Solved Questions

How to Use These Questions

These 34 questions run through the whole chapter in no fixed order. An instalment question can come straight after one on counting days, so before any working decide two things: is it SI or CI, and does it want the interest, the amount, the rate or the time? After that most answers take two or three lines.

Cover the answer and do each question on paper against a timer. 36 to 45 seconds is Tier 1 pace. If one takes you over a minute, find its fast route in Shortcuts & Speed Methods and do the question again the next day.

Roughly a third are simple interest and a third compound interest. The rest put the two together: CI − SI, SI and CI on the same sum, equal-amount splits and instalments.

The [Tier 2] questions need one more step than usual. Do them even if Tier 1 is your target, since Tier 1 asks easier versions of the same ideas.

Solved Examples

Question 1

Find the simple interest on ₹12,600 at 7% a year for 2 years 6 months.

Answer. One year's interest is 7% of 12,600 = ₹882. 2 years 6 months is 2.5 years, so the SI is 882 × 2.5 = ₹2,205.

Question 2

Find the compound interest on ₹18,500 for 2 years at 6% a year, compounded yearly.

Answer. Two years at 6% make 12 + 0.36 = 12.36%. 12.36% of 18,500 = ₹2,286.60. The SI would be ₹2,220, and the extra ₹66.60 is just 6% of the first year's interest of ₹1,110.

Question 3

On a sum, the SI for 2 years is ₹9,100 and the CI for 2 years at the same rate is ₹9,418.50. Find the rate and the sum.

Answer. One year's SI is ₹4,550. The CI is ₹318.50 more, and that extra is the interest on ₹4,550 for one year. So R = $\frac{318.50}{4550} \times 100$ = 7%, and P = 4,550 ÷ 0.07 = ₹65,000.

Question 4

₹5,750 amounts to ₹7,130 in 4 years at simple interest. Find the rate.

Answer. The interest is 7,130 − 5,750 = ₹1,380, so one year's interest is ₹345. R = $\frac{345}{5750} \times 100$ = 6%.

Question 5

Find the compound interest on ₹42,000 for 1.5 years at 20% a year, compounded half-yearly.

Answer. Half-yearly means 10% a half-year, and 1.5 years is 3 half-years. Three steps of 10% make 33.1% (the 10% table: 1,000 becomes 1,331). So the CI is 33.1% of 42,000 = ₹13,902.

Question 6

The difference between the compound and simple interest on a sum for 2 years at 4% is ₹34. Find the sum.

Answer. The difference is P × $\left(\frac{4}{100}\right)^2$ = $\frac{P}{625}$. So P = 34 × 625 = ₹21,250.

Question 7

In how many years will ₹9,800 earn ₹3,675 as simple interest at 12.5% a year?

Answer. One year's interest is 12.5% of 9,800, which is $\frac{1}{8}$ of it, ₹1,225. Then 3,675 ÷ 1,225 = 3 years.

Question 8

A sum at compound interest amounts to ₹10,000 in 2 years and to ₹12,500 in 3 years. Find the rate and the sum.

Answer. In the 3rd year ₹10,000 grew by ₹2,500, which is 25%. So R = 25%. 25% is $\frac{5}{4}$, so I go back two years: P = 10,000 × $\frac{16}{25}$ = ₹6,400.

Question 9

A loan of ₹25,500 at 4% compound interest is repaid in two equal yearly instalments. Find each instalment.

Answer. 4% is $\frac{26}{25}$. I call the instalment 676 units. Paid after 1 year it clears 650 units of loan, and paid after 2 years it clears 625, so the loan is 1,275 units. 1,275 units = ₹25,500, so 1 unit = ₹20 and the instalment = 676 × 20 = ₹13,520. Check: 25,500 grows to 26,520, less 13,520 leaves 13,000, which grows to exactly 13,520.

Question 10

₹9,125 is lent on 12 March and paid back on 4 August of the same year at 12% simple interest. Find the interest.

Answer. Days: 19 in March (13th to 31st), then 30 + 31 + 30 + 31 for April to July, and 4 in August. That makes 145 days. One year's interest is 12% of 9,125 = ₹1,095, which is ₹3 a day. So the SI is 145 × 3 = ₹435.

Question 11

Find the compound interest on ₹24,000 for 3 years if the rate is 5% in the first year, 8% in the second and 10% in the third.

Answer. I just multiply year by year: 24,000 × 1.05 = 25,200, then × 1.08 = 27,216, then × 1.1 = 29,937.60. So the CI is ₹5,937.60. Adding the rates to 23% gives ₹5,520, which is the SI and too small.

Question 12

₹51,000 is split into two parts. One is lent for 2 years and the other for 3 years, both at 12.5% compound interest, and both parts grow to the same amount. Find the parts.

Answer. The 3-year part gets one extra year of growth, so the 2-year part must be $\frac{9}{8}$ times the 3-year part. Parts = 9 : 8, which is 17 parts of ₹3,000. So it's ₹27,000 for 2 years and ₹24,000 for 3 years. Check: in its first year, ₹24,000 grows to ₹27,000, and from then on both parts grow side by side.

Question 13

A sum at simple interest amounts to ₹13,570 in 3 years and to ₹14,950 in 5 years. Find the sum and the rate.

Answer. The 2 years in between earned ₹1,380, so one year's interest is ₹690. P = 13,570 − 3 × 690 = ₹11,500, and R = $\frac{690}{11500} \times 100$ = 6%.

Question 14

A sum at compound interest becomes 8 times itself in 15 years. In how many years will it become 32 times itself?

Answer. 8 = $2^3$, so the sum doubles every 5 years. 32 = $2^5$, so it takes 5 × 5 = 25 years. Doing 15 × 32 ÷ 8 = 60 years is the trap.

Question 15

The simple interest on a sum for 3 years at 5% is ₹3,600. Find the compound interest on the same sum for 2 years at the same rate.

Answer. 3 years at 5% is 15% of P, so P = ₹24,000. The CI for 2 years is 10.25% of 24,000 = ₹2,460. Another way: one year's SI is ₹1,200, so 2 years' SI is ₹2,400, plus 5% of ₹1,200 = ₹60.

Question 16

Ravi borrows ₹9,200 at 1.5% a month simple interest. How much interest does he owe after 1 year 3 months?

Answer. 1 year 3 months is 15 months, and 15 × 1.5% = 22.5%. 22.5% of 9,200 = ₹2,070.

Question 17

Find the compound interest on ₹30,000 for 2 years 4 months at 10% a year, compounded yearly.

Answer. The 2 whole years go at CI: 30,000 × 1.21 = ₹36,300. The last 4 months are a third of a year, so they earn $\frac{10}{3}\%$ simple interest on ₹36,300, which is ₹1,210. The amount is ₹37,510, so the CI is ₹7,510.

Question 18 [Tier 2]

The difference between CI and SI on a sum at the same rate is ₹260 for 2 years and ₹806 for 3 years. Find the rate and the sum.

Answer. The 3-year difference ÷ the 2-year difference = $3 + \frac{R}{100}$. Here 806 ÷ 260 = 3.1, so R = 10%. At 10% the 2-year difference is 1% of P, so P = ₹26,000.

Question 19

The simple interest on a sum is $\frac{9}{25}$ of the sum, and the rate % equals the number of years. Find the rate.

Answer. The SI is 36% of P, so R × T = 36. Since R = T, $R^2$ = 36 and R = 6% (for 6 years).

Question 20

The compound interest on ₹4,000 for 2 years is ₹1,760. Find the rate.

Answer. The amount is ₹5,760, and $\frac{5760}{4000}$ = 1.44 = $1.2^2$. So R = 20%. Dividing 1,760 by 4,000 and then by 2 gives 22%, which is the SI way and wrong here.

Question 21 [Tier 2]

Meena borrows ₹35,000 at 10% compound interest. She pays ₹14,000 at the end of the 1st year and ₹14,000 at the end of the 2nd year. How much must she pay at the end of the 3rd year to clear the loan?

Answer. I run the loan year by year. Year 1: 35,000 grows to 38,500, less 14,000 leaves 24,500. Year 2: 24,500 grows to 26,950, less 14,000 leaves 12,950. Year 3: 12,950 grows to ₹14,245, and that is the last payment.

Question 22

₹63,000 is lent partly at 9% and partly at 13% simple interest. The total interest after 2 years is ₹13,020. How much was lent at 13%?

Answer. The yearly interest is ₹6,510, so the average rate is $\frac{6510}{63000} \times 100$ = $10\frac{1}{3}\%$. Alligation: 9% part : 13% part = $\left(13 - 10\frac{1}{3}\right) : \left(10\frac{1}{3} - 9\right)$ = $2\frac{2}{3} : 1\frac{1}{3}$ = 2 : 1. So ₹21,000 was lent at 13%. Check: 3,780 + 2,730 = 6,510.

Question 23

Find the compound interest on ₹3,20,000 for 6 months at 10% a year, compounded quarterly.

Answer. Quarterly means 2.5% a quarter, and 6 months is 2 quarters. Two steps of 2.5% make 5 + 0.0625 = 5.0625%. 5.0625% of 3,20,000 = ₹16,200.

Question 24

Find the difference between the compound interest and the simple interest on ₹37,500 for 3 years at 4% a year.

Answer. For 3 years the difference is $\frac{P R^2 (300 + R)}{100^3}$ = $\frac{37500 \times 16 \times 304}{1000000}$ = ₹182.40. Year by year it checks out: the CI is 12.4864% of 37,500 = ₹4,682.40 and the SI is ₹4,500.

Question 25

Arjun borrows ₹40,000 at 7.5% simple interest and lends the whole sum at 9.25% simple interest. What does he gain in 3 years?

Answer. He keeps 9.25 − 7.5 = 1.75% a year. 1.75% of 40,000 = ₹700 a year, so in 3 years he gains ₹2,100.

Question 26

A machine worth ₹2,50,000 loses 12% of its value every year. What is it worth after 2 years?

Answer. Each year it keeps 88% of its value, so after 2 years it's worth 2,50,000 × 0.88 × 0.88 = 2,50,000 × 0.7744 = ₹1,93,600. Taking 24% off in one go gives ₹1,90,000, which is wrong.

Question 27

Neha has ₹1,20,000 to invest for 2 years. Scheme A pays 10% simple interest and scheme B pays 9.5% compound interest. Which scheme pays more, and by how much?

Answer. A gives 20% of 1,20,000 = ₹24,000. B gives 19 + 0.9025 = 19.9025% of 1,20,000 = ₹23,883. So scheme A pays ₹117 more. B's interest on interest (₹1,083) doesn't make up for its lower rate.

Question 28

Karan borrows ₹25,000 at 8% simple interest. At the end of 2 years he pays back ₹12,000. How much does he owe at the end of 4 years?

Answer. After 2 years he owes 25,000 + 4,000 = ₹29,000. Paying ₹12,000 leaves ₹17,000, which now earns 8% for 2 more years: 17,000 + 2,720 = ₹19,720.

Question 29

The compound interest on a sum for 2 years at 12.5% is ₹6,800. Find the sum.

Answer. 12.5% is $\frac{9}{8}$, so in 2 years the sum becomes $\frac{81}{64}$ of itself and the CI is $\frac{17}{64}$ of it. P = 6,800 × $\frac{64}{17}$ = ₹25,600.

Question 30 [Tier 2]

A loan of ₹1,26,100 at 5% compound interest is repaid in three equal yearly instalments. Find each instalment.

Answer. 5% is $\frac{21}{20}$, so I call the instalment 9,261 units ($21^3$). Paid after 1, 2 and 3 years it clears 8,820, 8,400 and 8,000 units, so the loan is 25,220 units. 25,220 units = ₹1,26,100, so 1 unit = ₹5 and the instalment is ₹46,305. Check: 1,26,100 grows to 1,32,405, less 46,305 leaves 86,100; that grows to 90,405, less 46,305 leaves 44,100; and that grows to exactly 46,305.

Question 31

When the rate of simple interest falls from 9% to 7.5%, a man's yearly income from a deposit falls by ₹570. Find the deposit.

Answer. The fall of 1.5% of the deposit is ₹570, so the deposit is 570 × $\frac{100}{1.5}$ = ₹38,000.

Question 32 [Tier 2]

On ₹50,000 for 1 year, which pays more: 12% a year compounded half-yearly, or 12.3% a year compounded yearly? By how much?

Answer. Half-yearly at 12% is 6% twice, which is 12.36% for the year. That beats 12.3% by 0.06%, and 0.06% of 50,000 is ₹30. So 12% compounded half-yearly pays ₹30 more (₹6,180 against ₹6,150).

Question 33 [Tier 2]

Sunil borrows ₹10,500 at 12% simple interest and repays it in 3 equal yearly instalments, the first at the end of the first year. Find each instalment.

Answer. By the end of 3 years the loan has grown to 10,500 × 1.36 = ₹14,280. Each instalment earns interest only for the years left after it is paid: the first for 2 years (24%), the second for 1 year (12%) and the last for none. So x × (1.24 + 1.12 + 1) = 14,280, which gives 3.36x = 14,280 and x = ₹4,250.

Question 34

Find the compound interest on ₹7,000 for 3 years at 10% a year.

Answer. Three steps of 10% make 33.1%, from the 10% table (1,000 becomes 1,331). 33.1% of 7,000 = ₹2,317.