Chapter 8: Comparing Quantities

Find the ratio of:

a.  $Rs5$ to $50$ paise

b.  $15$ kg to $210$ g

c.  $9\text{ m}$ to $27\text{ cm}$

d.  $30$ days to $36$ hours

a.  $Rs5$ to $50$ paise

Now, converting rupees into paise.

Note: For finding the ratio, both the quantities

should be in the same units.

$\Rightarrow 5\times 100\text{ paise to }50\text{ paise}$   $\text{[}\because Rs1=100\text{ paise]}$

$\Rightarrow 500\text{ paise to }50\text{ paise}$

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

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

b.  $15$ kg to $210$ g

Now, converting kilograms into grams.

$\Rightarrow 15\times 1000\text{ g to }210\text{ g}$        $\text{[}\because 1\text{ kg}=1000\text{ g]}$

$\Rightarrow 15000\text{ g to }210\text{ g}$

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

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

c.  $9\text{ m}$ to $27\text{ cm}$

Now, converting metre into centimetre, we get

$\Rightarrow 9\times 100\text{ cm to }27\text{ cm}$        $\text{[}\because 1\text{ m}=100\text{ cm}]$

$\Rightarrow 900\text{ cm to }27\text{ cm}$

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

$\Rightarrow 30\times 24\text{ hours to }36\text{ hours}$        $[\because 1\text{ day}=24\text{ hours}]$

$\Rightarrow 720\text{ hours to }36\text{ hours}$

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

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

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

$\because 6$ students require $=3$ computers

$\therefore 1$ student requires $=\frac{3}{6}$ computers

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

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

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?

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

In Rajasthan $=\frac{570\text{ lakhs}}{3{\text{lakh km}}^2}$

$=190{\text{ people per km}}^2$

In U.P. $=\frac{1660\text{ lakhs}}{2{\text{lakh km}}^2}$

$=830{\text{ people per km}}^2$

(ii)          , Rajasthan is less populated.

Convert the given fractional numbers to per cents.

a.  $\frac{1}{8}$

b.  $\frac{5}{4}$

c.  $\frac{3}{40}$

d.  $\frac{2}{7}$

a.  $\frac{1}{8}=\frac{1}{8}\times 100\%$

   $=\frac{25}{2}\%$

   $=12.5\%$

b.  $\frac{5}{4}=\frac{5}{4}\times 100\%$

   $=5\times 25\%$

   $=125\%$

c.  $\frac{3}{40}=\frac{3}{40}\times 100\%$

     $=\frac{3}{2}\times 5\%=\frac{15}{2}\%$

     $=7.5\%$

d.  $\frac{2}{7}=\frac{2}{7}\times 100\%$

        $=\frac{200}{7}\%$

        $=28\frac{4}{7}\%$

Convert the given decimal fractions to per cents.

a.  $0.65$

b.  $2.1$

c.  $0.02$

d.  $12.35$

a.  $\begin{array}{l}0.65=\frac{65}{100}\times 100\%\\ =65\%\end{array}$ 

b.  $\begin{array}{l}2.1=\frac{21}{10}\times 100\%\\ =210\%\end{array}$

c.  $\begin{array}{l}0.02=\frac{2}{100}\times 100\%\\ =2\%\end{array}$

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

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

 

 

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

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

 

 

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

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

          $=60\%$

 

 

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

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

 

 

 

Find:

a.  $15\%\text{ of }250$

b.  $1\%\text{ of }1\text{ hour}$

c.  $20\%\text{ of Rs}.2500$

d.  $75\%\text{ of }1\text{ kg}$

a.  $15\%\text{ of }250=\frac{15}{100}\times 250$

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

b.   $1\%\text{ of }1\text{ hour}=1\%\text{ of }60\text{ minutes}$

$\begin{array}{c}=1\%\text{ of }(60\times 60)\text{ seconds}\\ =\frac{1}{100}\times 60\times 60\\ =6\times 6\\ =36\text{ seconds}\end{array}$

c.  $20\%\text{ of Rs}2500=\frac{20}{100}\times 2500$

$\begin{array}{c}=20\times 25\\ =Rs500\end{array}$

d.  $75\%\text{ of }1\text{ kg}=75\%\text{ of }1000\text{ g}$

$\begin{array}{c}=\frac{75}{100}\times 1000\\ =750\text{g}\\ =0.750\text{kg}\end{array}$

Find the whole quantity if

a.  $5\%\text{ of it is }600.$

b.  $12\%\text{ of it is Rs}1080.$

c.  $40\%\text{ of it is }500\text{ km}.$

d.  $70\%\text{ of it is }14\text{ minutes}.$

e.  $8\%\text{ of it is }40\text{ litres}.$

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

a.  $5\%\text{ of it is }600.$

$\begin{array}{l}\Rightarrow \frac{5}{100}\times x=600\\ \Rightarrow x=\frac{600\times 100}{5}=12,000\end{array}$

b.  $12\%\text{ of it is Rs}1080.$

$\begin{array}{l}\Rightarrow \frac{12}{100}\times x=1080\\ \Rightarrow x=\frac{1080\times 100}{12}=Rs9,000\end{array}$

c.  $40\%\text{ of it is }500\text{ km}.$

$\begin{array}{l}\Rightarrow \frac{40}{100}\times x=500\\ \Rightarrow x=\frac{500\times 100}{40}=1,250\text{km}\end{array}$

d.  $70\%\text{ of it is }14\text{ minutes}.$ 

$\begin{array}{l}\Rightarrow \frac{70}{100}\times x=14\\ \Rightarrow x=\frac{14\times 100}{70}=20\text{minutes}\end{array}$

e.  $8\%\text{ of it is }40\text{ litres}.$

$\begin{array}{l}\Rightarrow \frac{8}{100}\times x=40\\ \Rightarrow x=\frac{40\times 100}{8}=500\text{litres}\end{array}$

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

a.  $25\%$

b.  $150\%$

c.  $20\%$

d.  $5\%$

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)	$5\%$	$\frac{5}{100}$	$\frac{1}{20}$	$0.05$

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

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 $30\%$.

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?

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

$=40\%\text{ of }15000$

$=\frac{40}{100}\times 15000=6,000$

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

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

Let Meeta’s salary be $Rsx.$

Now, $10\%$ of Meeta’s salary $=Rs4000$

$\Rightarrow 10\%\text{ of }x=Rs4000$

$\Rightarrow \frac{10}{100}\times x=4000$

$\Rightarrow x=\frac{4000\times 100}{10}$

$\Rightarrow x=40,000$

Hence, Meeta’s salary is $Rs40,000.$

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

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 $=25\%\text{ of }20$

        $=\frac{25}{100}\times 20$

        $=5$

Hence, team won $5$ matches.

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 $Rs150$.

a.  Cost price of gardening shears $=Rs250$

Selling price of gardening shears $=Rs325$

Since, S.P. $>$ C.P.,

Hence, profit is earned on selling the item.

$\therefore \text{ Profit}=\text{S}\text{.P}\text{.}-\text{C}\text{.P}\text{.}=Rs325-Rs250=Rs75$

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

                            $=\frac{75}{250}\times 100=30\%$

Therefore, $\text{Profit earned}=Rs75$

$\text{and Profit}\%=30\%$

b.  Cost price of refrigerator $=Rs12,000$

Selling price of refrigerator $=Rs13,500$

Since, S.P. $>$ C.P.,

Hence,  profit is earned on selling the refrigerator.

$\begin{array}{l}\therefore \text{Profit}=\text{S}\text{.P}\text{.}-\text{ C}\text{.P}\text{.}=Rs13500-Rs12000\\ =Rs1,500\end{array}$

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

                            $=\frac{1500}{12000}\times 100=12.5\%$

Therefore, Profit $=Rs1,500\text{ and Profit}\%=12.5\%$

c.  Cost price of cupboard $=Rs2,500$

Selling price of cupboard $=Rs3,000$

Since, S.P. $>$ C.P.,

Hence, profit is earned on selling the cupboard.

$\therefore \text{ Profit}=\text{S}\text{.P}\text{.}-\text{C}\text{.P}\text{.}=Rs3,000-Rs2,500=Rs500$

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

                            $=\frac{500}{2500}\times 100=20\%$

Therefore, $\text{Profit}=Rs500$

$\text{and Profit}\%=20\%$

d.  Cost price of skirt $=Rs250$

Selling price of skirt $=Rs150$

Since, C.P. $>$ S.P.,

Hence, loss on selling skirt.

$\therefore \text{ Loss}=\text{C}\text{.P}\text{.}-\text{S}\text{.P}\text{.}=Rs250-Rs150=Rs100$

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

                            $=\frac{100}{250}\times 100=40\%$

Therefore, $\text{Profit}=Rs100$

$\text{and Profit}\%=40\%$

Convert each part of the ratio to percentage:

a.  $3:1$ 

b.  $2:3:5$ 

c.  $1:4$

d.  $1:2:5$

a.  $3:1$ 

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

Now, $\text{Fractional part will be}=\frac{3}{4}:\frac{1}{4}$

$\Rightarrow \text{ Percentage of given parts}=\frac{3}{4}\times 100:\frac{1}{4}\times 100$

$\Rightarrow \text{ Percentage of parts}=75\%:25\%$

b.  $2:3:5$ 

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

Now, $\text{Fractional part will be}=\frac{2}{10}:\frac{3}{10}:\frac{5}{10}$

$\Rightarrow \text{ Percentage of parts}=\frac{2}{10}\times 100:\frac{3}{10}\times 100:\frac{5}{10}\times 100$

$\Rightarrow \text{ Percentage of parts}=20\%:30\%:50\%$

c.  $1:4$

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

Now, $\text{Fractional part will be}=\frac{1}{5}:\frac{4}{5}$

$\Rightarrow \text{ Percentage of the given parts}=\frac{1}{5}\times 100:\frac{4}{5}\times 100$

$\Rightarrow \text{ Percentage of parts}=20\%:80\%$

d.  $1:2:5$

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

Now, $\text{Fractional part}=\frac{1}{8}:\frac{2}{8}:\frac{2}{8}$

$\Rightarrow \text{ Percentage of parts}=\frac{1}{8}\times 100:\frac{2}{8}\times 100:\frac{5}{8}\times 100$

$\Rightarrow \text{ Percentage of parts}=12.5\%:25\%:62.5\%$

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

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

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

$\begin{array}{c}\text{Percentage decrease}=\frac{\text{Decrease in population}}{\text{Original population}}\times 100\\ =\frac{500}{25000}\times 100\\ =2\%\end{array}$

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

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?

As per the question, price of the car increases from $Rs3,50,000\text{ to}$ $Rs3,70,000.$

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

Hence,

$\begin{array}{c}\text{Percentage increase}=\frac{\text{Increase in price}}{\text{ Original amount}}\times 100\\ =\frac{20000}{350000}\times 100=5\frac{5}{7}\%\end{array}$

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

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

The cost price of T.V. $=Rs10,000$

Profit per cent on T.V $=20\%$ 

Now, $\begin{array}{c}\text{Profit}=\frac{\text{Profit}\%}{100}\times \text{ C}\text{.P}\text{.}\\ =\frac{20}{100}\times 10000\\ =Rs2,000\end{array}$

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

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

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

Selling price of washing machine $=Rs13,500$

Loss percent $=20\%$ 

Let the cost price of washing machine be $\text{Rs }x.$

Since, $\text{Loss}=\text{Loss}\%\text{ of C}\text{.P}\text{.}$

$\Rightarrow \text{ Loss}=20\%\text{ of Rs}x=\frac{20}{100}\times x$

$=\frac{x}{5}$

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

$\Rightarrow 13500=x-\frac{x}{5}$

$\Rightarrow 13500=\frac{4x}{5}$

$\Rightarrow x=\frac{13500\times 5}{4}=Rs16,875$

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

(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?

(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 $=\frac{3}{25}\times 100=12\%$

(ii)          Quantity of Carbon in chalk stick $=3\text{ g}$

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

Then, $12\%\text{ of }x=3$

$\Rightarrow \frac{12}{100}\times x=3$

$\Rightarrow x=\frac{3\times 100}{12}=25\text{g}$

Hence, the weight of chalk stick is $25\text{ g}$.

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

The cost price of the book $=Rs275$

Loss per cent $=15\%$ 

$\begin{array}{l}\text{Loss}=\text{Loss}\%\text{ of C}\text{.P}\text{.}\\ =15\%\text{ of}Rs275\end{array}$

$=\frac{15}{100}\times 275=Rs41.25$

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

                         $=Rs233.75$

Hence, Amina sells the book for $Rs\mathrm{233.75.}$

Find the amount to be paid at the end of $3\text{ years}$ in each case:

a.  $\text{Principal}=Rs1,200\text{ at }12\%\text{ p}\text{.a}\text{.}$

b.  $\text{Principal}=Rs7,500\text{ at }5\%\text{ p}\text{.a}\text{.}$

a.  Given, Principal (P) $=Rs1,200,$

$\text{Rate }\left(\text{R}\right)$$=12\%$ p.a.,

Time (T) $=3$ years

$\begin{array}{c}\text{Simple Interest}=\frac{\text{P}\times \text{R}\times \text{T}}{100}\\ =\frac{1200\times 12\times 3}{100}\\ =Rs432\end{array}$

$\begin{array}{c}\text{Now, Amount}=\text{Principal}+\text{Simple Interest}\\ =Rs1200+Rs432\\ =Rs1,632\end{array}$

b.  Here, Principal (P) $=Rs7,500,$

$\text{Rate }\left(\text{R}\right)$$=5\%$ p.a.,

Time (T) $=3$ years

$\begin{array}{c}\text{Simple Interest}=\frac{\text{P}\times \text{R}\times \text{T}}{100}\\ =\frac{7500\times 5\times 3}{100}\\ =Rs1,125\end{array}$

$\begin{array}{c}\text{Now, Amount}=\text{Principal}+\text{Simple Interest}\\ =Rs7,500+Rs1,125\\ =Rs8,625\end{array}$

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

Given, Principal (P) $=Rs56,000,$ 

Simple Interest (S.I.) $=Rs280,$

 Time (T) $=2$ years

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

$\Rightarrow 280=\frac{56000\times \text{R}\times 2}{100}$

$\Rightarrow \text{ R}=\frac{280\times 100}{56000\times 2}$

$\Rightarrow \text{ R}=0.25\%$

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

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

Simple Interest $=Rs45,$ 

Rate (R) $=9\%$ p.a.,

Time (T) $=1$ years

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

$\Rightarrow 45=\frac{\text{P}\times 9\times 1}{100}$

$\Rightarrow \text{ P}=\frac{45\times 100}{9\times 1}$

$\Rightarrow \text{ P}=Rs500$

Hence, she has borrowed $Rs500$.