1. There is 60% increase in an amount in 6 years at simple interest. What will be the compound interest of Rs. 12,000 after 3 years at the same rate?
1. Rs. 2160
2. Rs. 3120
3. Rs. 3972
4. Rs. 6240
Answer & Explanation Answer:- 3 Explanation :- Let P = Rs. 100 Therefore, S.I.= Rs. 60 and T = 6 years. 2. The simple interest accrued on an amount of Rs. 25,000 at the end of three years is Rs. 7,500. What would be the compound interest accrued on the same amount at the same rate in the same period? Answer & Explanation Answer:- 2 Explanation :- P = Rs.25000 , S.I.= Rs. 7500 and T = 3 years. 3. A sum of money lent at compound interest for 2 years at 20% per annum would fetch Rs.482 more, if the interest was payable half yearly than if it was payable annually . The sum is Answer & Explanation Answer:- 2 Explanation :- Let sum be Rs.x 4. 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 & Explanation Answer:- 2 Explanation :- Let the sum be Rs. P. Therefore 5. The least number of complete years in which a sum of money put out at 20% compound interest will be more than doubled is: Answer:- 2 Explanation :- P(1+R/100)^n>2
R = (100*60)/(100*6) = 10%
Now , P= 12000 R= 10% T = 6 years
Therefore, Amount= 12000 [1 + (10/100)]^ 6
Amount = 15972
CI = Amount – Principal = 15972 – 12000 = Rs.3972
1. Rs. 7,750
2. Rs. 8,275
3. Rs. 8,500
4. Rs. 8,250
R = (7500*100)/(25000*3) = 10%
Now , P = 2500000 R= 10% T = 3 years
Therefore, Amount= 25000 [1 + (10/100)]^ 3
Amount = 25000(1331/1000)
Amount = 33275
CI = 33275- 25000 = Rs.3972 = Rs.8275
1. 10000
2. 20000
3. 40000
4. 50000
CI (compounded half yearly) = [x(1+(10/100)^4) -x]
CI = (4641/10000)x
CI (compounded yearly) = [x(120/100)^2) -x] =(11/25)x
(4641/10000)x – (11/25)x = 482
x=20000
1. 4360
2. 4460
3. 4560
4. 4660
Therefore
P(1+R/100)^3=6690…….…(1)
&
P(1+R/100)^6=10035………(2)
On dividing,
(1+R/100)^3 = 10025/6690 = 3/2.
Put This value of (1+R/100)^3 in Equation (1)
P*(3/2) = 6690
P = 6690*2/3 = Rs. 4460So , The sum is Rs.4460.
1. 3
2. 4
3. 5
4. 6
P(1+20/100)^n>2
(6/5)^n>2
6/5×6/5×6/5×6/5 > 2
P[(1+R/100)]^n > 2
P[(1+20/100)]^n > 2
(6/5)^n > 2
[(6/5) × (6/5) × (6/5) × (6/5)] > 2
Therefore n = 4