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 $=100$

Money deposited $=20\%$ of $100=20$

Remaining money $=100-20=80$

Money spended $=20\%$ of $80$

$=\frac{20}{100}\times 80=16$

Money Remained $=80-16=64$

$64\%=4800$

$100=\frac{4800}{64}\times 100=7500$

The cost of a flower vase got increased by $12\%.$ If the current cost is $Rs896$, what was its original cost?

Let the original cost of flower vase be $Rsx$

Now, the cost of flower vase is increased by $12\%$

So, $x+12\%$ of $x=Rs896$

$x+\frac{12}{100}x=896$

$\frac{112x}{100}=896$

$x=\frac{896\times 100}{112}=Rs800$

Hence, original cost of the flower vase in $Rs800$

Radhika borrowed $Rs12000$ from her friends. Out of which $Rs4000$ were borrowed at $18\%$ and the remaining at $15\%$ rate of interest per annum. What is the total interest after $3$ years?

For $1^{st}$ year interest, we have

${\text{P}}_1=Rs4000,{\text{R}}_1=18\%$ and ${\text{T}}_1=3\text{ yr}$

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

${\text{I}}_1=\frac{4000\times 18\times 3}{100}=Rs2160$

For $2^{nd}$ year interest, we have

${\text{P}}_2=12000-4000=Rs8000$

${\text{R}}_2=15\%$ and ${\text{T}}_2=3\text{yr}$

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

${\text{I}}_1=\frac{8000\times 15\times 3}{100}=Rs\text{3600}$

Hence, after $3\text{yr},$ total interest $={\text{I}}_1+{\text{I}}_2=2160+3600=Rs5760$

A man travelled $60\text{km}$ by car and $240\text{km}$ by train. Find what per cent of total journey did he travel by car and what per cent by train?

Distance travelled by car $=60\text{km}$

Distance travelled by train $=240\text{km}$

$\therefore$ Total journey $=60+240=300\text{km}$

Let $x\%$ of $300=60$

$\frac{x}{100}\times 300=60$

$x=\frac{60\times 100}{300}$

$x=20\%$

Let $y\%$ of $300=240$

$\frac{y}{100}\times 300=240$

$y=\frac{240\times 100}{300}$

Therefore, $20\%$ distance is travelled by the car and $80\%$ distance is travelled by the train.

By selling a chair for $Rs1440,$ a shopkeeper loses $10\%.$ At what price did he buy it?

We have, SP of the chair $=Rs1440$ and loss $=10\%$

Applying formula of Loss $\%$ we have,

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

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

$10=\frac{\text{CP}-1440}{\text{CP}}\times 100$

$\frac{10}{100}\text{CP}=\text{CP}-1440$

$\text{CP}-\frac{10}{100}\text{CP}=1440$

$\frac{9}{10}\text{CP}=1440$

$\text{CP}=\frac{1440\times 10}{9}$

$\text{CP}=\text{Rs }1600$

Hence, cost of the chair is $Rs1600$

Dhruvika invested money for a period from May $2006$ to April $2008$ at rate of $12\%$ per annum. If interest received by her is $Rs1620,$ find the money invested.

We have,

 $\text{I}=Rs1620$ and $Rs12\%$

Time $=$ from May $2006$ to April $2008=2\text{yr}$

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

$\text{P}=\frac{\text{I}\times 100}{\text{R}\times \text{T}}=\frac{1620\times 1000}{12\times 2}$

$\text{P}=Rs6750$

Hence, the invested money is $Rs6750$

A person wanted to sell a scooter at a loss of $25\%.$ But at the last moment he changed his mind and sold the scooter at a loss of $20\%.$ If the difference in the two SP’s is $Rs4000,$ then find the CP of the scooter.

Let CP of the scooter be $Rsx$

If he sells the scooter at a loss of $25\%,$ then

SP $=x-25\%$ of $x=x-\frac{25}{100}x=`\frac{75}{100}x$

& if he sells the scooter at a loss of $20\%,$ then

SP $=x-20\%$ of $x=x-\frac{20}{100}x=\frac{8}{100}x$

It is given that the difference in the two SP’s is $Rs4000$,

$\frac{80}{100}x-\frac{75}{100}x=4000$

$\frac{80x-75x}{100}=4000$

$\frac{5x}{100}=4000$

$x=\frac{4000\times 100}{5}=Rs80,000$

Hence, cost price of scooter is $Rs80,000$

The population of a village is $8000.$ Out of these, $80\%$ are literate and of these literate people, $40\%$ are women. Find the ratio of the number of literate women to the total population.

Given, total population of a village $=8000$

Literate people $=80\%$ of total population $=\frac{80}{100}\times 8000=6400$

Literate women $=40\%$ of total population $=\frac{40}{100}\times 6400=2560$

Ratio of literate women to total population $=2560:8000=\frac{2560}{320}:\frac{8000}{320}=8:25$

Hence, the ratio of literate women to total population is $8:25$.

In an entertainment programme, $250$ tickets of $Rs400$ and $500$ tickets of $Rs100$ were sold. If the entertainment tax is $40\%$ on ticket of $Rs400$ and $20\%$ on ticket of $`100,$ find how much entertainment tax was collected from the programme.

Given, $250$ tickets of $Rs400$ were sold,

Therefore, total amount received by selling these tickets $=250\times 400=Rs1,00,000$ 

Similarly, amount received by selling $500$ tickets of $Rs100=500\times 100=Rs50,000$.

As per the question, entertainment tax on $Rs400$ is $40\%$ and on $Rs100$ tickets is $20\%$ respectively.

So total entertainment tax collected $=40\%$ of total amount received by selling tickets of $Rs400+20\%$ of total amount received by selling tickets of $`100$

             $=40\%$ of $100000+20\%$ of $50000$

             $=\frac{40}{100}\times 100000+\frac{20}{100}\times 50000$

             $=40000+10000=Rs50000$

Hence, the total collected entertainment tax was $Rs50000$

Bhavya earns $Rs50,000$ per month and spends $80\%$ of it. Due to pay revision, her monthly income increases by $20\%$ but due to price rise, she has to spend $20\%$ more. Find her new savings.

 Bhavya earn per month $=Rs50000$

She spends per month $=80\%$ of $`50000$

$=\frac{80}{100}\times 50000=`40000$

Bhavya’s per month savings $=50000-40000=Rs10000$

Also, increment in monthly income $=20\%$ of $50000$

$=\frac{20}{100}\times 50000=Rs10000$

Bhavya’s new income $=50000+10000+Rs60000$

Increase expenditure $=20\%$ of $40000$

 $=\frac{20}{100}\times 40000=Rs8000$

So, Bhavya’s new expenditure $=40000+8000=Rs48000$

Now, Bhavya’s new savings $=60000-48000=Rs12000$

In an examination, there are three papers each of $100$ marks. A candidate obtained $53$ marks in the first and $75$ marks in the second paper. How many marks must the candidate obtain in the third paper to get an overall of $70$ per cent marks?

Let the marks of candidate in third paper be $x$ 

Then, total marks secured in all three papers will be $=53+75+x$

Total marks of three papers $=100+100+100=300$

Percentage of marks $=\frac{\text{Total marks secured}}{\text{Total makrs}}\times 100=\frac{53+75+x}{300}\times 100\%$

But it is given that, he obtained overall of $70\%$ marks.

$=\frac{53+75+x}{300}\times 100\%=70c$

$=\frac{128+x}{3}=70$

$=128+x=210$

$=x=210-218$

$x=82$

Hence, he needs to secure $82$ marks in the third paper to get an overall of $70\%$ marks.

Health Application

A doctor reports blood pressure in millimetres of mercury (mm Hg) as a ratio of systolic blood pressure to diastolic blood pressure (such as $140$ over $80$ ). Systolic pressure is measured when the heart beats, and diastolic pressure is measured when it rests. Refer to the table of blood pressure ranges for adults.

 

Blood Pressure Ranges
	Normal	Prehypertension	Hypertension (Very High)
Systolic Diastolic	Under $120\text{mm}$ Hg Under $80\text{mm}$ Hg	$120-139\text{mm}$ Hg $80-89\text{mm}$ Hg	$140\text{mm}$ Hg and above $90\text{mm}$ Hg and above

 

Manohar is a healthy $37$ years old man whose blood pressure is in the normal category.

a.  Calculate an approximate ratio of systolic to diastolic blood pressures in the normal range.

If Manohar’s systolic blood pressure is $102\text{mm}$ Hg, use the ratio from part (a) to predict his diastolic blood pressure.

Calculate ratio of average systolic to average diastolic blood pressure in the prehypertension category.

a.  Systolic blood pressure in the normal range $=120$ mm Hg

Diastolic blood pressure in the normal range $=80$ mm Hg

Approximate ratio of systolic to diastolic blood pressure

$\frac{\text{Systolic blood pressure in the normal range}}{\text{Diastolic blood pressure in the normal range}}$ $\begin{array}{l}=\frac{120}{80}\\ =\frac{3}{2}\end{array}$

Hence, approximate ratio is $3:2$

Manohar’s systolic blood pressure $=102$ mm Hg

Let diastolic blood pressure $=x$ mm Hg

As per the question,

$\frac{\text{Systolic blood pressure}}{\text{Diastolic blood pressure}}$ $=\frac{3}{2}$

$\frac{102}{x}=\frac{3}{2}$

$x=\frac{102\times 2}{3}=68$ mm Hg

Hence, Manohar’s diastolic blood pressure is $68$ mm Hg

Average systolic blood pressure in prehypertension category $=\frac{120+139}{2}=\frac{259}{2}$ mm Hg

Average diastolic blood pressure in prehypertension category $=\frac{80+89}{2}=\frac{169}{2}$ mm Hg

Hence, ratio of average systolic to average diastolic blood pressures

$=\frac{\text{Average systolic blood pressure}}{\text{average diastolic blood pressure}}$ $=\frac{\frac{259}{2}}{\frac{169}{2}}$$=\frac{259}{2}\times \frac{2}{169}=\frac{259}{169}$

Hence, required ratio is $259:169$

a.  Science Application: The king cobra can reach a length of $558cm$. This is only about $60$ per cent of the length of the largest reticulated python. Find the length of the largest reticulated python.

Physical Science Application: Unequal masses will not balance on a fulcrum if they are at equal distance from it; one side will go up and the other side will go down.

Unequal masses will balance when the following proportion is true:

$\frac{\text{Mass}1}{\text{Length}2}=\frac{\text{Mass}2}{\text{Length}1}$

Two children can be balanced on a see-saw when

$\frac{\text{Mass}1}{\text{Length}2}=\frac{\text{Mass}2}{\text{Length}1}.$ The child on the left and child on the right are balanced. What is the mass of the child on the right?

Life Science Application

A DNA model was built using the scale $2\text{cm}:0.0000001\text{ mm}.$ If the model of the DNA chain is $17\text{cm}$ long, what is the length of the actual chain?

a.  Length of the king cobra $=558$ cm

As per the question,

$60\%$ of length of reticulated python $=558$ cm

Length of reticulated python $=558\times \frac{100}{60}=930$ cm

It is given that, for balancing, $\frac{\text{Mass }1}{\text{Length }2}=\frac{\text{Mass }2}{\text{Length }1}$ 

According to the question,

Mass $1=24$ kg, Length $1=3$ m and Length $2=2$ m

$\frac{24}{2}=\frac{\text{Mass }2}{3}$

Mass $2=\frac{24\times 3}{2}=36$ kg

Let the length of the actual chain be $x$ mm.

Therefore, $\frac{2\text{ cm}}{0.0000001\text{ mm}}=\frac{17\text{ cm}}{x\text{ mm}}$

$x=\frac{17\times 0.0000001}{2}$

$=8.5\times 0.0000001=0.0000085$ mm

Language Application

Given below are few Mathematical terms.

Find

a.                The ratio of consonants to vowels in each of the terms.

b.               The percentage of consonants in each of the terms.

$\Rightarrow \text{ Ratio of consonants to vowels}= \frac{\text{Number of consonants}}{\text{Number of vowels}}$

a.  In term “Hypotenuse” $6$ consonants and $4$ vowels are there

$\therefore \frac{6}{4}=\frac{3}{2}$

$=3:2$

$\Rightarrow$ In term “Congruence” $6$ consonants and $4$ vowels are there

$\therefore \frac{6}{4}=\frac{3}{2}$

$=3:2$

$\Rightarrow$ In term “Perpendicular” $8$ consonants and $5$ vowels are there

$\therefore \frac{8}{5}$

$=8:5$

$\Rightarrow$ In term “Transversal” $8$ consonants and $3$ vowels are there

$\therefore \frac{8}{3}$

$=8:3$

$\Rightarrow$ In term “Correspondence” $9$ consonants and $5$ vowels are there

$\therefore \frac{9}{5}$

$=9:5$

Percentage of consonants $= \frac{\text{(Number of consonants)}}{\text{Total number of letters}}\times 100\%$

Total number of letters $=$ Number of consonants $+$ number of vowels

$\Rightarrow$ In term “Hypotenuse” total number of letters $=6+4=10$

$=\frac{6}{10} \times 100\%=60\%$

$\Rightarrow$ In term “Congruence” total number of letters $=6+4=10$

$=\frac{6}{10} \times 100\%=60\%$

$\Rightarrow$ In term “Perpendicular” total number of letters $=8+5=13$

$=\frac{8}{13}\times 100\%=61.53\%$

$\Rightarrow$ In term “Transversal” total number of letters $=8+3=11$

$=\frac{8}{11}\times 100\%=72.72\%$

$\Rightarrow$ In term “Correspondence” total number of letters $=9+5=14$

$=\frac{9}{14}\times 100\%=64.28\%$

What’s the Error? An analysis showed that $0.06$ per cent of the T-shirts made by one company were defective. A student says this is $6$ out of every $100.$ What is the student’s error?

As per the question

Defective T-shirts made by the company $=0.06\%$

                                                   $=\frac{0.06}{100}=\frac{6}{10000}$

According to the student, number of defective

T-shirts $=6$ out of every $100$

$=\frac{6}{100}$

Hence, student’s error is that, the defective T-shirts are $6$ out of every $1000$ 0 (not $100)$

What’s the Error? A student said that the ratios $\frac{3}{4}$ and $\frac{9}{16}$ were proportional. What error did the student make?

As per the definition or rule:

Two ratios a : b and c : d are said to be proportional, if $\frac{a}{b}=\frac{c}{d}$ or $ad=bc$ 

But in the given ratios $\frac{3}{4}$ and $\frac{9}{16},3\times 16\ne 4\times 9$

Hence, the ratios are not proportional. To make a ratio proportional to another ratio, we just simply multiply both numerator and denominator by same number. In our given case, student had multiplied numerator by $3$ and denominator by $4,$ which is incorrect.

What’s the Error? A clothing store charges $Rs1024$ for $4$ T-shirts. A student says that the unit price is $Rs25.6$ per T-shirt.

What is the error? What is the correct unit price?

By unitary method,

Cost of $4$ T-shirts $=Rs1024$

Cost of $1$ T-shirts $=\frac{1024}{4}=Rs256$

Hence, the correct unit price is $Rs256$

A tea merchant blends two varieties of tea in the ratio of $5:4.$ The cost of first variety is $Rs200$ per $kg$ and that of second variety is $Rs300$ per $kg$. If he sells the blended tea at the rate of $Rs275$ per $kg$, find out the percentage of her profit or loss.

 Ratio of blended two varieties of tea (green tea: lemon tea) $=5:4$

Cost of green tea $=Rs200$ per kg.

Cost of lemon tea $=\text{}Rs300$ per kg

SP of blended tea $=Rs275$ per kg

As per the ratio,

Let green tea be $5x$ kg and lemon tea be $4x$ kg.

Then, cost of green tea $=5x\times 200=Rs1000x$

Cost of lemon tea $=4x\times 300=Rs1200x$

Total CP $=1000x+1200x=Rs2200x$

Total quantity $=4x+5x=9x$ kg

So, for $9x$ kg

SP of blended tea $=275\times 9x=Rs2475x$

CP $<$ SP

So, there is profit on blended tea.

Profit $=$ SP $-$ CP

$=2475x-2200x=Rs275x$

Profit $\%=\frac{\text{profit}}{\text{CP}}=100\%=\frac{275x}{2200x}=100\%=\frac{275}{22}=12.5\%$

Hence, there is $12.5\%$ profit on blended tea (new variety)

A piece of cloth $5m$ long shrinks $10$ per cent on washing. How long will the cloth be after washing?

Given,

Length of shrink cloth $=10\%$ of $5$ m $=\frac{10}{100}\times 5=\frac{1}{2}$ m

Length of cloth after wash $=5-\frac{1}{2}=\frac{9}{2}=4.5$ m

Nancy obtained $426$ marks out of $600$ and the marks obtained by Rohit are $560$ out of $800.$ Whose performance is better?

Marks obtained by Nancy $=426$ out of $600$

Percentage marks $=\frac{426}{100}\times 100\%=71\%$

Rohit got marks $=560$ out of $800$

Percentage marks $=\frac{560}{800}\times 100\%=70\%$

Hence, Nancy’s performance is better, since she got $1\%$ more than Rohit.

A memorial trust donates $Rs5,00,000$ to a school, the interest on which is to be used for awarding $3$ scholarships to students obtaining first three positions in the school examination every year. If the donation earns an interest of $12$ per cent per annum and the values of the second and third scholarships are $Rs20,000$ and $Rs\text{15},000$ respectively, find out the value of the first scholarship.

Donation amount $=Rs500000$

Rate of interest for each year $=12\%$ per annum

Time period $=1$ yr

Interest received after $1$ yr $=\frac{500000\times 12\times 1}{100}=500\times 12=`60000$

Scholarship amount for $2^{nd}$ position $=`20000$

Scholarship amount for $3^{rd}$ position $=Rs15000$

Remaining amount for $1^{st}$ position student $=60000-\left(20000-15000\right)$

                   $=60000-35000=Rs25000$

Hence, scholarship amount for $1^{st}$ position student

Ambika got $99$ per cent marks in Mathematics, $76$ per cent marks in Hindi, $61$ per cent in English, $84$ per cent in Science, and $95\%$ in Social Science. If each subject carries $100$ marks, then find the percentage of marks obtained by Ambika in the aggregate of all the subjects.

 Each subject carries 100 marks

 Marks obtained by Ambika in,

Mathematics $=99$

Hindi $=76$

English $=61$

Science $=84$

Social science $=95$

Now, aggregate percentage of marks obtained $=\frac{\text{marks obtained by Ambika}}{\text{Total marks}}\times 100$

$=\frac{99+76+61+84+95}{5}\times 100=\frac{415}{5}=83\%$

What sum of money lent out at $16$ per cent per annum simple interest would produce $Rs9600$ as interest in $2$ years?

Here, $\text{I}=Rs9600,\text{T}=2$ yr and $\text{R}=16\%$

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

$\text{P}=\frac{\text{I}\times 100}{\text{R}\times \text{T}}=\frac{9600\times 100}{16\times 2}$

$\text{P}=Rs30000$

Hence, sum of money lent out at $16$ per cent per annum simple interest would be  $Rs30000$

Harish bought a gas-chullah for $Rs900$ and later sold it to Archana at a profit of $5$ per cent. Archana used it for a period of two years and later sold it to Babita at a loss of $20$ per cent. For how much did Babita get it?

 Harish bought the chullah for $Rs900$ and sold it to Archana at a profit of $5\%$

Cost price of chullah for Archana $=900+5\%$ of $900$

                   $=900+\frac{5}{100}\times 900=900+45=945$

Now, Archana sold it to Babita at a loss of $20\%$

Cost price of chullah for Babita $=945-20\%$ of $945$

                   $=945-\frac{20}{100}\times 945=945-189=Rs756$

Hence, Babita got chullah at $Rs756$

Match each of the entries in Column I with the appropriate entries in Column II:

	Column I	Column II
(i)	$3:5$	(A)	$Rs54$
(ii)	$2.5$	(B)	$Rs47$
(iii)	$100\%$	(C)	$Rs53$
(iv)	$\frac{2}{3}$	(D)	$Rs160$
(v)	$6\frac{1}{4}\%$	(E)	$60\%$
(vi)	$12.5\%$	(F)	$25\%$
(vii)	SP when CP $=Rs50$ and loss $=6\%$	(G)	$\frac{1}{16}$
(viii)	SP when CP $=Rs50$ and profit $=Rs4$	(H)	$250\%$
(ix)	Profit $\%$ when CP $=Rs40$ and SP $=Rs50$	(I)	$Rs159$
(x)	Profit $\%$ when CP $=Rs50$ and SP $=Rs60$	(J)	$66\frac{2}{3}\%$
(xi)	Interest when principal $=Rs\text{800}$, Rate of interest $=10\%$ per annum and period $=2$ years	(K)	$20\%$
(xii)	Amount when principal $=Rs150$, Rate of interest $=6\%$ per annum and period $=1$ year	(L)	$0.125$
		(M)	$3:2$
		(N)	$Rs164$
		(O)	$3:3$

 

(i)        Matches with (E)

Ratio given $=3:5$

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

(ii)    Matches with (H)

Given, $2.5$

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

(iii) Matches with (O)

Given, $100\%$

$\Rightarrow \text{ Ratio}=100\%:1$

$=\frac{100}{100}:1$

$\Rightarrow \text{ Ratio}=1:1$

$=1\times 3:1\times 3$

$=3:3$

(iv)  Matches with (J)

Given, $\frac{2}{3}$

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

(v)     Matches with (G)

Given, $6\frac{1}{4}\%=\frac{25}{4}\%$

For Fraction, $\frac{25}{4}\times \frac{1}{1000}=\frac{1}{16}$

(vi)  Matches with (L)

Given, $12.5\%$

For fraction, $\frac{125}{10}\times \frac{1}{100}=\frac{125}{1000}$

For decimal, $\frac{125}{1000}=0.125$

(vii)   Matches with (B)

Given, $\text{CP}=50,\text{Loss}\%=6\%$

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

$=\frac{(\text{CP}-\text{SP})}{\text{CP}}\times 100\%$

$\Rightarrow 6=\frac{50-\text{SP}}{50}\times 100$

$\text{SP}=47$

(viii)              Matches with (A)

Given, $\text{CP}=50,\text{ Profit}=4$

$\Rightarrow \text{ Profit}=\text{SP}-\text{CP}$

$\Rightarrow \text{ SP}=\text{Profit}+\text{CP}$

$=4+50=54$

(ix)  Matches with (F)

Given, $\text{CP}=40,\text{ SP}=50$

$\Rightarrow \text{ Profit}=\text{SP}-\text{CP}$

$=50-40=10$

$\text{Profit}\%=\frac{\text{Profit}}{\text{CP}}\times 100\%=\frac{10}{40}\times 100\%$

$\text{Profit}\%=25\%$

(x)     Matches with (K)

Given, $\text{CP}=50,\text{ SP}=60$

$\Rightarrow \text{ Profit}=\text{SP}-\text{CP}$

$=60-50=10$

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

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

(xi)  Matches with (D)

Given, $\text{P}=800,\text{ R}=10\%,\text{ T}=2\text{ year}$

$\Rightarrow \text{ I}=\frac{800\times 10\times 2}{100}$

$\Rightarrow \text{ I}=160$

(xii)   Matches with (l)

Given, $\text{P}=150,\text{ R}=6\%,\text{ T}=1\text{ year}$

Need to find out A

$\Rightarrow \text{ I}=\frac{150\times 6\times 1}{100}$

$=9$

$\Rightarrow \text{ A}=\text{P}+\text{I}$

$=159$

In a debate competition, the judges decide that $20$ per cent of the total marks would be given for accent and presentation. $60$ per cent of the rest are reserved for the subject matter and the rest are for rebuttal. If this means $8$ marks for rebuttal, then find the total marks.

Let $x$ be the total marks

Marks given for accent & presentation $=20\%$ of $x$ $=\frac{20}{100}\times x=\frac{x}{5}$

Remaining marks $=x-\frac{x}{5}=\frac{4x}{5}$

Marks reserved for subject matter $=60\%$ of rest marks $=\frac{60}{100}\times \frac{4x}{5}=\frac{12x}{25}$

Now, remaining marks $=\frac{4x}{5}-\frac{12x}{25}=\frac{20x-12x}{25}=\frac{8x}{25}$

According to question,

$\frac{8x}{25}=8$

$x=\frac{8\times 25}{8}=25$

Hence, total marks are $25.$

Divide $Rs10000$ in two parts so that the simple interest on the first part for $4$ years at $12$ per cent per annum may be equal to the simple interest on the second part for $4.5$ years at $16$ per cent per annum.

Given, money $=Rs10,000$

Now, we have divide $Rs10000$ in two parts such that SI on first part for $4$ yr at $12\%$ per annum may be equal to the SI on second part for $4.5$ yr at $16\%.$

Let first part = $x$

Then, second part $=Rs\left(10000-x\right)$

For $1^{st}$ part

$S{I}_1=\frac{P_1\times R_1\times T_1}{100}=\frac{x\times 12\times 4}{100}$

For $2^{nd}$ part,

${\text{P}}_2=Rs\left(1000-x\right),{\text{T}}_2=4.5$ yr, $R_2=16\%$

${\text{SI}}_2=\frac{{\text{P}}_2\times {\text{R}}_2\times {\text{T}}_2}{100}=\frac{\left(1000-x\right)\times 16\times 4.5}{100}$

Since ${\text{S}}_1={\text{S}}_2$

$\frac{x\times 12\times 4}{100}=\frac{\left(10000-x\right)\times 16\times 4.5}{100}$

$48x=\left(10000-x\right)\times 16\times 4.5$

$\frac{48x}{4.5\times 16}=\left(10000-x\right)$

$\frac{48x\times 10}{45\times 16}=1000-x$

$\frac{2}{3}x=10000-x$

$\frac{2}{3}x+x=10000$

$\frac{5x}{3}=10000$

$x=10000\times \frac{3}{5}=6000$

First part $=x=Rs6000$

Second part $=10000-x=10000-6000=Rs4000$

Hence, two parts of the sum are $Rs6000$ and $Rs4000$

$Rs9000$ becomes $Rs18000$ at simple interest in $8$ years. Find the rate per cent per annum.

Given, $\text{P}=Rs9000,\text{A}=Rs18000,\text{T}=8$ yr

We know that, $\text{A}=\text{P}+\text{I}$

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

$\text{R}=\frac{\text{I}\times 100}{\text{P}\times \text{T}}=\frac{9000\times 100}{9000\times 8}$

$\text{R}=12.5\%$

Hence, the rate of interest per annum is $12.5\%$

In how many years will the simple interest on a certain sum be $4.05$ times the principal at $13.5$ per cent per annum?

Let principal $=\text{P}$

$\text{R}=13.5\%$

$\text{I}=4.05$ times principal $=4.05\times \text{P}$

$\text{T}=?$

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

$4.05\times \text{P}=\frac{\text{P}\times 13.5\times \text{T}}{100}$

$\text{T}=\frac{405\text{ P}}{\text{P}\times 13.5}$

$=\frac{405\times 10}{135}=30$ yr

The simple interest on a certain sum for $8$ years at $12$ per cent per annum is $Rs3120$ more than the simple interest on the same sum for $5$ years at $14$ per cent per annum. Find the sum.

Given, ${\text{I}}_1-{\text{I}}_2=Rs3120$

${\text{T}}_1=8$ yr, ${\text{R}}_1=12\%$

${\text{T}}_2=5\text{ yr},{\text{R}}_2=14\%$

${\text{P}}_1={\text{P}}_2=\text{P}$

According to question,

${\text{I}}_1-{\text{I}}_2=3120$

$=\frac{{\text{P}}_1\times {\text{R}}_1\times {\text{T}}_1}{100}-\frac{{\text{P}}_2\times {\text{R}}_2\times {\text{T}}_2}{100}=3120$

$=\frac{\text{P}\times 12\times 8}{100}-\frac{\text{P}\times 14\times 5}{100}=3120$

$\frac{\text{P}}{100}\left[96-70\right]=3120\times 100$

$26\text{ P}=31200$

$\text{P}=\frac{31200}{26}=Rs12000$

Hence, sum is $Rs12000$

The simple interest on a certain sum for $2.5$ years at $12$ per cent per annum is $Rs300$ less than the simple interest on the same sum for $4.5$ years at $8$ per cent per annum. Find the sum.

Let the sum be $x$

${\text{P}}_1=x_1,={\text{R}}_1=12\%,{\text{T}}_1=2.5\text{ yr}=\frac{5}{2}\text{ yr}$                [given]

${\text{P}}_2=x_2,{\text{R}}_2=8\%,{\text{T}}_2=4.5\text{ yr}=\frac{9}{2}\text{ yr}$             [given]

According to question,

${\text{I}}_1-{\text{I}}_2=300$

$=\frac{{\text{P}}_2\times {\text{R}}_2\times {\text{T}}_2}{100}-\frac{{\text{P}}_1\times {\text{R}}_1\times {\text{T}}_1}{100}=300$

$=\frac{x\times 8\times 9}{2\times 100}-\frac{x\times 12\times 5}{2\times 100}=300$

$82x-60x=300\times 200$

$12x=60000$

$x=Rs5000$

Hence, sum is $Rs5000$

Designing a Healthy Diet

When you design your healthy diet, you want to make sure that you meet the dietary requirements to help you grow into a healthy adult.

As you plan your menu, follow the following guidelines

1.   Calculate your ideal weight as per your height from the table given at the end of this question.

2.   An active child should eat around $55.11$ calories for each kilogram desired weight.

3.   $55$ per cent of calories should come from carbohydrates. There are $4$ calories in each gram of carbohydrates.

4.   $15$ per cent of your calories should come from proteins. There are $4$ calories in each gram of proteins.

5.   $30$ per cent of your calories may come from fats. There are $9$ calories in each gram of fat.

Following is an example to design your own healthy diet.

Example

1.   Ideal weight $=40\text{kg}$.

2.   The number of calories needed $=40\times 55.11=2204.4$ 

3.   Calories that should come from carbohydrates

$=2204.4\times 0.55=1212.42$ calories.

Therefore, required quantity of carbohydrates

         $=\frac{1212.42}{4}=303.105g=300g.$ (approx).

4.   Calories that should come from proteins

$=2204.4\times 0.15=330.66$ calories.

Therefore, required quantity of protein

         $=\frac{330.66}{4}g=82.66g.$

5.   Calories that may come from fat $=2204.4\times 0.3$

$=661.3$ calories.

Therefore, required quantity of fat

         $=\frac{661.3}{9}g=73.47g.$

Answer the Given Questions

1.   Your ideal desired weight is __________ $kg$.

2.   The quantity of calories you need to eat is _______.

3.    The quantity of protein needed is ________ $g$.

4.   The quantity of fat required is ___________ $g$.

5.   The quantity of carbohydrates required is ________ $g$.

Ideal Height and Weight Proportion
Men			Women
Height		Weight	Height		Weight
Feet	cm	Kilograms	Feet	cm	Kilograms
$5\text{'}$	$152$	$48$	$4\text{'}7\text{'}\text{'}$	$140$	$34$
$5\text{'}1\text{'}\text{'}$	$155$	$51$	$4\text{'}8\text{'}\text{'}$	$142$	$36$
$5\text{'}2\text{'}\text{'}$	$157$	$54$	$4\text{'}9\text{'}\text{'}$	$145$	$39$
$5\text{'}3\text{'}\text{'}$	$160$	$56$	$4\text{'}1\text{'}\text{'}$	$147$	$41$
$5\text{'}4\text{'}\text{'}$	$163$	$59$	$4\text{'}11\text{'}\text{'}$	$150$	$43$
$5\text{'}5\text{'}\text{'}$	$165$	$62$	$5\text{'}$	$152$	$45$
$5\text{'}6\text{'}\text{'}$	$168$	$65$	$5\text{'}1\text{'}\text{'}$	$155$	$48$
$5\text{'}7\text{'}\text{'}$	$170$	$67$	$5\text{'}2\text{'}\text{'}$	$157$	$50$
$5\text{'}8\text{'}\text{'}$	$173$	$70$	$5\text{'}3\text{'}\text{'}$	$160$	$52$
$5\text{'}9\text{'}\text{'}$	$175$	$73$	$5\text{'}4\text{'}\text{'}$	$163$	$55$
$5\text{'}10\text{'}\text{'}$	$178$	$75$	$5\text{'}5\text{'}\text{'}$	$165$	$57$
$5\text{'}11\text{'}\text{'}$	$180$	$78$	$5\text{'}6\text{'}\text{'}$	$168$	$59$
$6\text{'}$	$183$	$81$	$5\text{'}7\text{'}\text{'}$	$170$	$61$
$6\text{'}1\text{'}\text{'}$	$185$	$84$	$5\text{'}8\text{'}\text{'}$	$173$	$64$
$6\text{'}2\text{'}\text{'}$	$188$	$86$	$5\text{'}9\text{'}\text{'}$	$175$	$66$
$6\text{'}3\text{'}\text{'}$	$191$	$89$	$5\text{'}10\text{'}\text{'}$	$178$	$68$
$6\text{'}4\text{'}\text{'}$	$193$	$92$	$5\text{'}11\text{'}\text{'}$	..	$70$

(i)             Let my height be $5\text{'}2\text{'}\text{'}$ ft.

Then, according to the table, my ideal weight $=54$ kg

(ii)         The quantity of calories needed $=54\times 55.11=2975.94$ calories

(iii)      Calorie that should come from proteins $=2975.94\times 0.15=446.391$ calories

Therefore, required of protein $=\frac{446.391}{4}=111.59775\text{ g}$

(iv)       Calories that may come from fat $=2975.94\times 0.3=892.782$ calories

Therefore, required quantity of fats $=\frac{892.782}{9}=99.198\text{ g}$

(v)          Calorie that should come from carbohydrate

$2975.94\times .55=1636.767$ 

Therefore required quantity of carbohydrate

$=\frac{1636.767}{4}=409.20\text{ g}$

$150$ students are studying English, Maths or both. $62$ per cent of students study English and $68$ per cent are studying Maths. How many students are studying both?

Total students $=150$

Students who study English $=62\%$ of $150=\frac{62}{100}\times 150=93$

Students who study Math $=68\%$ of $150=\frac{68}{100}\times 150=102$

Total students studying English & Maths $=93+102=195$

Students who study English & Maths both $=195-150=45$

Earth Science: The table lists the world’s $10$ largest deserts.

Largest Deserts in the World
Desert	Area (km ${}^2$ )
Sahara (Africa)	$8,800,000$
Gobi (Asia)	$1,300,000$
Australian Desert (Australia)	$1,250,000$
Arabian Desert (Asia)	$850,000$
Kalahari Desert (Africa)	$580,000$
Chihuahuan Desert (North America)	$370,000$
Takla Makan Desert (Asia)	$320,000$
Kara Kum (Asia)	$310,000$
Namib Desert (Africa)	$310,000$
Thar Desert (Asia)	$260,000$

 

b. What are the mean, median and mode of the areas listed?

c.  How many times the size of the Gobi Desert is the Namib Desert?

d. What percentage of the deserts listed are in Asia?

e.  What percentage of the total area of the deserts listed is in Asia?

f.  Mean $=$ $\frac{\text{Total area of all deserts}}{\text{Number of deserts}}$

$=\frac{14350000}{10}=1435000{\text{ km}}^2$

Median

$\begin{array}{l}=\frac{\left(\frac{n}{2}\right)\text{th}\text{term}+\left(\frac{n}{2}+1\right)\text{th}\text{term}}{2}\\ =\frac{\frac{10}{2}\text{th}\text{term}+\left(\frac{10}{2}+1\right)\text{th}\text{term}}{2}\end{array}$

$=\frac{5\text{th}\text{term}+6\text{th}\text{term}}{2}=\frac{580000+370000}{2}$

$=\frac{950000}{2}=475000{\text{ km}}^2$

Mode

$=$ Most frequent observation

$=310000{\text{ km}}^2$

Let the size of Gobi desert is $x$ times the Namib desert.

Gobi desert

$=x\times$ Namib desert

$=1300000=x\times 310000$

$\begin{array}{l}x=\frac{1300000}{310000}\\ =4.19\end{array}$

Hence, the size of Gobi Desert is $4.19$ times of Namib desert

Total number of desert $=10$

Number of deserts in Asia (Gobi, Arabian, Takla Makan, Kum, Thar) $=5$

Hence, percentage of deserts in Asia $=\frac{5}{10}\times 100\%=50\%$

Total area of all deserts $=14350000{\text{ km}}^2$

Total area of Asia’s deserts $\begin{array}{l}=1300000+8500000+320000+310000+260000\\ =3040000{\text{ km}}^2\end{array}$

Hence, percentage of the total area of the deserts listed in Asia $=\frac{\text{Total area of Asia}’\text{s deserts}}{\text{Total area of all deserts}}$

$\begin{array}{l}=\frac{3040000}{14350000}\times 100\%\\ =21.1\%\end{array}$

Geography Application: Earth’s total land area is about $148428950k{m}^2$. The land area of Asia is about $30$ per cent of this total. What is the approximate land area of Asia to the nearest square km?

Total land area of the earth $=148428950{\text{ km}}^2$

Land area of Asia $=30\%$ of land area of Earth

                   $=\frac{30}{100}\times 148428950=44528685{\text{ km}}^2$

The pieces of Tangrams have been rearranged to make the given shape.

By observing the given shape, answer the following questions:

a.    What percentage of total has been coloured?

                              i.  Red (R) $=$ _________

                          ii.  Blue (B) $=$ ________

                       iii.  Green (G) $=$ _______

b.   Check that the sum of all the percentages calculated above should be $100.$

c.    If we rearrange the same pieces to form some other shape, will the percentage of colours change?

a. Total coloured shape $=\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{16}+\frac{1}{16}+\frac{1}{4}+\frac{1}{4}=1$

i.               Red coloured shape $=\frac{1}{8}+\frac{1}{8}+\frac{1}{8}=\frac{1+1+1}{8}=\frac{3}{8}$

                   Hence, percentage of red coloured $=\frac{\text{Red}\text{coloured}\text{shape}}{\text{Total}\text{coloured}\text{shape}}\times 100$

$=\frac{\frac{3}{8}}{1}\times 100=\frac{3}{8}\times 100=37.5\%$

ii.           Blue coloured shape $=\frac{1}{4}+\frac{1}{4}=\frac{1+1}{4}$

                                         $=\frac{2}{4}=\frac{1}{2}$

Hence, Percentage of blue coloured $=\frac{\text{Blue}\text{coloured}\text{shape}}{\text{Total}\text{coloured}\text{shape}}\times 100$

$=\frac{\frac{1}{2}}{1}\times 100=\frac{1}{2}\times 100=50\%$

iii.        Green coloured shape $=\frac{1}{16}+\frac{1}{16}=\frac{1+1}{6}=\frac{2}{16}=\frac{1}{8}$

Hence, percentage of green coloured $=\frac{\text{Green}\text{coloured}\text{shape}}{\text{Total}\text{coloured}\text{shape}}\times 100$

$=\frac{\frac{1}{8}}{1}\times 100=\frac{1}{8}\times 100=12.5\%$

 

b. Sum of all percentages calculated $=$ Percentage of red coloured $+$ percentage of blue coloured $+$ percentage of green coloured $=37.5+50+12.5=100\%$

c. No, the percentage of colours will not change, because we just rearranging the parts and not changing the percentage of colours.