IMPORTANT TRICKS:

1. ONE YEAR OF SIMPLE INTEREST = ONE YEAR OF COMPOUND INTEREST.

2. THE AMOUNT AT THE END OF FIRST YEAR (OR PERIOD) WILL BECOME THE PRINCIPAL FOR THE SECOND YEAR(OR PERIOD) AND AMOUNT AT THE END OF SECOND YEAR (OR PERIOD) BECOMES THE PRINCIPAL FOR THIRD YEAR.

AMOUNT = PRINCIPAL + INTEREST

A=P(1+R/100)n

Where, A=Amount

P=principal

R =Rate percentage

n=number of years

For two years-

(1).CI/SI=(200+r)/200

(2).Difference between CI & SI =PR2/1002

For Three years-

Difference between CI & SI =PR2(300+R)/1003

If lent money is Rs. P and each instalments is Rs. R, then

P = x/(1+r/100) + x/(1+r/100)^2 + x / (1+r/100)^3 + ….

Important Question on Compound Interest:

1. The simple and compound interests on a sum of money for two years are Rs 2880 and Rs 3052.80 respectively. The Rate of interest per annum is..

(1) 10%

(2)12%

(3)15%

(4)18%

2. What annual instalment would clear a debt of Rs 4.24 lakhs in two years at the rate of 12% p.a. compound interest if interest is paid at end of year.

(1) 250880

(2)276440

(3)284220

(4)292660

3. Find the annual payment that will discharge a debt of Rs 15260 due in 4 years at 6% per annum simple interest( assume that each annual payment is made at the end of the year.)

(1) 3600

(2)3500

(3)3450

(4)3400

4. A person took a loan of Rs 6000 for 3 years, at 5% per annum compound interest. He repaid Rs 2100 in each of the first 2 years.The amount he should pay at the end of third year to clear all his debts is.

(1) 2425.5

(2)2552.5

(3)2635.5

(4)2745.5

5. If the difference between simple interest and compound interest on a certain sum of money for 3 years at 10% per annum is Rs 31, The sum is.

(1) 500

(2)750

(3)1000

(4)1250

Answers with Explanation:

1. (2) Since, CI/SI = (200+R)/200

3052.8/2880 = (1+R/200)

R = 12%

2. (1) 4.24 x 100000 = x/ (1+r/100) + x / (1+r/100)^2

= x/(1+12/100) + x/(1+12/100)^2

= 25x/28 + 252 x/282 = 1325x/784

X = 250880

3. (2) Let annual payment 100 Rs.

100 +(100+R) + (100 +2*R) + (100 +3*R) = 15260

100+ 106 + 112 + 118 = 15260

436 = 15260

100 = 15260×100 / 436 = 3500

4. (1) P = 6000

For 1sr year CI = 5% of 6000 = 300

Amount = 6000 + 300 = 6300

P for 2nd year = 63000 – 2100 = 4200

CI for 2nd year = 5% of 4200 = 210

Amount 2nd year = 4200 + 210 = 4410

P for 3rd year = 4410 – 2100 = 2310

CI for 3rd year = 5% of 2310 = 115.5

Required amount = 2310 + 115.5 = 2425.5

5. (3) Since difference = PR2 (300+R)

31 = Px102 (300+10)/ 1003

P = 1000

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