Aptitude - Compound Interest - Discussion
Discussion Forum : Compound Interest - General Questions (Q.No. 8)
8.
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: Option
Explanation:
| P |  |
1 + | 20 |  |
n | > 2P |  |
|
6 | |
n | > 2. |
| 100 | 5 |
| Now, | |
6 | x | 6 | x | 6 | x | 6 | |
> 2. |
| 5 | 5 | 5 | 5 |
So, n = 4 years.
Discussion:
26 comments Page 1 of 3.
Harini said:
2 years ago
Here they said that the money they put out in compound interest is greater than double of the money.
Let the money they put out in CI at a 20% rate of interest be P.
P(1+(20/100)^n > 2P,
(1+(20/100))^n > P,
(1.2)^n > 2.
taking log on both sides;
log(1.2)^n >log 2.
n> log 2/log 1.2.
n>3.8.
n = 4 yrs.
Let the money they put out in CI at a 20% rate of interest be P.
P(1+(20/100)^n > 2P,
(1+(20/100))^n > P,
(1.2)^n > 2.
taking log on both sides;
log(1.2)^n >log 2.
n> log 2/log 1.2.
n>3.8.
n = 4 yrs.
(16)
Tanish said:
3 years ago
@All.
When you divide 72 by ROI you will get an approximate no of years in which the principal will get double.
When you divide 72 by ROI you will get an approximate no of years in which the principal will get double.
(6)
Psit Kanpur said:
5 years ago
Can anyone explain about (6/5 * 6/5 * 6/5 * 6/5)>2?
(12)
Kanishk said:
5 years ago
Instead of trial and error we can use logarithm.
(1)
Sanjay Rajnedra Awate said:
7 years ago
For finding the time period in which a sum of money will double itself at R% rate of compound interest compounded annually, we generally use either of the following two formulas:
Time, T = 72 /R Years.
Time, T = 72 /R Years.
(4)
Rubi said:
7 years ago
Why we use amount formula instead of CISCO formula?
(4)
Kapil dev said:
8 years ago
Let principal be P.
P(1+R/100)^T>2P
P(1+20/100)^T>2P
(1+20/100)^T>2
(120/100)^T>2
1.2^T>2.
Now let's find out the minimum value of T for which the above equation becomes true.
If T = 1, 1.2T = 1.21 = 1.2
If T = 2, 1.2T = 1.22 * 1.4
If T = 3, 1.2T = 1.23 * 1.7
If T = 4, 1.2T = 1.24 * 2.07 which is greater than 2
Hence T = 4
i.e., the required number of years = 4.
P(1+R/100)^T>2P
P(1+20/100)^T>2P
(1+20/100)^T>2
(120/100)^T>2
1.2^T>2.
Now let's find out the minimum value of T for which the above equation becomes true.
If T = 1, 1.2T = 1.21 = 1.2
If T = 2, 1.2T = 1.22 * 1.4
If T = 3, 1.2T = 1.23 * 1.7
If T = 4, 1.2T = 1.24 * 2.07 which is greater than 2
Hence T = 4
i.e., the required number of years = 4.
(8)
Lokesh said:
8 years ago
Easy method.
How Will know n=72/r?
How Will know n=72/r?
Muhammad Zunnoorain said:
9 years ago
We can also solve it as;
2P = P(1.2)^n.
2 = 1.2^n.
Applying lag on both side.
Log(2) = n*log(1.2).
n = log(2)/log(1.2).
n = 3.8 to nearest year 4.
2P = P(1.2)^n.
2 = 1.2^n.
Applying lag on both side.
Log(2) = n*log(1.2).
n = log(2)/log(1.2).
n = 3.8 to nearest year 4.
(2)
Shubham said:
9 years ago
What is the least number of complete years in which sum will become more than double itself at 12% per annum?
Given me the answer with clear explanation.
Given me the answer with clear explanation.
(1)
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