Aptitude - Compound Interest - Discussion
Discussion Forum : Compound Interest - General Questions (Q.No. 5)
5.
The compound interest on Rs. 30,000 at 7% per annum is Rs. 4347. The period (in years) is:
Answer: Option
Explanation:
Amount = Rs. (30000 + 4347) = Rs. 34347.
Let the time be n years.
| Then, 30000 |  |
1 + | 7 |  |
n | = 34347 |
| 100 |
 |
|
107 | |
n | = | 34347 | = | 11449 | = | |
107 | |
2 |
| 100 | 30000 | 10000 | 100 |
n = 2 years.
Discussion:
46 comments Page 3 of 5.
France said:
9 years ago
107 * 107 is 11449 and 100 * 100 = 10000 but how is it equal to 34347/30000 ?
Anurag Nath said:
9 years ago
@Hitendra.
(107/100)n = 11449/10000.
=> 107 * 107/100 * 100 = 11449/10000.
=> (107/100)2 = 11449/10000 [Here n = 2].
Therefore n=2.
(107/100)n = 11449/10000.
=> 107 * 107/100 * 100 = 11449/10000.
=> (107/100)2 = 11449/10000 [Here n = 2].
Therefore n=2.
Dearlie Beloved said:
10 years ago
Shortcut through using ln:
t = ln (34 347/30 000)/ln (1.07).
t = 2 years.
t = ln (34 347/30 000)/ln (1.07).
t = 2 years.
Abhilok said:
10 years ago
2 times in 10 years.
2^3 = 8 times in 3*10 = 30 years.
Let power = n.
Given year = y.
n*y is the answer.
2^3 = 8 times in 3*10 = 30 years.
Let power = n.
Given year = y.
n*y is the answer.
(1)
Leo said:
10 years ago
Can anyone explain me the simplification of (107/100)2?
Bhabani said:
10 years ago
A sum becomes double in ten years in C.I. So in how many years the principal would be 8 times?
Trilok said:
10 years ago
11449/10000 = (107/100)^n , 1.1442 = (1.07)^n.
We can write as log 1.1442 = n log 1.07, n = log 1.1442/log 1.07 = 0.058501/0.029383 = 2 (as per scientific calculator).
We can write as log 1.1442 = n log 1.07, n = log 1.1442/log 1.07 = 0.058501/0.029383 = 2 (as per scientific calculator).
Isha sharma said:
1 decade ago
Guys this is simple divide numerator and denominator by 3 you will get your answer.
Digvijay said:
1 decade ago
Ya! it's not easy to find out. We have to try with value of t.
TEJAS said:
1 decade ago
Why 34347/30000?
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