Primary Mathematics Chapter 4 Operations On Fractions
Page 132 Exercise 4.10 Problem 1
Given: 90− page book is 50 pages.
What fraction of a 90− page book is 50 pages.
The fraction of a 90− page book in 50
pages= \(\frac{50}{90}\) or \(\frac{5}{9}\)
The fraction of a 90− page book in 50 pages are \(\frac{5}{9}\).
Page 132 Exercise 4.10 Problem 2
Given: Cameron has 40 toy cars.15 are battery operated.
What fraction of the toy cars are battery-operated
A fraction of toys are in battery
Operated = \(\frac{15}{40}\) or \(\frac{3}{8}\)
The fraction of toy cars in battery-operated is \(\frac{3}{8}\).
Page 132 Exercise 4.10 Problem 3
Given: Jim bought a packet of 60 stamps.24 were Canadian stamps.
What fraction of stamps were Canadian stamps.
The fraction of the stamps in Canadian Stamps is = \(\frac{24}{60}\) or \(\frac{2}{5}\).
The fraction of stamps in Canadian stamps are \(\frac{2}{5}\).
Page 133 Exercise 4.10 Problem 4
Given: Stewart had $25. He spent$5
What fraction of his money did he have left.
First of all, Stewart has $25 & He spent $5, then.
The money he left = $20
The fraction of his money he has left \(\frac{20}{25}\) or \(\frac{4}{5}\).
The fraction of the money he has left is \(\frac{4}{5}\)
Page 133 Exercise 4.10 Problem 5
Given: There are 100 children at a carnival,60 were boys.
Express the number of girls as a fraction of the total number of children at the carnival.
First of all, total children are 100 & boys are 60 , then girls are (100−60 = 40)
The fraction of number of girls are = \(\frac{240}{100}\) or \(\frac{2}{5}\).
The fraction of number of girls of the total number of children at the carnival is \(\frac{2}{5}\).
Page 134 Exercise 4.10 Problem 6
Given: Sara bought 40m of material.
She made 6 curtains from the material.
She used 2m to make each curtain.
What fraction of the material did she use for the 6 curtains.
First of all, if each curtain make 2m, then6 curtain make = 2 × 6
So,6 curtains make = 12m
The fraction of material for 6 curtains is
The fraction of material used for 6 curtains is \(\frac{1}{3}\)m.
Page 134 Exercise 4.10 Problem 7
In the question, we are asked how many mangoes Travis sold.
Given that, he had 160 mangoes and each mango costs $2
Use the equation mentioned in the Tip section and find the solution.
Given:
Each mango costs = $2
Total number of mangoes = 160
The total cost that he gained $ 240
We have
Total Cost = Number of mangoes×Cost of one mango
Number of mangoes \( =\frac{\text { Total Cost }}{\text { Cost of one mango }}\)
= \(\frac{240}{2}\)
= 120
Travis sells 120 mangoes for the cost of $240.
Page 134 Exercise 4.10 Problem 8
In the question, we are asked to find the fraction of his mangoes did he have left.
We found that how many mangoes he sold from exercise 8a.
So we can find how many mangoes left and can convert into fraction.
Number of mangoes that Travis sold =120
Number of mangoes left = 160 − 120
= 40
Fraction of mangoes that he has left
= \(\frac{40}{160}\)
= \(\frac{1}{4}\)
Fraction of his mangoes he has left = \(\frac{1}{4}\)
Page 135 Exercise 4.11 Problem 1
In the question, asked to color by given fraction from the set.
Find how many number of elements in the set.
Then find the fraction and color accordingly.
Given set contains 8 tortoises.
So, we have to find the given fraction of 8.
Given: Fraction =\(\frac{1}{2}\)
To find – \(\frac{1}{2}\) of 8
8 × \(\frac{1}{2}\) = 4
Therefore, color 4 Tortoises.
Color 4 tortoises.
Page 135 Exercise 4.11 Problem 2
In the question, asked to color by given fraction from the set.
Find how many number of elements in the set.
Then find the fraction and color accordingly.
Given set contains 9 sandwiches.
So, we have to find given fraction of 9.
Given: Fraction = \(\frac{1}{3}\)
To find – \(\frac{1}{3}\) of 9
9 × \(\frac{1}{3}\) = 3
Therefore, color 3 sandwiches.
Color 3 Sandwiches.
Page 135 Exercise 4.11 Problem 3
In the question, asked to color by a given fraction from the set.
Find how many number of elements in the set.
Then find the fraction and color accordingly.
The given set contains 15 socks.
So we have to find the given fraction of 15
Given: Fraction = \(\frac{2}{5}\)
To find – \(\frac{2}{5}\) of 15
15 × \(\frac{2}{5}\)
Therefore, Color 6 socks.
Color 6 socks.
Page 135 Exercise 4.11 Problem 4
In the question, asked to color by a given fraction from the set.
Find how many number of elements in the set.
Then find the fraction and color accordingly.
The given set contains 18 cakes.
We have to find a given fraction of 18.
Given: Fraction =\(\frac{5}{9}\)
To find – \(\frac{5}{9}\) of 18
18 × \(\frac{5}{9}\)= 10
Therefore, color 10 cakes.
Color 10 cakes.
Page 135 Exercise 4.11 Problem 5
In the question, we are asked to find the value of given problem.
Multiply the number by the fraction to get the solution.
We have to find
\(\frac{1}{2}\) of 12 = 12 × \(\frac{1}{2}\)
Simplify
\(\frac{1}{2}\) × 12 = 6
\(\frac{1}{2}\) of 12 = 6
Page 135 Exercise 4.11 Problem 6
In the question, we are asked to find the value of the given problem.
Multiply the number by the fraction to get the solution.
We have to find
\(\frac{1}{4}\) of 12 = 12 × \(\frac{1}{4}\)
Simplify
12 × \(\frac{1}{4}\) = 3
\(\frac{1}{4}\) of 12 = 3
Page 135 Exercise 4.11 Problem 7
In the question, we are asked to find the value of the given problem.
Multiply the number by the fraction to get the solution.
We have to find
\(\frac{1}{3}\) of 24 = 24 × \(\frac{1}{3}\)
Simplify
24 × \(\frac{1}{3}\) = 8
\(\frac{1}{3}\) of 24 = 8
Page 135 Exercise 4.11 Problem 8
In the question, we are asked to find the value of given problem.
Multiply the number by the fraction to get the solution.
We have to find
\(\frac{1}{6}\) of 36 = 36× \(\frac{1}{3}\)
Simplify
36× \(\frac{1}{6}\) = 6
\(\frac{1}{6}\) of 36 = 6
Page 136 Exercise 4.11 Problem 9
In the question, we are asked to find the value of given problem.
Multiply the number by the fraction to get the solution.
We have to find
\(\frac{1}{4}\) of 20 = 20 × \(\frac{1}{4}\)
= 5
To find –
\(\frac{3}{4}\) of 20 = 20× \(\frac{3}{4}\)
= 15
\(\frac{1}{4}\) of 20 = 5 , \(\frac{3}{4}\) of 20 = 15
Page 136 Exercise 4.11 Problem 10
In the question, we are asked to find the value of given problem.
Multiply the number by the fraction to get the solution.
We have to find
\(\frac{1}{5}\) of 25 = 25× \(\frac{1}{5}\)= 5
To find –
\(\frac{3}{5}\) of 25 = 25 × \(\frac{3}{5}\) = 15
\(\frac{1}{5}\) of 25 = 5 , \(\frac{3}{5}\) of 25 = 15
Page 136 Exercise 4.11 Problem 11
In the question, we are asked to find the value of given problem.
Multiply the number by the fraction to get the solution.
We have to find
\(\frac{1}{3}\) of 21 = 21× \(\frac{1}{3}\)= 7
To find –
\(\frac{2}{3}\) of 21 = 21 × \(\frac{2}{3}\) = 14
\(\frac{1}{3}\) of 21 = 7 , \(\frac{2}{3}\) of 21 = 14
Page 136 Exercise 4.11 Problem 12
In the question, we are asked to find the value of given problem.
Multiply the number by the fraction to get the solution.
We have to find
\(\frac{1}{10}\) of 30 = 30× \(\frac{1}{10}\)= 3
To find –
\(\frac{7}{10}\) of 30 = 30 × \(\frac{7}{10}\) = 21
\(\frac{1}{10}\) of 30 = 3 , \(\frac{7}{10}\) of 30 = 21
Page 137 Exercise 4.12 Problem 1
In the question, we are asked to find the value of given problem.
Using the rule mentioned in the Tip multiply the numbers with the fraction to get the solution.
We have to find
\(\frac{1}{4}\) of 8 = \(\frac{1}{4}\) × \(\frac{8}{1}\)
1. Multiply the numerators.
2. Multiply the denominators.
\(\frac{1}{4}\) of 8 = \(\frac{1×8}{4×1}\)
= \(\frac{8}{4}\)
Simplify
\(\frac{8}{4}\) = 2 , \(\frac{1}{4}\) of 8 = 2
Page 137 Exercise 4.12 Problem 2
In the question, we are asked to find the value of given problem.
Using the rule mentioned in the Tip multiply the numbers with the fraction to get the solution.
We have to find
\(\frac{1}{5}\) of 15 =\(\frac{1}{5}\) × \(\frac{15}{1}\)
1. Multiply the numerators.
2. Multiply the denominators.
\(\frac{1}{5}\) of 15 = \(\frac{1× 15}{5× 1}\)
= \(\frac{15}{5}\)
Simplifying
\(\frac{15}{5}\) = 3 , \(\frac{1}{5}\) of 15 = 3
Page 137 Exercise 4.12 Problem 3
In the question, we are asked to find the value of given problem.
Using the rule mentioned in the Tip multiply the numbers with the fraction to get the solution.
We have to find
\(\frac{1}{3}\) of 12 = \(\frac{1}{3}\) × \(\frac{12}{1}\)
1. Multiply the numerators.
2. Multiply the denominators.
\(\frac{1}{3}\) of 12 =\(\frac{1×12}{3× 1}\)
= \(\frac{12}{3}\)
Simplify
\(\frac{12}{3}\) = 4 , \(\frac{1}{3}\) of 12 = 4
Page 137 Exercise 4.12 Problem 4
In the question, we are asked to find the value of given problem.
Using the rule mentioned in the Tip multiply the numbers with the fraction to get the solution.
We have to find
\(\frac{1}{3}\) of 6 =\(\frac{6}{1}\) × \(\frac{1}{3}\)
1. Multiply the numerators.
2. Multiply the denominators.
\(\frac{1}{3}\) of 6 = \(\frac{6}{3}\)
Simplifying
\(\frac{6}{3}\) = 2
\(\frac{1}{3}\) of 6 = 2
Page 137 Exercise 4.12 Problem 5
In the question, we are asked to find the value of given problem.
Using the rule mentioned in the Tip multiply the numbers with the fraction to get the solution.
We have to find
\(\frac{2}{3}\) of 6= \(\frac{1}{3}\) of 6 =\(\frac{6}{1}\) × \(\frac{2}{3}\)
= \(\frac{12}{3}\)
1. Multiply the numerators.
2. Multiply the denominators.
Simplifying
\(\frac{12}{3}\) = 4
\(\frac{2}{3}\) of 6 = 4
Page 138 Exercise 4.12 Problem 6
In the question, we are asked to find the value of the given problem.
Using the rule mentioned in the Tip multiply the numbers with the fraction to get the solution
We have to find
\(\frac{1}{2}\) of 8= \(\frac{8}{1}\) × \(\frac{1}{2}\)
1. Multiply the numerators.
2. Multiply the denominators.
\(\frac{1}{2}\) of 8 = \(\frac{8}{2}\)
Simplifying
\(\frac{8}{2}\) = 4
\(\frac{1}{2}\) of 8 = 4
Page 138 Exercise 4.12 Problem 7
In the question, we are asked to find the value of the given problem.
Using the rule mentioned in the Tip multiply the numbers with the fraction to get the solution.
We have to find
\(\frac{1}{3}\) of 15 = \(\frac{1}{3}\) × \(\frac{15}{1}\)
1. Multiply the numerators.
2. Multiply the denominators.
\(\frac{1}{3}\) of 15= \(\frac{15}{3}\)
Simplifying
\(\frac{15}{3}\) = 5
\(\frac{1}{3}\) of 15 = 5
Page 138 Exercise 4.12 Problem 8
In the question, we are asked to find the value of the given problem.
Using the rule mentioned in the Tip multiply the numbers with the fraction to get the solution.
We have to find
\(\frac{1}{4}\) of 20 = \(\frac{20}{1}\) × \(\frac{1}{4}\)
1. Multiply the numerators.
2. Multiply the denominators.
\(\frac{1}{4}\) of 20= \(\frac{20}{4}\)
Simplifying
\(\frac{20}{4}\) = 5
\(\frac{20}{4}\) of 20 = 5
Page 138 Exercise 4.12 Problem 9
In the question, we are asked to find the value of given problem.
Using the rule mentioned in the Tip multiply the numbers with the fraction to get the solution.
We have to find
\(\frac{1}{6}\) of 18 = \(\frac{18}{1}\) × \(\frac{1}{6}\)
1. Multiply the numerators.
2. Multiply the denominators.
\(\frac{1}{6}\) of 18 = \(\frac{18}{6}\)
Simplifying
\(\frac{18}{6}\) = 3
\(\frac{1}{6}\) of 18 = 3
Page 139 Exercise 4.12 Problem 10
Given: The expression
\(\frac{2}{3}\) of 15 = \(\frac{2}{3}\) × 15
To find – The value of the given expression.
We multiply numerator with numerator and denominator with denominator and simplify.
First, we will convert 15 into a fraction using 1 as the denominator.
So, it can be written as ⇒ \(\frac{1}{15}\) .
Now, multiply the numerators and denominators.
\(\frac{2}{3}\) × \(\frac{15}{1}\)= \(\frac{2×15}{3×1}\)
Simplifying we get
\(\frac{30}{3}\) = 10
Therefore, the value of \(\frac{2}{3}\) of 15 is 10.
Page 139 Exercise 4.12 Problem 11
Given: The expression \(\frac{3}{4}\) of 20.
To find – The value of the given expression.
We multiply numerator with numerator and denominator with denominator and simplify.
We know that \(\frac{3}{4}\) of 20 can be written as \(\frac{3}{4}\) × 20.
First, we will convert 20 into a fraction using 1 as the denominator.
So, it can be written as ⇒ \(\frac{20}{1}\)
Now, multiply the numerators and denominators.
\(\frac{3}{4}\) × \(\frac{20}{1}\)= \(\frac{3×20}{4×1}\)
Simplifying we get
\(\frac{60}{4}\)= 15
Therefore, the value of \(\frac{3}{4}\) of 20 is 15.
Page 139 Exercise 4.12 Problem 12
Given: The expression \(\frac{4}{5}\) of 30.
To find – The value of the given expression.
We multiply numerator with numerator and denominator with denominator and simplify.
We know that \(\frac{4}{5}\) of 20 can be written as \(\frac{4}{5}\)× 30.
First, we will convert 20 into a fraction using 1 as the denominator.
So, it can be written as ⇒ \(\frac{30}{1}\)
Now, multiply the numerators and denominators.
\(\frac{4}{5}\) × \(\frac{30}{1}\)
= \(\frac{4×30}{5×1}\)
Simplifying we get
\(\frac{120}{5}\)= 24
Therefore, the value of \(\frac{4}{5}\) of 30 is 24.
Page 139 Exercise 4.12 Problem 13
Given: The expression \(\frac{5}{6}\) of 36.
To find – The value of the given expression.
We multiply numerator with numerator and denominator with denominator and simplify.
We know that \(\frac{5}{6}\) of 36 can be written as \(\frac{5}{6}\) × 36.
First, we will convert 20 into a fraction using 1 as the denominator.
So, it can be written as ⇒ \(\frac{36}{1}\)
Now, multiply the numerators and denominators.
\(\frac{5}{6}\) × \(\frac{36}{1}\) = \(\frac{5×36}{6×1}\)
Simplifying we get
\(\frac{180}{4}\)= 30
Therefore, the value of \(\frac{5}{6}\) of 36 is 30.
Page 140 Exercise 4.13 Problem 1
Given: Manfred had 25 picture cards. He gave \(\frac{2}{5}\) of them to his friends.
To find – The number of cards Manfred gave to his friends.
We find \(\frac{2}{5}\) of 25 cards.
We are given that Manfred gave \(\frac{2}{5}\) of them to his friends that implies
\(\frac{2}{5}\) of 25 = \(\frac{2}{5}\)×25
First, we will convert 25 into a fraction using 1 as the denominator.
So, it can be written as ⇒ \(\frac{25}{1}\)
Now, multiply the numerators and denominators.
\(\frac{2}{5}\)×\(\frac{25}{1}\) = \(\frac{2×25}{5×1}\)
Simplifying we get
\(\frac{50}{5}\) = 10
Therefore, Manfred gave 10 cards to his friends.
Page 140 Exercise 4.13 Problem 2
Given: Manfred had 25 picture cards. He gave \(\frac{2}{5}\) of them to his friends
To find – The number of cards left with Manfred.
We find \(\frac{2}{5}\) of 25 cards and subtract it from the total cards.
We are given that Manfred gave \(\frac{2}{5}\) of them to his friends that implies
\(\frac{2}{5}\) of 25 = \(\frac{2}{5}\) × 25
First, we will convert 25 into a fraction using 1 as the denominator.
So, it can be written as ⇒ \(\frac{25}{1}\)
Now, multiply the numerators and denominators.
\(\frac{2}{5}\)×\(\frac{25}{1}\) =\(\frac{2×25}{5×1}\)
Simplifying we get
\(\frac{50}{5}\) = 10
A number of remaining cards can be calculated by subtracting the cards he gave to his friends from the total number of cards he had.
25−10 = 15
Therefore, the number of cards left with Manfred is 15.
Page 140 Exercise 4.13 Problem 3
Given: Sharon had 40 dollars. She spent \(\frac{3}{8}\) of the money on a storybook.
To find – the cost of the storybook.
We find the money spent on the storybook which will be the cost of a storybook.
We are given that Sharon spent \(\frac{3}{8}\) of the money on a storybook that implies
\(\frac{3}{8}\) of 40 = \(\frac{3}{8}\)× 40
First, we will convert 40 into a fraction using 1 as the denominator.
So, it can be written as \(\frac{40}{1}\).
Now, multiply the numerators and denominators.
\(\frac{3}{8}\)×\(\frac{40}{1}\) = \(\frac{3×40}{8×1}\)
Simplifying we get
\(\frac{120}{8}\) = 15
Hence, the money spent on storybooks is 15.
Also, it is known that the cost of the storybook is equal to the expenditure on the storybook which is 15.
Therefore, the cost of the storybook is 15.
Page 140 Exercise 4.13 Problem 4
Given: Sharon had 40 dollars.
She spent \(\frac{3}{8}\) of the money on a storybook.
To find – The amount of money left with Sharon.
We find the money spent on the storybook and subtract it from the total cost.
We are given that Sharon spent \(\frac{3}{8}\) of the money on a storybook that implies
\(\frac{3}{8}\) of 40 = \(\frac{3}{8}\) × 40
First, we will convert 40 into a fraction using 1 as the denominator.
So, it can be written as \(\frac{40}{1}\).
Now, multiply the numerators and denominators.
\(\frac{3}{8}\)×\(\frac{40}{1}\) =\(\frac{3×40}{8×1}\)
Simplifying we get
\(\frac{120}{8}\) = 15
The remaining amount left with Sharon can be calculated by subtracting the amount spent on the storybook from the total money he had.
40−15 = 25
Therefore, 25 dollars are left with Sharon.
Page 141 Exercise 4.13 Problem 5
Given: There are 24 potatoes in a bag and Lily peeled \(\frac{2}{3}\)
To find – The number of potatoes Lily peeled.
We find the peeled potatoes by calculating \(\frac{2}{3}\) of 24 potatoes.
We are given that Lily peeled \(\frac{2}{3}\) of total potatoes that implies
\(\frac{2}{3}\) of = \(\frac{2}{3}\) × 24
First, we will convert 24 into a fraction using 1 as the denominator.
So, it can be written as ⇒ \(\frac{42}{1}\)
Now, multiply the numerators and denominators.
\(\frac{2}{3}\)×\(\frac{24}{1}\) =\(\frac{2×24}{3×1}\)
Simplifying we get
\(\frac{48}{3}\) = 16
Hence, the number of potatoes Lily peeled is 16.
Therefore, the number of potatoes Lily peeled is 16.
Page 141 Exercise 4.13 Problem 6
Given: Lindsay had 48 dollars. She spent \(\frac{3}{8}\) of the money.
To find – The amount of money she had spent.
We find the money spent by calculating \(\frac{3}{8}\)of 48 dollars.
We are given that Lindsay spent \(\frac{3}{8}\) of the money on a storybook that implies
\(\frac{3}{8}\) of 48 = \(\frac{3}{8}\) ×48
First, we will convert 48 into a fraction using 1 as the denominator.
So, it can be written as ⇒ \(\frac{48}{1}\)
Now, multiply the numerators and denominators.
\(\frac{3}{8}\)×\(\frac{48}{1}\)=\(\frac{3×48}{8×1}\)
Simplifying we get
\(\frac{144}{3}\) = 18
Hence, the money spent is 18.
Therefore, the amount of money Lindsay had spent is 18 dollars.
Page 142 Exercise 4.14 Problem 1
Given: Mandy had 25 dollars. She spent \(\frac{1}{5}\) of the money and saved the rest.
To find – The amount of money she saved.
We find the money spent and subtract it from the total amount.
We are given that Mandy spent \(\frac{1}{5}\) of the money that implies
\(\frac{1}{5}\) of 25 = \(\frac{1}{5}\)× 25
First, we will convert 25 into a fraction using 1 as the denominator.
So, it can be written as ⇒ \(\frac{25}{1}\).
Now, multiply the numerators and denominators.
\(\frac{1}{5}\)×\(\frac{25}{1}\)=\(\frac{1×25}{5×1}\)
Simplifying we get
\(\frac{25}{5}\)=5
Hence, the money spent is 5 dollars.
The remaining amount saved with Mandy can be calculated by subtracting the amount spent from the total money she had.
25−5 = 20
Therefore, the amount of money Mandy saved is 20 dollars.
Page 142 Exercise 4.14 Problem 2
Given: Matthew bought 45 oranges and used \(\frac{3}{5}\) of them to make orange juice.
To find – The number of oranges left with him.
We will subtract the used oranges from the total oranges.
We are given that Matthew used \(\frac{3}{5}\) of the total oranges that implies
\(\frac{3}{5}\) of 45 = \(\frac{3}{5}\)×45
First, we will convert 45 into a fraction using 1 as the denominator.
So, it can be written as ⇒ \(\frac{45}{1}\).
Now, multiply the numerators and denominators.
\(\frac{3}{5}\)×\(\frac{45}{1}\)=
\(\frac{3×45}{5×1}\)Simplifying we get
\(\frac{135}{5}\) = 27
Hence, the number of oranges used to make orange juice is 27.
The remaining oranges left with Matthew can be calculated by subtracting the number of oranges used from the total oranges he had.
45 − 27 = 18
Therefore, the number of oranges left with Matthew is 18 oranges.
Page 143 Exercise 4.14 Problem 3
Given: Nellie had 48 dollars and spent \(\frac{1}{4}\) of it on a calculator.
She also bought a book for 14 dollars.
To find – The amount of money she spent all together.
We add the amount spent on calculator and book.
We are given that Nellie spent \(\frac{1}{4}\)
of money she had on a calculator that implies
\(\frac{1}{4}\) of 45 = \(\frac{1}{4}\)×48
First, we will convert 48 into a fraction using 1 as the denominator.
So, it can be written as ⇒ \(\frac{48}{1}\)
Now, multiply the numerators and denominators.
\(\frac{1}{4}\)×\(\frac{48}{1}\)=\(\frac{1×48}{4×1}\)
Simplifying we get
\(\frac{48}{4}\)=12
Hence, the amount of money spent on a calculator is 12 dollars.
Also, we know that the amount spent on books is 14 dollars.
So, the total amount spent will be the amount spent on calculator + amount spent on book.
That is
12 + 14 = 26
Therefore, the amount of money she spent altogether is 26 dollars.
Page 144 Exercise 4.14 Problem 4
Given: There were 96 people on a board ship and \(\frac{1}{4}\)×\(\frac{48}{1}\)= of them were females.
To find – The number of males on the board ship.
We subtract the number of females from the total number of people.
We are given that \(\frac{1}{4}\) of total people were females. This implies that
\(\frac{1}{4}\) of = \(\frac{1}{4}\) × 96
First, we will convert 96 into a fraction using 1 as the denominator.
So, it can be written as ⇒ \(\frac{96}{1}\)
Now, multiply the numerators and denominators.
\(\frac{1}{4}\)×\(\frac{96}{1}\)=\(\frac{1×96}{4×1}\)
Simplifying we get
\(\frac{96}{4}\) = 24
Hence, the number of females is 24.
The number of males can be calculated by subtracting the number of females from the total number of people on the board ship.
This implies, Number of males
= 96−24
= 72
Page 145 Exercise 4.15 Problem 1
Given: Lynn spent \(\frac{7}{10}\) of the money she had.
She spent 42 dollars.
To find – The amount of money Lynn had at first.
We find the amount of money for 10 units.
We are given that Lynn spent \(\frac{7}{10}\) of the money she had which is 42 dollars.
This implies
7 units = 42 dollars
1 unit = 6 dollars
Now, the amount for 10 units will be the total amount.
⇒ 10 units = 6 × 10 dollars
⇒ 10 units = 60 dollars
Therefore, the amount of money Lynn had at first is 60 dollars.
Page 145 Exercise 4.15 Problem 2
Given: Lynn spent \(\frac{7}{10}\) of the money she had.
She spent 42 dollars.
To find – The amount of money Lynn saved.
We subtract the amount of money spent from the total money she had at first.
We are given that Lynn spent \(\frac{7}{10}\) of the money she had which is 42 dollars.
This implies
7 units = 42 dollars
1 unit = 6 dollars
Now, the amount for 10 units will be the total amount.
⇒ 10 units = 6 × 10 dollars
⇒ 10 units = 60 dollars
Hence, the amount of money Lynn had at first is 60 dollars.
We find the money she saved by subtracting money spent from the total amount of money.
Amount of money saved = 60 − 42 which is 18 dollars.
Therefore, the amount of money Lynn saved is 18 dollars.
Page 145 Exercise 4.15 Problem 3
Given: A group of children went for a school picnic and \(\frac{3}{7}\) of them were boys.
There were 18 boys
To find – The number of children presents there.
We find a number of children for 7 units.
We are given that \(\frac{3}{7}\) of the group of children were boys.
This implies
7 units =18 children
1 unit = \(\frac{18}{7}\) children
Now, the number of children for 7 units will be the total number.
⇒ 7 units = \(\frac{18}{7}\) × 7
⇒ 7 units = 18 children
Hence, the total number of children who went to the school picnic is 18 children.
Therefore, 18 children went to the school picnic.
Page 145 Exercise 4.15 Problem 4
Given: A group of children went for a school picnic and \(\frac{3}{7}\) of them were boys. There were 18 boys.
To find – The number of girls present there.
We find the number of girls by calculating the number of children for 4 units.
We are given that \(\frac{3}{7}\) of the group of children were boys.
This implies
3 units = 18 boys
1 unit = 6 boys
Now, the number of girls will be calculated by finding the number of children for 7−3=4 units.
⇒ 4 units = 6 × 4 girls
⇒ 7 units = 24 girls
Hence, the number of girls who went for the school picnic is 24 girls.
Therefore, the number of girls who went for the school picnic is 24 girls.
Page 146 Exercise 4.15 Problem 5
Given: Susan spent \(\frac{3}{10}\) of her money on a bag.
To find – The amount of money she had at first if the bag costs 9 dollars.
We find the total amount of money she had by calculating the amount for 10 units.
We are given that Susan spent \(\frac{3}{10}\) of her money on a bag which is 9 dollars.
This implies
3 units = 9 dollars
1 unit = 3 dollars
Now, the total amount of money she had is for 10 units.
⇒ 10 units = 10 × 3 dollars
⇒ 10 units = 30 dollars
Therefore, the amount of money she had at first if the bag costs 9 dollars is 30 dollars.
Page 147 Exercise 4.15 Problem 6
Given: John bought some stamps. He used \(\frac{3}{5}\)of them to make mail letters and had 12 stamps left.
To find – The number of stamps that he used.
We find the number of stamps he bought and then find \(\frac{3}{5}\) of them.
We are given that John used of the stamps and had 12 stamps left.
Remaining units are 5−3 = 2 units.
This implies
2units =12 stamps
1 unit =6 stamps
Now, the total number of stamps that she had is for 5 units.
⇒ 5 units = 6 × 5 stamps
⇒ 5 units = 30 stamps
We know that she used \(\frac{3}{5}\) of the total stamps.
Thus, the used number of stamps is \(\frac{3}{5}\) of 30 stamps.
Thus, used number of stamps is
⇒ \(\frac{3}{5}\)× 30 = \(\frac{3}{5}\) \(\frac{30}{1}\)
⇒ \(\frac{3×30}{5×1}\)= \(\frac{90}{5}\) 18 stamps.
Therefore, the number of stamps that he used is 18 stamps.