EXERCISE:2.3
Question 1
(i) $7 \times \frac{3}{5}=\frac{7 \times 3}{5}=\frac{21}{5}$
ii) $21 \times \frac{3}{14}=\frac{3 \times 3}{2}=\frac{9}{2}$
iii) $3 \frac{2}{5} \times 8=\frac{17}{5} \times 8=\frac{136}{5}$
iv) $5 \times 6 \frac{3}{4}= 5 \times \frac{27}{4}=\frac{135}{4}$
Question 2
i)$\frac{2}{3} \times 18=2 \times 6=12$
ii) $\frac{1}{2} \times 4 \frac{2}{9}=\frac{1}{7} \times \frac{38}{9}=\frac{19}{9}$
iii)$\frac{5}{8} \times 9 \frac{2}{3}=\frac{5}{8} \times \frac{29}{3}=\frac{145}{24}$
Question 3
i) $\frac{3}{7} \times \frac{5}{9}=\frac{5}{7 \times 3}=\frac{5}{21}$
ii) $\frac{2}{5} \times 5 \frac{1}{4}=\frac{2}{5} \times \frac{21}{4}=\frac{21}{10}$.
iii) $2 \frac{1}{3} \times 3 \frac{4}{21}=\frac{7}{3} \times \frac{109}{21}=\frac{109}{9}$
iv) $3 \frac{1}{6} \times 7 \frac{4}{23}=\frac{19}6 \times \frac{16555}{23}=\frac{19 \times 55}{2 \times 23}=\frac{1045}{46}$.
Question 4
i) $\frac{1}{3} \times 42=Rs 14$
ii) $\frac{3}{7} \times 5 \frac{1}{4}=\frac{3}{7} \times \frac{21}{4}=\frac{3 \times 3}{4}=\frac{9}{4} \mathrm{~kg}$
iii) $4 \frac{1}{2} \times 5 \frac{1}{2}=\frac{9}{2} \times \frac{11}{2}=\frac{99}{4} \cdot$ metres
Question 5
(i)$\frac{2}{2} \times \frac{3}{4} \quad ; \quad \frac{3}{8} \times \frac{3}{8}$
$=\frac{3}{7 \times 2} \quad ; \quad \frac{3 \times 1}{1 \times 8}$
$=\frac{3}{14} \quad ; \quad \frac{3}{8}$
$\frac{3}{14}=\frac{3 \times 4}{14 \times 4}=-\frac{12}{56} \quad ; \frac{3}{8}=\frac{3 \times 7}{8 \times 7}=\frac{21}{56} .$
$\therefore 21>12 .$
$\therefore \frac{3}{8}$ is greater than $\frac{3}{14}$
(ii) $\frac{1}{7} \times \frac{6}{7} ; \quad \frac{2}{3} \times \frac{3}{7}$
$=\frac{1 \times 3}{7} \quad ; \quad \frac{2 \times 1}{7}$
$=\frac{3}{7} \quad ; \frac{2}{7}$
As $\frac{3}{7},>\frac{2}{7} \quad(\because 3>2)$
Question 6
Given, 1 metre cloth costs Rs $31 \frac{3}{4}$ = Rs $\frac{127}{4}$
$5 \frac{1}{2}$ metres cloth costs = ?
Let it be equal = x
$5 \frac{1}{2}=\frac{5 \times 2+1}{2}=\frac{11}{2}$
$\frac{11 / 2}{1}=\frac{x}{(127 / 4)}$
$\frac{11}{2}=\frac{x \times 4}{127}$
$x=\frac{11 \times 127}{2 \times 4}$
X = $=\frac{1397}{8}$ = Rs $174 \frac{5}{8}$
$\therefore 5 \frac{1}{2}$ metre cloth costs is $174 \frac{5}{8}$
Question 7
Speed of a car = $\operatorname{105} \frac{1}{5} \mathrm{kmph}$
$=\frac{526}{5} \mathrm{kmph}$
Time taken = $3 \frac{3}{5}$ hours$=\frac{18}{5}$ hours
$\begin{aligned} \text { Distance } &=\text { speed } \times \text { Time } \\ &=\frac{526}{5} \times \frac{18}{5} \mathrm{~km} \end{aligned}$
Distance $=\frac{9468}{25} \mathrm{~km}$
Question 8
For 1 litre of petrol , the car runs 16km
For $2 \frac{3}{4}$ litres of petrol , the car runs ? km
Let the car runs be x km
$2 \frac{3}{4}=\frac{11}{4}$
i.e $\frac{x}{16}=\frac{11 / 4}{1}$
$\frac{x}{16}=\frac{11}{4}$
$x=\frac{11 \times 16}{4}$
$x=44 \mathrm{~km}$
Question 9
In one hour, Sushant reads $\frac{1}{3}$ part of book
Let In $2 \frac{1}{5}=\frac{11}{5}$ Hours, he reads x part of book
i.e
$\begin{aligned} \frac{x}{(1 / 3)} &=\frac{(11 / 5)}{(1)} \\ 3 x &=\frac{11}{5} \\ x &=\frac{11}{3 \times 5} \end{aligned}$
$x=\frac{11}{15}$ part of book
Question 10
Gold and copper ornament weigh = 52grams
Out of which, part of copper is $\frac{2}{13} \mathrm{th}$
Then the copper weighs -$\frac{2}{13}, \times 52$
= 8gms
Remaining Gold weighs = 52- 8
= 44grams
Question 11
Total no.of students in a class = 40
(i) Out of them, students like to study = $\frac{1}{5}th$ of total
$=\frac{1}{5} \times 40$
$=8$
(ii) Students like to study mathematics = $\frac{2}{5} th$ of Total
$=\frac{2}{5} \times 40$
=16
(iii) Fraction of total no . of students like to study
Science = $1-\left\{\frac{1}{5}+\frac{2}{5}\right\}$
$1-\left\{\frac{1+2}{5}\right\}$
=$1-\frac{3}{5}$
$=\frac{5-3}{5}$
$=\frac{2}{5}$
Fraction of students who like to study science = $\frac{2}{5} \th$ of total
Question 12
A rectangular sheet is having length = $12 \frac{1}{2}(m)=\frac{25}{2} c m=L$
Width B= $10 \frac{2}{3} \mathrm{~cm}=\frac{32}{3}$Cm
(i) We know perimeter of rectangle = $2 \times($ Length + width $)$
$=2 \times\left(\frac{25}{2}+\frac{32}{3}\right)$
$=2 \times\left(\frac{25 \times 3+32 x^{2}}{6}\right)$
$=2 \times\left(\frac{75+64}{6}\right)$
$=2 \times\left(\frac{139}{6}\right)$
∴ Perimeter of rectangle = $\frac{139}{3}$ cm
(ii) Area of rectangle = length $\times$ width
$=\frac{25}{2} \times \frac{32}{3}$
$=\frac{25 \times 16}{7 \times 1}$
$=\frac{400}{3}$
$=133 \frac{1}{3} \mathrm{~cm}^{2}$
∴ Area of rectangle = $133 \frac{1}{3} \mathrm{~cm}^{2}$.
Question 13
Part of students are girls = $\frac{25}{54}$
Then part of students are boys = $1-\frac{25}{54}$
$=\frac{54-25}{54}$
$=\frac{29}{54}$
Given number of boys is 2030
Let Total no. of students = x
Now, $\frac{29}{54} \times x=2030$
$x=\frac{2030 \times 54}{29}$
$x=3780$
∴ Total no. of students = 3780
No. of girls = Total -No of boys
= 3780 - 2030 = 1750
Question 14
Let number of trees in an archard = 'x'
Given banana trees are 148 in number
Also orange trees are $\frac{1}{5}$th total
Mango trees are $\frac{3}{13}$th total
Now, the part of banana trees= $1=\left\{\frac{1}{5}+\frac{3}{13}\right\}$
=$1-\frac{13+3 \times 5}{65}$
=$1-\frac{13+15}{65}$
$=1-\frac{28}{65}$
$=\frac{65-28}{65}$
$=\frac{37}{65}$
Now , as banana Trees = 148
ஃ $\frac{37}{65}$ th of total = 148
i.e
$\begin{aligned} \frac{37}{65} \times x &=148 \\ x &=\frac{148 \times 65}{3.7}=260 . \end{aligned}$
ஃ Total no. of trees in an archard = 260
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