Primary Mathematics Chapter 4 Operations On Fractions
Page 148 Exercise 4.16 Problem 1
Given: The expression \(\frac{5}{8}\) +\(\frac{3}{4}\)
Numerator and Denominator of \(\frac{3}{5}\) by 10
⇒ \(\frac{9 \times 5}{10 \times 5}-\frac{3 \times 10}{5 \times 10}\)
⇒ \(\frac{45}{50}-\frac{30}{50}\)
Since the Denominators are same, subtract the numerator
⇒ \(\frac{9 \times 5}{10 \times 5}\) – \(\frac{3 \times 10}{5 \times 10}\)
⇒ \(\frac{45}{50}\) – \(\frac{30}{50}\)
⇒ \(\frac{15}{50}\)
Now, simplify \(\frac{15 \div 5}{50 \div 5}=\frac{3}{10}\)
Hence, the solution of \(\frac{9}{10}-\frac{3}{5} \text { is } \frac{3}{10} \text {. }\)
Page 148 Exercise 4.16 Problem 2
Given: \(\frac{5}{8}\)−\(\frac{3}{4}\)
First, we, find the LCM of \(\frac{5}{8}\) and \(\frac{3}{4}\)
LCM of \(\frac{5}{8}\) is \(\frac{3}{4}\)
Now, we make the denominators of both fractions equal.
⇒ \(\frac{3}{4} \times \frac{2}{2}=\frac{6}{8}\)
Simplify the expression we have
⇒ \(\frac{5}{8}+ \frac{3}{4}=\frac{5}{8}+\frac{6}{8}\)
⇒ \(\frac{5+6}{8}=\frac{11}{8}\)
Therefore the solution for \(\frac{5}{8}+\frac{3}{4}=\frac{11}{8}\)
Page 148 Exercise 4.16 Problem 3
Given: 4\(\frac{5}{6}\)+ 3\(\frac{2}{3}\)
To Find – Find 4\(\frac{5}{6}\)+ 3\(\frac{2}{3}\)
To add fractions with unlike denominators, first, find the least common multiple of the two denominators and then add and simplify.
The LCM of 6 and 3 is 6. So, we need to find fractions equivalent to 4\(\frac{5}{6}\) and 3\(\frac{2}{3}\) which have 18 in the denominator.
Multiply the numerator and denominator of 4\(\frac{5}{6}\) by 3 , and multiply the numerator and denominator of 3\(\frac{2}{3}\) by 6.
⇒ 4\(\frac{5}{6}\)
⇒ \(\frac{6×4+5}{6}\)
⇒ \(\frac{29}{6}\)
Similarly 3\(\frac{2}{3}\)
⇒ \(\frac{3×3+2}{3}\)
⇒ \(\frac{11}{3}\)
Now, add the fractions
4\(\frac{5}{6}\) + 3\(\frac{2}{3}\)
⇒ \(\frac{29}{6}\) + \(\frac{11}{3}\)
⇒ \(\frac{29×3}{6×3}+\frac{11×6}{3×6}\)
⇒ \(\frac{87}{18}\) + \(\frac{66}{18}\)
Since the denominators are the same, add the numerators.
⇒ \(\frac{87}{18}\)+\(\frac{66}{18}\)
= \(\frac{87+66}{18}\)
= \(\frac{153}{18}\)
Now, simplify
⇒ \(\frac{153}{18}\)
= \(\frac{153÷9}{18÷9}\)
= \(\frac{17}{2}\)
Hence, the solution of 4\(\frac{5}{6}\) + 3\(\frac{2}{3}\) is \(\frac{17}{2}\).
Page 148 Exercise 4.16 Problem 4
Given: 5\(\frac{1}{4}\) – 2\(\frac{5}{12}\) =
To Find – Find 5\(\frac{1}{4}\) – 2\(\frac{5}{12}\)
To subtract fractions with unlike denominators, first, find the least common multiple of the two denominators and then subtract and simplify.
The LCM of 4 and 12 is 12. So, we need to find fractions equivalent to 5\(\frac{1}{4}\) and 2\(\frac{5}{12}\) which have 48 in the denominator.
Multiply the numerator and denominator of 5\(\frac{1}{4}\) by 12, and multiply the numerator and denominator of 2\(\frac{5}{12}\) by, 4.
⇒ 5\(\frac{1}{4}\)
= \(\frac{4×5+1}{4}\)
= \(\frac{21}{4}\)
Similarly
⇒ 2\(\frac{5}{12}\)
= \(\frac{12×2+5}{12}\)
= \(\frac{29}{12}\)
Now, add the fractions
⇒ 5\(\frac{1}{4}\)– 2\(\frac{5}{12}\)
= \(\frac{21}{4}\) – \(\frac{29}{12}\)
= \(\frac{21×12}{4×12}\) – \(\frac{29×4}{12×4}\)
= \(\frac{252}{48}\) – \(\frac{116}{48}\)
Since the denominators are the same, subtract the numerators.
⇒ \(\frac{252}{48}\) – \(\frac{116}{48}\)
= \(\frac{252-116}{48}\)
= \(\frac{136}{48}\)
Now, simplify
⇒ \(\frac{136}{48}\)
= \(\frac{136÷8}{48÷8}\)
= \(\frac{17}{6}\)
Hence, the solution of 5\(\frac{1}{4}\)− 2\(\frac{5}{12}\) is \(\frac{17}{6}\).
Page 149 Exercise 4.16 Problem 5
Given: ______− 6\(\frac{3}{4}\) = \(\frac{1}{2}\)
Take \(\frac{3}{4}\) to the other side and add to \(\frac{1}{2}\)
⇒ \(\frac{1}{2}+6 \frac{3}{4}\)
⇒ \(\frac{1}{2}+\frac{27}{4}\)
Take LCM of 2 and 4 to simplify
⇒ \(\frac{1}{2}+\frac{27}{4}\)
⇒ \(\frac{2+27}{4}\)
⇒ \(\frac{29}{4}\)
Hence the solution is \(\frac{29}{4}-6 \frac{3}{4}=\frac{1}{2}\)
Page 149 Exercise 4.16 Problem 6
Given: 3\(\frac{7}{8}\) +_____ = 5 \(\frac{1}{8}\)
Take 3\(\frac{7}{8}\) to the other side and subtract from 5 \(\frac{1}{8}\)
⇒ \(5 \frac{1}{8}-3 \frac{7}{8}\)
⇒ \(\frac{8 \times 5+1}{8}-\frac{8 \times 3+7}{8}\)
⇒ \(\frac{41}{8}-\frac{31}{8}\)
As the denominator is the same, simplify
⇒ \(\frac{41}{8}-\frac{31}{8}\)
⇒ \(\frac{10}{8} \text { or } \frac{5}{4}\)
Hence solution is [latex3 \frac{7}{8}+\frac{5}{4}=5 \frac{1}{8}][/latex]
Page 150 Exercise 4.16 Problem 7
Given: Rope A is 10\(\frac{2}{3}\)ft long and rope B is 8\(\frac{5}{6}\) ft long.
To Find – Find the sum of the length of the two ropes.
By adding the integers rule, find the sum of the length of the two ropes.
Simplify 10\(\frac{2}{3}\)ft
⇒ 10\(\frac{2}{3}\)
= \(\frac{10×3+2}{3}\)
= \(\frac{32}{3}\)
Similarly for 8\(\frac{5}{6}\) ft.
⇒ 8\(\frac{5}{6}\)
= \(\frac{6×8+5}{6}\)
= \(\frac{53}{6}\)
Add the length of the two ropes
⇒ \(\frac{32}{3}\)+\(\frac{53}{6}\)
Take LCM of 3 and 6, then add
⇒ \(\frac{32}{3}\)+\(\frac{53}{6}\)
= \(\frac{64}{6}\)+\(\frac{53}{6}\)
= \(\frac{64+53}{6}\)
= \(\frac{117}{6}\)
Or , \(\frac{39}{2}\) ft
Hence, the sum of the length of the two ropes is \(\frac{39}{2}\) ft.
Page 150 Exercise 4.16 Problem 8
Given: Rope A is 10\(\frac{2}{3}\) ft long and rope B is 8\(\frac{5}{6}\)
To Find – Find the difference in length of the two ropes.
By subtracting the integers rule, find the difference in length of the two ropes.
Simplify 10\(\frac{2}{3}\)ft
⇒ 10\(\frac{2}{3}\)
= \(\frac{10×3+2}{3}\)
= \(\frac{32}{3}\)
Similarly for 8\(\frac{5}{6}\) ft.
⇒ 8\(\frac{5}{6}\)
= \(\frac{6×8+5}{6}\)
= \(\frac{53}{6}\)
Subtract the length of the two ropes
⇒ \(\frac{32}{3}\) − \(\frac{53}{6}\)
Take LCM of 3 and 6, then subtract
⇒ \(\frac{32}{3}\)− \(\frac{53}{6}\)
= \(\frac{64}{6}\)−\(\frac{53}{6}\)
= \(\frac{64−53}{6}\)
= \(\frac{11}{6}\)
Hence, the difference in the length of the two ropes is \(\frac{11}{6}\).
Page 150 Exercise 4.16 Problem 9
Given: Brandy bought 3 yd of raffia. She used \(\frac{5}{6}\) yd to make a doll.
To Find – Find the length of the raffia left.
By subtracting the integers rule, find the length of the raffia left.
Given, Brandy bought 3 yd of raffia. She used \(\frac{5}{6}\) yd to make a doll.
Total raffia is 3yd . Out of which, \(\frac{5}{6}\) yd is used to make a doll. So, the length of the raffia left
⇒ (3-\(\frac{5}{6}\)) yd
Solve (3-\(\frac{5}{6}\)) yd
⇒ (3-\(\frac{5}{6}\)) yd
= (\(\frac{18-5}{6}\)) yd
= \(\frac{13}{6}\)
Hence, the length of the raffia left is \(\frac{13}{6}\).
Page 151 Exercise 4.16 Problem 10
Given: Total 10 number of fruits, out of which 3 number of apples, 1 number of pears, and 6 number of oranges.
To Find – What fraction of the fruits are oranges?
The total number of fruits is 10. Out of which, 6 are oranges.
So, a fraction of the fruits are oranges is \(\frac{6}{10}\) = \(\frac{3}{5}\)
Hence, \(\frac{3}{5}\) of the fruits are oranges.
Page 151 Exercise 4.16 Problem 11
Given: In a test, Matthew answered 32 out of 40 questions correctly.
To Find – What fraction of the questions did he answer correctly?
In a test, Matthew answered 32 out of 40 questions correctly.
So, a fraction of the questions did he answer correctly \(\frac{32}{40}\)
= \(\frac{4}{5}\)
Hence, a fraction of the questions he answered correctly is \(\frac{4}{5}\).
Page 151 Exercise 4.16 Problem 12
Given: Mrs. Reed bought 30 eggs.
She used 5 of them to bake a cake.
To Find – What fraction of the eggs did she have left?
Mrs. Reed bought 30 eggs.
She used 5 of them to bake a cake.
So, the remaining eggs she left were 30 − 5 = 2 eggs.
Thus, the fraction of the eggs she has left is \(\frac{25}{30}\) = \(\frac{5}{6}\).
Hence, the fraction of the eggs left with her is \(\frac{5}{6}\).
Page 152 Exercise 4.16 Problem 13
Given: Taylor made 98 sugar buns and 42 plain buns.
To Find – What fraction of the buns were plain buns?
The total number of buns Taylor has is 98 + 42 = 140 buns.
Thus, the fraction of the buns that were plain buns is \(\frac{42}{140}\) = \(\frac{21}{70}\).
Hence, \(\frac{21}{70}\) of the buns were plain buns.
Page 152 Exercise 4.16 Problem 14
Given: Rosa wants to tie 6 packages. She needs \(\frac{3}{5}\) yd of string for each package.
To Find – How many yards of the string must she buy?
Given, Rosa wants to tie 6 packages.
For each package, she needs \(\frac{3}{5}\) yd of string.
Hence , for 6 packages she needs 6 × \(\frac{3}{5}\)
= \(\frac{18}{5}\) yd of string.
Hence, she must buy \(\frac{18}{5}\) yd of string for 6 packages.
Page 152 Exercise 4.16 Problem 15
Given: The job had $24
He used \(\frac{7}{8}\) of it to buy a book.
To Find – What was the cost of the book?
Given, that Job had $24. He used \(\frac{7}{8}\) of $24 to buy a book.
So, the cost book is \(\frac{7}{8}\) × $24 = $21.
Hence, the cost of the book was $21.
Page 153 Exercise 4.16 Problem 16
Given: There are 60 roses. \(\frac{7}{10}\) of them are red roses.
The rest are yellow roses.
To Find – How many yellow roses are there?
Given
There are 60 roses. \(\frac{7}{10}\) of them are red roses.
So, red roses are
⇒ \(\frac{7}{10}\) × 60 = 42
Total of 60 roses, 42 roses are red.
Hence remaining roses are 60 − 42 = 18.
Therefore, there are 18 yellow roses.
Hence, there are 18 yellow roses.
Page 153 Exercise 4.16 Problem 17
Given: Ben had $35.
He spent \(\frac{2}{7}\) $35 on a pair of shoes.
To Find – How much money did he have left?
Given, Ben had $35. He spent \(\frac{2}{7}\) $35 on a pair of shoes.
Hence, money he had spent \(\frac{2}{7}\) × $35 = $10.
Therefore, he had left 35 −10 = $25.
Hence, Ben has $25 left.
Page 154 Exercise 4.16 Problem 18
Given: After spending \(\frac{3}{5}\) of his money on a tennis racket, Sean had $14 left.
To Find – How much did the tennis racket cost?
First, assume total money x and then find the value of x, for which he left $14
Then, substitute the value of x, in money he had spent that is \(\frac{3x}{5}\), and find the cost.
Let’s assume, Sean had total money x.
He spent \(\frac{3}{5}\) of his money x on a tennis racket.
So, he spent \(\frac{3x}{5}\) and left $14.
Therefore, the total money he had
⇒ x−\(\frac{3x}{5}\) =14
⇒ \(\frac{5x−3x}{5}\) = 14
⇒ \(\frac{2x}{5}\) = 14
⇒ 2x = 70
Or , x = 35
Hence, the total money he had was $35. He spent on tennis
⇒ \(\frac{3x}{5}\)
=\(\frac{3×35}{5}\)
= 21
Hence, the tennis racket cost $21.
Page 154 Exercise 4.16 Problem 19
Given: A computer costs $2290.
An oven costs \(\frac{1}{5}\) the cost of the computer.
To Find – How much more does the computer cost than the oven?
A computer costs $2290. An oven costs \(\frac{1}{5}\) the cost of the computer.
Hence, the oven cost \(\frac{1}{5}\) ×2290 = $458.
Therefore, the computer costs more than the oven is
$2290 − $458 = $1832
Hence, the computer costs $1832 more than the oven.