1) Rohit invests a sum of money at compound interest which doubles itself in 3 years, find in how many years will it amount to four times itself ?  Published on 22 Apr 16
a. 4
b. 6
c. 10
d. 14
e. None of these
Answer
Explanation

ANSWER: 6
Explanation: Case (I) When Time (T) = 3 years, principle P becomes 2P. Therefore, amount = 2P A = P (1 + R/100)^{T} 2P = P (1 + R/100)^{3} (1 + R/100)^{3} = 2 (1 + R/100) = ^{3}√2  (A) (III) Find T for which P becomes 4P When Time = X and Amount = 4P 4P = P (1 + R/100)^{X} (1 + R/100)^{X} = 4 (1 + R/100) = 4^{1/X}  (B) Equating (A) and (B) we get, 4^{1/X} = ^{3}√2 = 2^{1/3} 2^{2/X} = 2^{1/3} 2/X = 1/3 X = 6 years


2) What will be the difference between simple interest and compound interest @ 10% per annum on a sum of Rs. 1000 after 3 years?  Published on 06 Apr 16
a. Rs. 31
b. Rs. 32.10
c. Rs. 40.40
d. Rs. 64.10
e. None of these
Answer
Explanation

ANSWER: Rs. 31
Explanation: S.I. = Rs. [(1000 x 10 x 3) / 100] = Rs. 300 C.I. = Rs. [1000 x (1 + (10/100))^{3}  1000] = Rs. 331 Difference = Rs. (331  300) = Rs. 31


3) Find the compound interest on Rs. 25000 in 2 years at 4% per annum. The interest being compounded half yearly.  Published on 25 Mar 16
a. Rs 3570.250
b. Rs 2896.598
c. Rs 2060.804
d. Rs 1050.200
Answer
Explanation

ANSWER: Rs 2060.804
Explanation: Given: Principal = Rs. 25000, Rate = 4 % per half year, Time = 2 years = 4 half years Therefore, Amount = P( 1 + (R/2)/100)^{2n}
Amount = Rs. [25000*(1+ 2/100)^{4}] = Rs. (25000 * 51/50*51/50*51/50* 51/50) = Rs. 27060.804 Therefore, C.I. = Rs. ( 27060.804 – 25000) = Rs 2060.804


4) A sum of money becomes Rs 13,380 in 3 years and Rs 20,070 in 6 years at compound interest. The initial sum is?  Published on 12 Jun 15
a. Rs 9040
b. Rs 8900
c. Rs 8920
d. Rs 9160
Answer
Explanation

ANSWER: Rs 8920
Explanation: Use the formula Amount = Principal Amount(1 + R/100)^{n}
Where r is rate and t is time in years
=>13,380 = P(1+R/100)^{2} (1)
=>20,070 = P(1+R/100)^{6} (2)
=> 1.5 = (1+R/100)^{4}
=> P = Rs 8920

