McGraw Hill Math

Practice questions available in McGraw Hill Math Grade 6 Answer Key PDF Lesson 4.3 Absolute Value will engage students and is a great way of informal assessment.

McGraw-Hill Math Grade 6 Answer Key Lesson 4.3 Absolute Value

Exercises
Solve

Question 1.
|-3| = ____
Answer:
The absolute value of |-3| = 0.

Explanation:
The absolute value is the distance between a number and zero.
The distance between −3 and 0 is 3 .

Question 2.
|-26| = ____
Answer:
The absolute value  of |-26| = 26.

Explanation:
The absolute value is the distance between a number and zero.
The distance between -26 and 0 is 26. .

Question 3.
|423| = ____
Answer:
The absolute value of |423| = 423.

Explanation:
The absolute value is the distance between a number and zero.
The distance between 423  and 0 is 423 .

Question 4.
|-.7| = ____
Answer:
The absolute value of |-.7| = .7.

Explanation:
The absolute value is the distance between a number and zero.
The distance between -.7 and 0 = .7.

Question 5.
|-\(\frac{5}{6}\)| = ____
Answer:
The absolute value of |-\(\frac{5}{6}\)| = \(\frac{5}{6}\).

Explanation:
The absolute value is the distance between a number and zero.
The distance between –\(\frac{5}{6}\) and 0 = \(\frac{5}{6}\) .

Question 6.
|\(\frac{2}{3}\)| = ____
Answer:
The absolute value of |\(\frac{2}{3}\)| = \(\frac{2}{3}\).

Explanation:
The absolute value is the distance between a number and zero.
The distance between \(\frac{2}{3}\) and 0 =  \(\frac{2}{3}\).

Compare using <, >, or =.

Question 7.
|—6| ____ |—7|
Answer:
|—6| < |—7|

Explanation:
|—6| = 6.
|—7| = 7.
6 is lesser than 7.

Question 8.
|21| ___ |—21|
Answer:
|21| = |—21|

Explanation:
|21| = 21.
|—21| = 21.
|21| is equal to  |—21|.

Question 9.
|—8| ___ |—4|
Answer:
|—8| > |—4|

Explanation:
|—8| = 8.
|—4| = 4.
8 is greater than 4.

Question 10.
|—5| ___ |7|
Answer:
|—5| < |7|

Explanation:
|—5| = 5.
|7| = 7.
5 is lesser than 7.

Question 11.
|10| ______ |—12|
Answer:
|10| < |—12|

Explanation:
|10| = 10.
|—12| = 12.
10 is lesser than 12.

Question 12.
|423| ______ |425|
Answer:
|423| < |425|

Explanation:
|423| = 423.
|425| = 425.
423 is lesser than 425.

Question 13.
If Elena’s bank account balance is -$53.86, how much would she need to deposit in order to bring her balance to $10.00? ________
Answer:
$63.86 she needs to deposit in order to bring her balance to $10.00.

Explanation:
Amount of Elena’s bank account = -$53.86.
Amount of Elena’s needs to get the balance to $10.00:
-$53.86 + ?? = $10.
=> ?? = $10 + $53.86
=> ?? = $63.86.

Question 14.
If the temperature was 4 degrees lower than normal on Tuesday and 6 degrees higher than normal on Wednesday, which day’s temperature was more different than normal?
Answer:
Tuesday’s temperature was more different than normal because its decreasing than normal temperature.

Explanation:
Let the normal temperature be x.
Temperature on Tuesday = x – 4.
Temperature on Wednesday = x  + 6.

Question 15.
Vivian’s bank account balance is -$3.46. Evelyn’s bank account balance is -$2.98. Who has the greater debt? _______________________
Answer:
Amount of Vivian’s bank account has greater debt than Amount of Evelyn’s bank account which say his debt are more.

Explanation:
Amount of Vivian’s bank account = -$3.46.
Amount of Evelyn’s bank account = -$2.98.

Practice questions available in McGraw Hill Math Grade 6 Answer Key PDF Lesson 5.1 Plotting Ordered Pairs will engage students and is a great way of informal assessment.

McGraw-Hill Math Grade 6 Answer Key Lesson 5.1 Plotting Ordered Pairs

Exercises Give Ordered Pairs

Question 1.
Give the ordered pair for each point on the graph.

Answer:

Explanation:
Ordered pairs of points in the graph:
A = (1,2)
B = (3,-3)
C = (-5,-5)
D = (5,5)
E = (4,-5)
F = (1,1)
G = (-1,1)
H = (-4,4)
I = (8,8)
J = (-5,6)

Exercises
Plot Ordered Pairs

Plot the ordered pairs on the graph.

Question 1.
A(2, 3)
Answer:

Explanation:
A(2, 3) represents (x, y) axis points in the graph.

Question 2.
B(4, -4)
Answer:

Explanation:
B(4, -4)represents (x, y) axis points in the graph.

Question 3.
C(4, 4)
Answer:

Explanation:
C(4, 4) represents C(x, y) axis points in the graph.

Question 4.
D(-4, 4)
Answer:

Explanation:
D(-4, 4) represents D(x, y) axis points in the graph.

Question 5.
E(-4, 4)
Answer:

Explanation:
E(-4, 4) represents E(x, y) axis points in the graph.

Question 6.
F(5, 2)
Answer:

Explanation:
F(5, 2) represents F(x, y) axis points in the graph.

Question 7.
G(8, 2)
Answer:

Explanation:
G(8, 2) represents G(x, y) axis points in the graph.

Question 8.
H(8, 6)
Answer:

Explanation:
H(8, 6) represents H(x, y) axis points in the graph.

Question 9.
J(-5, 6)
Answer:

Explanation:
J(-5, 6) represents J(x, y) axis points in the graph.

Question 10.
K(3, -4)
Answer:

Explanation:
K(3, -4)represents K(x, y) axis points in the graph.

Question 11.
L(-2, -2)
Answer:

Explanation:
L(-2, -2) represents L(x, y) axis points in the graph.

Practice questions available in McGraw Hill Math Grade 6 Answer Key PDF Lesson 5.2 Distance will engage students and is a great way of informal assessment.

McGraw-Hill Math Grade 6 Answer Key Lesson 5.2 Distance

Exercises
Find The Distance Between The Points

Question 1.
(2, 5) and (2, 8) _________
Answer:

Explanation:
1. coordinates points = (2,5)
2. coordinates points = (2,8)
Distance between the two points = 3.

Question 2.
(-2, 1) and (-2, 3) _________
Answer:

Explanation:
1. coordinates points = (-2, 1)
2. coordinates points = (-2, 3)
Distance between the two points = 2.

Question 3.
(1, -4) and (1, 1) _________
Answer:

Explanation:
1. coordinates points = (1, -4)
2. coordinates points = (1, 1)
Distance between the two points = 5.

Question 4.
(3, 2) and (6, 2) _____
Answer:

Explanation:
1. coordinates points = (3, 2)
2. coordinates points = (6, 2)
Distance between the two points = 3.

Question 5.
(2, 3) and (-2, 3) _____
Answer:

Explanation:
1. coordinates points = (2,3)
2. coordinates points = (-2, 3)
Distance between the two points = 4.

Question 6.
(-4, -1) and (3, -1) _____
Answer:

Explanation:
1. coordinates points = (-4, -1)
2. coordinates points = (3, -1)
Distance between the two points = 7.

Practice questions available in McGraw Hill Math Grade 6 Answer Key PDF Lesson 6.1 Changing Improper Fractions to Mixed Numbers will engage students and is a great way of informal assessment.

McGraw-Hill Math Grade 6 Answer Key Lesson 6.1 Changing Improper Fractions to Mixed Numbers

Exercises
Change to Mixed Numbers

Question 1.
\(\frac{22}{7}\)
Answer:
Mixed fraction of \(\frac{22}{7}\) = 3\(\frac{1}{7}\)

Explanation:
\(\frac{22}{7}\) = 3 + \(\frac{1}{7}\)
= 3\(\frac{1}{7}\)

Question 2.
\(\frac{35}{4}\)
Answer:
Mixed fraction of \(\frac{35}{4}\) = 8\(\frac{3}{4}\)

Explanation:
\(\frac{35}{4}\) = 8 + \(\frac{3}{4}\)
= 8 \(\frac{3}{4}\)

Question 3.
\(\frac{73}{10}\)
Answer:
Mixed fraction of \(\frac{73}{10}\) = 7\(\frac{3}{10}\)

Explanation:
\(\frac{73}{10}\) = 7 + \(\frac{3}{10}\)
= 7\(\frac{3}{10}\)

Question 4.
\(\frac{47}{3}\)
Answer:
Mixed fraction of \(\frac{47}{3}\) = 15\(\frac{2}{3}\)

Explanation:
\(\frac{47}{3}\) = 15 + \(\frac{2}{3}\)
= 15\(\frac{2}{3}\)

Question 5.
\(\frac{87}{11}\)
Answer:
Mixed fraction of \(\frac{87}{11}\) = 7\(\frac{10}{11}\)

Explanation:
\(\frac{87}{11}\) = 7 + \(\frac{10}{11}\)
= 7\(\frac{10}{11}\)

Question 6.
\(\frac{35}{6}\)
Answer:
Mixed fraction of \(\frac{35}{6}\) = 5\(\frac{5}{6}\)

Explanation:
\(\frac{35}{6}\) = 5 + \(\frac{5}{6}\)
= 5\(\frac{5}{6}\)

Question 7.
\(\frac{26}{5}\)
Answer:
Mixed fraction of \(\frac{26}{5}\) = 5\(\frac{1}{5}\)

Explanation:
\(\frac{26}{5}\) = 5 + \(\frac{1}{5}\)
= 5\(\frac{1}{5}\)

Question 8.
\(\frac{111}{8}\)
Answer:
Mixed fraction of latex]\frac{111}{8}[/latex] = 13\(\frac{7}{8}\)

Explanation:
latex]\frac{111}{8}[/latex] = 13 + latex]\frac{7}{8}[/latex]
= 13\(\frac{7}{8}\)

Question 9.
\(\frac{32}{3}\)
Answer:
Mixed fraction of \(\frac{32}{3}\) = 10\(\frac{2}{3}\)

Explanation:
\(\frac{32}{3}\) = 10 + \(\frac{2}{3}\)
= 10\(\frac{2}{3}\)

Question 10.
\(\frac{66}{5}\)
Answer:
Mixed fraction of \(\frac{66}{5}\) = 13\(\frac{1}{5}\)

Explanation:
\(\frac{66}{5}\) = 13 + \(\frac{1}{5}\)
= 13\(\frac{1}{5}\)

Question 11.
\(\frac{211}{11}\)
Answer:
Mixed fraction of \(\frac{211}{11}\) = 19\(\frac{2}{11}\)

Explanation:
\(\frac{211}{11}\) = 19 + \(\frac{2}{11}\)
= 19\(\frac{2}{11}\)

Question 12.
\(\frac{21}{4}\)
Answer:
Mixed fraction of \(\frac{21}{4}\) = 5\(\frac{1}{4}\)

Explanation:
\(\frac{21}{4}\) = 5 + \(\frac{1}{4}\)
= 5\(\frac{1}{4}\)

Question 13.
\(\frac{78}{7}\)
Answer:
Mixed fraction of \(\frac{78}{7}\) = 11\(\frac{1}{7}\)

Explanation:
\(\frac{78}{7}\) = 11 + \(\frac{1}{7}\)
= 11\(\frac{1}{7}\)

Question 14.
\(\frac{82}{9}\)
Answer:
Mixed fraction of \(\frac{82}{9}\) = 9\(\frac{1}{9}\)

Explanation:
latex]\frac{82}{9}[/latex] = 9 + \(\frac{1}{9}\)
= 9\(\frac{1}{9}\)

Question 15.
\(\frac{13}{2}\)
Answer:
Mixed fraction of \(\frac{13}{2}\) = 6\(\frac{1}{2}\)

Explanation:
\(\frac{13}{2}\) = 6 + \(\frac{1}{2}\)
= 6\(\frac{1}{2}\)

Question 16.
\(\frac{67}{4}\)
Answer:
Mixed fraction of \(\frac{67}{4}\) = 16\(\frac{3}{4}\)

Explanation:
\(\frac{67}{4}\) = 16 + \(\frac{3}{4}\)
= 16\(\frac{3}{4}\)

Question 17.
\(\frac{11}{8}\)
Answer:
Mixed fraction of \(\frac{11}{8}\) = 1\(\frac{3}{8}\)

Explanation:
\(\frac{11}{8}\) = 1 + \(\frac{3}{8}\)
= 1\(\frac{3}{8}\)

Question 18.
\(\frac{43}{14}\)
Answer:
Mixed fraction of \(\frac{43}{14}\) = 3\(\frac{1}{14}\)

Explanation:
\(\frac{43}{14}\) = 3 + \(\frac{1}{14}\)
= 3\(\frac{1}{14}\)

Question 19.
\(\frac{137}{13}\)
Answer:
Mixed fraction of \(\frac{137}{13}\) = 10\(\frac{7}{13}\)

Explanation:
\(\frac{137}{13}\) = 10 + \(\frac{7}{13}\)
= 10\(\frac{7}{13}\)

Question 20.
\(\frac{45}{2}\)
Answer:
Mixed fraction of \(\frac{45}{2}\) = 22\(\frac{1}{2}\)

Explanation:
\(\frac{45}{2}\) = 22 + \(\frac{1}{2}\)
= 22\(\frac{1}{2}\)

Practice questions available in McGraw Hill Math Grade 6 Answer Key PDF Lesson 6.2 Changing Mixed Numbers to Improper Fractions will engage students and is a great way of informal assessment.

McGraw-Hill Math Grade 6 Answer Key Lesson 6.2 Changing Mixed Numbers to Improper Fractions

Exercises
Change to Improper Fractions

Question 1.
2\(\frac{2}{3}\)
Answer:
Improper fraction of 2\(\frac{2}{3}\) = \(\frac{8}{3}\)

Explanation:
2\(\frac{2}{3}\) = [(2 × 3) + 2] ÷ 3
= (6 + 2) ÷ 3
= \(\frac{8}{3}\)

Question 2.
5\(\frac{4}{7}\)
Answer:
Improper fraction of 5\(\frac{4}{7}\) = \(\frac{39}{7}\)

Explanation:
5\(\frac{4}{7}\) = [(5 × 7) + 4)] ÷ 7
= (35 + 4) ÷ 7
= \(\frac{39}{7}\)

Question 3.
21\(\frac{3}{5}\)
Answer:
Improper fraction of 21\(\frac{3}{5}\) = \(\frac{108}{5}\)

Explanation:
21\(\frac{3}{5}\) = [(21 × 5) + 3)] ÷ 5
= (105 + 3) ÷ 5
= \(\frac{108}{5}\)

Question 4.
5\(\frac{3}{8}\)
Answer:
Improper fraction of 5\(\frac{3}{8}\) =

Explanation:
5\(\frac{3}{8}\) = [(5 × 8) + 3] ÷ 8
= (40 + 3) ÷ 8
= \(\frac{43}{8}\)

Question 5.
22\(\frac{6}{7}\)
Answer:
Improper fraction of 22\(\frac{6}{7}\) = \(\frac{160}{7}\)

Explanation:
22\(\frac{6}{7}\) = [(22 × 7) + 6] ÷ 7
= (154 + 6) ÷ 7
= \(\frac{160}{7}\)

Question 6.
15\(\frac{4}{11}\)
Answer:
Improper fraction of 5\(\frac{4}{11}\) = \(\frac{59}{11}\)

Explanation:
5\(\frac{4}{11}\) = [(5 × 11) + 4] ÷ 11
= (55 + 4) ÷ 11
= \(\frac{59}{11}\)

Question 7.
13\(\frac{2}{3}\)
Answer:
Improper fraction of 13\(\frac{2}{3}\) = \(\frac{41}{3}\)

Explanation:
13\(\frac{2}{3}\) = [(13 × 3) + 2] ÷ 3
= (39 + 2) ÷ 3
= \(\frac{41}{3}\)

Question 8.
3\(\frac{11}{17}\)
Answer:
Improper fraction of 3\(\frac{11}{17}\) = \(\frac{62}{17}\)

Explanation:
3\(\frac{11}{17}\) = [(3 × 17) + 11] ÷ 17
= (51 + 11) ÷ 17
= \(\frac{62}{17}\)

Question 9.
2\(\frac{4}{19}\)
Answer:
Improper fraction of 2\(\frac{4}{19}\) = \(\frac{42}{19}\)

Explanation:
2\(\frac{4}{19}\) = [(2 × 19) + 4] ÷ 19
= (38 + 4) ÷ 19
= \(\frac{42}{19}\)

Question 10.
6\(\frac{3}{4}\)
Answer:
Improper fraction of 6\(\frac{3}{4}\) = \(\frac{27}{4}\)

Explanation:
6\(\frac{3}{4}\) = [(6 × 4) + 3] ÷ 4
= (24 + 3) ÷ 4
= \(\frac{27}{4}\)

Question 11.
1\(\frac{1}{51}\)
Answer:
Improper fraction of 1\(\frac{1}{51}\) = \(\frac{52}{51}\)

Explanation:
1\(\frac{1}{51}\) = [(1 × 51) + 1] ÷ 51
= (51 + 1) ÷ 51
= \(\frac{52}{51}\)

Question 12.
55\(\frac{1}{2}\)
Answer:
Improper fraction of 55\(\frac{1}{2}\) = \(\frac{111}{2}\)

Explanation:
55\(\frac{1}{2}\) = [(55 × 2) + 1] ÷ 2
= (110 + 1) ÷ 2
= \(\frac{111}{2}\)

Question 13.
10\(\frac{2}{23}\)
Answer:
Improper fraction of 10\(\frac{2}{23}\) = \(\frac{232}{23}\)

Explanation:
10\(\frac{2}{23}\) = [(10 × 23) + 2] ÷ 23
= (230 + 2) ÷ 23
= \(\frac{232}{23}\)

Question 14.
6\(\frac{4}{7}\)
Answer:
Improper fraction of 6\(\frac{4}{7}\) = \(\frac{46}{7}\)

Explanation:
6\(\frac{4}{7}\) = [(6 × 7) + 4] ÷ 4
= (42 + 4) ÷ 4
= \(\frac{46}{7}\)

Question 15.
13\(\frac{2}{7}\)
Answer:
Improper fraction of 13\(\frac{2}{7}\) = \(\frac{93}{7}\)

Explanation:
13\(\frac{2}{7}\) = [(13 × 7) + 2] ÷ 7
= [91 + 2) ÷ 7
= \(\frac{93}{7}\)

Question 16.
42\(\frac{1}{3}\)
Answer:
Improper fraction of 42\(\frac{1}{3}\) = \(\frac{127}{3}\)

Explanation:
42\(\frac{1}{3}\) = [(42 × 3) + 1] ÷ 3
= (126 + 1) ÷ 3
= \(\frac{127}{3}\)

Question 17.
5\(\frac{1}{19}\)
Answer:
Improper fraction of 5\(\frac{1}{19}\) = \(\frac{96}{19}\)

Explanation:
5\(\frac{1}{19}\) = [(5 × 19) + 1] ÷ 19
= (95 + 1) ÷ 19
= \(\frac{96}{19}\)

Question 18.
12\(\frac{2}{3}\)
Answer:
Improper fraction of 12\(\frac{2}{3}\) = \(\frac{38}{3}\)

Explanation:
12\(\frac{2}{3}\) = [(12 × 3) + 2] ÷ 3
= (36 + 2) ÷ 3
= \(\frac{38}{3}\)

Question 19.
2\(\frac{3}{4}\)
Answer:
Improper fraction of 2\(\frac{3}{4}\) = \(\frac{11}{4}\)

Explanation:
2\(\frac{3}{4}\) = [(2 × 4) + 3] ÷ 4
= (8 + 3) ÷ 4
= \(\frac{11}{4}\)

Question 20.
200\(\frac{33}{100}\)
Answer:
Improper fraction of 200\(\frac{33}{100}\) = \(\frac{20033}{100}\)

Explanation:
200\(\frac{33}{100}\) = [(200 × 100) + 33] ÷ 100
= (20000 + 33) ÷ 100
= \(\frac{20033}{100}\)

Practice questions available in McGraw Hill Math Grade 6 Answer Key PDF Lesson 25.1 Measures of Central Tendency (Mean, Median, Range) will engage students and is a great way of informal assessment.

McGraw-Hill Math Grade 6 Answer Key Lesson 25.1 Measures of Central Tendency (Mean, Median, Range)

Exercise

CALCULATE

Question 1.
What is the range for this set of numbers?
1, 8, 14, 22, 13, 19
Answer:
Given data is 1, 8, 14, 22, 13, 19
lowest number = 1
highest number = 22
range = high – low
range = 22 – 21
So, the range of the number is 21

Question 2.
What is the average for this set of numbers?
2, 2, 3, 3, 4, 6, 6, 10
Answer:
Given data is 2, 2, 3, 3, 4, 6, 6, 10
lowest number = 2
highest number = 10
Range = 10 – 2 = 8
So, the range of the number is 8.

Question 3.
What is the mode of this set of numbers?
2, 5, 10, 15, 6, 5, 6, 2, 5
Answer:
Given data is 2, 5, 10, 15, 6, 5, 6, 2, 5
2 – 2 times
5 – 3 times
6 – 2 times
10 – 1 time
15 – 1 time
The number 5 is repeated 3 times.
So, the mode is 5.

Question 4.
What is the median for this set of numbers?
3, 7, 2, 8, 9, 4, 6
Answer:
Given data 3, 7, 2, 8, 9, 4, 6
Set the numbers in the ascending or descending order.
2, 3, 4, 6, 7, 8, 9
The Median is the middle number.
Here 6 is in the middle.
So, the median is 6.

Question 5.
What is the range for this set of numbers?
2, 4, 9, 34, 35, 42, 15, 14
Answer:
Given data is 2, 4, 9, 34, 35, 42, 15, 14
Low number = 2
Highest number = 42
Range = 42 – 2
Range = 40
So, the range of the numbers is 40.

Question 6.
What is the mean for this set of numbers?
10, 20, 30, 69, 79, 89, 3
Answer:
Given data is 10, 20, 30, 69, 79, 89, 3
Mean = sum of observations/number of observations
Mean = (10 + 20 + 30 + 69 + 79 + 89 + 3)/7
Mean = 42.8
So, the mean of the set of numbers is 42.8

Question 7.
George earns tips of $10 for Sunday, $15 for Monday, and $35 for Tuesday. What is the average of his tips for the three days?
Answer:
Given data,
George earns tips of $10 for Sunday, $15 for Monday, and $35 for Tuesday.
Average = (10 + 15 + 35)/3
Average = 60/3
Average = 20
Therefore the average of his tips for the three days is $20.

Question 8.
If a group of 20 students in a class has an average age of 13 years, what would happen to the average if a student who was 12 years old was replaced with a student who was 14 years old?
Answer:
The average age of 20 students is 13.
sum/20 = 13
sum = 13 × 20
sum = 260
average = (12 + 14)/2 = 13
If 12 years old was replaced with a student who was 14 years old
average = (14 + 14)/2 = 14
So, the average becomes 14.

All the solutions provided in McGraw Hill Math Grade 1 Answer Key PDF Chapter 6 Lesson 1 Counting from 100 to 120 by 1s as per the latest syllabus guidelines.

McGraw-Hill Math Grade 1 Answer Key Chapter 6 Lesson 1 Counting from 100 to 120 by 1s

Count

Write the missing numbers.

Question 1.
Count from 100 to 103.
100 101 102 103
Explanation:
The given pattern is skip by ones
In skip by ones we add one to it

Question 2.
Count from 106 to 108.
___ ___ ____
Answer:
106 107 108
Explanation:
The given pattern is skip by ones
In skip by ones we add one to it

Question 3.
Count from 111 to 113.
___ ___ ____
Answer:
111 112 113
Explanation:
The given pattern is skip by ones
In skip by ones we add one to it

Question 4.
Count from 112 to 114.
___ ___ ____
Answer:
112 113 114
Explanation:
The given pattern is skip by ones
In skip by ones we add one to it

Question 5.
Count from 105 to 107.
___ ___ ____
Answer:
105 106 107
Explanation:
The given pattern is skip by ones
In skip by ones we add one to it

Question 6.
Count from 102 to 104.
___ ___ ____
Answer:
102 103 104
Explanation:
The given pattern is skip by ones
In skip by ones we add one to it

All the solutions provided in McGraw Hill Math Grade 1 Answer Key PDF Chapter 6 Lesson 2 Counting from 0 to 120 by 10s as per the latest syllabus guidelines.

McGraw-Hill Math Grade 1 Answer Key Chapter 6 Lesson 2 Counting from 0 to 120 by 10s

Count

Count by tens out loud. Write the missing numbers.

Question 1.
Count from 10 to 40 by 10s.
10 20 30 40
Explanation:
The given pattern is skip by tens
In skip by tens we have  to add ten to it

Question 2.
Count from 80 to 110 by 10s.
80 90 ___ ___
Answer:
80 90 100 110
Explanation:
The given pattern is skip by tens
In skip by tens we have to  add ten to it

Question 3.
Count from 20 to 50 by 10s.
____ ____ ___ ___
Answer:
20 30 40 50
Explanation:
The given pattern is skip by tens
In skip by tens we have to  add ten to it

Question 4.
Count from 80 to 110 by 10s.
___ ____ ___ ___
Answer:
80 90 100 110
Explanation:
The given pattern is skip by tens
In skip by tens we have to  add ten to it

Question 5.
Count from 0 to 30 by 10s.
___ ___ ___ ___
Answer:
0 10 20 30
Explanation:
The given pattern is skip by tens
In skip by tens we have to  add ten to it

Question 6.
Count from 50 to 80 by 10s.
___ ___ ____ ____
Answer:
50 60 70 80
Explanation:
The given pattern is skip by tens
In skip by tens we have to  add ten to it

All the solutions provided in McGraw Hill Math Grade 1 Answer Key PDF Chapter 6 Lesson 3 Counting from 0 to 100 by 5s as per the latest syllabus guidelines.

McGraw-Hill Math Grade 1 Answer Key Chapter 6 Lesson 3 Counting from 0 to 100 by 5s

Çount

Count by 5s. Write the numbers.

Question 1.
Count from 5 to 20 by 5s.
5 10 15 20
Explanation:
The given pattern is skip by fives
In skip by fives we have to add five to it

Question 2.
Count from 30 to 45 by 5s.
30 35 ____ ____
Answer:
30 35 40 45
Explanation:
The given pattern is skip by fives
In skip by fives we have to add five to it

Question 3.
Count from 55 to 70 by 5s.
55 60 ____ ___
Answer:
55 60 65 70
Explanation:
The given pattern is skip by fives
In skip by fives we have to add five to it

Question 4.
Count from 35 to 50 by 5s.
___ ___ ___ ____
Answer:
35 40 45 50
Explanation:
The given pattern is skip by fives
In skip by fives we have to add five to it

Question 5.
Count from 60 to 75 by 5s.
___ ___ ___ ____
Answer:
60 65 70 75
Explanation:
The given pattern is skip by fives
In skip by fives we have to add five to it

Question 6.
Count from 80 to 95 by 5s.
___ ___ ___ ____
Answer:
80 85 90 95
Explanation:
The given pattern is skip by fives
In skip by fives we have to add five to it

Practice the questions of McGraw Hill Math Grade 8 Answer Key PDF Lesson 6.9 Simple and Compound Interest to secure good marks & knowledge in the exams.

McGraw-Hill Math Grade 8 Answer Key Lesson 6.9 Simple and Compound Interest

Exercises Calculate

Question 1.
How much interest would you earn if you put $500 in a bank for 15 years and received simple interest of 8%?
Answer:
$ 600.00,

Explanation:
Simple Interest SI = \(\frac{P X R X T}{100}\),
P = Principle Amount,
R =  Rate of interest in %,
T = Time (Number of years),
Amount = Principle + Simple Interest,
SI = Amount – Principle,
P= $500,
R = 8%,
T = 15 yrs,
Simple Interest SI = \(\frac{P X R X T}{100}\),
= \(\frac{500 X 8 X 15}{100}\),
= 5 X 8 X 15,
= 600.

Question 2.
Calculate the simple interest on a bank account where you deposit $500 and earn 12% a year for 5 years.
Answer:
$300.00,

Explanation:
Simple Interest SI = \(\frac{P X R X T}{100}\),
P = Principle Amount,
R =  Rate of interest in %,
T = Time (Number of years),
Amount = Principle + Simple Interest,
SI = Amount – Principle,
P= $500,
R = 12%,
T = 5 yrs,
Simple Interest SI = \(\frac{P X R X T}{100}\),
= \(\frac{500 X 12 X 5}{100}\),
= 5 X 12 X 5,
= 300.

Question 3.
Calculate the ending balance of your savings account if you deposit $400 and earn simple interest of 7% for 5 years.
Answer:
$540.0,

Explanation:
Simple Interest SI = \(\frac{P X R X T}{100}\),
P = Principle Amount,
R =  Rate of interest in %,
T = Time (Number of years)
Amount = Principle + Simple Interest,
SI = Amount – Principle,
P= $400,
R = 7%,
T = 5 yrs,
Simple Interest SI = \(\frac{P X R X T}{100}\),
= \(\frac{400 X 7 X 5}{100}\),
= 4 X 7 X 5,
= 140
the ending balance = Principle + Interest,
A = 400 + 140 = 540.

Question 4.
Calculate the ending balance of your savings account if you deposited $1,000 and earned simple interest of 6% for 6 years.
Answer:
$1,360.00,

Explanation:
P = Principle Amount,
R =  Rate of interest in %,
T = Time (Number of years),
Amount = Principle + Simple Interest,
SI = Amount – Principle,
P= $1,000,
R = 6%,
T = 6 yrs,
Simple Interest SI = \(\frac{P X T X R}{100}\),
= \(\frac{1000 X 6 X 6}{100}\),
= 10 X 6 X 6,
= 360, The ending balance = Principle + Interest
A = $1,000 + $360 = $1,360.

Exercises Calculate

Question 5.
Calculate the interest earned over a 5-year period when you deposit $2,000 and earn compound interest of 8% per year.
Answer:
$938.66

Explanation:
Compound Interest CI = P [ 1 + \(\frac{R}{100}\) ]ⁿ – 1 ]
P = Principle Amount,
R =  Rate of interest in %,
T = Time (Number of years),
P= $1,000,
R = 8%,
n = 5 yrs,
Compound Interest CI = P [ 1 + \(\frac{R}{100}\) ]ⁿ – 1 ],
CI = 2000 [ 1 + \(\frac{8}{100}\)]⁵ – 1 ],
=2000 \(\frac{108}{100}\)⁵ – 1 ],
= 2000 X 0.469,
= $938.66.

Question 6.
How much interest would you earn if you put $500 in a bank for 20 years and received a compound interest rate of 4%?
Answer:
$595.56,

Explanation:
Compound Interest CI = P [ 1 + \(\frac{R}{100}\) ]ⁿ – 1 ],
P = Principle Amount,
R =  Rate of interest in %,
T = Time (Number of years),
P= $500,
R = 4%,
n = 20 yrs,
Compound Interest CI = P [ 1 + \(\frac{R}{100}\) ]ⁿ – 1 ],
CI = 500 [ 1 + \(\frac{4}{100}\)]²⁰ – 1 ],
=500 \(\frac{108}{100}\)⁵ – 1 ],
= 500 [ 2.19 – 1],
= 500 x 1.19,
= $595.56.

Question 7.
How much money would you owe if you borrowed $2,000 for 5 years, with a compound interest rate of 28%, and did not make any payments during that period?
Answer:
$6871.95,

Explanation:
Compound Interest CI = P [ 1 + \(\frac{R}{100}\) ]ⁿ – 1 ],
P = Principle Amount,
R =  Rate of interest in %,
T = Time (Number of years),
P= $2,000,
R = 28%,
n = 5 yrs,
Compound Interest CI = P [ 1 + \(\frac{R}{100}\) ]ⁿ – 1 ],
CI = 2000 [ 1 + \(\frac{28}{100}\)]⁵ – 1 ],
=2000 \(\frac{128}{100}\)⁵ – 1 ],
= 2000 X 3.435,
= $6871.95.

Question 8.
Is it better to receive compounded interest for 7 years at 12% on your balance of $500, or to receive the same rate of simple interest for 9 years on that same balance?
Answer:
7 years at 12%, Balance, Compound interest $1105.34
Simple interest $140.00,

Explanation:
Compound Interest CI = P [ 1 + \(\frac{R}{100}\) ]ⁿ – 1 ],
P= $500, R = 12%, n = 7 yrs, Compound Interest CI = P [ 1 + \(\frac{R}{100}\) ]ⁿ – 1 ], CI = 500 [ 1 + \(\frac{12}{100}\)]⁷ – 1 ],
= 500 \(\frac{112}{100}\)⁷ – 1 ] = 500[2.210 – 1] = 500 x 1.210,
= $605.34. The ending balance = Principle + Compound interest,
A = $500 + $605.34 = $1,105.34,
Amount = Principle + Simple Interest,
SI = Amount – Principle, P= $500, R = 12%,
T = 9 yrs,
Simple Interest SI = \(\frac{P X R X T}{100}\),
= \(\frac{500 X 12 X 9}{100}\) = 5 x 12 x 9 = 540.00,
The ending balance = Principle + Simple Interest,
A = $500 + $540 = $1,040.00.