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.
ROHIT said:
2 years ago
How did they get 10% as rate while calculating CI? Please explain the answer.
(5)
Vishwanath said:
3 years ago
@Keerti.
12000 * 331/1000 = 12 * 331/1 = 3972.
12000 * 331/1000 = 12 * 331/1 = 3972.
(5)
Kal el said:
3 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.
(4)
Vishwanath said:
3 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)
Keerthi said:
4 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)
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)
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)
Taraka said:
5 years ago
A = p+si
P(160/100) = p + (p*6*R).
R = 10%.
Next;
A = p+ci.
A = 12000 * (110/100) ^3
P = 12000.
Then substituting a and p then;
Ci = 3972.
P(160/100) = p + (p*6*R).
R = 10%.
Next;
A = p+ci.
A = 12000 * (110/100) ^3
P = 12000.
Then substituting a and p then;
Ci = 3972.
(2)
Harshit said:
4 years ago
How to solve (1 + 10/100) ^3?
Please explain it.
Please explain it.
(2)
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