INTEREST (Simple and Compound) - Q&A

EXERCISE 9(A)

1. Multiple Choice Type:

Choose the correct answer from the options given below.
(i) The interest on ₹483 for 2 years at 5% per annum is:
(a) ₹4,830
(b) ₹48.30
(c) ₹4.83
(d) ₹96.60

Solution:
Given: Principal (P) = ₹483, Time (T) = 2 years, Rate (R) = 5%
Simple Interest (I) = (P × R × T) / 100
I = (483 × 5 × 2) / 100
I = (483 × 10) / 100 = 4830 / 100 = ₹48.30
Answer: (b) ₹48.30

(ii) The simple interest on ₹8,490 at 5% and 73 days is:
(a) ₹849
(b) ₹84.90
(c) ₹8.49
(d) none of these

Solution:
Given: P = ₹8,490, R = 5%, T = 73 days
Convert days to years: T = 73/365 = 1/5 year
I = (P × R × T) / 100
I = (8490 × 5 × 1) / (100 × 5)
I = 8490 / 100 = ₹84.90
Answer: (b) ₹84.90

(iii) A sum of money, put at simple interest doubles itself in 8 years. The same sum will triple itself in:
(a) 16 years
(b) 12 years
(c) 24 years
(d) 18 years

Solution:
Case 1: Sum doubles in 8 years.
Interest = Principal (P).
P = (P × R × 8) / 100 ⇒ R = 100/8 = 12.5%
Case 2: Sum triples.
Interest = 2P.
2P = (P × 12.5 × T) / 100
2 = T / 8 ⇒ T = 16 years.
Answer: (a) 16 years

(iv) ₹5,000 earns ₹500 as simple interest in 2 years. Then the rate of interest is:
(a) 10%
(b) 5%
(c) 20%
(d) 2%

Solution:
R = (I × 100) / (P × T)
R = (500 × 100) / (5000 × 2)
R = 50000 / 10000 = 5%
Answer: (b) 5%

(v) ₹7,000 earns ₹1,400 as interest at 5% per annum. Then the time in this case is:
(a) 5 years
(b) 2 years
(c) 10 years
(d) 4 years

Solution:
T = (I × 100) / (P × R)
T = (1400 × 100) / (7000 × 5)
T = 140000 / 35000 = 4 years.
Answer: (d) 4 years

2. Find the interest and the amount on:
(i) ₹750 in 3 years 4 months at 10% per annum.

Solution:
Time = 3 years 4 months = 3 + 4/12 = 10/3 years.
I = (750 × 10 × 10) / (100 × 3) = 75000 / 300 = ₹250.
Amount = 750 + 250 = ₹1,000.
Answer: Interest = ₹250, Amount = ₹1,000

(ii) ₹5,000 at 8% per year from 23rd December 2011 to 29th July 2012.

Solution:
Days: Dec(8) + Jan(31) + Feb(29, leap) + Mar(31) + Apr(30) + May(31) + Jun(30) + Jul(29) = 219 days.
T = 219/365 = 3/5 year.
I = (5000 × 8 × 3) / (100 × 5) = 120000 / 500 = ₹240.
Amount = 5000 + 240 = ₹5,240.
Answer: Interest = ₹240, Amount = ₹5,240

(iii) ₹2,600 in 2 years 3 months at 1% per month.

Solution:
Rate = 1% per month = 12% per annum.
Time = 2 years 3 months = 2.25 years.
I = (2600 × 12 × 2.25) / 100 = ₹702.
Amount = 2600 + 702 = ₹3,302.
Answer: Interest = ₹702, Amount = ₹3,302

(iv) ₹4,000 in 1¹/₃ years at 2 paise per rupee per month.

Solution:
Time = 4/3 years.
Rate = 2% per month = 24% per annum.
I = (4000 × 24 × 4) / (100 × 3) = ₹1,280.
Amount = 4000 + 1280 = ₹5,280.
Answer: Interest = ₹1,280, Amount = ₹5,280

3. Rohit borrowed ₹24,000 at 7.5 percent per year. How much money will he pay at the end of 4 years to clear his debt?

Solution:
I = (24000 × 7.5 × 4) / 100 = 240 × 30 = ₹7,200.
Total to pay = 24000 + 7200 = ₹31,200.
Answer: ₹31,200

4. On what principal will the simple interest be ₹7,008 in 6 years 3 months at 5% per year?

Solution:
T = 6.25 years = 25/4 years.
P = (I × 100) / (R × T)
P = (7008 × 100 × 4) / (5 × 25) = 2803200 / 125 = ₹22,425.60
Answer: ₹22,425.60

5. Find the principal which will amount to ₹4,000 in 4 years at 6.25% per annum.

Solution:
Let Principal be P.
Amount = P + (P × 6.25 × 4)/100 = P + 0.25P = 1.25P
1.25P = 4000 ⇒ P = 3200.
Answer: ₹3,200

6. (i) At what rate per cent per annum will ₹630 produce an interest of ₹126 in 4 years?

Solution:
R = (126 × 100) / (630 × 4) = 12600 / 2520 = 5%.
Answer: 5%

(ii) At what rate percent per year will a sum double itself in 6¹/₄ years?

Solution:
I = P. T = 25/4 years.
R = (P × 100) / (P × 25/4) = 400/25 = 16%.
Answer: 16%

7. (i) In how many years will ₹950 produce ₹399 as simple interest at 7%?

Solution:
T = (399 × 100) / (950 × 7) = 39900 / 6650 = 6 years.
Answer: 6 years

(ii) Find the time in which ₹1,200 will amount to ₹1,536 at 3.5% per year.

Solution:
I = 1536 - 1200 = 336.
T = (336 × 100) / (1200 × 3.5) = 33600 / 4200 = 8 years.
Answer: 8 years

8. The simple interest on a certain sum of money is ³/₈ of the sum in 6¹/₄ years. Find the rate percent charged.

Solution:
I = 3/8 P. T = 25/4.
R = (3/8 P × 100) / (P × 25/4) = (300/8) × (4/25) = 1200/200 = 6%.
Answer: 6%

9. What sum of money borrowed on 24th May will amount to ₹10,210.20 on 17th October of the same year at 5 percent per annum simple interest?

Solution:
Days = 7(May)+30+31+31+30+17(Oct) = 146 days.
T = 146/365 = 2/5 year.
A = P(1 + RT/100) ⇒ 10210.20 = P(1 + (5×0.4)/100) = 1.02P
P = 10210.20 / 1.02 = 10010.
Answer: ₹10,010

10. In what time will the interest on a certain sum of money at 6% be ⁵/₈ of itself?

Solution:
I = 5/8 P.
T = (5/8 P × 100) / (P × 6) = 500 / 48 = 10 years 5 months.
Answer: 10 years 5 months

11. Ashok lent out ₹7,000 at 6% and ₹9,500 at 5%. Find his total income from the interest in 3 years.

Solution:
I1 = (7000 × 6 × 3)/100 = 1260.
I2 = (9500 × 5 × 3)/100 = 1425.
Total = 1260 + 1425 = 2685.
Answer: ₹2,685

12. Raj borrows ₹8,000; out of which ₹4,500 at 5% and the remaining at 6%. Find the total interest paid by him in 4 years.

Solution:
Remaining sum = 8000 - 4500 = 3500.
I1 = (4500 × 5 × 4)/100 = 900.
I2 = (3500 × 6 × 4)/100 = 840.
Total = 900 + 840 = 1740.
Answer: ₹1,740

EXERCISE 9(B)

1. Multiple Choice Type:

Choose the correct answer from the options given below.
(i) ₹5,000 is put in a bank at 5% simple interest. The amount at the end of 2 years will be:
(a) ₹5,250
(b) ₹5,500
(c) ₹5,500
(d) ₹4,500

Solution:
I = (5000 × 5 × 2)/100 = 500.
A = 5000 + 500 = 5500.
Answer: (b) ₹5,500

(ii) A sum of money triples itself in 20 years. The rate of interest is:
(a) 20%
(b) 10%
(c) 5%
(d) 15%

Solution:
I = 2P. T = 20.
R = (2P × 100)/(P × 20) = 10%.
Answer: (b) 10%

(iii) A sum of money earns simple interest equal to 0.5 times the sum in 10 years; the rate of interest per annum is :
(a) 20%
(b) 10%
(c) 5%
(d) none of these

Solution:
I = 0.5P. T = 10.
R = (0.5P × 100)/(P × 10) = 5%.
Answer: (c) 5%

(iv) A sum of ₹600 put at 5% S.I. amounts to ₹720 in:
(a) 3 years
(b) 4 years
(c) 5 years
(d) 6 years

Solution:
I = 120.
T = (120 × 100)/(600 × 5) = 4.
Answer: (b) 4 years

(v) Manoj invested ₹8,000 for 10 years at 10% p.a. simple interest. The amount at the end of 2 years will be:
(a) ₹9,600
(b) ₹9,800
(c) ₹16,000
(d) None of these

Solution:
T = 2 years (as asked).
I = (8000 × 10 × 2)/100 = 1600.
A = 9600.
Answer: (a) ₹9,600

2. If ₹3,750 amounts to ₹4,620 in 3 years at simple interest. Find:
(i) the rate of interest.
(ii) the amount of ₹7,500 in 5¹/₂ years at the same rate of interest.

Solution:
(i) I = 4620 - 3750 = 870.
R = (870 × 100)/(3750 × 3) = 7.73% (or 116/15 %).
(ii) I = (7500 × 116/15 × 5.5)/100 = 3190.
A = 7500 + 3190 = 10690.
Answer: Rate = 7.73%, Amount = ₹10,690

3. A sum of money, lent out at simple interest, doubles itself in 8 years. Find:
(i) the rate of interest.
(ii) in how many years will the sum become triple (three times) of itself at the same rate percent?

Solution:
(i) I = P. R = 100/8 = 12.5%.
(ii) I = 2P. T = (2P × 100)/(P × 12.5) = 16 years.
Answer: (i) 12.5% (ii) 16 years

4. Rupees 4,000 amounts to ₹5,000 in 8 years; in what time will ₹2,100 amount to ₹2,800 at the same rate?

Solution:
Case 1: I=1000. R = (1000 × 100)/(4000 × 8) = 3.125%.
Case 2: I=700. T = (700 × 100)/(2100 × 3.125) = 10.66 years = 10 years 8 months.
Answer: 10 years 8 months

5. What sum of money lent at 6.5% per annum will produce the same interest in 4 years as ₹7,500 produce in 6 years at 5% per annum?

Solution:
Target Interest = (7500 × 5 × 6)/100 = 2250.
P = (2250 × 100)/(6.5 × 4) ≈ 8653.85.
Answer: ₹8,653.85

6. A certain sum amounts to ₹3,825 in 4 years and to ₹4,050 in 6 years. Find the rate percent and the sum.

Solution:
Diff in 2 years = 225. 1 year interest = 112.5.
Sum = 3825 - (112.5 × 4) = 3375.
Rate = (112.5 × 100)/3375 = 3.33%.
Answer: Sum = ₹3,375, Rate = 3.33%

7. At what rate per cent of simple interest will the interest on ₹3,750 be one-fifth of itself in 4 years? What will it amount to in 15 years?

Solution:
I = 0.2P. R = (0.2P × 100)/(P × 4) = 5%.
For 15 years: I = (3750 × 5 × 15)/100 = 2812.5.
Amount = 6562.5.
Answer: Rate = 5%, Amount = ₹6,562.50

8. On what date will ₹1,950 lent on 5th January, 2011 amount to ₹2,125.50 at 5 per cent per annum simple interest?

Solution:
I = 175.5. T = (175.5 × 100)/(1950 × 5) = 1.8 years.
1.8 years = 1 year + 292 days.
Date: 23rd October 2012 (Leap year calculation involved).
Answer: 23rd October 2012

9. If the interest on ₹2,400 is more than the interest on ₹2,000 by ₹60 in 3 years at the same rate per cent, find the rate.

Solution:
Extra Principal = 400. Interest on it = 60.
R = (60 × 100)/(400 × 3) = 5%.
Answer: 5%

10. Divide ₹15,600 into two parts such that the interest on one at 5 percent for 5 years may be equal to that on the other at 4¹/₂ per cent for 6 years.

Solution:
25x = 27(15600-x) ⇒ 52x = 421200 ⇒ x = 8100.
Answer: ₹8,100 and ₹7,500

11. Simple interest on a certain sum is ¹⁶/₂₅ of the sum. Find the rate of interest and time, if both are numerically equal.

Solution:
16/25 = x²/100 ⇒ x² = 64 ⇒ x = 8.
Answer: Rate = 8%, Time = 8 years

12. Divide ₹9,000 into two parts in such a way that S.I. on one part at 16% p.a. and in 2 years is equal to the S.I. on the other part at 6% p.a. and in 3 years.

Solution:
32x = 18y ⇒ x/y = 9/16.
x = 9000 × 9/25 = 3240.
Answer: ₹3,240 and ₹5,760

EXERCISE 9(C)

1. Multiple Choice Type:

Choose the correct answer from the options given below.
(i) The C.I. on ₹1,000 at 20% per annum in 2 years is:
(a) ₹1,440
(b) ₹1,240
(c) ₹440
(d) ₹220

Solution:
A = 1000(1.2)² = 1440. CI = 440.
Answer: (c) ₹440

(ii) A sum of ₹2,000 is put at 10% compound interest. The amount at the end of 2 years will be:
(a) ₹2,400
(b) ₹2,420
(c) ₹2,420
(d) ₹4,840

Solution:
A = 2000(1.1)² = 2420.
Answer: (c) ₹2,420

(iii) The difference between C.I. and S.I. on ₹6,000 at 8% per annum in both the cases and in one year is :
(a) ₹480
(b) nothing
(c) ₹240
(d) none of these

Solution:
Difference in 1 year is 0.
Answer: (b) nothing

(iv) The difference between the C.I. in 1 year and compound interest in 2 years on ₹4,000 at 5% per annum is :
(a) ₹10
(b) ₹210
(c) ₹200
(d) ₹410

Solution:
CI(1yr) = 200. CI(2yrs) = 410.
Diff = 210.
Answer: (b) ₹210

2. A sum of ₹8,000 is invested for 2 years at 10% per annum compound interest. Calculate:
(i) interest for the first year.
(ii) principal for the second year.
(iii) interest for the second year.
(iv) final amount at the end of the second year.
(v) compound interest earned in 2 years.

Solution:
(i) 800 (ii) 8800 (iii) 880 (iv) 9680 (v) 1680.

3. A man borrowed ₹20,000 for 2 years at 8% per year compound interest. Calculate:
(i) the interest for the first year.
(ii) the interest for the second year.
(iii) the final amount at the end of the second year.
(iv) the compound interest for two years.

Solution:
(i) 1600 (ii) 1728 (iii) 23328 (iv) 3328.

4. Calculate the amount and the compound interest on ₹12,000 in 2 years at 10% per year.

Solution:
A = 12000(1.21) = 14520. CI = 2520.
Answer: Amount = ₹14,520, CI = ₹2,520

5. Calculate the amount and the compound interest on ₹10,000 in 3 years at 8% per annum.

Solution:
A = 10000(1.08)³ = 12597.12. CI = 2597.12.
Answer: Amount = ₹12,597.12, CI = ₹2,597.12

6. Calculate the compound interest on ₹5,000 in 2 years, if the rates of interest for successive years are 10% and 12% respectively.

Solution:
A = 5000(1.1)(1.12) = 6160. CI = 1160.
Answer: ₹1,160

7. Calculate the compound interest on ₹15,000 in 3 years; if the rates of interest for successive years are 6%, 8% and 10% respectively.

Solution:
A = 15000(1.06)(1.08)(1.1) = 18889.20. CI = 3889.20.
Answer: ₹3,889.20

8. Mohan borrowed ₹16,000 for 3 years at 5% per annum compound interest. Calculate the amount that Mohan would have to pay at the end of 3 years.

Solution:
A = 16000(1.05)³ = 18522.
Answer: ₹18,522

9. Rekha borrowed ₹40,000 for 3 years at 10% per annum compound interest. Calculate the interest paid by her for the second year.

Solution:
P2 = 44000. Interest = 4400.
Answer: ₹4,400

10. Calculate the compound interest for the second year on ₹15,000 invested for 5 years at 6% per annum.

Solution:
P2 = 15900. Interest = 954.
Answer: ₹954

11. A man invests ₹9,600 at 10% per annum compound interest for 3 years. Calculate:
(i) the interest for the first year.
(ii) the amount at the end of the first year.
(iii) the interest for the second year.
(iv) the interest for the third year.

Solution:
(i) 960 (ii) 10560 (iii) 1056 (iv) 1161.60.

12. A person invests ₹5,000 for two years at a certain rate of interest compounded annually. At the end of one year, this sum amounts to ₹5,600. Calculate:
(i) the rate of interest per annum.
(ii) the amount at the end of the second year.

Solution:
(i) Rate = (600/5000)*100 = 12%.
(ii) A = 5600(1.12) = 6272.
Answer: (i) 12% (ii) ₹6,272

13. Calculate the difference between the compound interest and the simple interest on ₹7,500 in two years and at 8% per annum.

Solution:
SI = 1200. CI = 1248. Diff = 48.
Answer: ₹48

14. Calculate the difference between the compound interest and the simple interest on ₹8,000 in three years at 10% per annum.

Solution:
SI = 2400. CI = 2648. Diff = 248.
Answer: ₹248

15. Rohit borrowed ₹40,000 for 2 years at 10% per annum C.I. and Manish borrowed the same sum for the same time at 10.5% per annum simple interest. Which of these two gives less interest and by how much?

Solution:
CI = 8400. SI = 8400. Difference is 0.
Answer: They are equal.

16. Mr. Sharma lends ₹24,000 at 13% p.a. simple interest and an equal sum at 12% p.a. compound interest. Find the total interest earned by Mr. Sharma in 2 years.

Solution:
SI = 6240. CI = 6105.6. Total = 12345.60.
Answer: ₹12,345.60

17. Peter borrows ₹12,000 for 2 years at 10% p.a. compound interest. He repays ₹8,000 at the end of first year. Find:
(i) the amount at the end of first year, before making the repayment.
(ii) the amount at the end of first year, after making the repayment.
(iii) the principal for the second year.
(iv) the amount to be paid at the end of the second year to clear the account.

Solution:
(i) 13200 (ii) 5200 (iii) 5200 (iv) 5720.

18. Gautam takes a loan of ₹16,000 for 2 years at 15% p.a. compound interest. He repays ₹9,000 at the end of first year. How much must he pay at the end of second year to clear the debt?

Solution:
A1 = 18400. Balance = 9400. A2 = 9400(1.15) = 10810.
Answer: ₹10,810

19. A certain sum of money, invested for 5 years at 8% p.a. simple interest, earns an interest of ₹12,000. Find:
(i) the sum of money.
(ii) the compound interest earned by this money in two years at 10% p.a. compound interest.

Solution:
(i) 12000 = 0.4P ⇒ P = 30000.
(ii) CI = 30000(1.21) - 30000 = 6300.
Answer: (i) ₹30,000 (ii) ₹6,300

20. Find the amount and the C.I. on ₹12,000 in one year at 10% per annum compounded half-yearly.

Solution:
Rate=5%, n=2. A = 12000(1.05)² = 13230. CI = 1230.
Answer: Amount = ₹13,230, CI = ₹1,230

21. Find the amount and the C.I. on ₹8,000 in 1¹/₂ years at 20% per year compounded half-yearly.

Solution:
Rate=10%, n=3. A = 8000(1.331) = 10648. CI = 2648.
Answer: Amount = ₹10,648, CI = ₹2,648

22. Find the amount and the compound interest on ₹24,000 for 2 years at 10% per annum compounded yearly.

Solution:
A = 24000(1.21) = 29040. CI = 5040.
Answer: Amount = ₹29,040, CI = ₹5,040

Test yourself

1. Multiple Choice Type:
(i) The S.I. on a certain sum in 3 years and at 8% per year is ₹720. The sum is :
(a) ₹6,000
(b) ₹9,000
(c) ₹3,000
(d) ₹4,000

Solution:
720 = 0.24P ⇒ P = 3000.
Answer: (c) ₹3,000

(ii) A sum of money becomes ⁵/₄ of itself in 5 years. The rate of interest is:
(a) 10%
(b) 5%
(c) 8%
(d) 15%

Solution:
I = 0.25P. T = 5. R = 5%.
Answer: (b) 5%

(iii) ³/₅ part of certain sum is lent at S.I. and the remaining is lent at C.I. If the rate of interest in both the cases is 20%. On the whole the total interest in 1 year is ₹1,000 then the sum is:
(a) ₹2,000
(b) ₹4,000
(c) ₹2,500
(d) ₹5,000

Solution:
SI and CI are same for 1 year. Total I = 20% of Sum. 1000 = 0.2S ⇒ S = 5000.
Answer: (d) ₹5,000

(iv) The amount of ₹1,000 invested for 2 years at 5% per annum compounded annually is:
(a) ₹1,100
(b) ₹1,102.50
(c) ₹1,200
(d) ₹8,000

Solution:
A = 1000(1.1025) = 1102.50.
Answer: (b) ₹1,102.50

(v) If the interest is compounded half-yearly, the time is:
(a) halved
(b) doubled
(c) tripled
(d) not changed

Solution:
Number of periods is doubled.
Answer: (b) doubled

2. Mohan lends ₹4,800 to John for 4¹/₂ years and ₹2,500 to Shyam for 6 years and receives a total sum of ₹2,196 as interest. Find the rate per cent per annum, provided it is the same in both the cases.

Solution:
I = [4800×4.5×R]/100 + [2500×6×R]/100 = 216R + 150R = 366R.
366R = 2196 ⇒ R = 6%.
Answer: 6%

3. John lent ₹2,550 to Mohan at 7.5 per cent per annum. If Mohan discharges the debt after 8 months by giving an old television and ₹1,422.50, find the price of the television.

Solution:
I = (2550 × 7.5 × 2/3)/100 = 127.50.
Total Debt = 2677.50.
Price = 2677.50 - 1422.50 = 1255.
Answer: ₹1,255

4. Divide ₹10,800 into two parts so that if one part is put at 18% per annum S.I. and the other part is put at 20% p.a. S.I. the total annual interest is ₹2,060.

Solution:
0.18x + 0.20(10800-x) = 2060 ⇒ -0.02x = -100 ⇒ x = 5000.
Answer: ₹5,000 and ₹5,800

5. Find the amount and the compound interest on ₹16,000 for 3 years at 5% per annum compounded annually.

Solution:
A = 16000(1.05)³ = 18522. CI = 2522.
Answer: Amount = ₹18,522, CI = ₹2,522

6. Find the amount and the compound interest on ₹20,000 for 1¹/₂ years at 10% per annum compounded half-yearly.

Solution:
R=5%, n=3. A = 20000(1.157625) = 23152.50.
Answer: Amount = ₹23,152.50, CI = ₹3,152.50

7. Find the amount and the compound interest on ₹32,000 for 1 year at 20% per annum compounded half-yearly.

Solution:
R=10%, n=2. A = 32000(1.21) = 38720.
Answer: Amount = ₹38,720, CI = ₹6,720

8. Find the amount and the compound interest on ₹4,000 in 2 years, if the rate of interest for first year is 10% and for the second year is 15%.

Solution:
A = 4000(1.1)(1.15) = 5060.
Answer: Amount = ₹5,060, CI = ₹1,060

9. Find the amount and the compound interest on ₹10,000 in 3 years, if the rates of interest for the successive years are 10%, 15% and 20% respectively.

Solution:
A = 10000(1.1)(1.15)(1.2) = 15180.
Answer: Amount = ₹15,180, CI = ₹5,180

10. A sum of money lent at simple interest amounts to ₹3,224 in 2 years and ₹4,160 in 5 years. Find the sum and the rate of interest.

Solution:
3yr diff = 936. 1yr = 312. P = 3224 - 624 = 2600. R = (312/2600)*100 = 12%.
Answer: Sum = ₹2,600, Rate = 12%

11. At what rate percent per annum compound interest will ₹5,000 amount to ₹5,832 in 2 years?

Solution:
5832/5000 = 1.1664 = (1.08)². Rate = 8%.
Answer: 8%

12. ₹16,000 invested at 10% p.a. compounded semi-annually amounts to ₹18,522. Find the time period of investment.

Solution:
R=5%. 1.05^n = 18522/16000 = 1.157625. n=3 half-years.
Answer: 1¹/₂ years