Percentage Questions - Percentage questions for class 6 - Quantitative Aptitude

$1)$ If $70\%$ of students in a school are boys and the number of girls is $504$. what is the number of boys?

Solution:

Boys $+$ Girls $=$ Total

$70\%+504$ $=100\%$

$504$ (girls) $=100\%-70%$

$504$ (girls) $=30\%$

$70\%$ (boys) $=\frac{504\times 70\%}{30\%}$

$70\%$ (boys) $=1176$

Number of boys $=1176$

$2)$ In an examination $70\%$ of the candidates passed in English, $80\%$ passed in Mathematics, and $10\%$ failed in both subjects. If $144$ candidates passed in both, the total number of candidates was?

Solution:

Pass in English $=70\%$

Pass in Math $=80\%$

Failed in both $=10\%$

Pass in both (English or Maths)  $=100%-10\%=90\%$

Pass in both subjects $=70\%+80\%-90\%$ $=60\%$

$144=60\%$

$60\%$ $=144$

$100\%$ $=\frac {144}{60}\times 100$

$100\%$ $=240$

Total number of candidates $=240$

2nd Method
        E           M           B
P     $70\%$    $80\%$     $90\%$
F     $30\%$    $20\%$     $10\%$

Pass in both $=70\%+80\%$ $-90\%$ $=60\%$

$60\%$ $=144$

$100\%$ $=\frac {144}{60}\times 100$

$100\%$ $=240$

Total no. of candidates $=240$

$3)$ A trader mixes $26$ kg of rice at  $\$20$ per kg with $30$ kg of rice of other variety at  $\$36$ per kg and sells the mixture at $\$30$ per kg. It is the profit percentage?
Solution:

Quantity of $v_{1}$ $=26kg$

Price of $v_{1}$ $=\$20/kg$

Cost of $v_{1}$ $=26\times 20$ $=\$520$

Quantity of $v_{2}$ $=30kg$

Price of $v_{2}$ $=\$36/kg$

Cost of $v_{2}$ $=30\times 36$ $=\$1080$

Quantity of mixture $=26+30$ $=56kg$

Cost price of mixture $=C.P$ of $v_{1}+C.P.$ of $v_{2}$

  $=520+1080$

The cost price of the mixture $=1600$

The selling price of mixture $=30/kg$

The total selling price of mixture $=30\times 56$ $=\$1680$

Profit $=S.P-C.P$

Profit $=1680-1600$ $=80$

Profit $=80$

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

Profit$\%$ $=\frac {80}{1600}\times 100$

Profit in percentage $=5\%$

$4)$ Solve the following
$a)$ How much $80\%$ of $40$ is greater than $4/5$ of $25$.
solution:

$80\%$ of $40-4/5$ of $25$

$=\frac {80}{100}\times 40-$ $\frac {4}{5}\times 25$

$=32-20$

$=12$

$b)$ How much $40\%$ of $30$ is less than $60\%$ of $50$.
solution:

$60\%$ of $50 - 40\%$ of $30$

$=\frac {60}{100}\times 50$ $-\frac {40}{100}\times 30$

$=30-12$

$=18$

c) What percent of $120$ is $90$?
solution:

1st Method

Let percent be $x\%$

$x\%$ of $120$ $=90$

$=\frac {x}{100}\times 120$ $=90$

$x=\frac {90\times 5}{6}$ $=75$

$x$ $=75\%$


2nd Method

Percentage $=\frac {part}{whole}\times 100$

Percentage $=\frac{90}{120}\times 100$

Percent $=75\%$

$5)$ If a car is sold $\$500$ at a profit of $17\%$ what would be the profit percentage if the car is sold for $\$470$
Solution:

Let the cost price $=100\%$

Profit $+$ C.P $=$Selling Price

$17\%+100\%$ $=\$500$

$117\%$ $=\$500$

$100\%$ $=\frac {50,000}{117}\times 100$

$100\%$ $=42735$

Cost price $=42735$

Profit$+$cost $=$sale price

Profit $=$sale price $-$ cost price

Profit $=50,000$ $-42735$

Profit $=4265$

Profit$\%$ $=\frac {profit}{C.P}\times 100$ $=\frac {4265}{42735}\times 100$

Profit $=9.98\%$ $=10\%$

$6)$ $b$ is what percent of $a$?

Solution:

$50$ is what percent of $200$?

$50$ $=x\%$ of $200$

$\frac {50}{200}$ $=\frac {x}{100}$

$x$ $=\frac {1}{4}\times 100$

$x=25$

$b$ $=x\%$ of $a$

$\frac {b}{a}$ $=\frac {x}{100}$

$x$ $=100\times \frac {b}{a}$

$7)$ $7\%$ of a number is $56$. Then what will be $25\%$ of that number?

Solution:

$1st$ method

$7\%$ $=56$

$1\%$ $=\frac {56}{7}$

$25\%$ $=8\times 25$

$25\%$ $=200$

$2nd$ method

$7\%$ of no. $=56$

$\frac {7}{100}\times x$ $=56$

$x$ $=\frac {56\times 100}{7}$

$x$ $=800$

$25\%$ of $800$

$=\frac {25}{100}\times 800$

$=200$

$8)$ The cost price of $20$ articles is same as the selling price of $x$ articles. If the profit is $25\%$ then what will be the value of $x$?

Solution:

Let the cost price of $1$ article is $100\%$

C.P of $20$ articles $=20\times 100\%$

S.P $=$Profit $+$ C.P

S.P $=25\%+100\%$

S.P $=125\%$

The selling price of $x$ articles $=125\%$ of $x$

According to question

$20\times 100\%$ $=125\%\times x$

$x$ $=\frac {20\times 100\%}{125\%}$ $=\frac {20\times 4}{5}$

$x$ $=16$

$9)$ After decreasing $24\%$ in the price of an article cost $\$912$.
Find the actual cost of the article.
Solution:

Let actual cost  $=100\%$

Decrease $=24\%$

Remaining $=100\%-24\%$

Remaining $=76\%$

$76\%$ of actual price $=912$

$100\%$ $=\frac {912}{76}\times$

$100\%$ $=\$1200$

$10)$ Ali sold a watch to Raheel at a gain of $5\%$ and Raheel sold it to Sarmand at a gain of $4\%$. If Sarmand paid $\$1092$ for it, the price paid by Ali is?

Solution:

Ali paid $=x$

$5\%$ of Ali's paid price $=5\%$ of $x$ $=\frac {5}{100} x$

Ali sold watch to Raheel $=x+\frac {5x}{100}$ $=\frac {100x+5x}{100}$ $=\frac {105x}{100}$

$4\%$ of Raheel's paid amount $=4\%$ of $\frac {105}{100}$

$=\frac {4}{100}\times \frac {21x}{20}$ $=\frac {1}{25}\times \frac{21x}{20}$ $=\frac {21x}{500}$

Raheel sold to Sarmand $=\frac {105x}{100}$ $+\frac {21x}{500}$ $=\frac {525x+21x}{500}$ $=\frac {546x}{500}$

Raheel sold a watch for $=\$1092$

$\frac {546x}{500}$ $=1092$

$x$ $=\frac {1092}{546}\times 500$ $x$ $=1000$

Ali paid $=\$1000$

$11)$ If the price of an item is increased by $20\%$. What percentage of the current price will be reduced to bring it back to its previous level?

Solution:

$∴$Restore$\%$ $=\frac {Change}{New\text{ }value}\times 100$

Old price $=100$

$20\%$ increment $=20$

New price $=100+20$ $=120$

Decrement $=20$

Decrement$\%$ $=\frac {20}{120}\times$ $=\frac {100}{6}$

Decrement$\%$ $=16.67\%$

$12)$ A number mistakenly divided by $5$ instead of being multiplied by $5$. Find the percentage change in results due to this mistake?

Solution:

Let no. be $=100$

Multiply by $5$ $=100\times$ $=500$

Divide by $5$ $=\frac{100}{5}$ $=20$

Change in result $=500-20$ $=480$

Percentage change in result $=\frac{480}{500}\times 100$

Percentage change in result $=96\%$

$13)$ $8$ kgs of solution contain $10\%$ sugar. If we further add $2$kgs of water in the solution, what will be the percentage of sugar?
Solution:

Solution $=8$ kg

Sugar $=10\%$ of solution

$10\%$ of $8$ kg

$\frac {10\times 8}{100}$ $=0.8$ kg

Water $=8-0.8$ $=7.2$ kg

Solution $=8$kg $+2$kg $=10$kg

Sugar $=0.8$ kg

$=\frac {0.8}{10}\times 100$

Sugar percentage $=8\%$

$14)$ The price of an onion increases by $25\%$. By what $\%$ should a housewife reduces consumption so that expenditure on onion remains the same?

solution:

Let price be

$\$100$ $=1$ kg

Increase $=25\%$

$\frac {100\times 25}{100}$ $=25$

$\$125$ $=1$ kg

$125$ $=1$ kg

$100$ $=\frac {1}{125}\times 100$ $=0.8$ kg

$d$ $=1-0.8$ $=0.2$

Decrease$\%$ $=\frac {0.2}{1}\times 100$

Decrease $=20\%$

$15)$ A number is increased by $10\%$ and then decreased by $10\%$. What will the number's net change be?
Solution:

Let the number $=100$

$10\%$ of $100$ $=\frac {10}{100}\times 100$ $=10$

$10\%$ increase in the number $=$ number $+10\%$ of a number

$=100+10$ $=110$

$10\%$ of $110$ $=\frac {10}{100}\times 11$

$10\%$ decrease in $110$ $=110-11$ $=99$

Number is decreased by $=100-99$ $=1$

Decrease$\%$ $=\frac {decrease}{number}\times 100$

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

Decrease $=1\%$

$16)$ A man spends $75\%$ of his income. His income increased by $20\%$ he increased his expenditure by $15\%$. His saving is increased by what percentage?

Solution:

Income $=100$

Spending $=75\%$ $=75$

Savings $=25$

Increase in salary $=20\%$ $=20$

Net Income $=100+20$ $=120$

Increase in spendings $=15\%$

$15\%$ of $75$ $=\frac {15\times 75}{100}$ $=11.25$

New spendings $=75+11.25$ $=86.25$

New savings $=$ new salary $-$ new spendings

$=120$ $-86.25$

New savings $=33.75$

Increase in saving $=33.75$ $-25$

$=8.75$

Percentage of increase in savings $=\frac {8.75}{25}\times 100$

Increase in savings $=35\%$

$17)$ A candidate who gets $30\%$ of total votes polled is defeated by $15000$. Find the no of votes winning candidate?
Solution:

Defeated candidate's votes $=30\%$

Winning candidate's votes $=70\%$

Candidate who got $30\%$ of votes is defeated by $15000$ votes

Defeated candidate's votes $+15000$ $=$ winning candidate's votes

$30\%$ $+15000$ $=70\%$

$15000$ $=40\%$

$1\%$ $=375$

$70\%$ $=26250$

Winning candidate's votes $=26250$

$18)$ Ali's salary is $25\%$ more than Ahmed's salary. How much Ahmed's salary is less than Ali's?
Solution:

Ahmed's salary $=100$

$25\%$ of Ahmed's salary $=\frac {25}{100}\times 100$ $=25$

Ali's salary =Ahmed's salary $+25$

$=100+25$ $=125$

Difference $=$Ali's salary$-$Ahmed's salary

$=125-100$

Difference $=25$

Difference$\%$ $=\frac{diff}{Ali's \text{ } salary}\times 100$ $=\frac {25}{125}\times$ $=20\%$

Ahmed's salary is $=20\%$ less than Ali's salary

$19)$ A team won $60\%$ of the games so far this season. If the team plays a total of $50$ games all season and wins $80\%$ of the remaining games. What number of games will the team win all season?
Solution:

Percentage $=\frac {part}{Total}\times 100$

$60$ $=\frac {part}{20}\times 100$

Part $=\frac {60\times 20}{100}$

Part $=12$

Remaining games $=50-20$ $=30$

percentage $=\frac {part}{total}\times 100$

$80$ $=\frac {P}{30}\times 100$

P $=\frac {80\times 30}{100}$

P $=24$

Total won by team in season $=12+24$ $=\frac {36}{50}$

$20)$ If A gets $25\%$ more than $B$ and $B$ gets $20\%$ more than $C$. What will be the share of $C$ from the sum $\$740$?
Solution:

Let $C's$ share $=x$

$B's$ share $=20\%$ more than $C's$ share

$B's$ share $=C's$ share share $+20\%$ of $C's$ share

$B's$ share $=x+\frac {20}{100}x$ $=x+\frac {1}{5}x$ $=\frac {5x+x}{5}$

$B's$ share $=\frac {6x}{5}$

$A's$ share $=25\%$ more than $B's$ share

$A's$ share $=B's$ share $+25\%$ of $B's$ share

$A's$ share $=\frac {6x}{5}+\frac {25}{100}\times \frac {6x}{5}$

$A's$ share $=\frac {6x}{5}+\frac {3x}{10}$

$A's$ share $=\frac {12x+3x}{10}$ $=\frac {15x}{10}$ $=\frac {3x}{2}$

$A's$ share $+B's$ share $+C's$ share $=740$

$\frac {3x}{2}+\frac {6x}{5}+\frac {x}{1}$ $=740$

$\frac {15x+12x+20x}{10}$ $=740$

$\frac {37x}{10}$ $=740$

$x$ $=\frac {740\times 10}{37}$

$x$ $=200$

$C's$ share $=200$

$21)$ Amir got $35\%$ marks and failed by $20$ marks Ali got $45\%$ marks and secure $20$ more marks than the minimum passing marks.
$a)$ What are passing marks?
$b)$ What is the percentage of passing marks?
solution:

let total marks $=x$

Amir's marks $=35\%$ of total marks $=35%$ of $x$

Amir's marks $=\frac {35\times x}{100}$ $=\frac {7x}{20}$

Amir got $20$ marks less than passing marks

so,

Passing marks $=\frac {7x}{20}+20$

Ali's marks $=45\%$ of total marks

Ali's marks $=\frac {45}{100}\times x$

Ali's marks $=\frac {9x}{20}$

Ali's marks $=$passing marks $+20$

$=\left( \frac {7x}{20}+20 \right)$

$\frac {9x}{20}$ $=\frac {7x}{20}$ $+20$ $+20$

$\frac {9x}{20}$ $-\frac {7x}{20}$ $=40$

$\frac {9x-7x}{20}$ $=40$

$2x$ $=40\times 20$

$x$ $=400$

Total marks $=400$

$a)$ What are passing marks?

Passing marks $=\frac {7x}{20}$ $+20$

Passing marks $=\frac {7\times 400}{20}$ $+20$

Passing marks $=140$ $+20$

Passing marks $=160$

$b)$ What is the percentage of passing marks?

Passing marks$\%$ $=\frac {passing\text{ } marks}{total\text{ }marks}\times 100$

Passing marks $=40\%$

$22)$ A coat is on sale for $\$120$ after a discount of $20\%$. What was the coat's original cost?
Solution:

let

Original Price $=100\%$

discount $=$Original price $-$ sale price

$100\%$ $-20\%$ $=120$

$80\%$ $=120$

$1\%$ $=\frac {120}{80}$

$100\%$ $=\frac {3}{2}\times 100$

$100\%$ $=150$

Original Price $=150$

$23)$ Ahmed earns a commission of $6\%$ of his total sales. How much must he sell to yield a commission of $\$135$?

Solution:

$6\%$ $=135$

$100\%$ $=\frac {135}{6}\times 100$

$100\%$ $=2250$

so,

Ahmed will get $\$135$ commission when he sells $\$2250$

$24)$ A man sells an article at a profit of $20\%$ if he had bought it at $20\%$ less and sold it for $\$5$ less he would have gained $25\%$. Determine the article's cost.
Solution:

Cost Price $=x$

Profit $=$Selling Price $-$Cost Price

$20\%$ of C.P $=$S.P $-x$

$\frac {20x}{100}$ $=$S.P $-x$

S.P $=x+$ $\frac {2x}{10}$ $=\frac {10x+2x}{10}$

S.P $=\frac {12x}{10}$

Profit $=$Selling Price $-$Cost Price

Profit $=x$ $-20\%$ of $x$

$25\%$ of C.P $=$S.P $-5$-$\left( x-\frac {20x}{100} \right)$

$\frac {25}{100}\times \frac {80x}{100}$ $=\frac {12x}{10}$ $-5$ $-\frac {100x-20}{100}$

$\frac {1}{5}x$ $=\frac {6x}{5}$ $-5$ $-\frac {80x}{100}$

$5$ $=\frac {6x}{5}$ $-\frac {4x}{5}-\frac {x}{5}$

$5$ $=\frac {6x-4x-x}{5}$

$5\times 5$ $=x$

$x$ $=5$

The cost price of the article $=25$

$25)$ In a certain store the profit is $320\%$ of cost. If the cost is increased by $25\%$ but the selling price remains constant what percentage of the selling price is the profit?
Solution:

Cost price $=100$

Profit $=320\%$ of $100$ $=320$

Selling price $=$C.P $+$profit

 $=100+320$

S.P $=420$

Increase in C.P $=25\%$ $=100+25$ $=125$

Selling Price $=420$

Profit $=$S.P $-$C.P

 $=420-125$

Profit $=295$

Profit$\%$ $=\frac {profit}{S.P}\times 100$

 $=\frac {295}{420}\times 100$

Profit$\%$ $=70%$

$26)$ Two numbers are $20\%$ and $50\%$ more than a third number.
What is the ratio of two numbers?
Solution:

let

$3rd$ no. $=x$

$1st$ no. $20\%$ more than $3rd$ no.

$1st$ $=x+20\%$ of $x$

 $=x$ $+\frac {20}{100}x$

$=x$ $+\frac {x}{5}$

$=\frac {5x+x}{5}$

$=\frac {6x}{5}$

$1st$ no. $=\frac {6x}{5}$

$2nd$ no. $=50\%$ more than $3rd$ no.

 $=x$ $+50\%$ of $x$

 $=x$ $+\frac {50}{100}\times x$

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

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

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

$2nd$ no. $=\frac {3x}{2}$

The ratio of $1st$ and $2nd$ no.

$\frac {6x}{5}$ $:\frac {3x}{2}$

$12:15$

$4:5$

$27)$ What will be the amount Ali has to pay after $3$ years if he secures a loan of $\$1000$ at $8\%$ simple interest?
Solution:

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

$P$ $=Principal$

$R$ $=Interest$ $Rate$

$T$ $=Time$

$P$ $=1000$; $R =8$; $T =3$

$S.I$ $=\frac {1000\times 8\times 3}{100}$ $=$240$

Amount $=\$1000+\$240$ $=\$1240$

$28)$ A car was sold at a loss of $10\%$. If it was sold for $\$70$ more there would have been a gain of $40\%$.
The cost price of the car was?
Solution

$10\%$ + $40\%$ $=70$

$50\%$ $=70$

$2\times 50\%$ $=2\times 70$

$100\%$ $=140$

$29)$ A single discount equivalent to a discount series of $20\%$, $10%$, and $25\%$ is
Solution:

A single discount equal to a series of discounts $r_1$, $r_2$,... is

$r$ $=\left[ 1-(1-r_1)(1-r_2)... \right]$

$r$ $=\left[ 1-(1-20\%)(1-10\%)(1-25\%) \right]$

$r$ $=\left[ 1-\left(1-\frac {20}{100}\right)\left(1-\frac {10}{100} \right)\left(1-\frac {25}{100} \right) \right]$

$r$ $=\left[ 1-\left(1-\frac {1}{5}\right)\left(1-\frac {1}{10}\right)\left(1-\frac {1}{4}\right) \right]$

$r$ $=\left[ 1-\left( \frac {5-1}{5}\right)\left( \frac{10-1}{10} \right)\left( \frac {4-1}{4}\right) \right]$

$r$ $=\left[ 1-\times\frac {4}{5}\times \frac {9}{10}\times\frac {3}{4} \right]$

$r$ $=\left[ 1-\frac {27}{50} \right]$

$r$ $=\frac {50-27}{50}$

$r$ $=\frac {23}{50}$

$r$ $=\frac {23\times 2}{50\times 2}$

$r$ $=\frac {46}{100}$

$r$ $=46\%$

$30)$ Ali's salary was decreased by $20\%$ and later increased by $20\%$. How much percent does he lose?
Solution:

Let Ali's original salary $=\$100$

New salary is $=120\%$ of $(80\%$ of $100)$

$=\frac {120\times 80\times 100}{100\times 100}$

New salary $=\$96$

So,

Ali's salary was decreased by $=4\%$

$31)$ In an examination $52\%$ of the candidates failed in English, $42\%$ failed in mathematics, $17\%$ failed in both. Calculate the proportion of students who passed both subjects.

Solution:

              E           M        Both

Fail      $52\%$    $42\%$   $17\%$

Pass   $48\%$    $58\%$    $83\%$

Pass$\%$ in both subjects $=48\%+58\%$ $-83\%$ $=23\%$

$32)$ Ali spends one-fifth of his salary on his children's education $20\%$ for household expenditures, $10\%$ for other miscellaneous expenses, and saves the rest. How much of his income does he save?
Solution:

$1/5$ $=20\%$

$%$ he saves

$=100\%$ $-(20+20+10)\%$

$=50\%$

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