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.
Abinash said:
2 decades ago
Dear friends, Can anybody tell me in brief how it is done please?
Sohi said:
1 decade ago
Let x=(6/5)^n
(6/5)^n >2....................(1)
if we put value of n=1 then x=1.2.........condtion 1 is false.
if we put value of n=2 then x=1.44.........condtion 1 is still false.
if we put value of n=3 then x=1.728.........condtion 1 is false.
if we put value of n=4 then x=2.07.........now condtion 1 is true.
Hence answer is 4.
Thanx.
(6/5)^n >2....................(1)
if we put value of n=1 then x=1.2.........condtion 1 is false.
if we put value of n=2 then x=1.44.........condtion 1 is still false.
if we put value of n=3 then x=1.728.........condtion 1 is false.
if we put value of n=4 then x=2.07.........now condtion 1 is true.
Hence answer is 4.
Thanx.
Kannan said:
1 decade ago
Hi friends,
In the explanation, in 1st step p is present.
But in 2nd step it is vanished where it is gone ?
In the explanation, in 1st step p is present.
But in 2nd step it is vanished where it is gone ?
Kailash said:
1 decade ago
Hello Kannan,
It got cancel (by dividing both side by P).
It got cancel (by dividing both side by P).
Sharvari said:
1 decade ago
Can someone explain in easier way?
Please.
Please.
Lakshmanan sp said:
1 decade ago
AS p(6/5)^n>2p.
Next step (6/5)^n>2.
Next step ,take n = 1, then 6/5 = 1.2 therefore (1.2)^1 = 1.2.
|||rly for n = 2(1.2)^2 = 1.44.
For n = 3(1.2)^3 = 1.44*1.2 = 1.7288.
For n = 4(1.2)^4 = 1.7288*1.2 = 2.07.
Hence minimum no.of years is 4.
Next step (6/5)^n>2.
Next step ,take n = 1, then 6/5 = 1.2 therefore (1.2)^1 = 1.2.
|||rly for n = 2(1.2)^2 = 1.44.
For n = 3(1.2)^3 = 1.44*1.2 = 1.7288.
For n = 4(1.2)^4 = 1.7288*1.2 = 2.07.
Hence minimum no.of years is 4.
Jahnavi said:
1 decade ago
Guys can you pls explain what is the first formula we are using there:
i.e., p(1 + 20/100)^n > 2p.
what is actually?
i.e., p(1 + 20/100)^n > 2p.
what is actually?
Narayana Rao Gartam said:
1 decade ago
Dear Jahnavi,
Amount = P(1+R/100)^n.
Sum of money is more than the double of Amount.
So, 2P < P(1+20/100)^n , Since R= 20%
P and P cancelled each other. So,
2< (1+20/100)^n.
2< (120/100)^n.
2<(6/5)^n.
Try n=1, 2, 3, 4, 5.
What least number can satisfy the 2<(6/5)^n.
Lets n=1 ( 2<1.2 so, wrong).
Lets n=2 ( 2<1.44 so, wrong).
Lets n=3 ( 2<1.728 so, wrong).
Lets n=4 ( 2<2.1036 so, right).
Answer is 4 years.
Amount = P(1+R/100)^n.
Sum of money is more than the double of Amount.
So, 2P < P(1+20/100)^n , Since R= 20%
P and P cancelled each other. So,
2< (1+20/100)^n.
2< (120/100)^n.
2<(6/5)^n.
Try n=1, 2, 3, 4, 5.
What least number can satisfy the 2<(6/5)^n.
Lets n=1 ( 2<1.2 so, wrong).
Lets n=2 ( 2<1.44 so, wrong).
Lets n=3 ( 2<1.728 so, wrong).
Lets n=4 ( 2<2.1036 so, right).
Answer is 4 years.
(1)
Swathi said:
1 decade ago
Hi,
We are using > 2. Can you explain it?
We are using > 2. Can you explain it?
Anjali said:
10 years ago
I am unable to understand why are taking amount formula in place of CI. Please explain it.
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