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Percentages questions for IPMAT, with solutions

10 free Percentages practice questions for IPMAT Indore, Rohtak and JIPMAT, 1 from past papers, each with the answer and a full worked solution. No sign-in. Try each one first, then open the solution. The lessons for this chapter are in Percentages on Learn.

1. Question 1Type the answerWritten by IPMATwayPercentage Applications

   A candidate needs $36\%$ of the maximum marks to pass. He scores $120$ marks and fails by $24$ marks. Find the maximum marks.

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   Answer: 400

   He scores $120$ and fails by $24$, so the pass mark is $120+24=144$. The pass mark is $36\%$ of the maximum: $0.36M=144$. So $M=\frac{144}{0.36}=400$. Check: $36\%$ of $400$ is $144$.

2. Question 2Type the answerWritten by IPMATwayPercentage Basics

   When $25\%$ of a number is subtracted from $210$, the result is the number itself. Find the number.

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   Answer: 168

   Let the number be $x$. Then $25\%$ of it is $\frac{x}{4}$. The condition says $210-\frac{x}{4}=x$. Add $\frac x4$ to both sides: $210=\frac{5x}{4}$. So $x=\frac{210\times4}{5}=168$. Check: $25\%$ of $168$ is $42$, and $210-42=168$.

3. Question 3Multiple choiceWritten by IPMATwayPercentage Basics

   Which of the following is the largest: $15\%$ of $80$, $80\%$ of $15$, $12\%$ of $100$, $25\%$ of $48$?

   1. A.$25\%$ of $48$
   2. B.$80\%$ of $15$
   3. C.all four are equal
   4. D.$12\%$ of $100$
   Show the answer and solution

   Answer: C. all four are equal

   Work out each one. $15\%$ of $80=\frac{15\times80}{100}=12$. $80\%$ of $15=\frac{80\times15}{100}=12$. $12\%$ of $100=12$. $25\%$ of $48=\frac{48}{4}=12$. All four equal $12$, so none is the largest. (In general $a\%$ of $b=b\%$ of $a$.)

4. Question 4Type the answerWritten by IPMATwayPercentage Applications

   In an election between two candidates, all votes were valid. The losing candidate got $35\%$ of the votes and lost by $360$ votes. How many votes were polled?

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   Answer: 1200

   All votes are valid, so the winner got $100-35=65\%$. The margin is $65\%-35\%=30\%$ of the votes polled. So $0.3N=360$ and $N=\frac{360}{0.3}=1200$. Check: $65\%$ of $1200=780$, $35\%=420$, difference $360$.

5. Question 5Type the answerWritten by IPMATwayPercentage Applications

   An alloy of $30$ kg contains $20\%$ copper. How many kg of pure copper must be added to make the copper content $25\%$?

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   Answer: 2

   Copper now: $20\%$ of $30=6$ kg. The other metals: $24$ kg, which do not change. After adding $x$ kg of copper, the other metals must be $75\%$ of the alloy: $24=0.75(30+x)$. So $30+x=32$ and $x=2$. Check: $\frac{8}{32}=25\%$.

6. Question 6Type the answerWritten by IPMATwayPercentage Basics

   A's income is $20\%$ more than B's, and B's income is $25\%$ less than C's. If C earns $32000, how much does A earn (in $)?

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   Answer: 28800

   Start from C, whose income is known: $\text{\textdollar}32000$. B earns $25\%$ less than C, i.e. $75\%$ of it: $0.75\times32000=24000$. A earns $20\%$ more than B, i.e. $120\%$ of B: $1.2\times24000=28800$. So A earns $\text{\textdollar}28800$.

7. Question 7Type the answerWritten by IPMATwayPercentage Change

   The length of a rectangle is increased by $20\%$ and its area increases by $50\%$. By what percentage is the breadth increased?

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   Answer: 25

   Area $=$ length $\times$ breadth. The length factor is $1.2$ and the area factor is $1.5$. So the breadth factor is $\frac{1.5}{1.2}=1.25$. The breadth goes up by $25\%$. Check: $1.2\times1.25=1.5$.

8. Question 8Multiple choiceWritten by IPMATwayPercentage Change

   In a town the literacy rate rose from $64\%$ to $72\%$, while the population grew by $25\%$. By what percentage did the number of literate people increase?

   1. A.$35\%$
   2. B.$33\%$
   3. C.$12\frac12\%$
   4. D.$40\frac{5}{8}\%$
   Show the answer and solution

   Answer: D. $40\frac{5}{8}\%$

   Literate people $=$ literacy rate $\times$ population. Take the old population as $P$. Old literates $=0.64P$. New population $=1.25P$ and new literates $=0.72\times1.25P=0.9P$. Ratio new/old $=\frac{0.9}{0.64}=1.40625$. So the number of literates rose by $40.625\%=40\frac58\%$.

9. Question 9Type the answerWritten by IPMATwayPercentage Change

   A number is increased by $25\%$, and the result is again increased by $25\%$. The final value exceeds the original number by $225$. Find the original number.

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   Answer: 400

   Two successive $25\%$ rises multiply: $1.25\times1.25=\frac{25}{16}$. So the final value is $\frac{25}{16}x$. It exceeds $x$ by $\frac{25}{16}x-x=\frac{9}{16}x$. This is $225$, so $x=225\times\frac{16}{9}=400$. Check: $400\to500\to625$, and $625-400=225$.

10. Question 10Multiple choiceAsked in IPMAT Indore 2024Percentage Applications

    A fruit seller had a certain number of apples, bananas and oranges at the start of the day. The number of bananas was $10$ more than the number of apples, and the total number of bananas and apples was a multiple of $11$. She was able to sell $70\%$ of apples, $60\%$ of bananas, and $50\%$ of oranges during the day. If she was able to sell $55\%$ of the fruits she had at the start of the day, then the minimum number of oranges she had at the start of the day was

    1. A.$220$
    2. B.$190$
    3. C.$210$
    4. D.$180$
    Show the answer and solution

    Answer: C. $210$

    Let apples $=a$, bananas $=a+10$, oranges $=o$. Sold: $0.7a+0.6(a+10)+0.5o=0.55(2a+10+o)$, which simplifies to $o=4a+10$. $70\%$ of $a$ must be whole, so $a$ is a multiple of 10; and $2a+10$ must be a multiple of 11. With $a=10k$: $20k+10\equiv0\pmod{11}\Rightarrow k=5$, so $a=50$ and $o=210$. (Check: sold $35+36+105=176=55\%$ of $320$.)