Question 1

Two equal amounts of money are deposited in two banks, each at 15% per annum for 3 1/ 2 years and 5 years. If the difference between their interests is Rs. 144, each sum is?

Accepted Answer

Let each sum be Rs. P. Then, [(P x 15 x 5) / 100] - [(P x 15 x 7) / 100] x 2 = 144 ⇒ 3P/4 -21P/40 = 144 ⇒ 9P/40 = 144 ∴ P = (144 x 40) / 9 =Rs. 640

Question 2

If the rate of interest rises from 6 1/2 to 8% a man's annual income increases by Rs. 4050. Find the capital.

Accepted Answer

Due to the rise in the rate of interest, annual income increases by Rs . (8 - 61/ 2) = Rs , 11/ 2 < sub> , when the capital is Rs . 100 Thus, the required capital = (100 x 2 x 4050) / 3 = Rs. 270000 <>

Question 3

The simple interest on a sum of money will be Rs. 600 after 10 years. If the principal is trebled after 5 years the total interest at the end of 10 years will be?

Accepted Answer

Let the sum be Rs. P, SI = Rs. 600, Time = 10 years ∴ Rate (600 x 100) / (P x 10)% S.I for first 5 years = Rs. (P x 5 x 6000)/(1000 x P) = Rs. 300 S.I for last 5 years = Rs. (3P x 5 x 6000)/(100 x P) = Rs. 900 Hence, total interest at the end of 10 years = 300 + 900 = Rs. 1200.

Question 4

A sum of Rs. 2540 is lent out into two parts. One at 12% and another one at 12 1/ 2 %. If the total annual income is Rs. 311.60, the money lent at 12% is?

Accepted Answer

Let money lent at 12% Rs. P Then, money lent at 121/2% = Rs. (2540 - P ) ∵ (P x 12 x 1)/100 + {(2540 - P) x 25/2 x 1}/100 = 311.60 ⇒ 3P/25 + 2540 - P/8 = 311.60 ⇒ 24P + 25(2540 - P) = 200 x 311.60 ∴ P = 63500 - 62320 = 1180

Question 5

A man lends Rs. 10000 in four parts. If he gets 8% on Rs. 2000, 7 1/ 2% on Rs . 4000 and 8 1/ 2% on Rs. 1400, What per cent must he get for the remainder if the average interest is 8.13%?

Accepted Answer

∵ [ (2000 x 8 x 1) /100] + [ (4000 x 15/2 x 1)/100] + [(1400 x 17/2 x 1) / 100] + [(2600 x R x 1)/100] = (10000 x 8.13 x 1) / 100 ⇒ 160 + 300 + 119 + 26R = 813 ⇒ 26R = 234 ⇒ R = 9%

Question 6

If the simple interest for 5 yr be equal to 40% of the principle, it will be equal to the principle after?

Accepted Answer

If interest is 40% of the principal then time = 5 years. So, when interest would be equal to 100% of the principal time would be = (100/40) x 5 years = 12.5 years = 12 yr 6 months

Question 7

What sum of money lent out at simple interest will amount to ₹ 3400 in 3 yr at 1% per month?

Accepted Answer

A = ₹ 3400 T = 3 yr , R = 1% per month = 12 % per annum Let the principle be ₹ P. SI = (PTR/100) = (P x 3 x 12)/100 = ₹ 36P/100 A = (P + SI) = P + 36P/100 = ₹ 136P/100 ⇒ 136P/100 = 3400 ⇒ P = (3400 x 100)/136 = 2500 ∴ Sum = ₹ 2500

Question 8

Simple interest on ₹ 1680 for 4 yr at 7 1/ 2 per annum is equal to the simple interest on ₹ 1200 at 7% per annum for a certain period of time. Time period of time is?

Accepted Answer

Let the required period of time be T yr. Then, (1680 x 4 x 15) /100 = (1200 x T x 7)/100 T = (1680 x 2 x 12)/(1200 x 7) = 6 yr Hence, the required time period is 6 yr.

Question 9

If a certain sum of money becomes double at simple interest in 12 yr. What would be the rate of interest per annum?

Accepted Answer

Let the amount be P, then amount after 12 yr = 2 P SI = 2P - P = P SI = (P x R x T)/ 100 P = (P x R x 12)/ 100 ⇒ R = 100/12 = 81/3%

Question 10

A sum of ₹ 4000 is lent out in two parts, one at 8% simple interest and other at 10% simple interest. If the annual interest is ₹ 352, the sum lent at 8% is

Accepted Answer

Let the money interest at 8% interest be ₹ P . Then, the money interest at 10% interest = ₹(4000 - P) According to the question, (P x 8 x 1)/100 + [(4000 - P) x 10 x 1]/100 = 352 ⇒ 8P + 40000 - 10P = 35200 ⇒ 40000 - 35200 = 2P ∴ P = 4800/2 = ₹ 2400

Question 11

A man invests ₹ 3000 at the rate of 5% per annum. How much more should he invest at the rate of 8%, so that he can earn a total of 6% per annum?

Accepted Answer

Let the extra money invested = ₹ P According to the question . (3000 x 5 x 1)/100 + (P x 8 x 1)/100 = [(3000 + P) x 6 x 1)]/100 ⇒ 15000 + 8P = 18000 + 6P ⇒ 2P = 18000 - 15000 ∴ P = 3000/2 = ₹ 1500

Question 12

When a bank reduce the rate of interest from 6 1/ 2% per annum to 5 1/ 2% per annum, depositor withdrew Rs. 600 and thereby his interest reduced by ₹ 73. Find the initial deposit.?

Accepted Answer

Initially , let P = ₹ A, R = 13/2% per annum and T = 1yr SI = PRT/100 = (A x 1 x 13/2)/100 = ₹ 13A/200 Now, new deposit = ₹ (A - 600), R = 11/2 % per annum and T = 1yr SI = PTR/100 = [(A - 600) x 1 x 11/2)/100] = ₹ 11(A - 600)/200 By the given condition , 13A/200 - 11(A - 600)/200 = 73 ⇒ 13A - 11(A - 600 ) = 200 x 73 ⇒ 2A = 200 x 73 - 600 x 11 ⇒ 2A = 14600 - 6600 ⇒ 2A = 8000 ∴ A = 4000 Hence, the initial investment is ₹ 4000.

Question 13

A farmer borrowed ₹ 4200 at 10% per annum. At the end of 5 yr he cleared his account by paying ₹ 5000 and a cow. The cost of the cow is?

Accepted Answer

Amount after 5 yr, 4200 + (4200 x 10 x 5)/100 = (4200 + 2100) = Rs 6300 ∴ Cost of the cow = (6300 - 5000) = ₹ 1300

Question 14

The simple interest on a sum of money will be ₹ 600 after 10 yr. It the principle is trebled after 5 yr, what will be the total interest at the end of tenth year?

Accepted Answer

Let the principle and rate of interest be P and R% Then, (P x R x 10)/100 = 600 ⇒ PR = 6000 Total interest at the end of year = 300 + (3P x R x 5)/100 = 300 + (15 x 6000)/100 = ₹ 1200

Question 15

Two equal sums of money were lent at simple interest at 10% per annum for 4 yr and 5 yr, respectively. If the difference in interest for two period was ₹ 220, then each sum is?

Accepted Answer

Let each sum be ₹ P then, (P x 10 x 5)/100 - (P x 10 x 4)/100 = 220 ⇒ P/10 = 220 ∴ P = 2200 Hence, each sum is ₹ 2200.

Question 16

A man borrowed ₹ 12000 for 4 yr at 7 3/ 4 per annum and a year later he again borrowed another ₹ 12000 for 3 yr at the same rate. How much should he pay at the end to settle the loans?

Accepted Answer

In the first case P = Rs. 12000, T = 4 yr and R = 31/4 % per annum SI = PTR/100 = (12000 x 4 x 31/4)/100 = ₹ 3720 A = (P + SI) = (12000 + 3720) = ₹ 15720 In the second case P = ₹ 12000, T = 3 yr and R = 31/4 % per annum SI = PTR/100 = (12000 x 3 x 31/4)/100 = ₹ 2790 A = (P + SI ) = (12000 + 2790) = ₹ 14790 ∴ Total memory to be paid at the end = (15720 + 14790) = ₹ 30510

Question 17

A man took a loan at simple interest at 7% in the first year with an increase of 0.5% in each subsequent year. He pays ₹ 3690 as interest after 5 yr . How much loan did he take?

Accepted Answer

Let the loan taken be ₹ P Then, [(P x 7 x 1)/100] + [(P x 7.5 x 1)/100] + [(P x 8 x 1)/100] + [(P x 8.5 x 1)/100] + [(P x 9 x 1)/100] = ₹ 3690 ⇒ P x (7 + 7.5 + 8 + 8.5 + 9)/100 = 3690 ⇒ (P x 40)/100 = 3690 ∴ P = (3690 x 100)/40 = ₹ 9225 Hence, the loan taken was ₹ 9225.

Question 18

What will be the difference in the compound interest on ₹ 50000 at 12% for one year, when the interest is paid yearly and half- yearly?

Accepted Answer

CI on ₹ 50000 at the rate of 12% for one year, when the interest is paid half-yearly = 50000(1 + 6/100)2 - 50000 = [50000 x (53/50) x (53/50)] - 50000 = ₹ 6180 CI when the interest is paid yearly = 50000(1 + 12/100)1 - 50000 = 50000(28/25 - 1) = 50000(3/25) = ₹ 6000 ∴ Required difference = CI - SI = 6180 - 6000 = ₹ 180

Question 19

Vikram borrowed ₹. 6450 at 5% simple interest repayable in 4 equal instalments. What will be the annual instalment payable by him?

Accepted Answer

Let the annual instalment be Rs. P. Amount of ₹ 100 after 4 yr = 100 + (100 x 5 x 4)/100 = ₹ 120 ∴ Present Value (PV) of ₹ 120 due after 4 yr = ₹ 100 Present Value (PV) of ₹ due after 4 yr = 100P/120 = 5P/6 Similary, PV of ₹ P due after 3 yr = 20P/23 PV of ₹ P due after 2 yr = 10P/11 PV of ₹ P due after 1 yr = 20P/21 ∴ 5P/6 + 20P/23 + 10P/11 + 20P/11 = 6450 ⇒ P = ₹ 1810

Question 20

Subbarao was approached by two neighbour for loan. He had ₹ 2540, a part of which he lent to one person at 12% interest per annum and the other part was lent to the second person at 12.5%. At the end of a year, Subbarao received ₹ 311.60 as interest…

Accepted Answer

Let the amount of money lent at 12% interest be ₹ P. ∴ Amount of money lent at 12.5% interest = ₹ (2540 - P) According to the question; (P x 12 x 1)/100 + [(2540 - P) x 12.5 x 1]/100 = 311.60 ⇒ -0.5P/100 + 31750/100 = 311.60 ⇒ 0.5P = 31750 - 31160 = 590 ∴ P = ₹ 1180

Simple Interest Questions