


8.
If the compound interest on a certain sum at 16 2/3% for 3 years is Rs. 1270, find the simple interest on the same sum at the same rate and for the same period.
9.
The difference between the compound interest and simple interest on a certain sum at 10% per annum for 2 years is Rs. 631. Find the sum.
10.
The difference between the compound interest and the simple interest accrued on an amount of Rs. 18,000 in 2 years was Rs. 405. What was the rate of interest p.c.p.a? [Bank P.O. 2003]
Answer & Solution
Answer: d) 15%
Solution: Let the rate be R% p.a. Then, [18000 [1 + (R/100)]^2  18000]  [(18000 * R * 2)/100] = 405 => 18000[[(100 + R)^2 / 10000]  1  (2R/100)] = 405 => 18000[((100 + R)^2  10000  200R)/10000] = 405 => (9/5)R^2 = 405 => R^2 = [(405 * 5)/9] = 225 => R = 15 Rate = 15% 11.
Did Rs. 1301 between A and B, so that the amount of A after 7 years is equal to the amount of B after 9 years, the interest being compounded at 4% per annum.
Answer & Solution
Answer: c) Rs. 676 and Rs. 625
Solution: Let the two parts be Rs. x and Rs. (1301  x). x[1 + (4/100)]^7 = (1301 x)[1 + (4/100)]^9 => x/(1301  x) =[1 + (4/100)]^2 = [(26/25) * (26/25)] => 625x = 676(1301  x) => 1301x = 676 * 1301 => x = 676. So, the two parts are Rs. 676 and Rs. (1301  676); i.e; Rs. 676 and Rs. 625 12.
A certain sum amounts to Rs. 7350 in 2 years and to Rs. 8575 in 3 years. Find the sum and rate percent.
13.
A sum of money amounts to Rs. 6690 after 3 years and to Rs. 10,035 after 6 years on compound interest. Find the sum.
Answer & Solution
Answer: d) Rs. 4460
Solution: Let the sum be Rs. P. Then, P[1 + (R/100)]^3 = 6690 ....... (i) P[1 + (R/100)]^6 = 10035 ....... (ii) On dividing, we get [1 + (R/100)]^3 = 10035/6690 = 3/2 Substituting this value in (i), we get: P * (3/2) = 6690 => P = [6690 * (2/3)] = 4460 Hence, the sum is Rs. 4460. 14.
A sum of money doubles itself at compound interest in 15 years. In how many years will it become eight times?
