Compound Interest ShortCut Tricks

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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