Aptitude - Compound Interest - Discussion
Discussion Forum : Compound Interest - General Questions (Q.No. 3)
3.
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?
Answer: Option
Explanation:
Let P = Rs. 100. Then, S.I. Rs. 60 and T = 6 years.
 R = |
 |
100 x 60 |  |
= 10% p.a. |
| 100 x 6 |
Now, P = Rs. 12000. T = 3 years and R = 10% p.a.
| C.I. |
|
|||||||||||||
|
||||||||||||||
| = 3972. |
Discussion:
94 comments Page 2 of 10.
Gaichalung kamei said:
2 years ago
160% increase..
160/100 = 8/5.
Let's assume 5 is the amount.
60% increase of 5 gives 8. There is a difference of 3 between 5 and 8. The given period is 6 yrs.
So, 6/3 = 2.
Now we can calculate the rate.
P × R × T/100 = 5 × R × 2/100 which gives 10%.
Now compound interest for 3 yrs with the same rate.
(11/10)³= 1331/1000.
The given principle is 12000. We get 12 as a multiplying factor.
now, CI = 331×12 = 3972.
160/100 = 8/5.
Let's assume 5 is the amount.
60% increase of 5 gives 8. There is a difference of 3 between 5 and 8. The given period is 6 yrs.
So, 6/3 = 2.
Now we can calculate the rate.
P × R × T/100 = 5 × R × 2/100 which gives 10%.
Now compound interest for 3 yrs with the same rate.
(11/10)³= 1331/1000.
The given principle is 12000. We get 12 as a multiplying factor.
now, CI = 331×12 = 3972.
(6)
Kal el said:
4 years ago
There is a difference between 'increase to' and 'increase by'.
So, please describe it clearly to understand.
So, please describe it clearly to understand.
(5)
Vishwanath said:
4 years ago
@Keerti.
12000 * 331/1000 = 12 * 331/1 = 3972.
12000 * 331/1000 = 12 * 331/1 = 3972.
(5)
Vishwanath said:
4 years ago
@Harshit.
First solve brackets, then multiply them into 3 times.
Then, You will get 1331/1000.
First solve brackets, then multiply them into 3 times.
Then, You will get 1331/1000.
(4)
Saleem said:
8 years ago
60%increase in 6 yrs.
i. e 10%every yr.
i. e R=10%.
Now A=P(1+10/100)^3=1200(1+1/10)^3=15972.
CI=15972-12000=3972.
i. e 10%every yr.
i. e R=10%.
Now A=P(1+10/100)^3=1200(1+1/10)^3=15972.
CI=15972-12000=3972.
(3)
Keerthi said:
5 years ago
1200*{(1+10/100)^3 - 1}, and next step how will become?
Especially- (1200*331/1000).
Please explain to me.
Especially- (1200*331/1000).
Please explain to me.
(3)
Sriram Manikanth said:
5 years ago
60 is interest 100 is the principle.
SI = pnr/100.
Therefor 60 = 100 * 6*r/100,
r = 10%.
SI = pnr/100.
Therefor 60 = 100 * 6*r/100,
r = 10%.
(3)
AKASH KADAM said:
8 months ago
For simple interest, 60% for the 6 years.
So, for one year it is - 10 % p.a.
Now the principle amount is- 12000.
We have to calculate the compound interest for 3 years.
On the 10% per annum rate.
So for the first year it's - 1200.
For the second year it's - 1320.
For the third year, it's - 1452.
Now add all of them - 1200 + 1320 + 1452.
= 3972.
So, for one year it is - 10 % p.a.
Now the principle amount is- 12000.
We have to calculate the compound interest for 3 years.
On the 10% per annum rate.
So for the first year it's - 1200.
For the second year it's - 1320.
For the third year, it's - 1452.
Now add all of them - 1200 + 1320 + 1452.
= 3972.
(3)
Harshit said:
5 years ago
How to solve (1 + 10/100) ^3?
Please explain it.
Please explain it.
(2)
Bhushan said:
7 years ago
SI for 6 yrs increase by 60%.
Means;
For 1st yr 10%.
1st yr , 10% of 12000 = 1200 - - - >(1)
Add 1200 ... ie. = 12000+1200= 13200.
For 2nd yr, 10% of 13200 = 1320 - - - > (2)
So, 13200+1320= 14520.
For 3rd yr, 10% of 14520 = 1452. - - - > (3)
So add the interest amount of all 3 equations ie....
1200 + 1320 + 1452 = 3972 is the right answer.
Means;
For 1st yr 10%.
1st yr , 10% of 12000 = 1200 - - - >(1)
Add 1200 ... ie. = 12000+1200= 13200.
For 2nd yr, 10% of 13200 = 1320 - - - > (2)
So, 13200+1320= 14520.
For 3rd yr, 10% of 14520 = 1452. - - - > (3)
So add the interest amount of all 3 equations ie....
1200 + 1320 + 1452 = 3972 is the right answer.
(2)
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