Simple Interest

Principal, Rate, Time, Interest and Amount

When you borrow money, you pay back more than you took. The extra money is the interest. Five words come up in every question.

• Principal (P): the money lent or borrowed, also called the sum.
• Rate (R): the interest on every ₹100 for one year, written as R% per annum (p.a.).
• Time (T): how long the money is kept, in years unless the question says otherwise.
• Interest (SI): the extra money earned or paid.
• Amount (A): principal + interest, the total paid back at the end.

Under simple interest, the interest is always worked out on the original principal, so it is the same every year. ₹100 at 8% earns ₹8 in the first year, ₹8 in the second, and ₹40 in 5 years.

$$\text{SI} = \frac{P \times R \times T}{100} \qquad A = P + \text{SI} = P \times \frac{100 + RT}{100}$$

Finding the interest. ₹7,500 at 9% for 4 years: SI = $\frac{7500 \times 9 \times 4}{100}$ = ₹2,700, and the amount is ₹10,200.

Finding P, R or T. Turn the formula around.

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

• ₹1,872 is earned in 3 years at 8%: P = $\frac{100 \times 1872}{8 \times 3}$ = ₹7,800.
• ₹6,400 earns ₹2,240 in 5 years: R = $\frac{100 \times 2240}{6400 \times 5}$ = 7%.
• ₹9,600 at 7.5% earns ₹2,520: T = $\frac{100 \times 2520}{9600 \times 7.5}$ = 3.5 years.

Think in R × T. In T years at R%, the interest is (R × T)% of the principal. 9% for 4 years is 36% of P, and 36% of ₹7,500 is ₹2,700. Most questions in this chapter get quicker once you see this one number.

Principal from the amount. "A sum amounts to ₹11,310 in 4 years at 7.5%." R × T = 30, so the amount is $\frac{13}{10}$ of P, and P = 11,310 × $\frac{10}{13}$ = ₹8,700. Check: 30% of 8,700 is 2,610, and 8,700 + 2,610 = 11,310.

[SSC Tip] Taking 30% off ₹11,310 gives ₹7,917, and that wrong figure is usually an option. The 30% was worked out on the principal, not on the amount, so divide by 1.3 instead of subtracting. It is the same mistake as finding the CP from the SP (Chapter 7).

Time in Months and Days, and Rates per Month

The rate is per year, so the time must be in years too.

Months. Divide by 12. ₹14,400 at 7.5% for 8 months: SI = $14400 \times \frac{7.5}{100} \times \frac{8}{12}$ = ₹720. Read "3 years 6 months" as 3.5 years, not 3.6.

Days. Divide by 365, even in a leap year, unless the question says otherwise. ₹36,500 at 8% for 73 days: SI = $36500 \times \frac{8}{100} \times \frac{73}{365}$ = ₹584.

Some day counts turn into neat fractions of a year.

Counting the days between two dates. Count one end only: leave out the day the money goes in and count the day it comes out. "₹18,250 is deposited on 12 March and withdrawn on 24 May of the same year, at 6%."

• March: 31 − 12 = 19 days
• April: 30 days
• May: 24 days
• Total = 73 days = $\frac{1}{5}$ year, so SI = 18,250 × 6% × $\frac{1}{5}$ = ₹219.

Keep the month lengths ready: 31 days for January, March, May, July, August, October and December; 30 for April, June, September and November; 28 for February (29 in a leap year).

Rate per month. Multiply by 12 to get the yearly rate. 1.25% per month is 15% per annum. ₹12,000 borrowed at 1.25% per month for 2.5 years: SI = 12,000 × 15% × 2.5 = ₹4,500. You can also stay in months: 1.25% × 30 months = 37.5%, and 37.5% of ₹12,000 is ₹4,500.

Paise per rupee. "2 paise per rupee per month" means ₹2 on every ₹100 each month. That is 2% per month, or 24% per annum. In the same way, "₹1.50 per ₹100 per month" is 18% per annum.

Rate per half-year. Under simple interest, 6% per half-year is simply 12% per annum, because nothing is added to the principal along the way.

Most wrong options in these questions come from a monthly rate used with a time in years, or from "2 years 4 months" read as 2.4 years.

Amount After Two Different Times, and Amounts at Two Rates

Two times, same sum. "A sum at simple interest amounts to ₹10,148 in 3 years and to ₹11,180 in 5 years. Find the sum and the rate."

The principal sits inside both amounts, so the difference between them is just the interest for the extra 2 years.

• Interest for 2 years = 11,180 − 10,148 = ₹1,032, so one year's interest = ₹516.
• Interest for 3 years = 3 × 516 = ₹1,548.
• Principal = 10,148 − 1,548 = ₹8,600.
• Rate = $\frac{516}{8600} \times 100$ = 6%.

As one formula, if the amounts are $A_1$ after $t_1$ years and $A_2$ after $t_2$ years,

$$P = \frac{A_1 t_2 - A_2 t_1}{t_2 - t_1}$$

Here: $\frac{10148 \times 5 - 11180 \times 3}{2} = \frac{50740 - 33540}{2} = 8600$.

The gap does not have to be whole years. If the amounts are given after 2 years and after 3.5 years, the difference is 1.5 years' interest.

Two rates, same time. "A sum amounts to ₹8,000 in 4 years at 7% and to ₹8,500 in 4 years at 9%." The extra 2% for 4 years is 8% of the principal, and that is the ₹500 gap. So P = 500 × $\frac{100}{8}$ = ₹6,250. Check: 6,250 × 1.28 = 8,000 and 6,250 × 1.36 = 8,500.

Rate raised or lowered. The same idea works without amounts. "If the rate were 2% higher, a sum would earn ₹432 more in 3 years." The extra is 2% × 3 = 6% of P, so P = 432 × $\frac{100}{6}$ = ₹7,200.

"When the rate falls from 9% to 7.5%, a man's yearly income from a deposit drops by ₹270." The 1.5% drop is worth ₹270, so the deposit is 270 × $\frac{100}{1.5}$ = ₹18,000.

[SSC Tip] In the two-times question, dividing the yearly interest of ₹516 by one of the amounts (₹10,148 or ₹11,180) gives a rate that looks fine but is wrong, and forgetting to halve ₹1,032 gives 12%. The rate is always on the principal, so find the principal first.

A Sum Becoming n Times Itself

If a sum becomes n times itself, the amount is nP and the interest is (n − 1)P. So the interest is (n − 1) × 100% of the principal:

$$R \times T = 100(n - 1)$$

• Doubles (n = 2): R × T = 100. A sum that doubles in 12.5 years is at 100 ÷ 12.5 = 8%.
• Trebles (n = 3): R × T = 200. At $6\frac{1}{4}\%$, a sum trebles in 200 ÷ 6.25 = 32 years.
• Becomes $\frac{7}{4}$ of itself in 15 years: the interest is $\frac{3}{4}$ of P, which is 75%, so R = 75 ÷ 15 = 5%.

Time for another multiple. The rate stays the same, so the time grows in step with the interest, (n − 1)P.

"A sum doubles in 15 years. In how many years will it become 4 times itself?" Doubling means an interest of 1P in 15 years. Becoming 4 times needs an interest of 3P, which takes 3 × 15 = 45 years.

"A sum becomes 3 times itself in 12 years. When will it become 7 times?" The interest goes from 2P to 6P, so the time goes from 12 to 12 × $\frac{6}{2}$ = 36 years.

$$\frac{T_2}{T_1} = \frac{n_2 - 1}{n_1 - 1}$$

Interest as a fraction or a percentage of the sum. "The SI on a sum is $\frac{7}{20}$ of the sum in 5 years." $\frac{7}{20}$ is 35%, which equals R × T, so R = 35 ÷ 5 = 7%. "In 8 years at 4.5%, the interest is what per cent of the sum?" Just R × T = 36%.

Rate equal to time. "The SI on a sum is $\frac{9}{16}$ of the sum, and the rate per cent equals the number of years." Then R × R = $\frac{9}{16} \times 100$ = 56.25, so R = 7.5% and the time is also 7.5 years.

Doubling in 15 years does not mean 4 times in 30 years under simple interest; that is how compound interest behaves. Under SI, count the interest: 1P, then 2P, then 3P, and the time goes up in the same steps.

Rates That Change Over the Years

Sometimes the rate changes partway through. Under simple interest each period is worked out on the same principal, so the percentages just add up:

$$\text{SI} = \frac{P(r_1 t_1 + r_2 t_2 + r_3 t_3)}{100}$$

Finding the interest. "₹19,000 is lent at 6% for the first 2 years, 8% for the next 3 years and 10% beyond 5 years. Find the interest for 9 years."

• First 2 years: 6 × 2 = 12%
• Next 3 years: 8 × 3 = 24%
• Years 6 to 9 (4 years): 10 × 4 = 40%
• Total = 76% of ₹19,000 = ₹14,440.

"Beyond 5 years" means only the years after the fifth, which is 4 years here, not 9. Using 10% for all 9 years is the usual slip.

The average rate is weighted by time. 76% over 9 years is about 8.44% a year, not the plain average (6 + 8 + 10) ÷ 3 = 8%. The plain average works only when every rate runs for the same number of years.

Finding the principal. "A sum earns ₹5,670 in 6 years: 5% for the first 2 years, 7% for the next 2 years and 9% for the rest." The total is 10 + 14 + 18 = 42%, so P = 5,670 × $\frac{100}{42}$ = ₹13,500.

Finding a missing rate. "₹9,000 is lent at 4% for 3 years and then at a new rate for 2 more years. The total interest is ₹2,340." The first 3 years earn 12% of 9,000 = ₹1,080, so the last 2 years earn ₹1,260, which is 14% of 9,000. The new rate is 14 ÷ 2 = 7%.

Parts of a Sum Lent at Different Rates

"₹24,000 is lent in two parts, one at 6% and the other at 10% simple interest. The total interest for a year is ₹2,016. Find the two parts."

Method 1: put it all at the lower rate first. At 6% the whole ₹24,000 would earn ₹1,440. The actual interest is ₹576 more, and every rupee moved to 10% earns 4 paise more. So the part at 10% is 576 ÷ 0.04 = ₹14,400, and the part at 6% is ₹9,600.

Method 2: an equation. Let x be at 6%. Then 0.06x + 0.10(24,000 − x) = 2,016, so 2,400 − 0.04x = 2,016 and x = 9,600.

Method 3: the average rate (alligation). The whole sum earns $\frac{2016}{24000} \times 100$ = 8.4% on average, and the two rates sit on either side of it:

$$\frac{\text{part at } 6\%}{\text{part at } 10\%} = \frac{10 - 8.4}{8.4 - 6} = \frac{1.6}{2.4} = \frac{2}{3}$$

So the parts are $\frac{2}{5}$ and $\frac{3}{5}$ of ₹24,000, that is ₹9,600 and ₹14,400. This is alligation (Chapter 6) with rates in place of prices. Check: 6% of 9,600 is 576 and 10% of 14,400 is 1,440, and 576 + 1,440 = 2,016.

Interest for more than one year. If the question gives the total interest for 3 years, divide by 3 first to get one year's interest, then use any method above.

Different sums for different times. "Ravi lends ₹4,500 to A for 3 years and ₹6,000 to B for 2 years at the same rate, and gets ₹2,040 as interest in all." Every rupee earns the same each year, so add principal × years: 4,500 × 3 + 6,000 × 2 = 13,500 + 12,000 = 25,500. Then R = $\frac{2040}{25500} \times 100$ = 8%.

Swapped rates. "If the two rates are swapped, what is the new yearly interest?" Each part gets each rate once across the two set-ups, so the two yearly interests add to (6 + 10)% of 24,000 = ₹3,840. After the swap the interest is 3,840 − 2,016 = ₹1,824. Check: 10% of 9,600 + 6% of 14,400 = 960 + 864 = 1,824.

[SSC Tip] The average rate is weighted by the money, so it is not (6 + 10) ÷ 2 = 8% unless the parts are equal. If your average rate does not fall between the two rates, you have forgotten to divide by the number of years.

Splitting a Sum for Equal Interest or Equal Amounts

Equal interest. If the parts earn the same interest, then $P_1 R_1 T_1 = P_2 R_2 T_2$. A part with a bigger R × T needs less money:

$$P_1 : P_2 = \frac{1}{R_1 T_1} : \frac{1}{R_2 T_2}$$

"Divide ₹20,000 into two parts so that the SI on the first at 5% for 6 years equals the SI on the second at 9% for 2 years." R × T is 30 for the first and 18 for the second, so $P_1 : P_2 = \frac{1}{30} : \frac{1}{18}$ = 18 : 30 = 3 : 5. The parts are ₹7,500 and ₹12,500. Check: 30% of 7,500 = 2,250 and 18% of 12,500 = 2,250.

Three parts, equal yearly interest. "₹26,000 is split into three parts lent at 4%, 6% and 8%, so that each part earns the same interest every year." The ratio is $\frac{1}{4} : \frac{1}{6} : \frac{1}{8}$ = 6 : 4 : 3 (multiply through by 24). The parts are ₹12,000, ₹8,000 and ₹6,000, and each earns ₹480 a year.

Equal amounts. Now principal plus interest must match: $P_1(100 + R_1 T_1) = P_2(100 + R_2 T_2)$, so

$$P_1 : P_2 = \frac{1}{100 + R_1 T_1} : \frac{1}{100 + R_2 T_2}$$

"A man divides ₹9,800 between his daughters aged 12 and 15, so that each gets the same amount when she turns 18. The money earns 5% simple interest." The younger one waits 6 years (100 + 30 = 130) and the elder waits 3 years (100 + 15 = 115). So younger : elder = $\frac{1}{130} : \frac{1}{115}$ = 115 : 130 = 23 : 26.

• One part = 9,800 ÷ 49 = ₹200.
• The younger gets 23 × 200 = ₹4,600 and the elder gets 26 × 200 = ₹5,200.
• Check: 4,600 × 1.3 = 5,980 and 5,200 × 1.15 = 5,980.

Which part is bigger? Money that stays out longer (or at a higher rate) grows more, so it must start smaller. If your answer gives the younger child the bigger share, the ratio is upside down.

For equal amounts, don't use the times alone (6 : 3) or the interests alone (30 : 15). The amount includes the principal, so the 100 has to be in both brackets.

Borrowing and Lending, Part Repayments and Comparing Loans

Borrowing at one rate and lending at another. The person keeps the difference in the rates on the same sum:

$$\text{gain per year} = \frac{P \times (r_2 - r_1)}{100}$$

"Meera borrows ₹45,000 at 6% and lends it at 8.5% simple interest. What does she gain in 3 years?" She gains 2.5% a year, so 7.5% in 3 years: 7.5% of 45,000 = ₹3,375.

Going backwards: "A man borrows at 7% and lends at 9.5%, and gains ₹600 a year." 2.5% of the sum is ₹600, so the sum is ₹24,000.

A part repaid midway. At the moment of a payment, work out what is owed (principal plus interest so far), take off the payment, and treat what is left as the new principal. SSC answers use this method.

"Vikram takes a loan of ₹20,000 at 8% simple interest. After 2 years he pays ₹9,200. How much must he pay at the end of 5 years (from the start) to clear the loan?"

• Owed after 2 years: 20,000 × 1.16 = ₹23,200.
• After the payment: 23,200 − 9,200 = ₹14,000. This is the new principal.
• 3 more years at 8% (24%): 14,000 × 1.24 = ₹17,360.

He pays 9,200 + 17,360 = ₹26,560 in all, so the loan cost him ₹6,560 in interest.

Comparing two deposits. When the sum is the same, compare R × T; the bigger one wins.

"₹50,000 can go into Bank A at 7.5% for 4 years or Bank B at 8% for 3.5 years." A gives 30% and B gives 28%, so A gives ₹1,000 more (₹15,000 against ₹14,000). The higher rate loses here, because its time is shorter.

Two loans at once. "Ankita borrows ₹30,000 at 9% for 2 years and ₹20,000 at 12% for 1.5 years." The interest is 18% of 30,000 + 18% of 20,000 = 5,400 + 3,600 = ₹9,000. Both loans cost 18% of their principal, so the total is simply 18% of ₹50,000.

Ratio of interests. For two sums, $\text{SI}_1 : \text{SI}_2 = P_1 R_1 T_1 : P_2 R_2 T_2$. ₹8,000 at 5% for 3 years and ₹6,000 at 8% for 2 years earn interest in the ratio 120,000 : 96,000 = 5 : 4.

Equal Annual Instalments under Simple Interest

A loan is often cleared by paying the same sum at the end of every year. Under simple interest, SSC's answers are worked on this basis: on the last day, what the lender is owed must equal what he has received, with each instalment carrying interest from the day it was paid.

Step by step. "A loan of ₹6,600 is to be cleared in 3 equal annual instalments at 10% simple interest. Find each instalment."

• What is owed at the end of 3 years: 6,600 × 1.3 = ₹8,580.
• The 1st instalment (x) is paid at the end of year 1, so it earns interest for 2 years and is worth 1.2x on the last day.
• The 2nd earns interest for 1 year (1.1x). The 3rd is paid on the last day (x).
• So 1.2x + 1.1x + x = 3.3x = 8,580, and x = ₹2,600.

The formula. For n instalments at R%, the instalments earn interest for n − 1, n − 2 and so on down to 0 years, which adds to $\frac{n(n - 1)}{2}$ years. So

$$\text{Amount due} = x\left(n + \frac{R}{100} \times \frac{n(n - 1)}{2}\right) \qquad x = \frac{100 \times \text{Amount due}}{100n + \frac{R n(n - 1)}{2}}$$

"A debt due in n years". Read this carefully. "What annual instalment will discharge a debt of ₹8,480 due in 4 years at 4%?" Here ₹8,480 is already the amount due at the end, so don't add interest to it. The denominator is 400 + $\frac{4 \times 4 \times 3}{2}$ = 400 + 24 = 424, so x = $\frac{848000}{424}$ = ₹2,000. Check: 2,000 × (1.12 + 1.08 + 1.04 + 1) = 2,000 × 4.24 = 8,480.

If the question instead says "₹8,480 is borrowed now", first find the amount due (8,480 × 1.16), then divide.

[Tier 2] Recurring deposit. A fixed sum is paid in every month, and each payment earns simple interest only for the months it stays in. For ₹1,500 a month for 12 months at 8%, the payments stay in for 12 months, 11 months and so on down to 1 month, which adds to 78 months. Interest = $1500 \times \frac{78}{12} \times \frac{8}{100}$ = ₹780, so the maturity value is 18,000 + 780 = ₹18,780. In general, the interest on x a month for n months is $x \times \frac{n(n + 1)}{2} \times \frac{R}{1200}$.

[SSC Tip] Dividing the amount due by the number of instalments (8,580 ÷ 3 = ₹2,860) ignores the interest the early instalments earn, and you will often see that figure among the choices.

Solved Examples

Question 1

Find the simple interest on ₹12,800 at 6.25% per annum for 3 years 6 months. (a) ₹2,880 (b) ₹2,800 (c) ₹15,600 (d) ₹2,400

Answer. 6 months is half a year, so the time is 3.5 years. SI = $\frac{12800 \times 6.25 \times 3.5}{100}$ = (b) ₹2,800. Reading the time as 3.6 years gives ₹2,880.

Question 2

A sum of money amounts to ₹13,260 in 5 years at 6% per annum simple interest. What is the sum? (a) ₹9,282 (b) ₹3,060 (c) ₹10,500 (d) ₹10,200

Answer. R × T = 6 × 5 = 30, so the amount is 130% of the sum. P = 13,260 × $\frac{100}{130}$ = (d) ₹10,200. Check: 30% of 10,200 is 3,060, and 10,200 + 3,060 = 13,260.

Question 3

At what rate per cent per annum will ₹14,000 give ₹1,260 as simple interest in 9 months? (a) 12% (b) 9% (c) 1% (d) 10%

Answer. 9 months is $\frac{9}{12} = \frac{3}{4}$ of a year. R = $\frac{100 \times 1260}{14000 \times \frac{3}{4}}$ = (a) 12%.

Question 4

Ravi deposited ₹21,900 in a bank on 19 January and withdrew it with interest on 2 April of the same year (not a leap year). The bank paid 10% per annum simple interest. How much did he get back? (a) ₹22,344 (b) ₹438 (c) ₹22,338 (d) ₹22,447.50

Answer. Count the days, leaving out 19 January: January 31 − 19 = 12, February 28, March 31, April 2. Total = 73 days = $\frac{1}{5}$ year. SI = 21,900 × 10% × $\frac{1}{5}$ = ₹438, so he gets 21,900 + 438 = (c) ₹22,338. Counting both 19 January and 2 April gives 74 days and ₹22,344.

Question 5

Meena borrowed ₹16,500 at 1.5 paise per rupee per month simple interest. How much interest does she pay after 2 years 4 months? (a) ₹6,930 (b) ₹577.50 (c) ₹7,128 (d) ₹5,940

Answer. 1.5 paise per rupee is 1.5% per month, which is 18% a year. 2 years 4 months is 28 months, so the interest is 1.5 × 28 = 42% of 16,500 = (a) ₹6,930.

Question 6

A sum of money at simple interest amounts to ₹11,780 in 3 years and to ₹13,300 in 5 years. What is the rate of interest? (a) 6.45% (b) 8% (c) 16% (d) 7.5%

Answer. The 2 extra years earn 13,300 − 11,780 = ₹1,520, so one year's interest is ₹760. Three years' interest is ₹2,280, so the sum is 11,780 − 2,280 = ₹9,500. Rate = $\frac{760}{9500} \times 100$ = (b) 8%.

Question 7

A sum was lent at simple interest for 3 years. Had it been lent at a rate 4% higher, it would have fetched ₹756 more. What is the sum? (a) ₹18,900 (b) ₹25,200 (c) ₹6,300 (d) ₹7,560

Answer. The extra 4% for 3 years is 12% of the sum, and that is ₹756. So the sum = 756 × $\frac{100}{12}$ = (c) ₹6,300. Check: 12% of 6,300 is 756.

Question 8

A sum of money doubles itself in 7 years at simple interest. In how many years will it become five times itself? (a) 35 years (b) 14 years (c) 21 years (d) 28 years

Answer. Doubling means the interest is 1P in 7 years. Becoming 5 times needs an interest of 4P, which takes 4 × 7 = (d) 28 years. 35 years is wrong: it multiplies 7 by 5, the amount, instead of by 4, the interest.

Question 9

The simple interest on a sum of money is $\frac{16}{25}$ of the sum, and the rate per cent is equal to the number of years. What is the rate of interest? (a) 8% (b) 6.4% (c) 64% (d) 16%

Answer. R × T = $\frac{16}{25} \times 100$ = 64, and R = T, so R × R = 64 and R = (a) 8%. The time is also 8 years.

Question 10

₹27,000 is lent at simple interest at 5% per annum for the first 3 years, 7% per annum for the next 4 years and 10% per annum beyond 7 years. What is the total interest at the end of 10 years? (a) ₹19,800 (b) ₹19,710 (c) ₹14,310 (d) ₹27,000

Answer. Add the percentages: 5 × 3 = 15, 7 × 4 = 28, and 10 × 3 = 30 for years 8 to 10. Total = 73% of 27,000 = (b) ₹19,710. Using 10% for all 10 years gives ₹27,000, which misreads 'beyond 7 years'.

Question 11

₹35,000 is divided into two parts. One part is lent at 8% and the other at 12% per annum simple interest. The total interest after 2 years is ₹6,440. How much was lent at 8%? (a) ₹10,500 (b) ₹17,500 (c) ₹24,500 (d) ₹21,000

Answer. Yearly interest = 6,440 ÷ 2 = ₹3,220, an average rate of $\frac{3220}{35000} \times 100$ = 9.2%. Part at 8% : part at 12% = (12 − 9.2) : (9.2 − 8) = 2.8 : 1.2 = 7 : 3. So the part at 8% is $\frac{7}{10}$ × 35,000 = (c) ₹24,500. Check: 24,500 at 8% for 2 years earns 3,920 and 10,500 at 12% for 2 years earns 2,520, a total of 6,440.

Question 12

Divide ₹21,000 into two parts so that the simple interest on the first part at 9% for 4 years equals the simple interest on the second part at 8% for 6 years. What is the first part? (a) ₹12,000 (b) ₹9,000 (c) ₹12,600 (d) ₹10,500

Answer. R × T is 36 for the first part and 48 for the second. Equal interest means $P_1 \times 36 = P_2 \times 48$, so $P_1 : P_2$ = 48 : 36 = 4 : 3. The first part is $\frac{4}{7}$ × 21,000 = (a) ₹12,000. Check: 36% of 12,000 = 4,320 and 48% of 9,000 = 4,320.

Question 13

A man divides ₹29,000 between his two sons, aged 11 and 14 years, so that each gets the same amount when he turns 20. The money earns 6% per annum simple interest. What is the younger son's share? (a) ₹15,400 (b) ₹14,500 (c) ₹11,600 (d) ₹13,600

Answer. The younger son's money stays 9 years (100 + 54 = 154) and the elder's stays 6 years (100 + 36 = 136). Equal amounts means younger : elder = $\frac{1}{154} : \frac{1}{136}$ = 136 : 154 = 68 : 77. One part = 29,000 ÷ 145 = ₹200, so the younger son gets 68 × 200 = (d) ₹13,600. Check: 13,600 × 1.54 = 20,944 and 15,400 × 1.36 = 20,944.

Question 14

Arjun borrowed ₹36,000 at 7% per annum simple interest and lent the whole sum at 9.5% per annum simple interest. What is his gain in 2.5 years? (a) ₹900 (b) ₹2,250 (c) ₹8,550 (d) ₹6,300

Answer. He gains 9.5 − 7 = 2.5% a year on ₹36,000, which is ₹900 a year. In 2.5 years: 900 × 2.5 = (b) ₹2,250. Don't stop at ₹900: that is the gain for one year only.

Question 15

Neha took a loan of ₹25,000 at 9% per annum simple interest. After 3 years she repaid ₹16,750. How much must she pay at the end of 5 years from the start to clear the loan? (a) ₹19,500 (b) ₹15,000 (c) ₹17,700 (d) ₹16,350

Answer. Owed after 3 years: 25,000 × 1.27 = ₹31,750. After paying ₹16,750, ₹15,000 is left, and this becomes the new principal. For the next 2 years at 9% (18%): 15,000 × 1.18 = (c) ₹17,700. Taking the payment off the 5-year amount (25,000 × 1.45 − 16,750) gives ₹19,500, but that keeps charging interest on the full ₹25,000 after part of it has been repaid.

Question 16

A loan of ₹10,600 is to be repaid in 2 equal annual instalments at 12% per annum simple interest. What is each instalment? (a) ₹6,200 (b) ₹6,572 (c) ₹5,936 (d) ₹5,300

Answer. Owed at the end of 2 years: 10,600 × 1.24 = ₹13,144. The first instalment x is paid a year early, so it earns 12% for a year and is worth 1.12x on the last day; the second is worth x. So 2.12x = 13,144 and x = (a) ₹6,200. Check: 6,200 × 1.12 + 6,200 = 6,944 + 6,200 = 13,144.

Question 17

What equal annual instalment will discharge a debt of ₹17,250 due in 3 years at 15% per annum simple interest? (a) ₹5,750 (b) ₹5,000 (c) ₹7,250 (d) ₹4,750

Answer. ₹17,250 is the amount due at the end, so no interest is added to it. The instalments earn interest for 2, 1 and 0 years: x(1.30 + 1.15 + 1) = 3.45x = 17,250, so x = (b) ₹5,000.

Question 18

Kavita has ₹60,000 to invest. Scheme A pays 8% per annum simple interest for 5 years, and Scheme B pays 9.5% per annum simple interest for 4 years. Which scheme gives more interest, and by how much? (a) Scheme B, by ₹1,200 (b) Scheme A, by ₹900 (c) Scheme A, by ₹1,200 (d) Both give the same

Answer. The sum is the same, so compare R × T. A gives 8 × 5 = 40% = ₹24,000 and B gives 9.5 × 4 = 38% = ₹22,800. So the answer is (c) Scheme A, by ₹1,200. The higher rate loses here because its time is a year shorter.

Question 19

A man invested a part of ₹40,000 at 6% and the rest at 9% per annum simple interest, and got ₹2,940 as interest in a year. Had he swapped the two rates, what would his yearly interest have been? (a) ₹2,940 (b) ₹3,000 (c) ₹3,180 (d) ₹3,060

Answer. Each part gets each rate once across the two set-ups, so the two yearly interests add to (6 + 9)% = 15% of 40,000 = ₹6,000. After the swap the interest is 6,000 − 2,940 = (d) ₹3,060. Check: 0.06x + 0.09(40,000 − x) = 2,940 gives x = 22,000 at 6% and 18,000 at 9%; swapped, 22,000 × 9% + 18,000 × 6% = 1,980 + 1,080 = 3,060.

Question 20

At a certain rate of simple interest, ₹7,500 amounts to ₹9,300 in 3 years. If the rate is increased by 2%, what will the same sum amount to in 3 years? (a) ₹9,750 (b) ₹9,450 (c) ₹9,486 (d) ₹9,900

Answer. The extra 2% for 3 years is 6% of ₹7,500 = ₹450, so the new amount is 9,300 + 450 = (a) ₹9,750. Check: the old interest is ₹1,800, which is 8% a year; at 10%, 7,500 × 1.3 = 9,750. ₹9,450 adds the extra 2% for one year only.

Question 21

[Tier 2] ₹25,300 is divided into three parts, which are lent at 10% per annum simple interest for 2, 3 and 4 years. The three amounts are equal. What is the part lent for 4 years? (a) ₹9,100 (b) ₹7,800 (c) ₹8,400 (d) ₹8,100

Answer. Equal amounts means $P_1 \times 120 = P_2 \times 130 = P_3 \times 140$, so $P_1 : P_2 : P_3 = \frac{1}{12} : \frac{1}{13} : \frac{1}{14}$. Multiply by 12 × 13 × 14 = 2,184 to get 182 : 168 : 156, which adds to 506. One part is 25,300 ÷ 506 = ₹50, so the 4-year part is 156 × 50 = (b) ₹7,800. Check: 9,100 × 1.2 = 8,400 × 1.3 = 7,800 × 1.4 = 10,920.