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Compound Interest [Without Using Formula] - Q&A


EXERCISE 2 (A)

1. Calculate the amount and the compound interest on:
(i) ₹ 3,500 at 10% per annum in 2 years
(ii) ₹ 6,000 in 3 years at 5% per year.

Answer:
(i) Principal (P) for 1st year = ₹ 3,500; Rate (R) = 10%
Interest for 1st year = (3500 × 10 × 1) / 100 = ₹ 350
Amount at end of 1st year = 3500 + 350 = ₹ 3,850
Principal for 2nd year = ₹ 3,850
Interest for 2nd year = (3850 × 10 × 1) / 100 = ₹ 385
Amount at end of 2nd year = 3850 + 385 = ₹ 4,235
Compound Interest = Final Amount - Original Principal = 4235 - 3500 = ₹ 735
Amount = ₹ 4,235

(ii) Principal (P) for 1st year = ₹ 6,000; Rate (R) = 5%
Interest for 1st year = (6000 × 5 × 1) / 100 = ₹ 300
Amount at end of 1st year = 6000 + 300 = ₹ 6,300
Principal for 2nd year = ₹ 6,300
Interest for 2nd year = (6300 × 5 × 1) / 100 = ₹ 315
Amount at end of 2nd year = 6300 + 315 = ₹ 6,615
Principal for 3rd year = ₹ 6,615
Interest for 3rd year = (6615 × 5 × 1) / 100 = ₹ 330.75
Amount at end of 3rd year = 6615 + 330.75 = ₹ 6,945.75
Compound Interest = 6945.75 - 6000 = ₹ 945.75
Amount = ₹ 6,945.75

2. Calculate the amount and the compound interest on:
(i) ₹ 8,000 in 2½ years at 15% per annum.
(ii) ₹ 20,000 in 2¼ years at 10% per annum.

Answer:
(i) For 1st year: P = ₹ 8,000, R = 15%
I = (8000 × 15 × 1) / 100 = ₹ 1,200
A = 8000 + 1200 = ₹ 9,200
For 2nd year: P = ₹ 9,200
I = (9200 × 15 × 1) / 100 = ₹ 1,380
A = 9200 + 1380 = ₹ 10,580
For next ½ year: P = ₹ 10,580, T = 1/2 year
I = (10580 × 15 × 1) / (100 × 2) = ₹ 793.50
Final Amount = 10580 + 793.50 = ₹ 11,373.50
Compound Interest = 11373.50 - 8000 = ₹ 3,373.50

(ii) For 1st year: P = ₹ 20,000, R = 10%
I = (20000 × 10 × 1) / 100 = ₹ 2,000
A = 20000 + 2000 = ₹ 22,000
For 2nd year: P = ₹ 22,000
I = (22000 × 10 × 1) / 100 = ₹ 2,200
A = 22000 + 2200 = ₹ 24,200
For next ¼ year: P = ₹ 24,200, T = 1/4 year
I = (24200 × 10 × 1) / (100 × 4) = 242000 / 400 = ₹ 605
Final Amount = 24200 + 605 = ₹ 24,805
Compound Interest = 24805 - 20000 = ₹ 4,805

3. Calculate the amount and the compound interest on:
(i) ₹ 4,600 in 2 years when the rates of interest of successive years are 10% and 12% respectively.
(ii) ₹ 16,000 in 3 years, when the rates of the interest for successive years are 10%, 14% and 15% respectively.

Answer:
(i) For 1st year: P = ₹ 4,600, R = 10%
I = (4600 × 10 × 1) / 100 = ₹ 460
A = 4600 + 460 = ₹ 5,060
For 2nd year: P = ₹ 5,060, R = 12%
I = (5060 × 12 × 1) / 100 = ₹ 607.20
Final Amount = 5060 + 607.20 = ₹ 5,667.20
Compound Interest = 5667.20 - 4600 = ₹ 1,067.20

(ii) For 1st year: P = ₹ 16,000, R = 10%
I = (16000 × 10 × 1) / 100 = ₹ 1,600
A = 16000 + 1600 = ₹ 17,600
For 2nd year: P = ₹ 17,600, R = 14%
I = (17600 × 14 × 1) / 100 = ₹ 2,464
A = 17600 + 2464 = ₹ 20,064
For 3rd year: P = ₹ 20,064, R = 15%
I = (20064 × 15 × 1) / 100 = ₹ 3,009.60
Final Amount = 20064 + 3009.60 = ₹ 23,073.60
Compound Interest = 23073.60 - 16000 = ₹ 7,073.60

4. Find the compound interest, correct to the nearest rupee, on ₹ 2,400 for 2½ years at 5 percent per annum.

Answer:
For 1st year: P = 2400, R = 5%
I = (2400 × 5) / 100 = 120
A = 2400 + 120 = 2520
For 2nd year: P = 2520, R = 5%
I = (2520 × 5) / 100 = 126
A = 2520 + 126 = 2646
For next ½ year: P = 2646, T = 1/2, R = 5%
I = (2646 × 5 × 1) / (100 × 2) = 66.15
Total CI = 120 + 126 + 66.15 = 312.15
Correct to nearest rupee = ₹ 312

5. Calculate the compound interest for the second year on ₹ 8,000/- invested for 3 years at 10% per annum.

Answer:
For 1st year: P = 8000, R = 10%
Interest = (8000 × 10) / 100 = 800
Amount = 8800
For 2nd year: P = 8800, R = 10%
Interest = (8800 × 10) / 100 = ₹ 880

6. A borrowed ₹ 2,500 from B at 12% per annum compound interest. After 2 years, A gave ₹ 2,936 and a watch to B to clear the account. Find the cost of the watch.

Answer:
P = 2500, R = 12%
Yr 1 Interest = (2500 × 12) / 100 = 300
Amount 1 = 2800
Yr 2 Interest = (2800 × 12) / 100 = 336
Amount 2 = 2800 + 336 = ₹ 3,136
Total amount to be paid = 3136
Amount paid = 2936 + Cost of watch
Cost of watch = 3136 - 2936 = ₹ 200

7. How much will ₹ 50,000 amount to in 3 years, compounded yearly, if the rates for the successive years are 6%, 8% and 10% respectively.

Answer:
Yr 1: P = 50000, R = 6%
I = 3000; A = 53000
Yr 2: P = 53000, R = 8%
I = (53000 × 8) / 100 = 4240; A = 53000 + 4240 = 57240
Yr 3: P = 57240, R = 10%
I = 5724; A = 57240 + 5724 = ₹ 62,964

8. Meenal lends ₹ 75,000 at C.I. for 3 years. If the rate of interest for the first two years is 15% per year and for the third year it is 16%, calculate the sum Meenal will get at the end of the third year.

Answer:
Yr 1: P = 75000, R = 15%
I = 11250; A = 86250
Yr 2: P = 86250, R = 15%
I = (86250 × 15) / 100 = 12937.50; A = 86250 + 12937.50 = 99187.50
Yr 3: P = 99187.50, R = 16%
I = (99187.50 × 16) / 100 = 15870; A = 99187.50 + 15870 = ₹ 1,15,057.50

9. Govind borrows ₹ 18,000 at 10% simple interest. He immediately invests the money borrowed at 10% compound interest compounded half-yearly. How much money does Govind gain in one year?

Answer:
Simple Interest paid by Govind = (18000 × 10 × 1) / 100 = ₹ 1800
Compound Interest received (Half-yearly):
1st Half-year: P = 18000, R = 10%, T = 1/2
I = (18000 × 10) / 200 = 900; A = 18900
2nd Half-year: P = 18900, R = 10%, T = 1/2
I = (18900 × 10) / 200 = 945; Total CI = 900 + 945 = 1845
Gain = CI Received - SI Paid = 1845 - 1800 = ₹ 45

10. Find the compound interest on ₹ 4,000 accrued in three years, when the rate of interest is 8% for the first year and 10% per year for the second and the third years.

Answer:
Yr 1: P = 4000, R = 8%
I = 320; A = 4320
Yr 2: P = 4320, R = 10%
I = 432; A = 4752
Yr 3: P = 4752, R = 10%
I = 475.20; A = 5227.20
Total CI = 5227.20 - 4000 = ₹ 1,227.20


EXERCISE 2 (B)

1. Calculate the difference between the simple interest and the compound interest on ₹ 4,000 in 2 years at 8% per annum compounded yearly.

Answer:
Simple Interest = (4000 × 8 × 2) / 100 = ₹ 640
Compound Interest:
Yr 1: I = (4000 × 8) / 100 = 320; A = 4320
Yr 2: I = (4320 × 8) / 100 = 345.60
Total CI = 320 + 345.60 = 665.60
Difference = 665.60 - 640 = ₹ 25.60

2. A man lends ₹ 12,500 at 12% for the first year, at 15% for the second year and at 18% for the third year. If the rates of interest are compounded yearly; find the difference between the C.I. of the first year and the compound interest for the third year.

Answer:
1st Year: P = 12500, R = 12%
CI (Yr 1) = (12500 × 12) / 100 = ₹ 1,500
Amount after Yr 1 = 12500 + 1500 = 14000
2nd Year: P = 14000, R = 15%
CI (Yr 2) = (14000 × 15) / 100 = 2100
Amount after Yr 2 = 14000 + 2100 = 16100
3rd Year: P = 16100, R = 18%
CI (Yr 3) = (16100 × 18) / 100 = ₹ 2,898
Difference = CI (Yr 3) - CI (Yr 1) = 2898 - 1500 = ₹ 1,398

3. A sum of money is lent at 8% per annum compound interest. If the interest for the second year exceeds that for the first year by ₹ 96, find the sum of money.

Answer:
Let Sum = P.
Interest 1st year (I1) = P × 8 / 100 = 0.08P
Interest 2nd year (I2) is calculated on Amount (P + 0.08P) = 1.08P.
I2 = 1.08P × 8 / 100 = 0.0864P
Given I2 - I1 = 96
0.0864P - 0.08P = 96
0.0064P = 96
P = 96 / 0.0064 = ₹ 15,000

4. A man borrows ₹ 6,000 at 5 percent C.I. per annum. If he repays ₹ 1,200 at the end of each year, find the amount of the loan outstanding at the beginning of the third year.

Answer:
Year 1: P = 6000, R = 5%
Interest = 300
Amount = 6300
Repaid = 1200
Balance = 6300 - 1200 = 5100
Year 2: P = 5100, R = 5%
Interest = 255
Amount = 5355
Repaid = 1200
Balance = 5355 - 1200 = 4155
Outstanding at beginning of 3rd year = ₹ 4,155

5. A man borrows ₹ 5,000 at 12 percent compound interest payable every six months. He repays ₹ 1,800 at the end of every six months. Calculate the third payment he has to make at the end of 18 months in order to clear the entire loan.

Answer:
Rate = 12% p.a. = 6% per 6 months.
1st Period (0-6 months): P = 5000
I = 5000 × 6 / 100 = 300; Amount = 5300
Paid = 1800; Balance = 3500
2nd Period (6-12 months): P = 3500
I = 3500 × 6 / 100 = 210; Amount = 3710
Paid = 1800; Balance = 1910
3rd Period (12-18 months): P = 1910
I = 1910 × 6 / 100 = 114.60; Amount = 2024.60
To clear loan, he pays = ₹ 2,024.60

6. On a certain sum of money, the difference between the compound interest for a year, payable half-yearly, and the simple interest for a year is ₹ 180/-. Find the sum lent out, if the rate of interest in both the cases is 10% per annum.

Answer:
Let Sum = P.
Simple Interest (1 yr) = P × 10 × 1 / 100 = 0.10P
Compound Interest (Half-yearly):
1st Half: I = P × 5 / 100 = 0.05P; A = 1.05P
2nd Half: I = 1.05P × 5 / 100 = 0.0525P
Total CI = 0.05P + 0.0525P = 0.1025P
Difference = 0.1025P - 0.10P = 0.0025P
0.0025P = 180
P = 180 / 0.0025 = ₹ 72,000

7. A manufacturer estimates that his machine depreciates by 15% of its value at the beginning of the year. Find the original value (cost) of the machine, if it depreciates by ₹ 5,355 during the second year.

Answer:
Let original value = x.
Depreciation Year 1 = 0.15x
Value beginning Year 2 = x - 0.15x = 0.85x
Depreciation Year 2 = 15% of 0.85x = 0.15 × 0.85x = 0.1275x
Given 0.1275x = 5355
x = 5355 / 0.1275 = ₹ 42,000

8. A man invests ₹ 5,600 at 14% per annum compound interest for 2 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, correct to the nearest rupee.

Answer:
(i) I (Yr 1) = (5600 × 14) / 100 = ₹ 784
(ii) Amount (End Yr 1) = 5600 + 784 = ₹ 6,384
(iii) I (Yr 2) = (6384 × 14) / 100 = 893.76. Nearest rupee = ₹ 894

9. (i) Find the difference between the compound interest of second year and the compound interest of third year on ₹ 48,000 invested for 5 years at 10% per annum compounded yearly.
(ii) A sum of ₹ 50,000 is invested for 8 years at compound interest, the rate of interest being 10%, 12%, 14% and 16% respectively for the first 4 consecutive years. Find the total of interests earned during the first and third years.

Answer:
(i) P = 48000, R = 10%
I (Yr 1) = 4800
P (Yr 2) = 52800
I (Yr 2) = (52800 × 10) / 100 = ₹ 5,280
P (Yr 3) = 52800 + 5280 = 58080
I (Yr 3) = (58080 × 10) / 100 = ₹ 5,808
Difference = 5808 - 5280 = ₹ 528

(ii) P = 50000.
Interest Yr 1 (R=10%) = 50000 × 10% = ₹ 5,000
Amount End Yr 1 = 55000.
Interest Yr 2 (R=12%) = 55000 × 12% = 6600.
Amount End Yr 2 = 55000 + 6600 = 61600.
Interest Yr 3 (R=14%) = 61600 × 14% = ₹ 8,624
Total Interest (1st & 3rd) = 5000 + 8624 = ₹ 13,624

10. A man saves ₹ 3,000 every year and invests it at the end of the year at 10% compound interest. Calculate the total amount of his savings at the end of the third year.

Answer:
End of Year 1: Saves 3000. Interest earned = 0. Total = 3000.
Year 2: Interest on 3000 = 300. Amount = 3300. Saves 3000. Total Principal = 6300.
Year 3: Interest on 6300 = 630. Amount = 6930. Saves 3000. Total Savings = 6930 + 3000 = ₹ 9,930

11. A man borrows ₹ 10,000 at 5% per annum compound interest. He repays 35% of the sum borrowed at the end of the first year and 42% of the sum borrowed at the end of the second year. How much must he pay at the end of the third year in order to clear the debt?

Answer:
P = 10000, R = 5%.
Yr 1: I = 500; A = 10500.
Repay = 35% of 10000 = 3500.
Balance = 10500 - 3500 = 7000.
Yr 2: I = 350; A = 7350.
Repay = 42% of 10000 = 4200.
Balance = 7350 - 4200 = 3150.
Yr 3: I = (3150 × 5) / 100 = 157.50.
Amount to clear debt = 3150 + 157.50 = ₹ 3,307.50

12. Mr. Mehta invested ₹ 8,000 every year at the beginning of the year, at 10% per annum compound interest. Calculate his total savings at the beginning of the third year.

Answer:
Start Year 1: Invests 8000.
End Year 1: I = 800; A = 8800.
Start Year 2: Invests 8000. Total P = 8800 + 8000 = 16800.
End Year 2: I = 1680; A = 18480.
Start Year 3: Invests 8000. Total Savings = 18480 + 8000 = ₹ 26,480

EXERCISE 2 (C)

1. A sum is invested at compound interest compounded yearly. If the interest for two successive years be ₹ 5,700 and ₹ 7,410, calculate the rate of interest.

Answer:
Difference in interest = 7410 - 5700 = 1710
This difference is the simple interest on the first year's interest.
Rate = (1710 × 100) / (5700 × 1) = 30%

2. A certain sum of money is put at compound interest, compounded half-yearly. If the interest for two successive half-years are ₹ 650 and ₹ 760.50; find the rate of interest.

Answer:
Difference = 760.50 - 650 = 110.50
Rate per half-year = (110.50 × 100) / 650 = 17%
Rate per annum = 17% × 2 = 34%

3. A certain sum amounts to ₹ 5,292 in two years and ₹ 5,556.60 in three years, interest being compounded annually. Find:
(i) the rate of interest
(ii) the original sum.

Answer:
(i) Interest for 3rd year = Amount(3 yrs) - Amount(2 yrs) = 5556.60 - 5292 = 264.60
Rate = (264.60 × 100) / 5292 = 5%
(ii) Let Sum = P.
Amount 2 yrs = P(1 + 5/100)(1 + 5/100) = P(1.05)(1.05) = 1.1025P
1.1025P = 5292
P = 5292 / 1.1025 = ₹ 4,800

4. The compound interest, calculated yearly, on a certain sum of money for the second year is ₹ 1,089 and for the third year is ₹ 1,197.90. Calculate the rate of interest and the sum of money.

Answer:
Difference = 1197.90 - 1089 = 108.90
Rate = (108.90 × 100) / 1089 = 10%
CI for 2nd year = 1089.
Let Sum = P.
CI for 1st year = P × 10 / 100 = 0.1P
CI for 2nd year = CI for 1st year + Interest on CI of 1st year
1089 = 0.1P + (0.1P × 10 / 100) = 0.1P + 0.01P = 0.11P
P = 1089 / 0.11 = ₹ 9,900

5. Mohit invests ₹ 8,000 for 3 years at a certain rate of interest, compounded annually. At the end of one year it amounts to ₹ 9,440. Calculate:
(i) the rate of interest per annum.
(ii) the amount at the end of the second year.
(iii) the interest accrued in the third year.

Answer:
(i) Interest 1st yr = 9440 - 8000 = 1440.
Rate = (1440 × 100) / 8000 = 18%
(ii) P for 2nd yr = 9440.
Interest 2nd yr = 9440 × 18 / 100 = 1699.20
Amount 2nd yr = 9440 + 1699.20 = ₹ 11,139.20
(iii) P for 3rd yr = 11139.20
Interest 3rd yr = 11139.20 × 18 / 100 = ₹ 2,005.056 (approx ₹ 2005)

6. Geeta borrowed ₹ 15,000 for 18 months at a certain rate of interest compounded semi-annually. If at the end of six months it amounted to ₹ 15,600; calculate:
(i) the rate of interest per annum.
(ii) the total amount of money that Geeta must pay at the end of 18 months in order to clear the account.

Answer:
(i) Interest for 6 months = 15600 - 15000 = 600.
Rate for 6 months = (600 × 100) / 15000 = 4%
Rate per annum = 8%
(ii) Amount after 6 months = 15600.
Interest next 6 months (2nd period) = 15600 × 4% = 624. Amount = 16224.
Interest next 6 months (3rd period) = 16224 × 4% = 648.96.
Total Amount = 16224 + 648.96 = ₹ 16,872.96

7. Ramesh invests ₹ 12,800 for three years at the rate of 10% per annum compound interest. Find:
(i) the sum due to Ramesh at the end of the first year.
(ii) the interest he earns for the second year.
(iii) the total amount due to him at the end of the third year.

Answer:
(i) I (Yr 1) = 12800 × 10% = 1280. Sum Due = 14080.
(ii) I (Yr 2) = 14080 × 10% = ₹ 1,408.
(iii) Amount (End Yr 2) = 14080 + 1408 = 15488.
I (Yr 3) = 15488 × 10% = 1548.80.
Total Amount = 15488 + 1548.80 = ₹ 17,036.80

8. The compound interest, calculated yearly, on a certain sum of money for the second year is ₹ 864 and for the third year is ₹ 933.12. Calculate the rate of interest and the compound interest on the same sum and at the same rate, for the fourth year.

Answer:
Difference = 933.12 - 864 = 69.12
Rate = (69.12 × 100) / 864 = 8%
CI for 4th year = CI for 3rd year + Interest on it
CI for 4th year = 933.12 + (933.12 × 8 / 100) = 933.12 + 74.65 = ₹ 1,007.77

9. A sum of money placed out at compound interest amounts to ₹ 20,160 in 3 years and to ₹ 24,192 in 4 years. Calculate:
(i) the rate of interest.
(ii) amount in 2 years and
(iii) amount in 5 years.

Answer:
(i) Interest for 4th year = 24192 - 20160 = 4032.
Rate = (4032 × 100) / 20160 = 20%
(ii) Amount 3 yrs = Amount 2 yrs × 1.20
Amount 2 yrs = 20160 / 1.20 = ₹ 16,800
(iii) Amount 5 yrs = Amount 4 yrs × 1.20
Amount 5 yrs = 24192 × 1.20 = ₹ 29,030.40

10. ₹ 8,000 is lent out at 7% compound interest for 2 years. At the end of the first year ₹ 3,560 are returned. Calculate:
(i) the interest paid for the second year.
(ii) the total interest paid in two years
(iii) the total amount of money paid in two years to clear the debt.

Answer:
Yr 1: P = 8000, R = 7%.
Interest 1 = 560. Amount = 8560.
Returned = 3560. Balance = 5000.
(i) Interest 2 = 5000 × 7% = ₹ 350
(ii) Total Interest = 560 + 350 = ₹ 910
(iii) Amount at end of 2nd year = 5000 + 350 = 5350.
Total Paid = 3560 (Yr 1) + 5350 (Yr 2) = ₹ 8,910

11. The cost of a machine depreciated by ₹ 4,000 during the first year and by ₹ 3,600 during the second year. Calculate:
(i) the rate of depreciation.
(ii) the original cost of the machine.
(iii) its cost at the end of the third year.

Answer:
(i) Difference = 4000 - 3600 = 400.
Rate = (400 × 100) / 4000 = 10%
(ii) Dep Yr 1 = 10% of Cost = 4000.
Cost = 4000 × 100 / 10 = ₹ 40,000
(iii) Cost start Yr 2 = 36000. Dep Yr 2 = 3600. Cost start Yr 3 = 32400.
Dep Yr 3 = 32400 × 10% = 3240.
Cost end Yr 3 = 32400 - 3240 = ₹ 29,160

12. The cost of a machine is ₹ 32,000. Its value depreciates at the rate of 5% every year. Find the total depreciation in its value by the end of 2 years.

Answer:
Yr 1 Dep = 32000 × 5% = 1600. Value = 30400.
Yr 2 Dep = 30400 × 5% = 1520.
Total Depreciation = 1600 + 1520 = ₹ 3,120

13. Find the sum, invested at 10% compounded annually, on which the interest for the third year exceeds the interest of the first year by ₹ 252.

Answer:
Let sum = P.
I1 = 0.10P
I2 = (P + 0.10P) × 10% = 0.11P
I3 = (1.10P + 0.11P) × 10% = 1.21P × 10% = 0.121P
Diff = I3 - I1 = 0.121P - 0.10P = 0.021P
0.021P = 252
P = 252 / 0.021 = ₹ 12,000

14. A man borrows ₹ 10,000 at 10% compound interest compounded yearly. At the end of each year, he pays back 30% of the sum borrowed. How much money is left unpaid just after the second year?

Answer:
Sum borrowed = 10000. 30% of sum = 3000.
Yr 1: I = 1000; A = 11000. Pays 3000. Bal = 8000.
Yr 2: I = 800; A = 8800. Pays 3000. Bal = ₹ 5,800

15. A man borrows ₹ 10,000 at 10% compound interest compounded yearly. At the end of each year, he pays back 20% of the amount for that year. How much money is left unpaid just after the second year?

Answer:
Yr 1: I = 1000; A = 11000.
Pays 20% of 11000 = 2200.
Bal = 11000 - 2200 = 8800.
Yr 2: I = 880; A = 9680.
Pays 20% of 9680 = 1936.
Bal = 9680 - 1936 = ₹ 7,744

EXERCISE 2 (D)

1. What sum will amount to ₹ 6,593.40 in 2 years at C.I., if the rates are 10 percent and 11 percent for the two successive years ?

Answer:
Let sum = P.
A1 = P(1.10) = 1.1P
A2 = 1.1P(1.11) = 1.221P
1.221P = 6593.40
P = 6593.40 / 1.221 = ₹ 5,400

2. The value of a machine depreciated by 10% per year during the first two years and 15% per year during the third year. Express the total depreciation of the machine, as percent, during the three years

Answer:
Let Value = 100.
Yr 1 Dep = 10; Val = 90.
Yr 2 Dep = 9; Val = 81.
Yr 3 Dep = 15% of 81 = 12.15; Val = 68.85.
Total Depreciation = 100 - 68.85 = 31.15%

3. Rachna borrows ₹ 12,000 at 10 per cent per annum interest compounded half-yearly. She repays ₹ 4,000 at the end of every six months. Calculate the third payment she has to make at the end of 18 months in order to clear the entire loan.

Answer:
Rate = 10% p.a. = 5% per 6 months.
Period 1: 12000 + 5% = 12600. Pays 4000. Bal = 8600.
Period 2: 8600 + 5% (430) = 9030. Pays 4000. Bal = 5030.
Period 3: 5030 + 5% (251.50) = 5281.50.
Third Payment = ₹ 5,281.50

4. On a certain sum of money, invested at the rate of 10 percent per annum compounded annually, the interest for the first year plus the interest for the third year is ₹ 2,652. Find the sum.

Answer:
Let sum = P.
I1 = 0.10P.
I2 = 0.11P.
I3 = (P + 0.10P + 0.11P) × 10% = 1.21P × 10% = 0.121P.
I1 + I3 = 0.10P + 0.121P = 0.221P.
0.221P = 2652.
P = 2652 / 0.221 = ₹ 12,000

5. During every financial year, the value of a machine depreciates by 12%. Find the original cost of a machine which depreciates by ₹ 2,640 during the second financial year of its purchase.

Answer:
Let Cost = x.
Dep Yr 1 = 0.12x. Val = 0.88x.
Dep Yr 2 = 0.88x × 12% = 0.1056x.
0.1056x = 2640.
x = 2640 / 0.1056 = ₹ 25,000

6. Find the sum on which the difference between the simple interest and the compound interest at the rate of 8% per annum compounded annually be ₹ 64 in 2 years.

Answer:
Diff = SI on Interest of 1st year = I1 × 8% = 64.
I1 = 64 / 0.08 = 800.
I1 = P × 8% = 800.
P = 800 / 0.08 = ₹ 10,000

7. A sum of ₹ 13,500 is invested at 16% per annum compound interest for 5 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, correct to the nearest rupee.

Answer:
(i) I1 = 13500 × 16 / 100 = ₹ 2,160
(ii) A1 = 13500 + 2160 = ₹ 15,660
(iii) I2 = 15660 × 16 / 100 = 2505.60. Nearest rupee = ₹ 2,506

8. Saurabh invests ₹ 48,000 for 7 years at 10% per annum compound interest. Calculate:
(i) the interest for the first year.
(ii) the amount at the end of the second year.
(iii) the interest for the third year.

Answer:
(i) I1 = 48000 × 10% = ₹ 4,800
(ii) A1 = 52800. I2 = 5280. A2 = 52800 + 5280 = ₹ 58,080
(iii) I3 = 58080 × 10% = ₹ 5,808

9. Ashok borrowed ₹ 12,000 at some rate per cent compound interest. After a year, he paid back ₹ 4,000. If compound interest for the second year be ₹ 920, find:
(i) the rate of interest charged
(ii) the amount of debt at the end of the second year.

Answer:
(i) Let rate = r.
A1 = 12000(1+r) - 4000.
I2 = [12000(1+r) - 4000] × r = 920.
This simplifies to a quadratic. Trying r = 10% (0.1):
A1 = 12000(1.1) - 4000 = 13200 - 4000 = 9200.
I2 = 9200 × 0.1 = 920. Matches.
Rate = 10%
(ii) Debt at end of 2nd year = Balance after Yr 1 + I2 = 9200 + 920 = ₹ 10,120

10. On a certain sum of money, lent out at C.I., interests for first, second and third years are ₹ 1,500; ₹ 1,725 and ₹ 2,070 respectively. Find the rate of interest for the (i) second year (ii) third year.

Answer:
(i) Rate for 2nd year = (Difference between I2 and I1) / I1 × 100
Rate = (1725 - 1500) / 1500 × 100 = 225 / 1500 × 100 = 15%
(ii) Rate for 3rd year = (Difference between I3 and I2) / I2 × 100
Rate = (2070 - 1725) / 1725 × 100 = 345 / 1725 × 100 = 20%

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Quick Review Flashcards - Click to flip and test your knowledge!
Question
What term describes the money borrowed from a bank or agency for a specified period?
Answer
Principal
Question
What is the extra money paid to a lender for using their money called?
Answer
Interest
Question
What is the total money paid back to a lender at the end of a specified period called?
Answer
Amount
Question
State the fundamental formula for calculating the total Amount ($A$) using Principal ($P$) and Interest ($I$).
Answer
$A = P + I$
Question
Interest that is calculated on the original principal throughout the entire loan period is known as _____ interest.
Answer
Simple
Question
What is the mathematical formula for Simple Interest ($I$)?
Answer
$I = \frac{P \times R \times T}{100}$
Question
If a financial problem mentions 'interest' without any further description, which type of interest is always implied?
Answer
Simple Interest
Question
In Compound Interest, what happens to the interest due at the end of a fixed period?
Answer
It is added to the principal to form the new principal for the next period.
Question
How is the total Compound Interest ($C.I.$) determined using the final amount and the original principal?
Answer
$C.I. = \text{Final Amount} - \text{Original Principal}$
Question
For the first year of a loan, how does Compound Interest compare to Simple Interest given the same principal and rate?
Answer
They are equal ($C.I. = S.I.$)
Question
From the second year onwards, why is Compound Interest higher than Simple Interest for the same sum and rate?
Answer
Because the interest is calculated on a principal that grows each year.
Question
What is the specific name for the time interval after which interest is added to the principal and the principal changes?
Answer
Conversion period
Question
If interest is compounded half-yearly, what is the length of the conversion period?
Answer
Six months
Question
In Compound Interest calculations, the amount at the end of the first year becomes the _____ for the second year.
Answer
Principal
Question
Calculate the Simple Interest on ₹ $1,000$ at $10\%$ per annum for 3 years.
Answer
₹ $300$
Question
If the Compound Interest for the 1st, 2nd, and 3rd years are ₹ $100$, ₹ $110$, and ₹ $121$ respectively, what is the total $C.I.$ for 3 years?
Answer
₹ $331$
Question
When interest is compounded half-yearly, what value for Time ($T$) is used to calculate interest for each conversion period?
Answer
$T = \frac{1}{2}$ year
Question
If ₹ $8,000$ is lent at $5\%$ compound interest for 2 years, what is the interest for the first year?
Answer
₹ $400$
Question
What is the amount at the end of the first year for ₹ $8,000$ at $5\%$ compound interest?
Answer
₹ $8,400$
Question
Calculate the interest for the second year if the principal at the start of that year is ₹ $8,400$ and the rate is $5\%$.
Answer
₹ $420$
Question
What is the total Compound Interest on ₹ $16,000$ in 3 years if the successive annual rates are $10\%$, $12\%$, and $15\%$?
Answer
₹ $6,668.80$
Question
To calculate C.I. for $2 \frac{1}{2}$ years compounded annually, what time value is used for the interest calculation of the final period?
Answer
$T = \frac{1}{2}$ year
Question
Find the difference between C.I. and S.I. on ₹ $4,000$ at $8\%$ per annum for 2 years.
Answer
₹ $25.60$
Question
What is the formula for finding the Rate of interest ($R$) when the C.I. of two consecutive periods is known?
Answer
$R = \frac{\text{Difference in C.I.} \times 100}{\text{C.I. of preceding period} \times \text{Time}}$
Question
What is the formula for finding the Rate of interest ($R$) when the Amounts ($A$) of two consecutive periods are known?
Answer
$R = \frac{\text{Difference in Amounts} \times 100}{\text{Preceding Amount} \times \text{Time}}$
Question
Term: Depreciation
Answer
Definition: The reduction in the value of an asset (like a machine) over time due to use or age.
Question
How is the value of a machine at the beginning of the second year calculated if it depreciates by $10\%$ in the first year?
Answer
$\text{Original Cost} - (10\% \text{ of Original Cost})$
Question
In a repayment problem, if a borrower pays back ₹ $2,500$ at the end of the first year, how is the principal for the second year determined?
Answer
$\text{Amount at the end of 1st year} - ₹ 2,500$
Question
If the Simple Interest on a sum for 3 years is ₹ $600$, what is the Simple Interest for just the first year?
Answer
₹ $200$
Question
For any conversion period after the first, why is the Compound Interest always more than the C.I. of the previous period?
Answer
Because the interest is calculated on a principal that has increased by the previous period's interest.
Question
If the C.I. of the 1st year is $x$, what is the C.I. for the 2nd year at rate $r\%$?
Answer
$x + (r\% \text{ of } x)$
Question
The difference between the compound interests for any two consecutive conversion periods is the interest on the _____ of the preceding period.
Answer
Interest
Question
The difference between the amounts of any two consecutive conversion periods is the interest on the _____ of the preceding period.
Answer
Amount
Question
If an asset worth ₹ $100$ depreciates by $10\%$ in the first year and $10\%$ of the remaining value in the second year, what is its value after 2 years?
Answer
₹ $81$
Question
If interest is compounded half-yearly, how many conversion periods are there in 1.5 years?
Answer
Three
Question
Find the interest on ₹ $10,000$ for $\frac{1}{2}$ year at $8\%$ per annum.
Answer
₹ $400$
Question
If the amount after 2 years at C.I. is ₹ $6,272$ and after 3 years is ₹ $7,024.64$, what is the interest for the 3rd year?
Answer
₹ $752.64$
Question
In Example 9, what assumed principal ($P$) is used to solve the problem by the unitary method?
Answer
₹ $100$
Question
If a machine depreciates by $10\%$ per annum and the depreciation during the second year is ₹ $9$, what was the value at the start of the second year?
Answer
₹ $90$
Question
If the amount at the end of the 2nd year is ₹ $13,860$ and an additional ₹ $6,000$ is invested at the start of the 3rd year, what is the principal for the 3rd year?
Answer
₹ $19,860$
Question
Find the interest on ₹ $6,000$ for 1 year at $10\%$ per annum.
Answer
₹ $600$
Question
If a sum grows to ₹ $22,400$ after 1 year and ₹ $8,400$ is repaid, what is the new principal for the 2nd year?
Answer
₹ $14,000$
Question
Find the interest on ₹ $5,900$ for $\frac{1}{2}$ year at $10\%$ per annum.
Answer
₹ $295$
Question
If the difference between C.I. of the 1st and 3rd year is ₹ $0.5125$ for a principal of ₹ $100$, what is the principal for a difference of ₹ $61.50$?
Answer
₹ $12,000$
Question
What is the Simple Interest on ₹ $46,875$ at $4\%$ for 1 year?
Answer
₹ $1,875$
Question
If the C.I. for two successive years is ₹ $2,700$ and ₹ $2,880$, calculate the rate of interest.
Answer
$6 \frac{2}{3} \%$
Question
How does the Simple Interest on a fixed sum at a constant rate change over multiple years?
Answer
It remains the same every year.
Question
If the S.I. for 3 years is ₹ $600$ and the C.I. for 2 years is ₹ $410$ on the same sum and rate, what is the interest for the 1st year?
Answer
₹ $200$ (for both S.I. and C.I.)
Question
In half-yearly compounding at $10\%$ p.a., what is the interest on ₹ $8,000$ for the first six months?
Answer
₹ $400$
Question
If a machine depreciates by $15\%$ of its value each year, and its initial value is ₹ $V$, what is the depreciation amount in the first year?
Answer
$0.15 V$
Question
If the principal for the 2nd year is ₹ $14,000$ and the rate is $12\%$, what is the interest for the 2nd year?
Answer
₹ $1,680$
Question
Calculate the amount for the 3rd year if the principal is ₹ $19,712$ and the rate is $15\%$.
Answer
₹ $22,668.80$
Question
What is the C.I. for the 3rd year in Problem 19 if the interest for the 2nd year is ₹ $1,056$ and the rate is $10\%$?
Answer
₹ $1,161.60$
Question
If the C.I. for the 5th year is ₹ $665.50$ and the rate is $10\%$, what is the C.I. for the 6th year?
Answer
₹ $732.05$
Question
In Example 18, if the amount in 5 years is ₹ $8,100$ and in 6 years is ₹ $8,748$, what is the interest for the 6th year?
Answer
₹ $648$
Question
What is the rate percent in Example 18 if the interest is ₹ $648$ on a sum of ₹ $8,100$?
Answer
$8\%$
Question
If a machine's original cost is ₹ $100$ and it depreciates by $10\%$ annually, what is its value at the beginning of the 3rd year?
Answer
₹ $81$
Question
For ₹ $6,000$ at $10\%$ C.I., find the interest for the second year.
Answer
₹ $660$
Question
If the principal at the start of the final half-year is ₹ $7,260$ and the rate is $10\%$, what is the interest for that half-year?
Answer
₹ $363$
Question
In Example 11, find the interest for the 3rd year on ₹ $50,700$ at $4\%$.
Answer
₹ $2,028$