGregor pours \(\frac{3}{8}\) pound of dirt in each of his 4 flower pots. How much dirt does Mr. McGregor use to fill the 4 pots?
Answer:
(3/8) x 4 = 12/8 = 3/2 or 1(1/2)
McGregor needs 1(1/2) pounds of dirt to fill the 4 pots.
Explanation:
Mr. McGregor pours 3/8 pound of dirt in each of his 4 flower pots. Multiply 3/8 with 4 the product is 3/2. The fraction 3/2 in mixed fraction form is 1(1/2). McGregor needs 1(1/2) pounds of dirt to fill the 4 pots.

Lesson Check

Fill in the bubble completely to show your answer.

Question 9.
Which multiplication problem does the model represent?

Answer:

7/8 x 2 = 7/4
So, option B is correct.
Explanation:
In the above image we can observe two 1-whole fraction strips side-by-side. The 8 fraction strips all with the same denominator that fit exactly under the two wholes. So, drawn a circle for 7/8 of 2 on the model given. The number sentence is (7/8) × 2 = 7/4. So, the multiplication (7/8) x 2 represents the above model.

Question 10.
Which multiplication problem does the model represent?

Answer:

Option A is correct.
Explanation:
In the above image we can observe 3 circles. Each circle is shaded with 5 parts out 12 parts. Multiply (5/12) with 3 the product is 5/4. The multiplication problem (5/12) x 3 represents the above model.

Grade 5 Go Math Answer Key Lesson 6.2 Question 11.
Marianne is completing a 4-mile route for charIty Every \(\frac{1}{10}\) mile is marked along the route. For each mile, she runs \(\frac{7}{10}\) mile and walks \(\frac{3}{10}\) mile. How many miles does Marianne run?
(A) 1\(\frac{1}{10}\) miles
(B) 2\(\frac{4}{5}\) miles
(C) 1\(\frac{1}{5}\) miles
(D) 2\(\frac{2}{5}\) miles
Answer:

4 x (7/10) = 14/5 = 2(4/5)
Marianne runs 2(4/5) miles.
So, option B is correct.
Explanation:
Marianne is completing a 4-mile route for charity. Every 1/10 mile is marked along the route. For each mile, she runs 7/10 mile and walks 3/10 mile. Multiply 4 miles with 7/10 the product is 14/5. The fraction 14/5 in mixed fraction is 2(4/5). Marianne runs 2(4/5) miles. So, draw a circle to option B.

Question 12.
Terrance runs 5 miles each week. His brother runs \(\frac{5}{6}\) the distance Terrance runs in one week. How far does Terrance’s brother run in one week?
(A) 3\(\frac{1}{3}\) miles
(B) 4\(\frac{1}{6}\) miles
(C) 5\(\frac{1}{6}\) miles
(D) 4 miles
Answer:

5 x (5/6) = 25/6 = 4(1/6)
Terrance’s brother run 4(1/6) miles in one week.
So, option B is correct.
Explanation:
Terrance runs 5 miles each week. His brother runs 5/6 the distance Terrance runs in one week. Multiply 5 miles with 5/6 the product is 25/6. The fraction form of 25/6 in mixed fraction is 4(1/6). Terrance’s brother run 4(1/6) miles in one week. So, draw a circle to option B.

Question 13.
Multi-Step Colton’s recipe makes 2 dozen brownies. His recipe calls for \(\frac{7}{8}\) cup of vegetable oil. How much oil will Colton need to make 6 dozen brownies?
(A) 2\(\frac{5}{8}\) cups
(B) 1\(\frac{3}{4}\) cups
(C) 5\(\frac{1}{4}\) cups
(D) 3\(\frac{1}{2}\) cups
Answer:

3 x 7/8 = 21/8 = 2(5/8)
Colton need 2(5/8) cups of oil to make 6 dozen brownies.
So, option A is correct.
Explanation:
Colton’s recipe makes 2 dozen brownies. His recipe calls for 7/8 cup of vegetable oil. Multiply 3 with 7/8 the product is 21/8. The fraction 21/8 in mixed fraction form is 2(5/8). Colton need 2(5/8) cups of oil to make 6 dozen brownies. So, draw a circle to option A.

Texas Go Math Grade 5 Lesson 6.2 Answer Key Question 14.
Multi-Step Kiesha brought 3 loaves of cornbread to a football party. \(\frac{5}{12}\) of each loaf was eaten. If Kiesha gave 1 whole loaf of the leftover bread to the party hosts, what part of the loaf did she have left to take home?
(A) \(\frac{3}{4}\)
(B) \(\frac{1}{4}\)
(C) \(\frac{1}{2}\)
(D) \(\frac{7}{12}\)
Answer:

1- (5/12) = 7/12
3 x (7/12) = 7/4
(7/4) – 1 = 3/4
Kiesha left 3/4 part of a loaf to take home.
So, option A is correct.
Explanation:
Kiesha brought 3 loaves of cornbread to a football party. 5/12 of each loaf was eaten. Kiesha gave 1 whole loaf of the leftover bread to the party hosts. First subtract 5/12 from 1 the difference is 7/12. Multiply 3 loaves with 7/12 the product is 7/4. Subtract 1 from 7/4 the difference is 3/4. Kiesha left 3/4 part of a loaf to take home. So, draw a circle to option A.

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 Unit 1 Assessment Answer Key.

Texas Go Math Grade 5 Unit 1 Assessment Answer Key

Vocabulary

Choose the best term from the box.

Vocabulary
Associative Property
Commutative Property
inverse operations

Question 1.
The _________ states that changing the grouping of factors does not change the product. (p. 5)
Answer:
Commutative Property,

Explanation:
Commutative property is applicable only for addition and multiplication processes.
Thus, it means we can change the position or swap the numbers when adding or
multiplying any two numbers. This is one of the major properties of integers.
For example: 1+2 = 2+1 and 2 x 3 = 3 x 2.
therefore,  the ___Commutative Property__ states that changing the grouping of
factors does not change the product.

5th Grade Go Math Unit 1 Assessment Answer Key Question 2.
Addition and subtraction are ___________. (p. 41)
Answer:
Inverse operations,

Explanation:
Addition and subtraction are the inverse operations of each other.
Simply, this means that they are the opposite.
We can undo an addition through subtraction and
we can undo a subtraction through addition.

Concepts and Skills

Compare. Write <, >, or =. (TEKS 5.2.B)

Question 3.
6.35 0.695
Answer:
6.35 > 0.695,

Explanation:
Given to compare between 6.35 and 0.695
as 6.35 is greater than 0.695 therefore
6.35 > 0.695.

Question 4.
0.02 0.020
Answer:
0.02 = 0.020,

Explanation:
Given to compare between 0.02 and 0.020 as
0.02 is equal to 0.020 therefore 0.02 = 0.020.

Texas Go Math Grade 5 Unit 1 Answer Key Question 5.
0.132 0.2
Answer:
0.132 < 0.2,

Explanation:
Given to compare between 0.132 and 0.2
as 0.132 is less than 0.2 therefore 0.132 < 0.2.

Estimate. Then solve. (TEKS 5.3.A. 5.3.B, 5.3.C)

Question 6.
Estimate: _____9,000_______

Answer:

8760, Estimate is 9,000,

Explanation:
Given to multiply 24 with 365 we get 8760 as shown above,
as per estimation 24 ≈ 25 and 365 ≈ 360 we get 25 X  360 = 9,000.

Question 7.
Estimate: ______30______
616 ÷ 22
Answer:
22)616(28
44_
      176
      176
0

28, Estimate is 30,

Explanation:
Given to find 616 ÷ 22 we get 28 as
per estimation 616 ≈ 600 and 22 ≈ 20,
now we divide 600 ÷ 20 = 30.

Question 8.
Estimate: ____250_________
5,184 ÷ 18
Answer:
18)5184(288
     36
     158
     144
       144
       144
0
288, Estimate is 250

Explanation:
Given to find 5,184 ÷ 18 we get 288 as
per estimation 5,184 ≈ 5,200 and  ≈ 20,
now we divide 5000 ÷ 20 = 250.

Use models or strategies to find the product. Show your work. (TEKS 5.3.D, 5.3.E)

Grade 5 Unit 1 Assessment Answer Key Texas Go Math Question 9.
0.05 × 1.32
Answer:

0.05 X 1.32 = 0.066,

Explanation:
Given to find 0.05 X 1.32 if we multiply
      11
1.32
X0.05
 0.0660 the result 0.0660 shaded region is
shown in the graph.

Question 10.
23 × 5.28
Answer:
23 × 5.28 = 121.44,

Explanation:
Upon multiplying 23 X 5.28 we get
2
23
X5.28
001.84
004.60
115.00
11
121.44

Question 11.
4.2 × 14.85
Answer:
4.2 × 14.85 = 62.37,

Explanation:
Given to find 4.2 X 14.85 we get
4.2
14.85
00.21
03.36
16.80
42.00
11
62.37

Use models or strategies to find the quotient. Show your work. (TEKS 5.3.F, 5.3.G)

Question 12.
3.6 ÷ 4
Answer:
3.6 ÷ 4 = 0.9,

Explanation:
4)3.6(0.9
   3.6
0
Upon dividing 3.6 ÷ 4 we get 0.9.

Texas Go Math Grade 5 Answer Key Pdf Unit 1 Question 13.
16.24 ÷ 29
Answer:
16.24 ÷ 29 = 0.56,

Explanation:
29)16.24(0.56
       14.50
         1.74
         1.74
0   
Upon dividing 16.24 ÷ 29 we get 0.56.

Question 14.
96.72 ÷ 62
Answer:
96.72 ÷ 62 = 1.56,

Explanation:
62)96.72(1.56
     62
     34.72
     31.00
        3.72
        3.72
         0
Fill in the bubble completely to show your answer.

Question 15.
Kaya’s score in the gymnastics competition is 15.4 when rounded to the nearest tenth.
Which of the following is her actual score? (TEKS 5.2.C)
(A) 15.333
(B) 15.496
(C) 15.395
(D) 15.349
Answer:
Kaya’s actual score is (C) 15.395,

Explanation:
Given Kaya’s score in the gymnastics competition is 15.4,
when rounded to the nearest tenth her actual score out of
15.333, 15.496, 15.395, 15.349 would be 15.395 as we see
15.4 is near to and in between 15.333<15.349<15.395 and 15.496.

Question 16.
A bakery uses 1,750 kilograms of flour to make 1,000 loaves of bread.
How much flour is needed to make 10 loaves? (TEKS 5.3.G)
(A) 17.5 kilograms
(B) 175,000 kilograms
(C) 1.75 kilograms
(D) 175 kilograms
Answer:
(A) 17.5 kilograms,

Explanation:
Given a bakery uses 1,750 kilograms of flour to make 1,000 loaves of bread
now flour needed to make 1 loaves is 1,750 kilograms ÷ 1,000 = 1.75 kilograms,
flour needed to make 10 loaves is 1.75 kilograms X 10 = 17.5 kilograms,
therefore matches with (A) 17.5 kilograms.

Question 17.
Maxine paints a mural that is 4.65 meters long.
The width of the mural is 0.8 times the length.
Maxine increases the width by another 0.5 meters.
How wide is the mural? (TEKS 5.3.E, 5.3.K)
(A) 3.77 meters
(B) 37.2 meters
(C) 4.22 meters
(D) 3.72 meters
Answer:
(D) 3.72 meters,

Explanation:
Given Maxine paints a mural that is 4.65 meters long.
The width of the mural is 0.8 times the length.
So width is 4.65 X 0.8 = 3.72 meters, if Maxine increases
the width by another 0.5 meters then wide the mural is
3.72 X 0.5 meters = 3.72 meters + 1.86 meters = 3.72 meters
which matches with (D).

5th Grade Unit 1 Math Test Texas Go Math Question 18.
Juan uses the model below to solve a problem.
Which of the following equations matches Juan’s model? (TEKS 5.3.F)

(A) 0.4 × 32 = 12.8
(B) 1.28 ÷ 4 = 0.32
(C) 0.32 ÷ 4 = 0.08
(D) 0.32 + 4 = 4.32
Answer:
(A) 0.4 × 32 = 12.8,

Explanation:
Given Juan uses the 2 graph models of 10 X 10 above to solve
the problem the first graph if we count has 32 boxes of 0.4 each,
and the result  is 12.8 boxes therefore the equation is
(A) 0.4 × 32 = 12.8.

Question 19.
The price of a shirt is $26.50. The matching shorts are 0.9 times
the price of the shirt. If Li wants to buy the shirt and the shorts,
how much money will he need? (TEKS 5.3.E, 5.3.K)
(A) $238.50
(B) $35.50
(C) $23.85
(D) $50.35
Answer:
(D) $50.35,

Explanation:
Given the price of a shirt is $26.50. The matching shorts are 0.9 times
the price of the shirt. So price of the shirt is $26.50 X 0.9 = $23.85,
If Li wants to buy the shirt and the shorts the price is
$26.50 + $23.85 = $50.35 which matches with (D) above.

Question 20.
Ali’s times for the four laps of the race are: 15.36 seconds, 15.95 seconds,
17.83 seconds, and 18.25 seconds. About how long did Ali take to
complete the whole race? (TEKS 5.3.A)
(A) 47 seconds
(B) 18 seconds
(C) 15 seconds
(D) 67 seconds
Answer:
(D) 67 seconds,

Explanation:
Given Ali’s times for the four laps of the race are: 15.36 seconds, 15.95 seconds,
17.83 seconds, and 18.25 seconds. Long did Ali took to complete the whole race
is 15.36 + 15.95 + 17.83 + 18.25 = 67.39 seconds therefore whole is 67 seconds
matches with (D).

Question 21.
Goran wants to build a square picture frame with sides that are 5.25 inches long.
Natalie wants to build a square sandbox and needs 11 times the amount of wood
that Goran needs to build his frame. They have 4 pieces of wood that are each 65.5 inches long.
How much wood will they have left over after making the frame and sandbox? (TEKS 5.3.E, 5.3.K)
(A) 10 inches
(B) 2.5 inches
(C) 7.75 inches
(D) Not here
Answer:
(A) 10 inches,

Explanation:
Given Goran wants to build a square picture frame with sides that are
5.25 inches long so picture frame requires is 4 X 5.25 inches = 21 inches,
Natalie build’s a square sandbox that needs 11 times the amount of wood that
Goran needs to build his frame is 11 X 21 = 231 inches,
They have 4 pieces of wood that are each 65.5 inches long,
so wood they have is 4 X 65.5 = 262 inches,
Both Goran and Natalie needs 21 + 231 = 252 inches,
Wood will they have left over after making the frame and sandbox is
262 inches – 252 inches = 10 inches matches with (A).

Texas Go Math Grade 5 Unit 1 Assessment Answers Key Question 22.
Mustafa buys 6 cans of beans. Each can contain 12.6 ounces of beans.
Mustafa uses 0.7 of the beans in a stew and the rest of the beans for tacos.
How many ounces does he use for the tacos? (TEKS 5.3.E, 5.3.K)
(A) 21 ounces
(B) 75.60 ounces
(C) 52.92 ounces
(D) 22.68 ounces
Answer:
(C) 52.92 ounces,

Explanation:
Given Mustafa buys 6 cans of beans. Each can contains 12.6 ounces of beans.
So Mustafa has 6 X 12.6 = 75.6 ounces,
Mustafa uses 0.7 of the beans in a stew and the rest of the beans for tacos.
So Mustafa uses 75.6 X 0.7 = 52.92 ounces for the tacos matches with (C).

Question 23.
The scale at a butcher shop shows the weight of the meat as 5.363 pounds.
The butcher rounds the weight to the nearest hundredth.
Which of the following shows the new number in expanded form? (TEKS 5.2.A, 5.2.C)
(A) 5 + 0.3 + 0.06 + 0.003
(B) 5 + 0.3 + 0.07
(C) 5 + 0.3 + 0.06
(D) 5 + 3 + 6 + 3
Answer:
(A) 5 + 0.3 + 0.06 + 0.003,

Explanation:
Given the scale at a butcher shop shows the weight of the meat as 5.363 pounds.
The butcher rounds the weight to the nearest hundredth.
So the new number in expanded form is
5 X 1 + 3 X 0.1 + 6 X 0.01 + 3 X 0.003 =
5 + 0.3 + 0.06 + 0.03 matches with (A).

Question 24.
The table shows the times recorded by the top 3 swimmers in the 100 meter race.
What is the value of the digit 6 in the fastest recorded time? (TEKS 5.2.A, 5.2.B)

(A) 0.06
(B) 0.006
(C) 0.6
(D) 6
Answer:
(C) 0.6,

Explanation:
The table above showed the times recorded by the top
3 swimmers in the 100 meter race are as 51.695 seconds,
51.563 seconds and 51.536 seconds among the three the
fastest recorded time is 51.695 seconds and the value of the
digit 6 in the fastest recorded time 6 X 0.1 = 0.6 matches with (C).

Question 25.
Tickets to the school play cost $3.65 for children and $5.65 for adults.
Sonal buys tickets for 3 children and 2 adults. How much money should
she get back if she gives the cashier $50? (TEKS 5.3.E, 5.3.K)
(A) $39.05
(B) $27.75
(C) $26
(D) $38.70
Answer:
(B) $27.75,

Explanation:
Given Tickets to the school play cost $3.65 for children and $5.65 for adults.
Sonal buys tickets for 3 children and 2 adults, for 3 children the cost will be
3 X $3.65 = $10.95 and for 2 adults it will be 2 X$5.65 = $11.3,
So Sonal buys tickets of cost $10.95 + $11.3 = $22.25 in total.
Now money should she get back if she gives the cashier $50 is
$50 – $22.25 = $27.75 which matches with (B).

Texas Go Math 5th Grade Unit 1 Assessment Answers Question 26.
The prices for different beverages and snacks at a snack stand in a park
are shown on the table. Emily spent $8.11 on park snacks for her
friends and herself. Make a list of the items she may have purchased.
Justify the amount spent. (TEKS 5.3.K )

Answer:
Given amount spent by Emily is $8.11 and the actual amount spent by Emily which is approximately equal to $7.71,

Explanation:
Given the prices for different beverages and snacks at a snack stand in a park
are shown in the table. Emily spent $8.11 on park snacks for her
friends and herself. and made a list of the items she would have purchased,
so upon adding the amount purchased we get
$0.89 + $1.29 +$1.78 + $2.50 + $1.25 = $7.71 approximately to $8.11.

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 5.8 Answer Key Use Properties of Addition.

Texas Go Math Grade 5 Lesson 5.8 Answer Key Use Properties of Addition

Unlock the Problem

Jane and her family are driving to Big Lagoon State Park. On the first day, they travel \(\frac{1}{3}\) of the total distance. On the second day, they travel \(\frac{1}{3}\) of the total distance in the morning and then \(\frac{1}{6}\) of the total distance in the afternoon. How much of the total distance has Jane’s family driven by the end of the second day?

Use the Associative Property

Day 1 + Day 2

So, Jane’s family has driven _________ of the total distance by the end of the second day.
Answer:

So, Jane’s family has driven \(\frac{4}{6}\) of the total distance by the end of the second day.

Math Talk
Mathematical Processes

Explain Why grouping the fractions differently makes it easier to find the sum.
Answer:
Because, Adding 2 fraction makes very tough
so, grouping the fractions differently makes it easier to find the sum.

Example

Use the Commutative Property and the Associative Property.

Answer:

Try This!

Subtraction is not commutative or associative. When you subtract, perform operations in parentheses first. Then subtract from left to right.
a. \(\frac{7}{8}\) – \(\frac{1}{2}\) – \(\frac{1}{8}\) = __________ – \(\frac{1}{8}\) = ____________
Answer:
\(\frac{7}{8}\) – \(\frac{1}{2}\) – \(\frac{1}{8}\) =\(\frac{3}{8}\) – \(\frac{1}{8}\) = \(\frac{2}{8}\)

b. (\(\frac{7}{8}\) – \(\frac{1}{2}\)) – \(\frac{1}{8}\) = ____________ – \(\frac{1}{8}\) = ____________
Answer:
(\(\frac{7}{8}\) – \(\frac{1}{2}\)) – \(\frac{1}{8}\) = ____________ – \(\frac{1}{8}\) = ____________

c. \(\frac{7}{8}\) – (\(\frac{1}{2}\) – \(\frac{1}{8}\)) = \(\frac{3}{8}\) –\(\frac{1}{8}\)= \(\frac{2}{8}\)
Answer:
(\(\frac{7}{8}\) – \(\frac{1}{2}\)) – \(\frac{1}{8}\) = \(\frac{7}{8}\) – \(\frac{1}{8}\) = \(\frac{6}{8}\)

Explain how you can use your answers to conclude that subtraction is not associative.
Answer:
The equation one and two shows that by using associative property addition gives the same answer
but two and three shows that by using associative property subtraction gives the different answer

Share and Show

Use the properties and mental math to solve. Write your answer in the simplest form.

Question 1.
(2\(\frac{5}{8}\) + \(\frac{5}{6}\)) + 1\(\frac{1}{8}\)
Answer:
(\(\frac{21}{8}\) + \(\frac{5}{6}\)) + \(\frac{9}{8}\)
(\(\frac{63}{24}\) + \(\frac{40}{24}\)) + \(\frac{27}{8}\)
\(\frac{103}{24}\) =
\(\frac{130}{24}\)
Explanation:
Written the number sentence to represent the problem.
Used the Associative Property to group fractions with equal denominators together.
Used mental math to add the fractions with
equal denominators.
Written equivalent fractions with equal denominators and then added

Go Math Lesson 5.8 5th Grade Properties of Addition Question 2.
\(\frac{5}{12}\) + (\(\frac{5}{12}\) + \(\frac{3}{4}\))
Answer:
(\(\frac{5}{12}\) + \(\frac{9}{12}\)) + \(\frac{5}{12}\)
\(\frac{5+5+9}{12}\) =
\(\frac{19}{12}\)
Explanation:
Written the number sentence to represent the problem.
Used the Associative Property to group fractions with equal denominators together.
Used mental math to add the fractions with
equal denominators.
Written equivalent fractions with equal denominators and then added

Question 3.
(3\(\frac{1}{4}\) + 2\(\frac{5}{6}\)) + 1\(\frac{3}{4}\)
Answer:
(\(\frac{39}{12}\) + \(\frac{34}{12}\)) + \(\frac{7}{4}\)
\(\frac{94}{12}\)
Explanation:
Written the number sentence to represent the problem.
Used the Associative Property to group fractions with equal denominators together.
Used mental math to add the fractions with
equal denominators.
Written equivalent fractions with equal denominators and then added

Problem Solving

Use the properties and mental math to solve. Write your answer in simplest form.

Question 4.
(\(\frac{2}{7}\) + \(\frac{1}{3}\)) + \(\frac{2}{3}\)
Answer:
(\(\frac{2}{7}\) + \(\frac{1}{3}\)) + \(\frac{2}{3}\)
(\(\frac{6}{21}\) + \(\frac{7}{21}\)) + \(\frac{2}{3}\)
(\(\frac{13}{21}\) + \(\frac{2}{3}\))
\(\frac{13 +14}{21}\) =
\(\frac{27}{21}\)
Explanation:
Written the number sentence to represent the problem.
Used the Associative Property to group fractions with equal denominators together.
Used mental math to add the fractions with
equal denominators.
Written equivalent fractions with equal denominators and then added

Question 5.
(\(\frac{1}{5}\) + \(\frac{1}{2}\)) + \(\frac{2}{5}\)
Answer:
(\(\frac{1}{5}\) + \(\frac{1}{2}\)) + \(\frac{2}{5}\)
(\(\frac{2}{10}\) + \(\frac{5}{10}\)) + \(\frac{4}{10}\)
\(\frac{11}{10}\)
Explanation:
Written the number sentence to represent the problem.
Used the Associative Property to group fractions with equal denominators together.
Used mental math to add the fractions with
equal denominators.
Written equivalent fractions with equal denominators and then added

5th Grade Math Properties Lesson 5.8 Answer Key Question 6.
(\(\frac{1}{6}\) + \(\frac{3}{7}\)) + \(\frac{2}{7}\)
Answer:
(\(\frac{1}{6}\) + \(\frac{3}{7}\)) + \(\frac{2}{7}\)
(\(\frac{7}{42}\) + \(\frac{24}{42}\)) + \(\frac{18}{42}\)
\(\frac{49}{42}\)
Explanation:
Written the number sentence to represent the problem.
Used the Associative Property to group fractions with equal denominators together.
Used mental math to add the fractions with
equal denominators.
Written equivalent fractions with equal denominators and then added

Question 7.
Explain why grouping the fractions differently makes it easier to find the sum.
Answer: easy method
Explanation:
Written the number sentence to represent the problem.
Used the Associative Property to group fractions with equal denominators together.
Used mental math to add the fractions with
equal denominators.
Written equivalent fractions with equal denominators and then added by this process it becomes easy

Problem Solving

Use the map to solve 8-9.

Question 8.
H.O.T. Multi-Step On one afternoon, Mario walks from his house to the library. That evening, Mario walks from the library to the mall, and then to Kyle’s house. Describe how you can use the properties to find how far Mario walks.

Answer:
(\(\frac{8}{5}\) + \(\frac{2}{5}\)) + \(\frac{4}{5}\)
\(\frac{8+2+4}{5}\)
\(\frac{14}{5}\)
Explanation:
Written the number sentence to represent the problem.
Used the Associative Property to group fractions with equal denominators together.
Used mental math to add the fractions with
equal denominators.
Written equivalent fractions with equal denominators and then added by this process it becomes easy

Question 9.
H.O.T. Pose a Problem Write and solve a new problem that uses the distances between four locations.
Answer: On one evening, Mario walks from his mall to the school. That evening, Mario walks from the school to the sports complex,  Describe how you can use the properties to find how far Mario walks.
(\(\frac{2}{5}\) + \(\frac{2}{3}\))
\(\frac{6+10}{15}\)
\(\frac{16}{15}\)

Daily Assessment Task

Fill in the bubble completely to show your answer.

Question 10.
During a scavenger hunt, Ben’s team completed four tasks in the following times: 2\(\frac{1}{3}\) hours, 1\(\frac{1}{2}\) hours, 1\(\frac{1}{3}\) hours, and 1\(\frac{1}{3}\) hours. How long did it take Ben’s team to complete the scavenger hunt?
(A) 6\(\frac{1}{2}\) hours
(B) 5\(\frac{1}{2}\) hours
(C) 6 hours
(D) 5 hours
Answer: A
2\(\frac{1}{3}\) hours+1\(\frac{1}{2}\) hours+ 1\(\frac{1}{3}\) hours+1\(\frac{1}{3}\) hours.
\(\frac{14+9+8+8}{6}\) hours+\(\frac{39}{6}\)

Question 11.
Use Symbols Elijah wants to add (2\(\frac{3}{5}\) + 8\(\frac{1}{6}\)) + 5\(\frac{1}{5}\). He rewrites the problem as (8\(\frac{1}{6}\) + 2\(\frac{3}{5}\)) + 5\(\frac{1}{5}\). Then he uses the Associative Property to rewrite the problem. Which shows his next step?
(A) 8\(\frac{1}{6}\) + (2\(\frac{3}{5}\) + 5\(\frac{1}{5}\))
(B) (8\(\frac{1}{6}\)) (2\(\frac{3}{5}\) + 5\(\frac{1}{5}\))
(C) (8\(\frac{1}{6}\) + 2\(\frac{3}{5}\) + 5\(\frac{1}{5}\))
(D) (2\(\frac{3}{8}\) + 8\(\frac{1}{6}\)) (5\(\frac{1}{5}\))
Answer: A
Explanation:
His next step by using the property is
8\(\frac{1}{6}\) + (2\(\frac{3}{5}\) + 5\(\frac{1}{5}\))

Go Math 5th Grade Properties of Addition  Question 12.
Multi-Step Glen finds the sum of (3\(\frac{3}{10}\) + 4\(\frac{1}{3}\)) + 2\(\frac{1}{10}\) and Ana finds the sum of (4\(\frac{1}{3}\) + 3\(\frac{3}{10}\)) + 2\(\frac{1}{10}\). What is the total sum of their answers?
(A) 9\(\frac{11}{15}\)
(B) 9\(\frac{1}{15}\)
(C) 18\(\frac{7}{15}\)
(D) 19\(\frac{7}{15}\)
Answer: D
Explanation:
9\(\frac{11}{15}\) + 9\(\frac{11}{15}\)
19\(\frac{7}{15}\)

Texas Test Prep

Question 13.
Use the properties and mental math to solve.
\(\frac{11}{12}\) – (\(\frac{2}{3}\) – \(\frac{1}{12}\))
(A) \(\frac{7}{12}\)
(B) \(\frac{1}{3}\)
(C) \(\frac{1}{4}\)
(D) \(\frac{1}{6}\)
Answer: D
Explanation:
\(\frac{11}{12}\) – (\(\frac{2}{3}\) – \(\frac{1}{12}\))
\(\frac{1}{6}\)

Texas Go Math Grade 5 Lesson 5.8 Homework and Practice Answer Key

Use the properties and mental math to solve. Write your answer in the simplest form.

Question 1.
(\(\frac{3}{7}\) + \(\frac{2}{3}\)) + \(\frac{1}{3}\) ___________
Answer:
(\(\frac{3}{7}\) + \(\frac{2}{3}\)) + \(\frac{1}{3}\)
(\(\frac{9}{21}\) + \(\frac{14}{21}\)) + \(\frac{1}{2}\)
(\(\frac{23}{21}\) + \(\frac{1}{2}\))
\(\frac{23 + 21}{21}\) =
\(\frac{44}{21}\)
Explanation:
Written the number sentence to represent the problem.
Used the Associative Property to group fractions with equal denominators together.
Used mental math to add the fractions with
equal denominators.
Written equivalent fractions with equal denominators and then added

Question 2.
\(\frac{4}{7}\) + (\(\frac{1}{6}\) + \(\frac{2}{7}\)) ____________
Answer:
(\(\frac{4}{7}\) + \(\frac{1}{6}\)) + \(\frac{2}{7}\)
(\(\frac{7}{42}\) + \(\frac{2}{42}\)) + \(\frac{4}{7}\)
\(\frac{24 + 9}{42}\) =
\(\frac{33}{42}\)
Explanation:
Written the number sentence to represent the problem.
Used the Associative Property to group fractions with equal denominators together.
Used mental math to add the fractions with
equal denominators.
Written equivalent fractions with equal denominators and then added

Question 3.
(\(\frac{4}{5}\) + \(\frac{1}{2}\)) + \(\frac{2}{5}\) _____________
Answer:
(\(\frac{4}{5}\) + \(\frac{1}{2}\)) + \(\frac{2}{5}\)
(\(\frac{8 + 5}{10}\) + \(\frac{2}{5}\))
\(\frac{13 + 4}{10}\) =
\(\frac{17}{10}\)
Explanation:
Written the number sentence to represent the problem.
Used the Associative Property to group fractions with equal denominators together.
Used mental math to add the fractions with
equal denominators.
Written equivalent fractions with equal denominators and then added

Question 4.
(3\(\frac{5}{8}\) + \(\frac{1}{6}\)) + 2\(\frac{1}{8}\) ____________
Answer:
(3\(\frac{5}{8}\) + \(\frac{1}{6}\)) + 2\(\frac{1}{8}\)
(\(\frac{29}{8}\) + \(\frac{1}{6}\)) + \(\frac{17}{8}\)
\(\frac{87 +51 + 4 }{24}\) =
\(\frac{142}{24}\)
Explanation:
Written the number sentence to represent the problem.
Used the Associative Property to group fractions with equal denominators together.
Used mental math to add the fractions with
equal denominators.
Written equivalent fractions with equal denominators and then added

Go Math Grade 5 Answer Key Lesson 5.8 Question 5.
2\(\frac{1}{6}\) + (2\(\frac{1}{4}\) + 1\(\frac{1}{6}\)) ______________
Answer:
2\(\frac{1}{6}\) + (2\(\frac{1}{4}\) + 1\(\frac{1}{6}\))
(\(\frac{9}{4}\) + \(\frac{7}{6}\)) + \(\frac{13}{6}\)
\(\frac{26+27+14}{12}\) =
\(\frac{67}{12}\)
Explanation:
Written the number sentence to represent the problem.
Used the Associative Property to group fractions with equal denominators together.
Used mental math to add the fractions with
equal denominators.
Written equivalent fractions with equal denominators and then added

Question 6.
(\(\frac{1}{7}\) + \(\frac{1}{6}\)) + \(\frac{3}{7}\) ______________
Answer:
(\(\frac{1}{7}\) + \(\frac{1}{6}\)) + \(\frac{3}{7}\)
(\(\frac{6+7+18}{4}\) = \(\frac{31}{42}\))

Explanation:
Written the number sentence to represent the problem.
Used the Associative Property to group fractions with equal denominators together.
Used mental math to add the fractions with
equal denominators.
Written equivalent fractions with equal denominators and then added

Question 7.
(\(\frac{3}{4}\) + \(\frac{7}{12}\)) + \(\frac{5}{12}\) ____________
Answer:
(\(\frac{3}{4}\) + \(\frac{7}{12}\)) + \(\frac{5}{12}\)
(\(\frac{9+7+5}{12}\)
\(\frac{21}{12}\)

Explanation:
Written the number sentence to represent the problem.
Used the Associative Property to group fractions with equal denominators together.
Used mental math to add the fractions with
equal denominators.
Written equivalent fractions with equal denominators and then added

Question 8.
\(\frac{3}{5}\) + (2\(\frac{1}{5}\) + 1\(\frac{1}{4}\)) ______________
Answer:
\(\frac{3}{5}\) + (2\(\frac{1}{5}\) + 1\(\frac{1}{4}\))
(\(\frac{3}{5}\) + \(\frac{11}{5}\)) + \(\frac{5}{4}\)
\(\frac{12+44+25}{20}\) =
\(\frac{81}{20}\)
Explanation:
Written the number sentence to represent the problem.
Used the Associative Property to group fractions with equal denominators together.
Used mental math to add the fractions with
equal denominators.
Written equivalent fractions with equal denominators and then added

Question 9.
(3\(\frac{1}{2}\) + 1\(\frac{3}{5}\)) + 1\(\frac{1}{2}\) _____________
Answer:
(3\(\frac{1}{2}\) + 1\(\frac{3}{5}\)) + 1\(\frac{1}{2}\)
(\(\frac{7}{2}\) + \(\frac{8}{5}\)) + \(\frac{3}{2}\)
\(\frac{35+16+15}{10}\) =
\(\frac{66}{10}\)
Explanation:
Written the number sentence to represent the problem.
Used the Associative Property to group fractions with equal denominators together.
Used mental math to add the fractions with
equal denominators.
Written equivalent fractions with equal denominators and then added

Question 10.
Write a word problem for the addends \(\frac{7}{10}\), \(\frac{1}{5}\) and \(\frac{9}{10}\). Explain how you solved your problem.
Answer:
\(\frac{7}{10}\) + \(\frac{1}{5}\) + \(\frac{9}{10}\)
\(\frac{7+2+9}{10}\) =
\(\frac{18}{10}\)
Explanation:
Written the number sentence to represent the problem.
Used the Associative Property to group fractions with equal denominators together.
Used mental math to add the fractions with
equal denominators.
Written equivalent fractions with equal denominators and then added

Problem Solving

Question 11.
Last night, Halley spent \(\frac{4}{5}\) hour on math homework, \(\frac{1}{2}\) hour on science homework, and \(\frac{3}{5}\) hour on social studies homework. How long did Halley spend on homework last night?
Answer: \(\frac{19}{10}\)
Explanation
\(\frac{4}{5}\) + \(\frac{1}{2}\) + \(\frac{3}{5}\)
\(\frac{8+5+6}{10}\)
\(\frac{19}{10}\)

Question 12.
The rainfall totals at the airport for the last three months were 2\(\frac{1}{6}\) inches, 1\(\frac{2}{3}\) inches, and 1\(\frac{1}{6}\) inches. What was the total rainfall for the last three months?
Answer: \(\frac{90}{18}\)
Explanation:
2\(\frac{1}{6}\) + 1\(\frac{2}{3}\) +1\(\frac{1}{6}\)
\(\frac{39+30+21}{18}\)
\(\frac{90}{18}\)

Lesson Check

Fill in the bubble completely to show your answer.

Question 13.
Shelly volunteered at the pet shelter 2\(\frac{1}{2}\) hours in June, 1\(\frac{2}{3}\) hours in July, and the same amount of hours in August as in June. How many hours did Shelly volunteer this summer?
(A) 4\(\frac{1}{6}\) hours
(B) 5 hours
(C) 6\(\frac{2}{3}\) hours
(D) 5\(\frac{2}{3}\) hours
Answer: C
Explanation:
2\(\frac{1}{2}\) + 2\(\frac{1}{2}\) + 1\(\frac{2}{3}\)
\(\frac{40}{6}\)
6\(\frac{2}{3}\) hours

Go Math Grouping Property of Addition 5th Grade Question 14.
Erica uses a recipe for cookies that calls for 2\(\frac{3}{4}\) cups of flour and 1\(\frac{1}{2}\) cups of sugar. Erica doubles the recipe and adds the flour and sugar to a mixing bowl. What is the amount of dry ingredients in the bowl?
(A) 5\(\frac{3}{4}\) cups
(B) 8\(\frac{1}{2}\) cups
(C) 4\(\frac{1}{4}\) cups
(D) 9\(\frac{1}{2}\) cups
Answer: B
Explanation:
2\(\frac{3}{4}\) + 1\(\frac{1}{2}\)
\(\frac{34}{8}\) x 2
\(\frac{68}{2}\)
8\(\frac{1}{2}\) cups

Question 15.
Stefan wants to add (3\(\frac{5}{7}\) + 1\(\frac{1}{5}\)) + 2\(\frac{1}{7}\). He uses the Commutative Property to rewrite the problem. Which shows this step?
(A) (3\(\frac{5}{7}\) + 1\(\frac{1}{5}\) + 2\(\frac{1}{7}\))
(B) 3\(\frac{5}{7}\) + (1\(\frac{1}{5}\) + 2\(\frac{1}{7}\))
(C) (1\(\frac{1}{5}\) + 3\(\frac{5}{7}\)) + 2\(\frac{1}{7}\)
(D) (2\(\frac{1}{7}\) + 1\(\frac{1}{5}\)) + 3\(\frac{5}{7}\)
Answer: B
(3\(\frac{5}{7}\) + 1\(\frac{1}{5}\)) + 2\(\frac{1}{7}\)
3\(\frac{5}{7}\) + (1\(\frac{1}{5}\) + 2\(\frac{1}{7}\))

Question 16.
Habib is making a frame for a rectangular picture that is 2\(\frac{1}{4}\) feet by 3\(\frac{3}{8}\) feet. How many feet of wood trim should Habib buy to make the frame?
(A) 6\(\frac{3}{4}\) feet
(B) 10\(\frac{1}{4}\) feet
(C) 11\(\frac{1}{4}\) feet
(D) 5\(\frac{5}{7}\) feet
Answer:

Question 17.
Multi-Step Juliana used 10\(\frac{1}{2}\) yards of yarn to make three yarn dolls. She used 4\(\frac{1}{2}\) yards of yarn for the first doll and 2\(\frac{1}{5}\) yards for the second doll. How much yarn did Juliana use for the third doll?
(A) 3\(\frac{4}{5}\) yards
(B) 2\(\frac{4}{5}\) yards
(C) 3\(\frac{1}{5}\) yards
(D) 2\(\frac{1}{5}\) yards
Answer: A
Explanation:
4\(\frac{1}{2}\) + 2\(\frac{1}{5}\)
\(\frac{67}{10}\) + 10\(\frac{1}{2}\)
3\(\frac{4}{5}\) yards

Question 18.
Multi-Step Yuan finds the sum of \(\frac{3}{10}\) + (\(\frac{1}{5}\) + \(\frac{1}{10}\)). Then he adds \(\frac{1}{15}\) to the sum. What is Yuan’s final sum?
(A) \(\frac{3}{5}\)
(B) \(\frac{1}{5}\)
(C) \(\frac{1}{3}\)
(D) \(\frac{2}{3}\)
Answer: D
Explanation:
\(\frac{3}{10}\) + (\(\frac{1}{5}\) + \(\frac{1}{10}\))
\(\frac{3+2+1}{10}\)
\(\frac{6}{10}\) + \(\frac{1}{15}\)
\(\frac{75}{150}\)
\(\frac{2}{3}\)