Chapter 8: Comparing Quantities

## Exercise 1: (Multiple Choice Questions and Answers 1-23)

In questions 1 to 23, there are four options, out of which one is correct. Write the correct one.

# Question: 1

$20%$ of $700\text{\hspace{0.17em}}\text{m}$ is

a.    $560\text{\hspace{0.17em}}\text{m}$

b.   $70\text{\hspace{0.17em}}\text{m}$

c.    $210\text{\hspace{0.17em}}\text{m}$

d.   $140\text{\hspace{0.17em}}\text{m}$

## Solution

(d)

$20%$ of $\begin{array}{l}700\text{\hspace{0.17em}}\text{m}=\frac{20}{100}×700\\ =20×7\\ =140\text{\hspace{0.17em}}\text{m}\end{array}$

Therefore, $20%$ of $700\text{\hspace{0.17em}}\text{m}$ is $140\text{\hspace{0.17em}}\text{m}$.

# Question: 2

Gayatri’s income is $\text{Rs}\text{\hspace{0.17em}}1,60,000$ per year. She pays $15%$ of this as house rent and $10%$ of the remainder on her child’s education. The money left with her is

a.    $\text{Rs}\text{\hspace{0.17em}}136000$

b.   $\text{Rs}\text{\hspace{0.17em}}120000$

c.    $\text{Rs}\text{\hspace{0.17em}}122400$

d.   $\text{Rs}\text{\hspace{0.17em}}14000$

## Solution

(c)

Given, Gayatri’s income $=\text{Rs}\text{\hspace{0.17em}}160000$

Money paid as house rent = $15%$ of $160000$ $=\frac{15}{100}×160000=$ $\text{Rs}\text{\hspace{0.17em}}24000$

Remaining amount $=160000-24000=$ $\text{Rs}\text{\hspace{0.17em}}136000$

Money spent on child’s education $=10%$ of $136000$ $=\frac{10}{100}×136000=$ $\text{Rs}\text{\hspace{0.17em}}13600$

Money left with Gayatri $=136000-13600=\text{\hspace{0.17em}}\text{Rs}\text{\hspace{0.17em}}122400$

# Question: 3

The ratio of Fatima’s income to her savings is $4:1$. The percentage of money saved by her is

a.    $20%$

b.   $25%$

c.    $40%$

d.   $80%$

## Solution

(a)

Given ratio of income to savings of Fatima $=4:1$

Here, we need to find out the percentage of money saved by Fatima.

Let the income $=4x$ and her savings $=x$

Percentage of money saved by Fatima $=\frac{\text{savings}}{\text{income}+\text{savings}}×100%$

$\begin{array}{l}=\frac{x}{4x+x}×100\\ =\frac{x}{5x}×100%\\ =\frac{1}{5}×100%\\ =20%\end{array}$

# Question: 4

$0.07$ is equal to

a.    $70%$

b.   $7%$

c.    $0.7%$

d.   $0.07%$

## Solution

(b)

$0.07=\frac{7}{100}×100%=7%$

# Question: 5

In a scout camp, $40%$ of the scouts were from Gujarat State and $20%$ of these were from Ahmedabad. The percentage of scouts in the camp from Ahmedabad is:

a.    $25$

b.   $32.5$

c.    $8$

d.   $50$

## Solution

(c)

Let the number of scouts in scout camp $=100$

The number of scouts from Gujarat $=40%$ of $100$ $=\frac{40}{100}×100=40$

Number of scouts from Ahmedabad $=20%$ of $40$ $=\frac{20}{100}×40=8$

$=\frac{8}{100}×100%=8%$

# Question: 6

What percent of $\text{Rs}\text{\hspace{0.17em}}4500$ is $\text{Rs}\text{\hspace{0.17em}}9000$?

a.    $200$

b.   $\frac{1}{2}$

c.    $2$

d.   $50$

## Solution

(a)

Let the percentage be $x%$

$=200$

Option (a) is the correct one

# Question: 7

$5.2$ is equal to

a.    $50%$

b.   $5.2%$

c.    $520%$

d.   $0.52%$

## Solution

(c)

We have, $5.2$

In percentage, $\frac{52}{10}×100%=520%$

# Question: 8

The ratio $3:8$ is equal to

a.    $3.75%$

b.   $37.5%$

c.    $0.375%$

d.   $267%$

## Solution

(b)

Given, ratio $=3:8$

In percentage, $\frac{3}{8}×100%=37.5%$

# Question: 9

$225%$ is equal to

a.    $9:4$

b.   $4:9$

c.    $3:2$

d.   $2:3$

## Solution

(a)

We have, $225%$ in fraction, $225×\frac{1}{100}=\frac{225}{100}=\frac{9}{4}$

$\therefore$ Required ratio $=9:4$

# Question: 10

A bicycle is purchased for $\text{Rs}1800$ and is sold at a profit of $12%$. Its selling price is

a.    $\text{Rs}1584$

b.   $\text{Rs}2016$

c.    $\text{Rs}1788$

d.   $\text{Rs}1812$

## Solution

(b)

Given, cost price of bicycle $=\text{Rs}\text{\hspace{0.17em}}1800$ and profit $=12%$

As we know, profit percentage can be calculated as,

Profit $%=\frac{\text{Profit}}{\text{CP}}×100$

$12=\frac{\text{Profit}}{1800}×100$

Profit $=12×18=\text{\hspace{0.17em}}\text{Rs}\text{\hspace{0.17em}}216$

SP $=$ CP $+$ Profit $=1800+216=\text{\hspace{0.17em}}\text{Rs}\text{\hspace{0.17em}}\text{2016}$

Hence, selling price of bicycle is $\text{Rs}\text{\hspace{0.17em}}\text{\hspace{0.17em}}2016$

# Question: 11

A cricket bat was purchased for $\text{Rs}\text{\hspace{0.17em}}800$ and was sold for $\text{Rs}1600$. Then profit earned is

a.    $100%$

b.   $64%$

c.    $50%$

d.   $60%$

## Solution

(a)

Given, cost price of bat $=\text{\hspace{0.17em}}\text{Rs}800$

Selling price of bat $=\text{\hspace{0.17em}}\text{Rs}1600$

Profit $=$ SP $-$ CP

We know that profit percentage is given as,

Profit $%=\frac{\text{Profit}}{\text{CP}}×100=\frac{800}{800}×100=100%$

Hence, profit earned is $100%$

# Question: 12

A farmer bought a buffalo for $\text{Rs}44000$ and a cow for $\text{Rs}18000$. He sold the buffalo at a loss of $5%$ but made a profit of $10%$ on the cow. The net result of the transaction is

a.    loss of $\text{Rs}\text{\hspace{0.17em}}200$

b.   profit of $\text{Rs}400$

c.    loss of $\text{Rs}400$

d.   profit of $\text{Rs}\text{\hspace{0.17em}}200$

## Solution

(c)

CP of buffalo $=\text{\hspace{0.17em}}\text{Rs}44,000$

Loss $%=5%$

Loss $%=\frac{\text{Loss}}{\text{cp}}×100$

$5=\frac{\text{Loss}}{44000}×100$

Loss $=5×440=\text{\hspace{0.17em}}\text{Rs}2200$

SP $=$ CP $-$ Loss $=44000-2200=\text{\hspace{0.17em}}\text{Rs}41800$

CP of cow $=\text{\hspace{0.17em}}\text{Rs}18000$

Profit $%=10%$

Profit $%=\frac{\text{Profit}}{\text{CP}}×100$

$10=\frac{\text{Profit}}{18000}×10$

Profit $=\text{\hspace{0.17em}}\text{Rs}1800$

SP $=$ CP $+$ Profit $=18000+1800=\text{\hspace{0.17em}}\text{Rs}19800$

Total CP of buffalo and cow $=44000+18000=\text{\hspace{0.17em}}\text{Rs}62000$

Total SP of buffalo and cow $=41800+19800=\text{\hspace{0.17em}}\text{Rs}61600$

Net loss $=$ CP $-$ SP $=62000-61600=\text{\hspace{0.17em}}\text{Rs}400$

# Question: 13

If Mohan’s income is $25%$ more than Raman’s income, then Raman’s income is less than Mohan’s income by

a.    $25%$

b.   $28%$

c.    $20%$

d.   $75%$

## Solution

(c)

Let the Raman’s income be $x$.

Mohan’s income is $25%$ more than Raman’s income.

Then, Mohan’s income  $=x+25%$ of $x$

$=x+\frac{25}{100}x=x\left(1+\frac{25}{100}\right)$

$=x\left(\frac{100+25}{100}\right)=\frac{125}{100}x$

Percentage of Raman’s income less than Mohan’s income

$=\frac{\frac{125}{100}x-x}{\frac{125}{100}x}×100=\frac{\frac{125-100x}{100}}{\frac{125}{100}x}×100$

$=\frac{25x}{125x}×100=20%$

# Question: 14

The interest on $\text{Rs}30000$ for $3$ years at the rate of $15%$ per annum is

a.    $\text{Rs}\text{\hspace{0.17em}}4500$

b.   $\text{Rs}\text{\hspace{0.17em}}9000$

c.    $\text{Rs}\text{\hspace{0.17em}}18000$

d.   $\text{Rs}\text{\hspace{0.17em}}13500$

## Solution

(d)

Given, $\text{P}=\text{\hspace{0.17em}}\text{Rs}\text{\hspace{0.17em}}30000,\text{\hspace{0.17em}}\text{T}=3\text{\hspace{0.17em}}\text{yr,}\text{\hspace{0.17em}}\text{R}=15%$

We know that, $\text{I}=\frac{\text{P}×\text{R}×\text{T}}{100}=\frac{30000×15×3}{100}=\text{\hspace{0.17em}}\text{Rs}\text{\hspace{0.17em}}13500$

# Question: 15

Amount received on $\text{Rs}3000$ for $2$ years at the rate of $11%$ per annum is

a.    $\text{Rs}2340$

b.   $\text{Rs}3660$

c.    $\text{Rs}4320$

d.   $\text{Rs}3330$

## Solution

(b)

Given, $\text{P}=\text{\hspace{0.17em}}\text{Rs}\text{\hspace{0.17em}}3000,\text{\hspace{0.17em}}\text{T}=2\text{\hspace{0.17em}}\text{yr,}\text{\hspace{0.17em}}\text{R}=11%$

$\text{I}=\frac{\text{P}×\text{R}×\text{T}}{100}=\frac{3000×11×2}{100}=\text{\hspace{0.17em}}\text{Rs}\text{\hspace{0.17em}}660$

Now, amount $\text{(A)}=\text{P}+\text{I}$ $=3000+660=\text{\hspace{0.17em}}\text{Rs}\text{\hspace{0.17em}}3660$

# Question: 16

Interest on $\text{Rs}\text{\hspace{0.17em}}12000$ for $1$ month at the rate of $10%$ per annum is

a.    $\text{Rs}\text{\hspace{0.17em}}1200$

b.   $\text{Rs}\text{\hspace{0.17em}}600$

c.    $\text{Rs}\text{\hspace{0.17em}}100$

d.   $\text{Rs}\text{\hspace{0.17em}}12100$

## Solution

(c)

Given,

$\begin{array}{l}\text{P}=\text{\hspace{0.17em}}\text{Rs}\text{\hspace{0.17em}}12000,\text{\hspace{0.17em}}\\ \text{T}=1\text{\hspace{0.17em}}\text{month}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{1}{12}\text{yr}\text{\hspace{0.17em}}\\ \text{R}=10%\end{array}$

$\begin{array}{l}=\frac{\text{P}×\text{R}×\text{T}}{100}\\ =\frac{12000×10×1}{100×12}\\ =\text{\hspace{0.17em}}\text{Rs}\text{\hspace{0.17em}}100\end{array}$

# Question: 17

Rajni and Mohini deposited $\text{Rs}\text{\hspace{0.17em}}3000$ and $\text{Rs}\text{\hspace{0.17em}}4000$ in a company at the rate of $10%$ per annum for $3$ years and $2\frac{1}{2}$ years respectively. The difference of the amounts received by them will be

a.    $\text{Rs}\text{\hspace{0.17em}}100$

b.   $\text{Rs}\text{\hspace{0.17em}}1000$

c.    $\text{Rs}\text{\hspace{0.17em}}900$

d.   $\text{Rs}\text{\hspace{0.17em}}1100$

## Solution

(d)

Interest received by Rajni after 3 years, $\text{P}=\text{\hspace{0.17em}}\text{Rs}\text{\hspace{0.17em}}3600,\text{\hspace{0.17em}}\text{R}=10%\text{\hspace{0.17em}}\text{per}\text{\hspace{0.17em}}\text{annum,T}=3\text{\hspace{0.17em}}\text{yr}$

$�$ I $=\frac{\text{P}×\text{R}×\text{T}}{100}=\frac{3000×10×3}{100}=\text{\hspace{0.17em}}\text{Rs}\text{\hspace{0.17em}}900$

And total amount received will be, $\text{A}=\text{P}+\text{I}=3000+900=\text{\hspace{0.17em}}Rs\text{\hspace{0.17em}}3900$

Interest received by Mohini after $2\frac{1}{2}\text{yr}$ years, $\text{P}=\text{\hspace{0.17em}}Rs\text{\hspace{0.17em}}4000,\text{\hspace{0.17em}}\text{R}=10%\text{\hspace{0.17em}}\text{per}\text{\hspace{0.17em}}\text{annum,}\text{\hspace{0.17em}}\text{T}=2\frac{1}{2}\text{yr}\text{\hspace{0.17em}}\text{=}\frac{5}{2}\text{yr}$

$\therefore$ I $=\frac{\text{P}×\text{R}×\text{T}}{100}=\frac{4000×10×5}{100×2}=\text{\hspace{0.17em}}Rs\text{\hspace{0.17em}}1000$

And total amount received will be,

$\text{A}=\text{P}+\text{I}=4000+1000=5000$

Difference in amounts $=5000-3900=\text{\hspace{0.17em}}Rs\text{\hspace{0.17em}}1100$

# Question: 18

If $90%$ of $x$ is $315\text{\hspace{0.17em}}km,$ then the value of $x$

a.    $325\text{\hspace{0.17em}}km$

b.   $350\text{\hspace{0.17em}}km$

c.    $405\text{\hspace{0.17em}}km$

d.   $340\text{\hspace{0.17em}}km$

## Solution

(b)

Given,

$90%$ of $x$$=315\text{\hspace{0.17em}}km$

$\begin{array}{l}⇒\frac{90}{100}×x=315\\ ⇒x=\frac{315×100}{90}\end{array}$

$x=350\text{\hspace{0.17em}}km$

# Question: 19

On selling an article for $Rs329,$ a dealer lost $6%$. The cost price of the article is

a.    $Rs310.37$

b.   $Rs348.74$

c.    $Rs335$

d.   $Rs350$

## Solution

(d)

Given, SP of an article is $Rs\text{\hspace{0.17em}}329$ and loss per cent $=6%$

We know that, loss per cent $=\frac{\text{Loss}}{\text{CP}}×100$

$6=\frac{\text{CP}-\text{SP}}{\text{CP}}×100$

$\frac{6}{100}\text{CP}=\text{CP}-\text{SP}$

$\text{CP}-\frac{6}{100}\text{CP}=329$

$\frac{94}{100}\text{CP}=329$

$\text{CP}=\frac{329×100}{94}$

$\text{CP}=\text{\hspace{0.17em}}Rs\text{\hspace{0.17em}}350$

# Question: 20

$\frac{25%\text{\hspace{0.17em}}of\text{\hspace{0.17em}}50%\text{\hspace{0.17em}}of\text{\hspace{0.17em}}100%}{25×50}$ is equal to

a.    $1.1%$

b.   $0.1%$

c.    $0.01%$

d.   $1%$

## Solution

We have, $\frac{25%\text{\hspace{0.17em}}of\text{\hspace{0.17em}}50%\text{\hspace{0.17em}}of\text{\hspace{0.17em}}100%}{25×50}=\frac{25}{100}×\frac{50}{100}×\frac{100}{100}×\frac{1}{25×50}$

$\frac{125000}{1250×100×100×100}=\frac{1}{10000}=0.0001$

In percentage, $0.0001×100%=0.01%$

# Question: 21

The sum which will earn a simple interest of $Rs\text{\hspace{0.17em}}126$ in $2$ years at $14%$ per annum is

a.    $Rs\text{\hspace{0.17em}}394$

b.   $Rs\text{\hspace{0.17em}}395$

c.    $Rs\text{\hspace{0.17em}}450$

d.   $Rs\text{\hspace{0.17em}}540$

## Solution

(c)

Given, $\text{I}=\text{\hspace{0.17em}}Rs\text{\hspace{0.17em}}126,\text{\hspace{0.17em}}\text{R}=14%$ and $\text{T}=2\text{\hspace{0.17em}}\text{yr}$

$\text{I}=\frac{\text{P}×\text{R}×\text{T}}{100}$

$\text{I}=126=\frac{\text{P}×14×2}{100}$

$126×100=\text{P}×14×2$

$\text{P}=\frac{12600}{14×2}$

$\text{P}=\text{\hspace{0.17em}}Rs450$

# Question: 22

The per cent that represents the unshaded region in the figure. a.    $75%$

b.   $50%$

c.    $40%$

d.   $60%$

## Solution

(c)

Total parts $=10×10=100$

Shaded parts $=60$

Percent of shaded parts $=\frac{60}{100}×100%=60%$

Then, percent of un-shaded parts $=100-60=40%$

Hence, the per cent that represents the un-shaded region is $40%$

# Question: 23

The per cent that represents the shaded region in the figure is a.    $36%$

b.   $64%$

c.    $27%$

d.   $48%$

## Solution

(a)

Total parts $=10×10=100$

Shaded parts $=36$

Percent of shaded parts $=\frac{36}{100}×100%=36%$

Hence, the per cent that represents the shaded region is $36%$

In each of the questions 24 to 59, fill in the blanks to make the statements true.

# Question: 24

$2:3=$ ______ $%$

## Solution

Given ratio $=2:3$

In percentage $=\frac{2}{3}×100%=66\frac{2}{3}%$

# Question: 25

$18\frac{3}{4}%=$ _______: ______

## Solution

Given percentage $=18\frac{3}{4}%=\frac{75}{4}%$

In fraction, $\frac{75}{4}×\frac{1}{100}=\frac{3}{16}$

Ratio $=3:16$

# Question: 26

$30%$ of $Rs\text{\hspace{0.17em}}360=$ _____.

## Solution

We have, $30%$ of $Rs\text{\hspace{0.17em}}360=\frac{30}{100}×360=\text{\hspace{0.17em}}Rs\text{\hspace{0.17em}}108$

# Question: 27

$120%$ of $50\text{\hspace{0.17em}}\text{km}=$ _____.

## Solution

We have, $120%$ of $50\text{\hspace{0.17em}}\text{km}=\frac{120}{100}×50\text{\hspace{0.17em}}\text{km}=60\text{\hspace{0.17em}}\text{km}$

# Question: 28

$2.5=$ _____ $%$

## Solution

We have, $2.5$

In percentage, $2.5×100%=250%$

# Question: 29

$\frac{8}{5}=$ _____ $%$

## Solution

We have, $\frac{8}{5}$

In percentage, $\frac{8}{5}×100%=160%$

# Question: 30

A ________ with its denominator $100$ is called a per cent

## Solution

A fraction with its denominator $100$ is called a per cent.

# Question: 31

$15\text{\hspace{0.17em}}kg$ is ______ $%$ of $50\text{\hspace{0.17em}}kg.$

## Solution

Let $x%$ of $50\text{\hspace{0.17em}}kg$ be $15\text{\hspace{0.17em}}kg$

Then, $\frac{x}{100}×50=15$

$\frac{x}{2}=15$

$x=15×2=30%$

Hence, $15\text{\hspace{0.17em}}kg$ is $30%$ of $50\text{\hspace{0.17em}}kg$

# Question: 32

Weight of Nikhil increased from $60\text{\hspace{0.17em}}\text{kg}$ to $66\text{\hspace{0.17em}}\text{kg}$. Then, the increase in weight is _____ $%$.

## Solution

Given, Initial weight of Nikhil $=60\text{\hspace{0.17em}}kg$

After increase in weight, weight became $=66\text{\hspace{0.17em}}\text{kg}$ Increase in weight $=66-60=6\text{\hspace{0.17em}}\text{kg}$

$\therefore$ Percentage increase of weight

# Question: 33

In a class of $50$ students, $8%$ were absent on one day. The number of students present on that day was ________.

## Solution

Given, Total number of students in the class $=50$

Absent percentage on one day $=8%$

Percentage of students present on that day $=100-8=92%$

$\therefore$ Number of students present on that day $=92%$ of $50$

$=\frac{92}{100}×50=\frac{92}{2}=46$

So, the number of students present on that day, was $46.$

# Question: 34

Savitri obtained $440$ marks out of $500$ in an examination. She secured _______ $%$ marks in the examination.

## Solution

Marks obtained by Savitri out of $500=440$

Percentage of marks obtained $=\frac{440}{500}×100%=88%$

Hence, Savitri secured $88%$ marks in the examination

# Question: 35

Out of a total deposit of $Rs\text{\hspace{0.17em}}1500$ in her bank account, Abida withdrew $40%$ of the deposit. Now the balance in her account is ______.

## Solution

Total deposit $=\text{\hspace{0.17em}}Rs\text{\hspace{0.17em}}1500$

Amount withdrawn $=40%$ of $Rs\text{\hspace{0.17em}}1500$

$=\frac{40}{100}×1500=\text{\hspace{0.17em}}Rs\text{\hspace{0.17em}}600$

$\therefore$ Balance in the account $=1500-600=\text{\hspace{0.17em}}Rs\text{\hspace{0.17em}}900$

# Question: 36

________ is $50%$ more than $60$.

## Solution

Let number be $x.$

It is given that $x$ is $50%$ more than $60.$

Therefore, according to question

$x=60+50%$ of $60.$

$\begin{array}{l}=60+\frac{50}{100}×60\\ =60+30=90\end{array}$

# Question: 37

John sells a bat for $Rs\text{\hspace{0.17em}}75$ and suffers a loss of $Rs\text{\hspace{0.17em}}8$. The cost price of the bat is ________.

## Solution

Given, SP of bat $=\text{\hspace{0.17em}}Rs\text{\hspace{0.17em}}75$ and loss $=\text{\hspace{0.17em}}Rs\text{\hspace{0.17em}}8$

We know that, CP $=$ SP $+$ Loss

$=75+8=\text{\hspace{0.17em}}Rs\text{\hspace{0.17em}}83$

Hence, cost price of the bat is $Rs\text{\hspace{0.17em}}83.$

# Question: 38

If the price of sugar is decreased by $20%,$ then the new price of $3kg$ sugar originally costing $Rs\text{\hspace{0.17em}}120$ will be ________.

## Solution

Original price of $3\text{\hspace{0.17em}}kg$ sugar $=\text{\hspace{0.17em}}Rs\text{\hspace{0.17em}}120$

Given that, price of sugar is decreased by $20%.$

So, new price of sugar $=$ Original price $-20%$ of original price

$=120-20%$ of $120$

$\begin{array}{l}=120-\frac{20}{100}×120\\ =120-24=\text{\hspace{0.17em}}Rs\text{\hspace{0.17em}}96\end{array}$

# Question: 39

Mohini bought a cow for $Rs\text{\hspace{0.17em}}9000$ and sold it at a loss of $Rs\text{\hspace{0.17em}}900$. The selling price of the cow is ________.

## Solution

Given, CP of cow $=\text{\hspace{0.17em}}Rs\text{\hspace{0.17em}}9000$ and loss $=\text{\hspace{0.17em}}Rs\text{\hspace{0.17em}}900$

We know that, SP $=$ CP $-$ Loss $=9000-900=\text{\hspace{0.17em}}Rs\text{\hspace{0.17em}}8100.$

Hence, the selling price of the cow is $Rs\text{\hspace{0.17em}}8100$

# Question: 40

Devangi buys a chair for $Rs700$ and sells it for $Rs\text{\hspace{0.17em}}750$. She earns a profit of ________ $%$ in the transaction.

## Solution

Given, CP of a chair $=\text{\hspace{0.17em}}Rs700$ and SP of a chair $=\text{\hspace{0.17em}}Rs750$

Since, SP $>$ CP

$\therefore$ Profit $=$ SP $-$ CP $=\text{\hspace{0.17em}}Rs\text{\hspace{0.17em}}\left(750-700\right)=\text{\hspace{0.17em}}Rs\text{\hspace{0.17em}}50$

Now, Profit $%=\frac{\text{Profit}}{\text{CP}}×100=\frac{50}{700}×100=7\frac{1}{7}%$

Hence, Devangi’s profit is $7\frac{1}{7}%.$

# Question: 41

Sonal bought a bed sheet for $Rs400$ and sold it for $Rs440$. Her ________ $%$ is ________.

## Solution

Given, CP of a bed sheet $=\text{\hspace{0.17em}}Rs400$ and SP of a bed sheet $=\text{\hspace{0.17em}}Rs440$

Since, SP $>$ CP

$\therefore$ Profit $=$ SP $-$ CP $=\text{\hspace{0.17em}}Rs\left(440-400\right)=\text{\hspace{0.17em}}Rs40$

Now, profit $%=\frac{\text{Profit}}{\text{CP}}×100=\frac{40}{400}×100=10%$

Hence, Sonal’s profit is $10%.$

# Question: 42

Nasim bought a pen for $Rs\text{\hspace{0.17em}}60$ and sold it for $Rs\text{\hspace{0.17em}}54$. His ________ $%$ is ________.

## Solution

Given, CP of a Pen $=\text{\hspace{0.17em}}Rs\text{\hspace{0.17em}}60$ and SP of a pen $=\text{\hspace{0.17em}}Rs\text{\hspace{0.17em}}54$

Since, SP $<$ CP

$\therefore$ Loss $=$ CP $-$ SP $=\text{\hspace{0.17em}}Rs\left(60-54\right)=\text{\hspace{0.17em}}Rs\text{\hspace{0.17em}}6$

Now, Loss $%=\frac{\text{Loss}}{\text{CP}}×100=\frac{6}{60}×100=10%$

Hence, Nasim’s loss is $10%.$

# Question: 43

Aahuti purchased a house for $Rs50,59,700$ and spent $Rs40300$ on its repairs. To make a profit of $5%$, she should sell the house for $Rs$ ________.

## Solution

Given, CP of house $=\text{\hspace{0.17em}}Rs5059700$

And amount spent on repairing $=\text{\hspace{0.17em}}Rs40300$

So, total CP of house $=\text{\hspace{0.17em}}Rs\left(5059700+40300\right)=\text{\hspace{0.17em}}Rs5100000$

Profit $%=\frac{\text{Profit}}{\text{CP}}×100%$

$5=\frac{\text{SP}-\text{CP}}{\text{CP}}×100$

$5=\frac{\text{SP}-5100000}{5100000}×100$

$\frac{5×5100000}{100}\text{=SP}-5100000$

$SP=5100000+255000$

$SP=\text{\hspace{0.17em}}Rs5355000$

# Question: 44

If $20$ lemons are bought for $Rs10$ and sold at $5$ for three rupees, then ________ in the transaction is ________ $%$.

## Solution

CP of $20$ lemons $=\text{\hspace{0.17em}}Rs\text{\hspace{0.17em}}10$

By applying unitary method,

If SP of $5$ lemons is $Rs\text{\hspace{0.17em}}3$

Then, SP of $1$ lemon $=\text{\hspace{0.17em}}\frac{3}{5}$

SP of $20$ lemons $=\text{\hspace{0.17em}}\text{\hspace{0.17em}}\frac{3}{5}×20=12$

Now, CP $=\text{\hspace{0.17em}}Rs\text{\hspace{0.17em}}10$ and SP $=\text{\hspace{0.17em}}Rs\text{\hspace{0.17em}}12$

Since, SP $>$ CP

Profit $=$ SP $-$ CP

$=\text{\hspace{0.17em}}Rs\text{\hspace{0.17em}}\left(12-10\right)=\text{\hspace{0.17em}}Rs\text{\hspace{0.17em}}2$

Now, profit $%=\frac{\text{Profit}}{\text{CP}}×100=\frac{2}{10}×100=20%$

Hence, profit is $20%$

# Question: 45

Narain bought $120$ oranges at $Rs\text{\hspace{0.17em}}4$ each. He sold $60%$ of the oranges at $Rs\text{\hspace{0.17em}}5$ each and the remaining at $Rs\text{\hspace{0.17em}}3.50$ each. His ________ is ________ $%$.

## Solution

CP of $1$ orange $=\text{\hspace{0.17em}}Rs4$ (given)

CP of $120$ oranges $=4×120=\text{\hspace{0.17em}}Rs480$

Now, $60%$ of $120$ oranges $=\frac{60}{100}×120=72$

$\therefore$ SP of $72$ oranges $=72×5=\text{\hspace{0.17em}}Rs\text{\hspace{0.17em}}360$

And SP of remaining oranges $=\left(120-72\right)×3.50=48×3.50=\text{\hspace{0.17em}}Rs\text{\hspace{0.17em}}168$

Total SP of $120$ oranges $=360+168=\text{\hspace{0.17em}}Rs\text{\hspace{0.17em}}528$

Since, SP $>$ CP

Profit $=$ SP $-$ CP $=$ $Rs\left(528-480\right)=\text{\hspace{0.17em}}Rs48$

Now,

Profit per cent

$=\frac{\text{Profit}}{\text{CP}}×100%=\frac{48}{480}×100=10%$

Hence, profit is $10%$

# Question: 46

A fruit seller purchased $20\text{\hspace{0.17em}}kg$ of apples at $Rs\text{\hspace{0.17em}}50\text{\hspace{0.17em}}per\text{\hspace{0.17em}}kg$. Out of these, $5%$ of the apples were found to be rotten. If he sells the remaining apples at $Rs\text{\hspace{0.17em}}60\text{\hspace{0.17em}}per\text{\hspace{0.17em}}kg$, then his _________is _________ $%$.

## Solution

Price for per kg apples $=\text{\hspace{0.17em}}Rs\text{\hspace{0.17em}}\text{\hspace{0.17em}}50$

Total weight of purchased apples $=20\text{\hspace{0.17em}}\text{kg}$

Since, $5%$ were rotten, so weight of good apples $=20\text{\hspace{0.17em}}\text{kg}-5%$ of $20\text{\hspace{0.17em}}\text{kg}$ (rotten)

$=20-\frac{5}{100}×20=20-1=19\text{\hspace{0.17em}}\text{kg}$

Also, he sells $19\text{\hspace{0.17em}}\text{kg}$ apples at $Rs\text{\hspace{0.17em}}60$ per kg

Total SP $=$ $19×60=\text{\hspace{0.17em}}Rs\text{\hspace{0.17em}}1140$

Cost price of $20\text{\hspace{0.17em}}kg$ apples $=20×50=\text{\hspace{0.17em}}Rs\text{\hspace{0.17em}}1000$

Profit $=$ SP $-$ CP

Now, Profit $%=\frac{\text{Profit}}{\text{CP}}×100%=\frac{140}{1000}×100%=\frac{140}{10}=14%$

So, his profit is $14%.$

# Question: 47

Interest on $Rs\text{\hspace{0.17em}}3000$ at $10%$ per annum for a period of $3$ years is ________.

## Solution

Given, $\text{P}=\text{\hspace{0.17em}}Rs\text{\hspace{0.17em}}3000,\text{\hspace{0.17em}}\text{R}=10%$ and $\text{T}=3\text{\hspace{0.17em}}yr$

$\text{I}=\frac{\text{P}×\text{R}×\text{T}}{100}$

$=\frac{3000×10×3}{100}=\text{\hspace{0.17em}}Rs\text{\hspace{0.17em}}900$

Hence, interest is $Rs\text{\hspace{0.17em}}900$

# Question: 48

Amount obtained by depositing $Rs\text{\hspace{0.17em}}20,000$ at $8%$ per annum for six months is ________.

## Solution

Amount deposited $=\text{\hspace{0.17em}}Rs\text{\hspace{0.17em}}20,000$

Rate of interest $=8%$

Time period $=6$ months $=\frac{6}{12}\text{\hspace{0.17em}}\text{yr}=\frac{1}{2}\text{yr}$

$\text{I}=\frac{\text{P}×\text{R}×\text{T}}{100}=\frac{20000×8×\frac{1}{2}}{100}=\frac{20000×8}{200}=100×8=\text{\hspace{0.17em}}Rs\text{\hspace{0.17em}}800$

Amount received $=$ Principal $+$ Interest $=20000+800=\text{\hspace{0.17em}}Rs\text{\hspace{0.17em}}20800$

# Question: 49

Interest on $Rs\text{\hspace{0.17em}}12500$ at $18%$ per annum for a period of $2$ years and $4$ months is ________.

## Solution

Given, $\text{P}=\text{\hspace{0.17em}}Rs\text{\hspace{0.17em}}12500$ and $\text{R}=18%$

$T=2\text{\hspace{0.17em}}yr4$ months $=\left(2+\frac{4}{12}\right)yr=\left(2+\frac{1}{3}\right)yr=\frac{7}{3}yr$

We know that, $\text{I}=\frac{\text{P}×\text{R}×\text{T}}{100}=\frac{12500×18×7}{3×100}=\text{\hspace{0.17em}}Rs\text{\hspace{0.17em}}5250$

# Question: 50

$25\text{\hspace{0.17em}}ml$ is _________ per cent of $5$ litres.

## Solution

Let $25\text{\hspace{0.17em}}ml$ be $x%$ of $5L$

Then, $25\text{\hspace{0.17em}}ml=x%$ of $5l$

$25=\frac{x}{100}×5×1000$ [Converting litre into millilitre]

$\frac{25×100}{5×1000}=x$

$x=0.5$

Hence, $25\text{\hspace{0.17em}}ml$ is $0.5%$ of $5l.$

# Question: 51

If A is increased by $20%,$ it equals B. If B is decreased by $50%,$ it equals C. Then __________ $%$ of A is equal to C.

## Solution

Given, if $\text{A}$ is increased by $20%,$ then it is equal to $\text{B}\text{.}$

$\therefore$$\text{A}+20%$ of $\text{A}=\text{B}$

$⇒\text{A}\left(1+\frac{20}{100}\right)=\text{B}$

$⇒\frac{120}{100}\text{A}=\text{B}$

$⇒\text{B}=\frac{6}{5}\text{A}$               (i)

If B is decreased by $50%,$ then it is equal to C.

$\text{B}-50%$ of $\text{B}=\text{C}$

$⇒\text{B}\left(1-\frac{50}{100}\right)=\text{C}$

$⇒\text{B}×\frac{50}{100}=\text{C}$

$⇒\frac{1}{2}\text{B}=\text{C}$

$⇒\text{B}=2\text{C}$          (ii)

On comparing eq. (i) and (ii), we get

$\frac{6}{5}\text{A}=2\text{C}$

$⇒\frac{\text{A}}{\text{C}}=\frac{10}{6}$

$⇒\frac{\text{C}}{\text{A}}=\frac{6}{10}$

$⇒\text{C}=\frac{3}{5}\text{A}$

In percentage $\frac{\text{C}}{\text{A}}×100=\frac{\frac{3}{5}\text{A}}{\text{A}}×100%$

$=\frac{3}{5}×100%=60%$

Hence, $60%$ of A is equal to C.

# Question: 52

Interest $=\frac{P×R×T}{100},$ where

$T$ is ____________

$R%$ is ____________ and

$P$ is ____________.

## Solution

Here, T is time period, $\text{R}%$ is rate of interest and P is Principal.

# Question: 53

The difference of interest for $2$ years and $3$ years on a sum of $Rs\text{\hspace{0.17em}}2100$ at $8%$ per annum is _________.

## Solution

Given, $\text{P}=\text{\hspace{0.17em}}Rs\text{\hspace{0.17em}}2100$ and $\text{R}=8%$

For, $\text{T}=2\text{\hspace{0.17em}}\text{yr}$

$\text{I}=\frac{P×R×T}{100}=\frac{2100×8×2}{100}=\text{\hspace{0.17em}}Rs\text{\hspace{0.17em}}336$

For $3\text{yr}\text{.}$

$\text{I}=\frac{P×R×T}{100}=\frac{2100×8×3}{100}=\text{\hspace{0.17em}}Rs\text{\hspace{0.17em}}504$

Difference between both interests $=504-336=\text{\hspace{0.17em}}Rs\text{\hspace{0.17em}}168$

# Question: 54

To convert a fraction into a per cent, we _________ it by $100$.

## Solution

To convert a fraction into a per cent, we multiple it by $100.$

# Question: 55

To convert a decimal into a per cent, we shift the decimal point two places to the _________.

## Solution

To convert a decimal into a per cent, we shift the decimal point into places to the right.

# Question: 56

The _________ of interest on a sum of $Rs\text{\hspace{0.17em}}2000$ at the rate of $6%$ per annum for $1\frac{1}{2}$ years and $2$ years is $Rs420$.

## Solution

Given, $\text{P}=\text{\hspace{0.17em}}Rs\text{ }\text{\hspace{0.17em}}2000$ and $\text{R}=6%$ $1\frac{1}{2}\text{yr}\text{\hspace{0.17em}}\text{=}\frac{3}{2}\text{yr,}$

Now, $\text{I}=\frac{P×R×T}{100}=\frac{2000×6×3}{100×2}=\text{\hspace{0.17em}}Rs180$

For $2\text{yr,}$

$\text{I}=\frac{P×R×T}{100}=\frac{2000×6×2}{100}=\text{\hspace{0.17em}}Rs240$

The sum of both interests is $Rs\left(180+240\right)=\text{\hspace{0.17em}}Rs420$

# Question: 57

When converted into percentage, the value of $6.5$ is _________ than $100%$.

## Solution

In percentage, $6.5×100%=650%$

When converted into percentage, the value of 6.5 is more than $100%.$

In questions 58 and 59 copy each number line. Fill in the blanks so that each mark on the number line is labelled with a per cent, a fraction and a decimal. Write all fractions in lowest terms.

# Question: 58 ## Solution

Percentage $=$ Fraction $×100$

Decimal $=\frac{\text{Percentage}}{100}$

Now, according to the formula, we have

 Percentage Fraction Decimal $0%$ $0$ $0$ $10%$ $\frac{1}{10}$ $0.1$ $20%$ $\frac{1}{5}$ $0.2$ $30%$ $\frac{3}{10}$ $0.3$ $40%$ $\frac{2}{5}$ $0.4$ $50%$ $\frac{1}{2}$ $0.5$ $60%$ $\frac{3}{5}$ $0.6$ $70%$ $\frac{7}{10}$ $0.7$ $80%$ $\frac{4}{5}$ $0.8$ $90%$ $\frac{9}{10}$ $0.9$ $100%$ $1$ $1$

# Question: 59 ## Solution

Percentage $=$ Fraction $×100$

Decimal $=\frac{\text{Percentage}}{100}$

Now, according to the formula, we have

 Percentage Fraction Decimal $0%$ $0$ $0$ $12.5%$ $\frac{1}{8}$ $0.125$ $25%$ $\frac{1}{4}$ $0.25$ $37.5%$ $\frac{3}{8}$ $0.375$ $50%$ $\frac{1}{2}$ $0.5$ $62.5%$ $\frac{5}{8}$ $0.625$ $75%$ $\frac{3}{4}$ $0.75$ $87.5%$ $\frac{7}{8}$ $0.875$ $100%$ $1$ $1$

True/False

In questions $60$ to $79,$ state whether the statements are true of False.

# Question: 60

$\frac{2}{3}=66\frac{2}{3}%.$

## Solution

True

Given fraction $=\frac{2}{3}$

In percentage, $\frac{2}{3}×100%=\frac{200}{3}=66\frac{2}{3}%$

# Question: 61

When an improper fraction is converted into percentage then the answer can also be less than $100$.

## Solution

False

Let’s consider, an improper fraction $\frac{12}{5}$

In percentage, $\frac{12}{5}×100%=240%$

Hence, when an improper fraction is converted into percentage, then the answer is always greater than $100.$

# Question: 62

$8$ hours is $50%$ of $4$ days.

## Solution

False

Let $8\text{\hspace{0.17em}}\text{hr}$ be $x%$ of $4$ days

Then, $8\text{\hspace{0.17em}}\text{hr}=x%$ of $4$ days

$⇒8=\frac{x}{100}×4×24$

$⇒\frac{8×100}{4×24}=x$

$⇒x=\frac{25}{3}=8\frac{1}{3}$

Hence, $8\text{\hspace{0.17em}}\text{hr}$ is  $8\frac{1}{3}%$ of $4$ days

# Question: 63

The interest on $Rs\text{\hspace{0.17em}}350$ at $5%$ per annum for $73$ days is $Rs\text{\hspace{0.17em}}35$.

## Solution

False

Given, $\text{P}=\text{\hspace{0.17em}}Rs\text{\hspace{0.17em}}350,\text{\hspace{0.17em}}\text{R}=5%$

And $\text{T}=73$ days $=\frac{73}{365}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{yr}$

$\therefore \text{I=}\frac{\text{P}×\text{R}×\text{T}}{100}$

$⇒\text{I}=\frac{350×5×73}{100×365}=\frac{127750}{36500}$

$⇒\text{I}=\text{\hspace{0.17em}}Rs\text{\hspace{0.17em}}3.5$

# Question: 64

The simple interest on a sum of $Rs\text{P}$ for T years at $\text{R}%$ per annum is given by the formula: Simple Interest = $\frac{\text{T}×\text{P}×\text{R}}{100}.$

## Solution

True

Simple Interest $=\frac{\text{P}×\text{R}×\text{T}}{100}$

It can also be written as, SI $=\frac{\text{T}×\text{P}×\text{R}}{100}$

# Question: 65

$75%=\frac{4}{3}.$

## Solution

False

In fraction, $75%$ can be written as

$\frac{75}{100}=\frac{3}{4}$

$\therefore 75%=\frac{3}{4}$

# Question: 66

$12%$ of $120$ is $100$.

## Solution

False

$12%$ of $120=\frac{12}{100}×120=\frac{1440}{100}=14.4$

Therefore, $12%$ of $120$ is $14.4$

# Question: 67

If Ankita obtains $336$ marks out of $600,$ then percentage of marks obtained by her is $33.6.$

## Solution

False

Marks obtained by Ankita out of $600=336$

Percentage of marks $=\frac{336}{600}×100%=56%$

Hence, Ankita got $56%$ marks.

# Question: 68

$0.018$ is equivalent to $8%.$

## Solution

False

In percentage, $0.018$ can be written as,

$0.018×100\text{\hspace{0.17em}}=\text{\hspace{0.17em}}1.8%$

Hence, $0.018$ is equivalent to $1.8%$

# Question: 69

$50%$ of $Rs\text{\hspace{0.17em}}50$ is $Rs\text{\hspace{0.17em}}25.$

## Solution

True

Since, $50%$ of $Rs\text{\hspace{0.17em}}50=\frac{50}{100}×50=\text{\hspace{0.17em}}Rs\text{\hspace{0.17em}}25$

Hence, $50%$ of $Rs\text{\hspace{0.17em}}50$ is $Rs\text{\hspace{0.17em}}25.$

# Question: 70

$250\text{\hspace{0.17em}}cm$ is $4%$ of $1\text{\hspace{0.17em}}km.$

## Solution

False

$250\text{\hspace{0.17em}}\text{cm}=\frac{250}{100}=2.5\text{\hspace{0.17em}}\text{m}$

Now, $4%$ of $1\text{\hspace{0.17em}}\text{km}=\frac{4}{100}×1000\text{\hspace{0.17em}}m$

$=40\text{\hspace{0.17em}}\text{m}$

Hence, $250\text{\hspace{0.17em}}\text{cm}\ne 4%$ of $1\text{\hspace{0.17em}}\text{km}$

# Question: 71

Out of $600$ students of a school, $126$ go for a picnic. The percentage of students that did not go for the picnic is $75.$

## Solution

False

Total number of  students in school $=600$

Number of students who went for picnic $=126$

$\therefore$ Number of students who did not go for picnic $=600-126=474$

Percentage of students who did not go for picnic, $\frac{474}{600}×100%=79%$

Therefore, $79%$ of students did not go for picnic.

# Question: 72

By selling a book for $Rs\text{\hspace{0.17em}}50$, a shopkeeper suffers a loss of $10%$. The cost price of the book is $Rs\text{\hspace{0.17em}}60$.

## Solution

False

SP of book $=\text{\hspace{0.17em}}Rs\text{\hspace{0.17em}}50$

Loss per cent on book $=10%$

Loss per cent $=\frac{\text{Loss}}{\text{CP}}×100%$

Loss per cent $=\frac{\text{CP}-\text{SP}}{\text{CP}}×100%$

$\therefore 10=\frac{\text{CP}-50}{\text{CP}}×100%$

$⇒10\text{\hspace{0.17em}}\text{CP}=100\text{\hspace{0.17em}}\text{CP}-5000$

$⇒10\text{\hspace{0.17em}}\text{CP}=5000$

$⇒\text{CP}=\frac{5000}{90}$

$\therefore \text{CP}=\text{\hspace{0.17em}}Rs\text{\hspace{0.17em}}55.55$

Hence, the cost price of the book is $Rs55.55$

# Question: 73

If a chair is bought for $Rs\text{\hspace{0.17em}}2000$ and is sold at a gain of $10%,$ then selling price of the chair is $Rs\text{\hspace{0.17em}}2010$.

## Solution

False

Given,  and Profit per cent $=10%$

Profit per cent $=\frac{\text{Profit}}{\text{CP}}×100$

Profit $=\frac{\text{SP}-\text{CP}}{\text{CP}}×100$

10 $=\frac{\text{SP}-2000}{2000}×100$

$\frac{\left(10×2000\right)}{100}=\text{SP}-2000$

$200=\text{SP}-2000$

$\text{SP}=\text{\hspace{0.17em}}Rs\text{\hspace{0.17em}}2200$

Hence, SP of chair is $Rs\text{\hspace{0.17em}}2200$

# Question: 74

If a bicycle was bought for $Rs\text{\hspace{0.17em}}650$ and sold for $Rs\text{\hspace{0.17em}}585$, then the percentage of profit is $10$.

## Solution

False

Given, CP of bicycle $=\text{\hspace{0.17em}}Rs\text{\hspace{0.17em}}650$ and SP of bicycle $=\text{\hspace{0.17em}}Rs\text{\hspace{0.17em}}585$

Since, CP $>$ SP

Loss $=$ CP $-$ SP $=650-585=\text{\hspace{0.17em}}Rs\text{\hspace{0.17em}}65$

Now, Loss per cent

$=\frac{\text{Loss}}{\text{CP}}×100=\frac{65}{650}×100=10%$

Hence, Loss per cent is $10%$

# Question: 75

Sushma sold her watch for $Rs\text{\hspace{0.17em}}3320$ at a gain of $Rs\text{\hspace{0.17em}}320$. For earning a gain of $10%$ she should have sold the watch for $Rs\text{\hspace{0.17em}}3300.$

## Solution

True

Given, SP of watch $=Rs\text{\hspace{0.17em}}3320$

Profit on SP $=Rs\text{\hspace{0.17em}}320$

CP $=$ SP $-$ Profit $=3320-320=\text{\hspace{0.17em}}Rs\text{\hspace{0.17em}}3000$

Now, for earning profit $10%$, we need to find new SP.

Profit per cent $=\frac{\text{Profit}}{\text{CP}}×100$

Profit $=\frac{\text{SP}-\text{CP}}{\text{CP}}×100$

Profit per cent $=\frac{\text{SP}-\text{CP}}{\text{CP}}×100$

$10=\frac{\text{SP}-3000}{3000}×100$

$300=\text{SP}-3000$

$\text{SP}=\text{\hspace{0.17em}}Rs\text{\hspace{0.17em}}3300$

# Question: 76

Interest on $Rs\text{\hspace{0.17em}}1200$ for $1\frac{1}{2}$ years at the rate of $15%$ per annum is $Rs\text{\hspace{0.17em}}180.$

## Solution

False

Given,  and $\text{R}=15%$

$\therefore \text{I}=\frac{\text{P}×\text{R}×\text{T}}{100}$

$⇒\text{I}=\frac{1200×15×3}{100×2}=\text{\hspace{0.17em}}Rs\text{\hspace{0.17em}}\text{\hspace{0.17em}}270$

So, interest is $Rs\text{\hspace{0.17em}}270$

# Question: 77

Amount received after depositing $Rs\text{\hspace{0.17em}}800$ for a period of $3$ years at the rate of $12%$ per annum is $Rs\text{\hspace{0.17em}}896.$

## Solution

False

Given, $\text{P}=\text{\hspace{0.17em}}Rs\text{\hspace{0.17em}}800,\text{\hspace{0.17em}}\text{T}=3\text{\hspace{0.17em}}\text{yr}$ and $\text{R}=12%$

$\therefore \text{I}=\frac{\text{P}×\text{R}×\text{T}}{100}$

$\text{I}=\frac{800×12×3}{100}=\text{\hspace{0.17em}}Rs\text{\hspace{0.17em}}288$

So, interest is $Rs\text{\hspace{0.17em}}288$

Hence, the amount received will be

A = P+ I

= $800+288=1088$

# Question: 78

$Rs\text{\hspace{0.17em}}6400$ were lent to Feroz and Rashmi at $15%$ per annum for $3\frac{1}{2}$ and $5$ years respectively. The difference in the interest paid by them is $Rs\text{\hspace{0.17em}}150.$

## Solution

False

As per the question, Feroz borrowed $Rs\text{\hspace{0.17em}}6400$ for  at $15%$ per annum.

Here,  ${\text{P}}_{1}=\text{\hspace{0.17em}}Rs\text{\hspace{0.17em}}6400,\text{\hspace{0.17em}}{\text{T}}_{1}=3\frac{1}{2}\text{yr}=\frac{7}{2}\text{yr}$ and ${\text{R}}_{1}=15%$

$\therefore {\text{I}}_{1}=\frac{{\text{P}}_{1}×{\text{R}}_{1}×{\text{T}}_{1}}{100}$

$\text{I}=\frac{6400×15×7}{100×2}=\text{\hspace{0.17em}}Rs\text{\hspace{0.17em}}3360$

Rashmi borrowed $Rs\text{\hspace{0.17em}}6400$ for $5\text{yr}$ at $15%$

Here, ${\text{P}}_{2}=\text{\hspace{0.17em}}Rs\text{\hspace{0.17em}}6400,\text{\hspace{0.17em}}{\text{R}}_{2}=15%\text{\hspace{0.17em}}\text{and}\text{\hspace{0.17em}}{\text{T}}_{2}=5\text{\hspace{0.17em}}\text{yr}$

$\therefore {\text{I}}_{2}=\frac{{\text{P}}_{2}×{\text{R}}_{2}×{\text{T}}_{2}}{100}$

${\text{I}}_{2}=\frac{6400×15×5}{100}=\text{\hspace{0.17em}}Rs4800$

Difference between interests $=4800-3360=\text{\hspace{0.17em}}Rs\text{\hspace{0.17em}}1440$

Hence, the difference in interest, paid by them is $Rs\text{\hspace{0.17em}}1440$

# Question: 79

A vendor purchased $720$ lemons at $Rs\text{\hspace{0.17em}}120$ per hundred. $10%$ of the lemons were found rotten which he sold at $Rs\text{\hspace{0.17em}}50$ per hundred. If he sells the remaining lemons at $Rs\text{\hspace{0.17em}}125$ per hundred, then his profit will be $16%$.

## Solution

False

As per the question, cost price of $100$ lemons $=\text{\hspace{0.17em}}Rs\text{\hspace{0.17em}}120$

Cost price of $1$ lemon $=\text{\hspace{0.17em}}Rs\text{\hspace{0.17em}}\frac{120}{100}$

And cost price of $720$ lemons $Rs\text{\hspace{0.17em}}25.6$

Now, according to question, $10%$ of the lemons were rotten.

$\therefore 10%$ of $720$ lemons $=\frac{10}{100}×720=72$ lemons

Selling price of $100$ rotten lemons $=\text{\hspace{0.17em}}Rs50$

Selling price of $1$ rotten lemon $=\text{\hspace{0.17em}}Rs\frac{50}{100}$

And selling price of $72$ rotten lemons $=\frac{50}{100}×72=\text{\hspace{0.17em}}Rs\text{\hspace{0.17em}}36$

Also, selling price of good lemons $=\text{\hspace{0.17em}}Rs125$ per hundred

& selling price of $\left(720-72\right)$ good lemons $=\frac{125}{100}×\left(720-72\right)=\text{\hspace{0.17em}}Rs\text{\hspace{0.17em}}810$

Now, total selling price of $720$ lemons $=36+810=\text{\hspace{0.17em}}Rs846$

As, SP $<$ CP

Therefore, vendor will bear loss.

# Question: 80

Find the value of $x$ if

a.  $8%$ of $Rs\text{\hspace{0.17em}}x$ is $Rs\text{\hspace{0.17em}}100$

$32%$ of $x\text{\hspace{0.17em}}\text{kg}$ is $400\text{\hspace{0.17em}}\text{kg}$

$35%$ of $Rs\text{\hspace{0.17em}}x$ is $Rs\text{\hspace{0.17em}}280$

$45%$ of marks $x$ is $Rs\text{\hspace{0.17em}}405.$

## Solution

a.  $8%$ of $Rs\text{\hspace{0.17em}}x$ is $Rs\text{\hspace{0.17em}}100$                 (given)

$\therefore \frac{8}{100}×x=100$

$⇒x=\frac{100×100}{8}=\text{\hspace{0.17em}}Rs\text{\hspace{0.17em}}1250$

$32%$ of $x\text{\hspace{0.17em}}\text{kg}$ is $400\text{\hspace{0.17em}}\text{kg}$                   (given)

$\therefore \frac{32}{100}×x=400$

$⇒x=\frac{400×100}{32}=\text{\hspace{0.17em}}Rs\text{\hspace{0.17em}}1250\text{\hspace{0.17em}}\text{kg}$

$35%$ of $Rs\text{\hspace{0.17em}}x$ is $Rs\text{\hspace{0.17em}}280$                  (given)

$\therefore \frac{35}{100}×x=280$

$⇒x=\frac{280×100}{35}=\text{\hspace{0.17em}}Rs800$

$45%$ of marks $x$ is     $Rs405$  (given)

$\therefore \frac{45}{100}×x=405$

$⇒x=\frac{405×100}{45}=\text{\hspace{0.17em}}Rs\text{\hspace{0.17em}}900$

# Question: 81

Imagine that, a $10×10$ grid has value $300$ and that this value is divided evenly among the small squares. In other words, each small square is worth $3.$ Use a new grid for each part of this problem and label each grid “Value : $300$ ”. a.  Shade $25%$ of the grid. What is $25%$ of $300?$ Compare the two answers.

What is the value of $25$ squares?

Shade $17%$ of the grid. What is $17%$ of $300?$ Compare the two answers.

What is the value of $\frac{1}{10}$ of the grid?

## Solution

a.  We have to shade $25%$ of the grid i.e., $\left(\frac{1}{4}\right)$ th part of grid $\left(\frac{1}{4}\right)$ th part of grid covers $25$ squares. Since, one square = $3$
So, total value of $25$ such squares $=25×3=75$
Now, $25%$ of $300=\frac{25}{100}×300=25×3=75$ Hence, the above $2$ values are equal

Value of $25$ squares $=25×3=75$

$17%$ of the grid means $17$ squares. So we will shade $17$ squares Total value these $17$ squares $=17×3=51$

Now, $17%$ of $300=\frac{17}{100}×300=17×3=51$

Hence, the above two values are equal.

Value of $1$  grid $=300$

Value of $\frac{1}{10}$ of the grid $\begin{array}{l}=\frac{1}{100}×300\\ =30\end{array}$

# Question: 82

Express $\frac{1}{6}$ as a per cent.

## Solution

In percentage, $\frac{1}{6}$ can be written as

$\begin{array}{l}=\frac{1}{6}×100%\\ =\frac{50}{3}=16.6%\end{array}$

# Question: 83

Express $\frac{9}{40}$ as a per cent.

## Solution

In percentage, $\frac{9}{40}$ can be written as

$\begin{array}{l}=\frac{9}{40}×100%\\ =22\frac{1}{2}%\end{array}$

# Question: 84

Express $\frac{1}{100}$ as a per cent.

## Solution

In percentage, $\frac{1}{100}$ can be written as $\frac{1}{100}×100%=1%$

# Question: 85

Express $80%$ as fraction in its lowest term.

## Solution

In fraction, $80%$ can be written as

$\begin{array}{l}80×\frac{1}{100}=\frac{80}{100}\\ =\frac{4}{5}\end{array}$

# Question: 86

Express $33\frac{1}{3}%$ as a ratio in the lowest term.

## Solution

$33\frac{1}{3}%$ can be written as $\frac{100}{3}%$

For ratio in lowest term, $\frac{100}{3}%$ can be written as $\begin{array}{l}=\frac{100}{3}%:1\\ =\frac{100}{3}×\frac{1}{100}:1\end{array}$

$\begin{array}{l}=\frac{1}{3}:1\\ =\frac{1}{3}×3:1×3\\ =1:3\end{array}$

# Question: 87

Express $16\frac{2}{3}%$ as a ratio in the lowest form.

## Solution

Given, $16\frac{2}{3}%=\frac{50}{3}%$

For ratio in lowest term,

$\begin{array}{l}=\frac{50}{3}%:1\\ =\frac{50}{3}×\frac{1}{100}:1\end{array}$

$\begin{array}{l}=\frac{1}{6}:1\\ =\frac{1}{6}×6:1×6\\ =1:6\end{array}$

# Question: 88

Express $150%$ as a ratio in the lowest form.

## Solution

For the ratio in lowest term

$\begin{array}{l}150%:1\\ =\frac{150}{100}:1\\ =\frac{3}{2}:1\\ =\frac{3}{2}×2:1×2\\ =3:2\end{array}$

# Question: 89

Sachin and Sanjana are calculating $23%$ of $800.$

Now calculate $52%$ of $700$ using both the ways described above. Which way do you find easier?

## Solution

First way,

$52%$ of $700=\left(1%\text{\hspace{0.17em}}of\text{\hspace{0.17em}}700\right)×52$

$=\left(\frac{1}{100}×700\right)×52$

$=7×52=364$

Second way,

$52%$ of $700$ $\begin{array}{l}=\frac{52}{100}×700\\ =0.52×700\\ =364\end{array}$

Hence, second way is the easier to find the percentage.

# Question: 90

Write $0.089$ as a percent.

## Solution

In percentage, $0.089$ can be written as,
$0.089×100%=8.9%$    [to convert in per cent, multiply by $100$ ]

# Question: 91

Write $1.56$ as a percent.

## Solution

In percentage, $1.56$ can be written as, $1.56×100%=156%$

# Question: 92

What is $15%$ of $20?$

## Solution

We need to find, $15%$ of $20$, which can be written as,

$=\frac{15}{100}×20=3$

# Question: 93

What is $800%$ of $800?$

## Solution

We need to find $800%$ of $800$, which can be written as,

$=\frac{800}{100}×800=6400$

# Question: 94

What is $100%$ of $500?$

## Solution

We need to find $100%$ of $500$, which can be written as,

$=\frac{100}{100}×500=500$

# Question: 95

What per cent of $1$ hour is $30$ minutes?

## Solution

Let $x%$ of $1$ h be $30$ min.

Then, $\frac{x}{100}×1\text{\hspace{0.17em}}\text{h}=30$ min.

$\frac{x}{100}×60$ min $=30$ min

$x=\frac{30×100}{60}$

$x=50%$

Therefore, $50%$ of $1$ hr. is $30$ min.

# Question: 96

What per cent of $1$ day is $1$ minute?

## Solution

Let $x%$ of $1$ day is $1$ min

Then, $\frac{x}{100}×1$ day $=1$ min

$\frac{x}{100}×24\text{\hspace{0.17em}}$ h $=1$ min

$\frac{x}{100}×1440$ min $=1$ min

$x=\frac{100}{1440}=\frac{10}{144}$

$x=0.069%$

Therefore, $0.069%$ of $1$ day is $1$ min

# Question: 97

What per cent of $1\text{\hspace{0.17em}}km$ is $1000$ metres?

## Solution

Let $x%$ of $1$ km is $1000$ m

Then, $\frac{x}{100}×1$ km $=1000$ m

$\frac{x}{100}×1000$ m $=1000$ m

$x×10=1000$

$x=100%$

Therefore, $100%$ of $1$ km is $1000$ m

# Question: 98

Find out $8%$ of $25\text{\hspace{0.17em}}\text{kg}.$

## Solution

We have to find, $8%$ of $25$ kg, which can be written as,

$=\frac{8}{100}×25=2$ kg

Hence, $8%$ of $25\text{\hspace{0.17em}}\text{kg}$ is 2 kg

# Question: 99

What percent of $Rs80$ is $Rs100?$

## Solution

Let $x%\text{\hspace{0.17em}}Rs\text{\hspace{0.17em}}80$ is $Rs\text{\hspace{0.17em}}100$

Then, $\frac{x}{100}×80=100$

$x=\frac{100×10}{8}=\text{\hspace{0.17em}}Rs\text{\hspace{0.17em}}125%$

Hence, $125%$ of $Rs\text{\hspace{0.17em}}80$ is $Rs\text{\hspace{0.17em}}100$

# Question: 100

$45%$ of the population of a town are men and $40%$ are women. What is the percentage of children?

## Solution

Given, percentage of men in town $=45%$

Percentage of women in town $=40%$

So, percentage of children in town will be $=100-45-40=15%$

Hence, $15%$ of the population of a town are children

# Question: 101

The strength of a school is $2000.$ If $40%$ of the students are girls then how many boys are there in the school?

## Solution

According to the question,

The strength of school $=2000$

Percentage of girls in school $=40%$

Percentage of boys in school $=100-40=60%$

Number of boys in school $=60%$ of 2000, i.e $\begin{array}{l}=2000=\frac{60}{100}×2000\\ =60×20=1200\end{array}$

Hence, number of boys in school are $1200.$

# Question: 102

Chalk contains $10%$ calcium, $3%$ carbon and $12%$ oxygen. Find the amount of carbon and calcium (in grams) in $2\frac{1}{2}\text{\hspace{0.17em}}\text{kg}$ of chalk.

## Solution

Given, percentage of calcium in chalk $=10%$

Percentage of carbon in chalk $=3%$

Percentage of oxygen in chalk $=12%$

Weight of chalk $=2\frac{1}{2}\text{\hspace{0.17em}}\text{kg}=\frac{5}{2}\text{\hspace{0.17em}}\text{kg}=2.5\text{\hspace{0.17em}}\text{kg}=2.5×1000\text{gm}=2500\text{gm}$

Amount of carbon in chalk $=3%$ of $2500g$

$=\frac{3}{100}×2500=25×3=75g$

Amount of calcium in chalk $=10%$ of $2500\text{g}$

$=\frac{10}{100}×2500=10×25=250\text{g}$

Therefore, amount of carbon & calcium are $75\text{g}$ and $250\text{g}$ respectively.

# Question: 103

$800\text{\hspace{0.17em}}\text{kg}$ of mortar consists of $55%$ sand, $33%$ cement and rest lime. What is the mass of lime in mortar?

## Solution

Given, percentage of sand in mortar $=55%$

Percentage of cement in mortar $=33%$

So, percentage of lime in mortar $=100-55-33=100-88=12%$

$\therefore$ Weight of mortar

$\therefore$ Mass of lime in mortar $=12%$ of

Therefore, weight of lime in mortar is

# Question: 104

In a furniture shop, $24$ tables were bought at the rate of $Rs450$ per table. The shopkeeper sold $16$ of them at the rate of $Rs\text{\hspace{0.17em}}600$ per table and the remaining at the rate of $400$ per table. Find her gain or loss percent.

## Solution

Buying cost of the tables $=24×450=10800=$ C.P

Selling cost of the tables $=\left(16×600\right)+\left(8×400\right)$

$=9600+3200=12800=$ S.P

Gain or loss $=$ S.P $>$ C.P,

Therefore,

Gain $=$ S.P $-$ C.P $=12800-10800=2000$

Gain $%=\left(\frac{\text{gain}}{\text{CP}}×100\right)%=\left(\frac{2000}{10800}×100\right)%=18.51$

# Question: 105

Medha deposited $20%$ of her money in a bank. After spending $20%$ of the remainder, she has $Rs4800$ left with her. How much did she originally have?

## Solution

Let the total money Medha had $=100$