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Simple and Compound Interest questions for IPMAT, with solutions

10 free Simple and Compound Interest 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 Simple and Compound Interest on Learn.

1. Question 1Type the answerWritten by IPMATwayGrowth and Depreciation

   The population of a village is $8000$. It increases by $10\%$ in the first year and decreases by $10\%$ in the second year. What is the population after $2$ years?

   Show the answer and solution

   Answer: 7920

   After a $10\%$ rise the population is $8000\times1.1=8800$. The $10\%$ fall acts on $8800$: $880$. So the population is $8800-880=7920$. Overall it is $1.1\times0.9=0.99$ of the start, a net fall of $1\%$.

2. Question 2Type the answerWritten by IPMATwayCompound Interest

   Find the amount (in ₹) on ₹$16000$ at $20\%$ per annum for $1\frac12$ years, compounded half-yearly.

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   Answer: 21296

   Half-yearly compounding at $20\%$ p.a. means $10\%$ per half-year. $1\frac12$ years is 3 half-years. The amount is $16000\times1.1^3=16000\times1.331$. So it is ₹$21296$.

3. Question 3Type the answerWritten by IPMATwaySimple Interest

   A sum lent at $12\%$ per annum simple interest for $3$ years earns ₹$1800$ more than when it is lent at $10\%$ per annum for the same period. Find the sum (in ₹).

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   Answer: 30000

   The extra money comes only from the extra rate: $12\%-10\%=2\%$ per year. Over 3 years that is $2\times3=6\%$ of the sum. So $\frac{6}{100}P=1800$. Hence $P=30000$.

4. Question 4Multiple choiceWritten by IPMATwaySimple Interest

   The simple interest on ₹$4000$ is ₹$1000$. If the number of years equals the rate of interest per annum, the rate is

   1. A.$25\%$
   2. B.$12.5\%$
   3. C.$5\%$
   4. D.$10\%$
   Show the answer and solution

   Answer: C. $5\%$

   Put $T=R$ in $\text{SI}=\frac{PRT}{100}$. Then $\frac{4000\times R\times R}{100}=1000$, i.e. $40R^2=1000$. So $R^2=25$ and $R=5$. The rate is $5\%$ (and the time is 5 years).

5. Question 5Multiple choiceAsked in IPMAT Indore 2023Compound Interest

   If the difference between compound interest and simple interest for a certain amount of money invested for 3 years at an annual interest rate of $10\%$ is INR 527, then the amount invested in INR is

   1. A.$170000$
   2. B.$15000$
   3. C.$150000$
   4. D.$17000$
   Show the answer and solution

   Answer: D. $17000$

   For 3 years the CI on $P$ at $10\%$ is $P(1.1^3-1)=0.331P$. The SI is $3\times0.1P=0.3P$. The difference is $0.331P-0.3P=0.031P$. So $0.031P=527$. Hence $P=\frac{527}{0.031}=17000$.

6. Question 6Type the answerWritten by IPMATwayCompound Interest

   Equal sums are lent for $3$ years, one at $30\%$ per annum simple interest and the other at $30\%$ per annum compound interest (compounded annually). By what percentage is the compound interest greater than the simple interest?

   Show the answer and solution

   Answer: 33

   Take $P=100$. SI for 3 years at $30\%$ is $3\times30=90$. CI is $100\times1.3^3-100=219.7-100=119.7$. The excess is $119.7-90=29.7$. As a percentage of the SI: $\frac{29.7}{90}\times100=33\%$.

7. Question 7Multiple choiceWritten by IPMATwayCompound Interest

   The difference between compound interest and simple interest on a sum is ₹$320$ for $2$ years and ₹$992$ for $3$ years, at the same rate. The sum (in ₹) is

   1. A.₹$320000$
   2. B.₹$99200$
   3. C.₹$32000$
   4. D.₹$3200$
   Show the answer and solution

   Answer: C. ₹$32000$

   For 2 years, CI $-$ SI $=P r^2$ (with $r$ as a fraction). For 3 years, CI $-$ SI $=P r^2(3+r)$. Dividing: $\frac{992}{320}=3.1=3+r$, so $r=0.1$, i.e. $10\%$. Then $P\times0.01=320$ gives $P=32000$.

8. Question 8Multiple choiceWritten by IPMATwaySimple Interest

   Likhit earns $x\%$ on the first ₹$2000$ of his savings and $y\%$ on the rest. He earns ₹$700$ on savings of ₹$4000$ and ₹$900$ on savings of ₹$5000$. Find $x$.

   1. A.$15$
   2. B.$17.5$
   3. C.$18$
   4. D.$20$
   Show the answer and solution

   Answer: A. $15$

   Going from 4000 to 5000 adds 1000 at $y\%$ and adds $200$ of income, so $y=20$. With 4000: $\frac{2000x}{100}+\frac{2000\times20}{100}=700$, i.e. $20x=300$ and $x=15$.

9. Question 9Multiple choiceWritten by IPMATwayGrowth and Depreciation

   The population of a town is $20{,}000$. It grows $10\%$ in the first year, $10\%$ in the second year, and then falls $10\%$ in the third year. What is the population after three years?

   1. A.$21{,}780$
   2. B.$20{,}000$
   3. C.$22{,}000$
   4. D.$24{,}200$
   Show the answer and solution

   Answer: A. $21{,}780$

   Each year's change applies to the previous year's population. After two years: $20000\times1.1\times1.1=24200$. The third year loses $10\%$ of $24200$, i.e. $2420$, leaving $21780$ (multiplier $1.089$). Adding the percentages ($+10\%$) gives $22000$, which is wrong because the base changes; stopping at year two gives $24200$; and $20000$ wrongly assumes equal rises and falls cancel.

10. Question 10Multiple choiceWritten by IPMATwayGrowth and Depreciation

    A machine bought for Rs $80{,}000$ loses $25\%$ of its value each year (reducing balance). After $2$ years it is sold for Rs $36{,}000$. By how much is the sale price below its book value at that time?

    1. A.Rs $4{,}000$
    2. B.Rs $9{,}000$
    3. C.Rs $44{,}000$
    4. D.Rs $2{,}250$
    Show the answer and solution

    Answer: B. Rs $9{,}000$

    Reducing-balance depreciation multiplies the value by $0.75$ each year: after two years the book value is $80000\times0.75\times0.75=45000$. The sale price is Rs $36{,}000$, so it is $45000-36000=9000$ below book value. Flat depreciation would give a book value of $40000$ and a shortfall of $4000$, comparing with the cost gives $44000$, and depreciating for $3$ years gives a book value of $33750$, a gap of $2250$.