Chapter 8: Comparing Quantities

# Question: 1

Find the ratio of:

a.  $Rs5$ to $50$ paise

b.  $15$ kg to $210$ g

c.   to

d.  $30$ days to $36$ hours

## Solution

a.  $Rs5$ to $50$ paise

Now, converting rupees into paise.

Note: For finding the ratio, both the quantities

should be in the same units.

Thus, the ratio is $=\frac{500}{50}$

$=\frac{10}{1}=10:1$

b.  $15$ kg to $210$ g

Now, converting kilograms into grams.

Thus, the ratio is $=\frac{15000}{210}$

$=\frac{500}{7}=500:7$

c.   to

Now, converting metre into centimetre, we get

Thus, the ratio is $=\frac{900}{27}$

$=\frac{100}{3}=100:3$

d.  $30$ days to $36$ hours

Now, converting days into hours, we get

Thus, the ratio is $=\frac{720}{36}$

$=\frac{20}{1}=20:1$

# Question: 2

In a computer lab, there are $3$ computers for every $6$ students. How many computers will be needed for $24$ students?

## Solution

students require $=3$ computers

student requires $=\frac{3}{6}$ computers

students require $=\frac{3}{6}×24=12$ computers

Hence, $12$ computers will be needed for $24$ students.

# Question: 3

Population of Rajasthan $=570$ lakhs and population of U.P. $=1660$ lakhs.Area of Rajasthan $=3$ lakh ${\text{km}}^{2}$ and area of $\text{U}\text{.P}\text{.}=2$ lakh ${\text{km}}^{2}$.

(i)             How many people are there per ${\text{km}}^{2}$ in both states?

(ii)          Which state is less populated?

## Solution

(i)             People residing per ${\text{km}}^{2}=\frac{\text{Population}}{\text{Area}}$

In Rajasthan

In U.P.

(ii)          , Rajasthan is less populated.

# Question: 1

Convert the given fractional numbers to per cents.

a.  $\frac{1}{8}$

b.  $\frac{5}{4}$

c.  $\frac{3}{40}$

d.  $\frac{2}{7}$

## Solution

a.  $\frac{1}{8}=\frac{1}{8}×100%$

$=\frac{25}{2}%$

$=12.5%$

b.  $\frac{5}{4}=\frac{5}{4}×100%$

$=5×25%$

$=125%$

c.  $\frac{3}{40}=\frac{3}{40}×100%$

$=\frac{3}{2}×5%=\frac{15}{2}%$

$=7.5%$

d.  $\frac{2}{7}=\frac{2}{7}×100%$

$=\frac{200}{7}%$

$=28\frac{4}{7}%$

# Question: 2

Convert the given decimal fractions to per cents.

a.  $0.65$

b.  $2.1$

c.  $0.02$

d.  $12.35$

## Solution

a.  $\begin{array}{l}0.65=\frac{65}{100}×100%\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=65%\end{array}$

b.  $\begin{array}{l}2.1=\frac{21}{10}×100%\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=210%\end{array}$

c.  $\begin{array}{l}0.02=\frac{2}{100}×100%\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=2%\end{array}$

d.  $\begin{array}{l}12.35=\frac{1235}{100}×100%\\ =1235%\end{array}$

# Question: 3

Estimate what part of the figures is coloured and hence find the per cent which is coloured.

## Solution

(i)             Coloured part of the given figure $=\frac{1}{4}$

$\therefore$ Per cent of coloured part $=\frac{1}{4}×100%=25%$

(ii)          Coloured part of the given figure $=\frac{3}{5}$

$\therefore$ Per cent of coloured part from the given figure $=\frac{3}{5}×100%$

$=60%$

(iii)      Coloured part of the given figure $=\frac{3}{8}$

$\therefore$ Per cent of coloured part from the given figure $\begin{array}{l}=\frac{3}{8}×100%=\frac{3}{2}×25%\\ =37.5%\end{array}$

# Question: 4

Find:

a.

b.

c.

d.

## Solution

a.

$\begin{array}{c}=15×2.5\\ =37.5\end{array}$

b.

c.

$\begin{array}{c}=20×25\\ =\text{\hspace{0.17em}}Rs500\end{array}$

d.

$\begin{array}{c}=\frac{75}{100}×1000\\ =750\text{\hspace{0.17em}}\text{g}\text{\hspace{0.17em}}\\ =0.750\text{\hspace{0.17em}}\text{kg}\end{array}$

# Question: 5

Find the whole quantity if

a.

b.

c.

d.

e.

## Solution

Let the whole quantity be $x$ in given questions:

a.

b.

c.

d.

e.

# Question: 6

Convert given per cents to decimal fractions and also to fractions in simplest forms:

a.  $25%$

b.  $150%$

c.  $20%$

d.  $5%$

## Solution

 S. No. Per cents Fractions Simplest form Decimal form (a) $25%$ $\frac{25}{100}$ $\frac{1}{4}$ $0.25$ (b) $150%$ $\frac{150}{100}$ $\frac{3}{2}$ $1.5$ (c) $20%$ $\frac{20}{100}$ $\frac{1}{5}$ $0.2$ (d) $5%$ $\frac{5}{100}$ $\frac{1}{20}$ $0.05$

# Question: 7

In a city, $30%$ are females, $40%$ are males and remaining are children. What per cent are children?

## Solution

Given:

Percentage of females in the city $=30%$

Percentage of males in the city $=40%$

Total percentage of females and males in the city $=30+40=70%$

Percentage of children

$=$ $100%$$-$ Percentage of males and females

$=100%-70%$

$=30%$

Hence, percentage of children in the city is $30%$.

# Question: 8

Out of $15,000$ voters in a constituency, $60%$ voted. Find the percentage of voters who did not vote. Can you now find how many actually did not vote?

## Solution

Total voters in the constituency $=15,000$

Percentage of voters who voted  $=60%$

Percentage of voters who did not vote $=100-60=40%$

Actual voters, who did not vote

$=\frac{40}{100}×15000=6,000$

Hence, number of voters who did not vote = $6,000$.

# Question: 9

Meeta saves $Rs4000$ from her salary. If this is $10%$ of her salary. What is her salary?

## Solution

Let Meeta’s salary be $Rs\text{\hspace{0.17em}}x.$

Now, $10%$ of Meeta’s salary $=\text{\hspace{0.17em}}Rs4000$

Hence, Meeta’s salary is $Rs40,000.$

# Question: 10

A local cricket team played $20$ matches in one season. It won $25%$ of them. How many matches did they win?

## Solution

Number of matches played by local cricket team $=20$

Percentage of matches won by the local cricket team $=25%$

Total matches won by the local cricket team

$=5$

Hence, team won $5$ matches.

# Question: 1

Tell what is the profit or loss in the following transactions. Also find profit per cent or loss per cent in each case.

a.  Gardening shears bought for $Rs250$ and sold for $Rs325$.

b.  A refrigerator bought for $Rs12,000$ and sold at $Rs13,500$.

c.  A cupboard bought for $Rs2,500$ and sold at $Rs3,000.$

d.  A skirt bought for $Rs250$ and sold at $Rs150$.

## Solution

a.  Cost price of gardening shears $=\text{\hspace{0.17em}}Rs250$

Selling price of gardening shears $=\text{\hspace{0.17em}}Rs325$

Since, S.P. $>$ C.P.,

Hence, profit is earned on selling the item.

Now $\text{Profit}%=\frac{\text{Profit}}{\text{C}\text{.P}\text{.}}×100$

$=\frac{75}{250}×100=30%$

Therefore,

b.  Cost price of refrigerator $=\text{\hspace{0.17em}}Rs12,000$

Selling price of refrigerator $=\text{\hspace{0.17em}}Rs13,500$

Since, S.P. $>$ C.P.,

Hence,  profit is earned on selling the refrigerator.

Now $\text{Profit}%=\frac{\text{Profit}}{\text{C}\text{.P}\text{.}}×100$

$=\frac{1500}{12000}×100=12.5%$

Therefore, Profit

c.  Cost price of cupboard $=\text{\hspace{0.17em}}Rs2,500$

Selling price of cupboard $=\text{\hspace{0.17em}}Rs3,000$

Since, S.P. $>$ C.P.,

Hence, profit is earned on selling the cupboard.

Now $\text{Profit}%=\frac{\text{Profit}}{\text{C}\text{.P}\text{.}}×100$

$=\frac{500}{2500}×100=20%$

Therefore,

d.  Cost price of skirt $=\text{\hspace{0.17em}}Rs250$

Selling price of skirt $=\text{\hspace{0.17em}}Rs150$

Since, C.P. $>$ S.P.,

Hence, loss on selling skirt.

Now $\text{Loss}%=\frac{\text{Loss}}{\text{C}\text{.P}\text{.}}×100$

$=\frac{100}{250}×100=40%$

Therefore,

# Question: 2

Convert each part of the ratio to percentage:

a.  $3:1$

b.  $2:3:5$

c.  $1:4$

d.

## Solution

a.  $3:1$

Total $\text{part}=3+1=4$

Now,

b.  $2:3:5$

Total $\text{part}=2+3+5=10$

Now,

c.  $1:4$

Total $\text{part}=1+4=5$

Now,

d.

Total $\text{part}=1+2+5=8$

Now,

# Question: 3

The population of a city decreased from $25,000$ to $24,500$. Find the percentage decrease.

## Solution

As per the question, population of a city decreases from $25,000$ to $24,500$.

Decrease in population $=25,000-24,500=500$

Hence, the percentage decrease in population is $2%$.

# Question: 4

Arun bought a car for $Rs3,50,000.$ The next year, the price went upto $Rs3,70,000$. What was the Percentage of price increase?

## Solution

As per the question, price of the car increases from  $\text{\hspace{0.17em}}Rs3,70,000.$

Change in price of the car $\begin{array}{l}=\text{\hspace{0.17em}}Rs3,70,000-\text{\hspace{0.17em}}Rs3,50,000\\ =\text{\hspace{0.17em}}Rs20,000.\end{array}$

Hence,

Hence, the percentage of price increased is $5\frac{5}{7}%.$

# Question: 5

I buy a T.V. for $Rs10,000$ and sell it at a profit of $20%$. How much money do I get for it?

## Solution

The cost price of T.V. $=\text{\hspace{0.17em}}Rs10,000$

Profit per cent on T.V $=20%$

Now,

Selling $\text{price}=\text{C}\text{.P}\text{.}+\text{\hspace{0.17em}}\text{Profit}=\text{\hspace{0.17em}}Rs10,000+\text{\hspace{0.17em}}Rs2,000$ $=\text{\hspace{0.17em}}Rs12,000$

Hence, he gets $Rs12,000$ on selling his T.V.

# Question: 6

Juhi sells a washing machine for $Rs13,500.$ She loses $20%$ in the bargain. What was the price at which she bought it?

## Solution

Selling price of washing machine $=\text{\hspace{0.17em}}Rs13,500$

Loss percent $=20%$

Let the cost price of washing machine be

Since,

$=\frac{x}{5}$

Therefore, $\text{S}\text{.P}\text{.}=\text{C}\text{.P}\text{.}-\text{Loss}$

Hence, the cost price of washing machine is $Rs16,875.$

# Question: 7

(i)             Chalk contains calcium, carbon and oxygen in the ratio $10:3:12.$ Find the percentage of carbon in chalk.

(ii)          If in a stick of chalk, carbon is $3\text{g}$, what is the weight of the chalk stick?

## Solution

(i)             Given ratio of calcium, carbon and oxygen in chalk $=10:3:12$

Total part $=10+3+12=25$

Part of Carbon $=\frac{3}{25}$

Percentage of Carbon part in chalk $=\frac{3}{25}×100=12%$

(ii)          Quantity of Carbon in chalk stick

Let the weight of chalk be $x\text{\hspace{0.17em}}\text{g}.$

Then,

Hence, the weight of chalk stick is .

# Question: 8

Amina buys a book for $Rs275$ and sells it at a loss of $15%$. How much does she sell it for?

## Solution

The cost price of the book $=\text{\hspace{0.17em}}Rs275$

Loss per cent $=15%$

$=\frac{15}{100}×275=\text{\hspace{0.17em}}Rs41.25$

Therefore, $\begin{array}{l}\text{S}\text{.P}\text{.}=\text{C}\text{.P}\text{.}-\text{Loss}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\text{\hspace{0.17em}}Rs275-\text{\hspace{0.17em}}Rs41.25\end{array}$

$=\text{\hspace{0.17em}}Rs233.75$

Hence, Amina sells the book for $Rs233.75.$

# Question: 9

Find the amount to be paid at the end of  in each case:

a.

b.

## Solution

a.  Given, Principal (P)

$=12%$ p.a.,

Time (T) $=3$ years

b.  Here, Principal (P)

$=5%$ p.a.,

Time (T) $=3$ years

# Question: 10

What rate gives $Rs280$ as interest on a sum of $Rs56,000$ in $2$ years?

## Solution

Given, Principal (P) $=\text{\hspace{0.17em}}Rs56,000,$

Simple Interest (S.I.) $=\text{\hspace{0.17em}}Rs280,$

Time (T) $=2$ years

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

Hence, the rate of interest on sum is $0.25%$.

# Question: 11

If Meena gives an interest of $Rs45$ for one year at $9%$ rate p.a. What is the sum she has borrowed?

## Solution

Simple Interest $=\text{\hspace{0.17em}}Rs45,$

Rate (R) $=9%$ p.a.,

Time (T) $=1$ years

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

Hence, she has borrowed $Rs500$.