Chapter 8: Comparing Quantities

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

$20\%$ of $700\text{m}$ is

a.    $560\text{m}$

b.   $70\text{m}$

c.    $210\text{m}$

d.   $140\text{m}$

(d)

 $20\%$ of $\begin{array}{l}700\text{m}=\frac{20}{100}\times 700\\ =20\times 7\\ =140\text{m}\end{array}$ 

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

Gayatri’s income is $\text{Rs}1,60,000$ 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.    $\text{Rs}136000$

b.   $\text{Rs}120000$

c.    $\text{Rs}122400$

d.   $\text{Rs}14000$

(c)

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

Money paid as house rent = $15\%$ of $160000$ $=\frac{15}{100}\times 160000=$ $\text{Rs}24000$ 

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

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

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

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.   $80\%$

(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 $=\frac{\text{savings}}{\text{income}+\text{savings}}\times 100\%$

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

$0.07$ is equal to

a.    $70\%$

b.   $7\%$

c.    $0.7\%$

d.   $0.07\%$

(b)

 $0.07=\frac{7}{100}\times 100\%=7\%$

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.   $50$

(c)

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

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

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

Percentage of scouts from Ahmedabad

$=\frac{\text{Number of scouts}\text{from}\text{Ahmedabad}}{\text{Total number}\text{scouts}}\times 100\%$

$=\frac{8}{100}\times 100\%=8\%$

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

a.    $200$

b.   $\frac{1}{2}$

c.    $2$

d.   $50$

(a)

Let the percentage be $x\%$

$\Rightarrow x\%\text{ of }4500=9000$

$\Rightarrow \frac{x}{100}\times 4500=9000$

$\Rightarrow 45x=9000$

$\Rightarrow x=\frac{9000}{45}$$=200$

 

Option (a) is the correct one

$5.2$ is equal to

a.    $50\%$

b.   $5.2\%$

c.    $520\%$

d.   $0.52\%$

(c)

We have, $5.2$

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

The ratio $3:8$ is equal to

a.    $3.75\%$

b.   $37.5\%$

c.    $0.375\%$

d.   $267\%$

(b)

Given, ratio $=3:8$

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

$225\%$ is equal to

a.    $9:4$

b.   $4:9$

c.    $3:2$

d.   $2:3$

(a)

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

$\therefore$ Required ratio $=9:4$

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.   $\text{Rs}1812$

(b)

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

As we know, profit percentage can be calculated as,  

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

$12=\frac{\text{Profit}}{1800}\times 100$

Profit $=12\times 18=\text{Rs}216$

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

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

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

a.    $100\%$

b.   $64\%$

c.    $50\%$

d.   $60\%$

(a)

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

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

Profit $=$ SP $-$ CP $=\text{Rs}\left(1600-800\right)=\text{Rs }800$

We know that profit percentage is given as,

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

Hence, profit earned is $100\%$

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 $\text{Rs}200$

b.   profit of $\text{Rs}400$

c.    loss of $\text{Rs}400$

d.   profit of $\text{Rs}200$

(c)

 CP of buffalo $=\text{Rs}44,000$ 

Loss $\%=5\%$

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

$5=\frac{\text{Loss}}{44000}\times 100$

Loss $=5\times 440=\text{Rs}2200$

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

 CP of cow $=\text{Rs}18000$

Profit $\%=10\%$

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

$10=\frac{\text{Profit}}{18000}\times 10$

Profit $=\text{Rs}1800$

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

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

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

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

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.   $75\%$

(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

 

$=\frac{\text{Mohan's}\text{income - Raman's}\text{income}}{\text{Mohan's}\text{income}}\times 100\%$

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

$=\frac{25x}{125x}\times 100=20\%$

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

a.    $\text{Rs}4500$

b.   $\text{Rs}9000$

c.    $\text{Rs}18000$

d.   $\text{Rs}13500$

(d)

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

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

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.   $\text{Rs}3330$

(b)

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

$\text{I}=\frac{\text{P}\times \text{R}\times \text{T}}{100}=\frac{3000\times 11\times 2}{100}=\text{Rs}660$

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

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

a.    $\text{Rs}1200$

b.   $\text{Rs}600$

c.    $\text{Rs}100$

d.   $\text{Rs}12100$

(c)

Given,

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

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

Rajni and Mohini deposited $\text{Rs}3000$ and $\text{Rs}4000$ 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.    $\text{Rs}100$

b.   $\text{Rs}1000$

c.    $\text{Rs}900$

d.   $\text{Rs}1100$

(d)

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

$�$ I $=\frac{\text{P}\times \text{R}\times \text{T}}{100}=\frac{3000\times 10\times 3}{100}=\text{Rs}900$

And total amount received will be, $\text{A}=\text{P}+\text{I}=3000+900=Rs3900$

Interest received by Mohini after $2\frac{1}{2}\text{yr}$ years, $\text{P}=Rs4000,\text{R}=10\%\text{per}\text{annum,}\text{T}=2\frac{1}{2}\text{yr}\text{=}\frac{5}{2}\text{yr}$

$\therefore$ I $=\frac{\text{P}\times \text{R}\times \text{T}}{100}=\frac{4000\times 10\times 5}{100\times 2}=Rs1000$

And total amount received will be,

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

Difference in amounts $=5000-3900=Rs1100$

If $90\%$ of $x$ is $315km,$ then the value of $x$ 

a.    $325km$

b.   $350km$

c.    $405km$

d.   $340km$

(b)

Given,

$90\%$ of $x$$=315km$

 

$\begin{array}{l}\Rightarrow \frac{90}{100}\times x=315\\ \Rightarrow x=\frac{315\times 100}{90}\end{array}$

$x=350km$

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.   $Rs350$

(d)

Given, SP of an article is $Rs329$ and loss per cent $=6\%$

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

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

                         $6\text{ CP}=100\left(\text{CP}-\text{SP}\right)$

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

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

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

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

                         $\text{CP}=Rs350$

$\frac{25\%of50\%of100\%}{25\times 50}$ is equal to

a.    $1.1\%$

b.   $0.1\%$

c.    $0.01\%$

d.   $1\%$

We have, $\frac{25\%of50\%of100\%}{25\times 50}=\frac{25}{100}\times \frac{50}{100}\times \frac{100}{100}\times \frac{1}{25\times 50}$

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

In percentage, $0.0001\times 100\%=0.01\%$

The sum which will earn a simple interest of $Rs126$ in $2$ years at $14\%$ per annum is

a.    $Rs394$

b.   $Rs395$

c.    $Rs450$

d.   $Rs540$

(c)

Given, $\text{I}=Rs126,\text{R}=14\%$ and $\text{T}=2\text{yr}$

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

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

$126\times 100=\text{P}\times 14\times 2$

$\text{P}=\frac{12600}{14\times 2}$

$\text{P}=Rs450$

The per cent that represents the unshaded region in the figure.

a.    $75\%$

b.   $50\%$

c.    $40\%$

d.   $60\%$

(c)

 Total parts $=10\times 10=100$

Shaded parts $=60$

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

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

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

The per cent that represents the shaded region in the figure is

a.    $36\%$

b.   $64\%$

c.    $27\%$

d.   $48\%$

(a)

 Total parts $=10\times 10=100$

Shaded parts $=36$

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

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.

$2:3=$ ______ $\%$

Given ratio $=2:3$

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

$18\frac{3}{4}\%=$ _______: ______

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

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

Ratio $=3:16$ 

$30\%$ of $Rs360=$ _____.

We have, $30\%$ of $Rs360=\frac{30}{100}\times 360=Rs108$

$120\%$ of $50\text{km}=$ _____.

We have, $120\%$ of $50\text{km}=\frac{120}{100}\times 50\text{km}=60\text{km}$

$2.5=$ _____ $\%$

We have, $2.5$

In percentage, $2.5\times 100\%=250\%$

$\frac{8}{5}=$ _____ $\%$

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

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

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

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

$15kg$ is ______ $\%$ of $50kg.$

Let $x\%$ of $50kg$ be $15kg$

Then, $\frac{x}{100}\times 50=15$

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

$x=15\times 2=30\%$

Hence, $15kg$ is $30\%$ of $50kg$

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

Given, Initial weight of Nikhil $=60kg$

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

$\therefore$ Percentage increase of weight

$\frac{\text{increase in weight}}{\text{initial weight}}\times 100\%=\frac{6}{60}\times 100\%=\frac{60}{6}\%=10\%$

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

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 $50$

 $=\frac{92}{100}\times 50=\frac{92}{2}=46$

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

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

Marks obtained by Savitri out of $500=440$

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

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

Out of a total deposit of $Rs1500$ in her bank account, Abida withdrew $40\%$ of the deposit. Now the balance in her account is ______.

Total deposit $=Rs1500$

Amount withdrawn $=40\%$ of $Rs1500$

 $=\frac{40}{100}\times 1500=Rs600$

$\therefore$ Balance in the account $=1500-600=Rs900$

________ is $50\%$ more than $60$.

Let number be $x.$

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

Therefore, according to question

$x=60+50\%$ of $60.$ 

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

John sells a bat for $Rs75$ and suffers a loss of $Rs8$. The cost price of the bat is ________.

Given, SP of bat $=Rs75$ and loss $=Rs8$

We know that, CP $=$ SP $+$ Loss

$=75+8=Rs83$

Hence, cost price of the bat is $Rs83.$

If the price of sugar is decreased by $20\%,$ then the new price of $3kg$ sugar originally costing $Rs120$ will be ________.

Original price of $3kg$ sugar $=Rs120$

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

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

$=120-20\%$ of $120$

$\begin{array}{l}=120-\frac{20}{100}\times 120\\ =120-24=Rs96\end{array}$

Mohini bought a cow for $Rs9000$ and sold it at a loss of $Rs900$. The selling price of the cow is ________.

Given, CP of cow $=Rs9000$ and loss $=Rs900$

We know that, SP $=$ CP $-$ Loss $=9000-900=Rs8100.$

Hence, the selling price of the cow is $Rs8100$

Devangi buys a chair for $Rs700$ and sells it for $Rs750$. She earns a profit of ________ $\%$ in the transaction.

Given, CP of a chair $=Rs700$ and SP of a chair $=Rs750$

Since, SP $>$ CP

$\therefore$ Profit $=$ SP $-$ CP $=Rs\left(750-700\right)=Rs50$

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

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

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

Given, CP of a bed sheet $=Rs400$ and SP of a bed sheet $=Rs440$

Since, SP $>$ CP

$\therefore$ Profit $=$ SP $-$ CP $=Rs\left(440-400\right)=Rs40$

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

Hence, Sonal’s profit is $10\%.$

Nasim bought a pen for $Rs60$ and sold it for $Rs54$. His ________ $\%$ is ________.

Given, CP of a Pen $=Rs60$ and SP of a pen $=Rs54$

Since, SP $<$ CP

$\therefore$ Loss $=$ CP $-$ SP $=Rs\left(60-54\right)=Rs6$

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

Hence, Nasim’s loss is $10\%.$

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 $Rs$ ________.

Given, CP of house $=Rs5059700$

And amount spent on repairing $=Rs40300$

So, total CP of house $=Rs\left(5059700+40300\right)=Rs5100000$

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

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

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

 

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

$SP=5100000+255000$

$SP=Rs5355000$

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

CP of $20$ lemons $=Rs10$

By applying unitary method,

If SP of $5$ lemons is $Rs3$

Then, SP of $1$ lemon $=\frac{3}{5}$

SP of $20$ lemons $=\frac{3}{5}\times 20=12$

Now, CP $=Rs10$ and SP $=Rs12$

Since, SP $>$ CP

Profit $=$ SP $-$ CP

$=Rs\left(12-10\right)=Rs2$

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

Hence, profit is $20\%$

Narain bought $120$ oranges at $Rs4$ each. He sold $60\%$ of the oranges at $Rs5$ each and the remaining at $Rs3.50$ each. His ________ is ________ $\%$.

CP of $1$ orange $=Rs4$ (given)

CP of $120$ oranges $=4\times 120=Rs480$

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

$\therefore$ SP of $72$ oranges $=72\times 5=Rs360$

And SP of remaining oranges $=\left(120-72\right)\times 3.50=48\times 3.50=Rs168$

Total SP of $120$ oranges $=360+168=Rs528$

Since, SP $>$ CP

Profit $=$ SP $-$ CP $=$ $Rs\left(528-480\right)=Rs48$

Now,

Profit per cent

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

Hence, profit is $10\%$

A fruit seller purchased $20kg$ of apples at $Rs50perkg$. Out of these, $5\%$ of the apples were found to be rotten. If he sells the remaining apples at $Rs60perkg$, then his _________is _________ $\%$.

Price for per kg apples $=Rs50$

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

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

$=20-\frac{5}{100}\times 20=20-1=19\text{kg}$

Also, he sells $19\text{kg}$ apples at $Rs60$ per kg

Total SP $=$ $19\times 60=Rs1140$

Cost price of $20kg$ apples $=20\times 50=Rs1000$

Profit $=$ SP $-$ CP $=1140-1000=\text{Rs }140$

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

So, his profit is $14\%.$

Interest on $Rs3000$ at $10\%$ per annum for a period of $3$ years is ________.

Given, $\text{P}=Rs3000,\text{R}=10\%$ and $\text{T}=3yr$

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

$=\frac{3000\times 10\times 3}{100}=Rs900$

Hence, interest is $Rs900$

Amount obtained by depositing $Rs20,000$ at $8\%$ per annum for six months is ________.

 Amount deposited $=Rs20,000$

Rate of interest $=8\%$

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

$\text{I}=\frac{\text{P}\times \text{R}\times \text{T}}{100}=\frac{20000\times 8\times \frac{1}{2}}{100}=\frac{20000\times 8}{200}=100\times 8=Rs800$

Amount received $=$ Principal $+$ Interest $=20000+800=Rs20800$

Interest on $Rs12500$ at $18\%$ per annum for a period of $2$ years and $4$ months is ________.

Given, $\text{P}=Rs12500$ and $\text{R}=18\%$

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

We know that, $\text{I}=\frac{\text{P}\times \text{R}\times \text{T}}{100}=\frac{12500\times 18\times 7}{3\times 100}=Rs5250$

$25ml$ is _________ per cent of $5$ litres.

Let $25ml$ be $x\%$ of $5L$

Then, $25ml=x\%$ of $5l$

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

$\frac{25\times 100}{5\times 1000}=x$

$x=0.5$

Hence, $25ml$ is $0.5\%$ of $5l.$

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.

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

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

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

$\Rightarrow \frac{120}{100}\text{A}=\text{B}$

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

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

$\text{B}-50\%$ of $\text{B}=\text{C}$

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

$\Rightarrow \text{B}\times \frac{50}{100}=\text{C}$

$\Rightarrow \frac{1}{2}\text{B}=\text{C}$

$\Rightarrow \text{B}=2\text{C}$          (ii)

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

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

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

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

$\Rightarrow \text{C}=\frac{3}{5}\text{A}$

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

                   $=\frac{3}{5}\times 100\%=60\%$

Hence, $60\%$ of A is equal to C.

Interest $=\frac{P\times R\times T}{100},$ where

$T$ is ____________

$R\%$ is ____________ and

$P$ is ____________.

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

The difference of interest for $2$ years and $3$ years on a sum of $Rs2100$ at $8\%$ per annum is _________.

Given, $\text{P}=Rs2100$ and $\text{R}=8\%$

For, $\text{T}=2\text{yr}$

 

$\text{I}=\frac{P\times R\times T}{100}=\frac{2100\times 8\times 2}{100}=Rs336$

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

 

$\text{I}=\frac{P\times R\times T}{100}=\frac{2100\times 8\times 3}{100}=Rs504$

Difference between both interests $=504-336=Rs168$

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

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

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

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

The _________ of interest on a sum of $Rs2000$ at the rate of $6\%$ per annum for $1\frac{1}{2}$ years and $2$ years is $Rs420$.

Given, $\text{P}=Rs2000$ and $\text{R}=6\%$ $1\frac{1}{2}\text{yr}\text{=}\frac{3}{2}\text{yr,}$

Now, $\text{I}=\frac{P\times R\times T}{100}=\frac{2000\times 6\times 3}{100\times 2}=Rs180$

For $2\text{yr,}$

$\text{I}=\frac{P\times R\times T}{100}=\frac{2000\times 6\times 2}{100}=Rs240$

The sum of both interests is $Rs\left(180+240\right)=Rs420$

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

In percentage, $6.5\times 100\%=650\%$

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.

Percentage $=$ Fraction $\times 100$

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

Now, according to the formula, we have

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

 

Percentage $=$ Fraction $\times 100$

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.

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

True

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

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

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

False

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

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

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

$8$ hours is $50\%$ of $4$ days.

False

Let $8\text{hr}$ be $x\%$ of $4$ days

Then, $8\text{hr}=x\%$ of $4$ days

$\Rightarrow 8=\frac{x}{100}\times 4\times 24$

$\Rightarrow \frac{8\times 100}{4\times 24}=x$

$\Rightarrow x=\frac{25}{3}=8\frac{1}{3}$

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

The interest on $Rs350$ at $5\%$ per annum for $73$ days is $Rs35$.

False

Given, $\text{P}=Rs350,\text{R}=5\%$

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

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

 

$\Rightarrow \text{I}=\frac{350\times 5\times 73}{100\times 365}=\frac{127750}{36500}$

$\Rightarrow \text{I}=Rs3.5$

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

True

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

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

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

False

In fraction, $75\%$ can be written as

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

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

$12\%$ of $120$ is $100$.

False

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

Therefore, $12\%$ of $120$ is $14.4$

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

False

Marks obtained by Ankita out of $600=336$

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

Hence, Ankita got $56\%$ marks.

$0.018$ is equivalent to $8\%.$

False

In percentage, $0.018$ can be written as,

 $0.018\times 100=1.8\%$

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

$50\%$ of $Rs50$ is $Rs25.$

True

Since, $50\%$ of $Rs50=\frac{50}{100}\times 50=Rs25$

Hence, $50\%$ of $Rs50$ is $Rs25.$

$250cm$ is $4\%$ of $1km.$

False

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

Now, $4\%$ of $1\text{km}=\frac{4}{100}\times 1000m$

                         $=40\text{m}$

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

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

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, $\frac{474}{600}\times 100\%=79\%$

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

By selling a book for $Rs50$, a shopkeeper suffers a loss of $10\%$. The cost price of the book is $Rs60$.

False

 SP of book $=Rs50$  

Loss per cent on book $=10\%$

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

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

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

$\Rightarrow 10\text{CP}=100\text{CP}-5000$

$\Rightarrow 10\text{CP}=5000$

$\Rightarrow \text{CP}=\frac{5000}{90}$

$\therefore \text{CP}=Rs55.55$

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

If a chair is bought for $Rs2000$ and is sold at a gain of $10\%,$ then selling price of the chair is $Rs2010$.

False

Given, $\text{CP of the chair}=Rs2000$ and Profit per cent $=10\%$

 

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

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

                            10 $=\frac{\text{SP}-2000}{2000}\times 100$

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

$200=\text{SP}-2000$

$\text{SP}=Rs2200$

Hence, SP of chair is $Rs2200$

If a bicycle was bought for $Rs650$ and sold for $Rs585$, then the percentage of profit is $10$.

False

Given, CP of bicycle $=Rs650$ and SP of bicycle $=Rs585$

Since, CP $>$ SP

Loss $=$ CP $-$ SP $=650-585=Rs65$

Now, Loss per cent

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

Hence, Loss per cent is $10\%$

Sushma sold her watch for $Rs3320$ at a gain of $Rs320$. For earning a gain of $10\%$ she should have sold the watch for $Rs3300.$

True

Given, SP of watch $=Rs3320$ 

Profit on SP $=Rs320$

CP $=$ SP $-$ Profit $=3320-320=Rs3000$

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

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

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

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

 

$10=\frac{\text{SP}-3000}{3000}\times 100$

$300=\text{SP}-3000$

$\text{SP}=Rs3300$

Interest on $Rs1200$ for $1\frac{1}{2}$ years at the rate of $15\%$ per annum is $Rs180.$

False

Given, $\text{P}=Rs1200,\text{ T}=1\frac{1}{2}\text{yr}=\frac{3}{2}\text{yr}$ and $\text{R}=15\%$

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

$\Rightarrow \text{I}=\frac{1200\times 15\times 3}{100\times 2}=Rs270$

So, interest is $Rs270$

Amount received after depositing $Rs800$ for a period of $3$ years at the rate of $12\%$ per annum is $Rs896.$

False

Given, $\text{P}=Rs800,\text{T}=3\text{yr}$ and $\text{R}=12\%$

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

$\text{I}=\frac{800\times 12\times 3}{100}=Rs288$

So, interest is $Rs288$

Hence, the amount received will be

A = P+ I

= $800+288=1088$ 

$Rs6400$ 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 $Rs150.$

False

As per the question, Feroz borrowed $Rs6400$ for $3\frac{1}{2}\text{ yr}$ at $15\%$ per annum.

Here,  ${\text{P}}_1=Rs6400,{\text{T}}_1=3\frac{1}{2}\text{yr}=\frac{7}{2}\text{yr}$ and ${\text{R}}_1=15\%$

$\therefore {\text{I}}_1=\frac{{\text{P}}_1\times {\text{R}}_1\times {\text{T}}_1}{100}$

$\text{I}=\frac{6400\times 15\times 7}{100\times 2}=Rs3360$

Rashmi borrowed $Rs6400$ for $5\text{yr}$ at $15\%$

Here, ${\text{P}}_2=Rs6400,{\text{R}}_2=15\%\text{and}{\text{T}}_2=5\text{yr}$

$\therefore {\text{I}}_2=\frac{{\text{P}}_2\times {\text{R}}_2\times {\text{T}}_2}{100}$

${\text{I}}_2=\frac{6400\times 15\times 5}{100}=Rs4800$

Difference between interests $=4800-3360=Rs1440$

Hence, the difference in interest, paid by them is $Rs1440$

A vendor purchased $720$ lemons at $Rs120$ per hundred. $10\%$ of the lemons were found rotten which he sold at $Rs50$ per hundred. If he sells the remaining lemons at $Rs125$ per hundred, then his profit will be $16\%$.

False

As per the question, cost price of $100$ lemons $=Rs120$

Cost price of $1$ lemon $=Rs\frac{120}{100}$

And cost price of $720$ lemons $Rs25.6$

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

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

 

Selling price of $100$ rotten lemons $=Rs50$

Selling price of $1$ rotten lemon $=Rs\frac{50}{100}$

And selling price of $72$ rotten lemons $=\frac{50}{100}\times 72=Rs36$

Also, selling price of good lemons $=Rs125$ per hundred

& selling price of $\left(720-72\right)$ good lemons $=\frac{125}{100}\times \left(720-72\right)=Rs810$

Now, total selling price of $720$ lemons $=36+810=Rs846$

 As, SP $<$ CP

Therefore, vendor will bear loss.

Find the value of $x$ if

a.  $8\%$ of $Rsx$ is $Rs100$

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

$35\%$ of $Rsx$ is $Rs280$

$45\%$ of marks $x$ is $Rs405.$

a.  $8\%$ of $Rsx$ is $Rs100$                 (given)

 

$\therefore \frac{8}{100}\times x=100$

$\Rightarrow x=\frac{100\times 100}{8}=Rs1250$

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

 

$\therefore \frac{32}{100}\times x=400$

 

$\Rightarrow x=\frac{400\times 100}{32}=Rs1250\text{kg}$

$35\%$ of $Rsx$ is $Rs280$                  (given)

 

$\therefore \frac{35}{100}\times x=280$

 

$\Rightarrow x=\frac{280\times 100}{35}=Rs800$

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

 

$\therefore \frac{45}{100}\times x=405$

$\Rightarrow x=\frac{405\times 100}{45}=Rs900$ 

Imagine that, a $10\times 10$ 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?

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 $=25\times 3=75$
Now, $25\%$ of $300=\frac{25}{100}\times 300=25\times 3=75$

Hence, the above $2$ values are equal

Value of $25$ squares $=25\times 3=75$

$17\%$ of the grid means $17$ squares. So we will shade $17$ squares

Total value these $17$ squares $=17\times 3=51$

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

Hence, the above two values are equal.

 

Value of $1$  grid $=300$

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

 

Express $\frac{1}{6}$ as a per cent.

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

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

Express $\frac{9}{40}$ as a per cent.

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

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

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

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

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

In fraction, $80\%$ can be written as

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

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

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

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

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

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

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

For ratio in lowest term,

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

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

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

For the ratio in lowest term

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

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?

First way,

$52\%$ of $700=\left(1\%of700\right)\times 52$

             $=\left(\frac{1}{100}\times 700\right)\times 52$

            

             $=7\times 52=364$

Second way,

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

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

Write $0.089$ as a percent.

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

Write $1.56$ as a percent.

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

What is $15\%$ of $20?$

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

$=\frac{15}{100}\times 20=3$

What is $800\%$ of $800?$

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

$=\frac{800}{100}\times 800=6400$

What is $100\%$ of $500?$

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

$=\frac{100}{100}\times 500=500$

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

Let $x\%$ of $1$ h be $30$ min.

Then, $\frac{x}{100}\times 1\text{h}=30$ min.

$\frac{x}{100}\times 60$ min $=30$ min

$x=\frac{30\times 100}{60}$

$x=50\%$

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

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

Let $x\%$ of $1$ day is $1$ min

Then, $\frac{x}{100}\times 1$ day $=1$ min

$\frac{x}{100}\times 24$ h $=1$ min

$\frac{x}{100}\times 1440$ min $=1$ min

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

$x=0.069\%$

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

What per cent of $1km$ is $1000$ metres?

Let $x\%$ of $1$ km is $1000$ m

Then, $\frac{x}{100}\times 1$ km $=1000$ m

$\frac{x}{100}\times 1000$ m $=1000$ m

$x\times 10=1000$

$x=100\%$

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

Find out $8\%$ of $25\text{kg}.$

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

$=\frac{8}{100}\times 25=2$ kg

Hence, $8\%$ of $25\text{kg}$ is 2 kg

What percent of $Rs80$ is $Rs100?$

Let $x\%Rs80$ is $Rs100$

Then, $\frac{x}{100}\times 80=100$

$x=\frac{100\times 10}{8}=Rs125\%$

Hence, $125\%$ of $Rs80$ is $Rs100$

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

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

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

 

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 $\begin{array}{l}=2000=\frac{60}{100}\times 2000\\ =60\times 20=1200\end{array}$

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

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

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

Percentage of carbon in chalk $=3\%$

Percentage of oxygen in chalk $=12\%$

Weight of chalk $=2\frac{1}{2}\text{kg}=\frac{5}{2}\text{kg}=2.5\text{kg}=2.5\times 1000\text{gm}=2500\text{gm}$

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

$=\frac{3}{100}\times 2500=25\times 3=75g$

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

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

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

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

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 $=800\text{ kg}$

$\therefore$ Mass of lime in mortar $=12\%$ of $800\text{ kg}=\frac{12}{100}\times 800=12\times 8=96\text{ kg}$

Therefore, weight of lime in mortar is $96\text{ kg}.$

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 $Rs600$ per table and the remaining at the rate of $400$ per table. Find her gain or loss percent.

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

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

                                $=9600+3200=12800=$ S.P

Gain or loss $=$ S.P $>$ C.P,

Therefore,

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

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

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?

Let the total money Medha had $=100$