The Compound Interest on an investment in 6 years is $400

rajatchopra1994 You missed the logic i guess.

On P investment, we 're getting 400 bucks interest in 6 years, so we'll get the same interest in the next 6 years. Total is 400+400=800 bucks.
Whatever extra we're getting is the interest on the interest of first 6 years.

Dear nick1816 ,

I already understood the explanation.
For others, it will be confusing to understand. That's why I have mentioned it.
Hope you will get my Point.

Regards,
Rajat Chopra

rajatchopra1994 bro if someone will have a doubt, he/she can ask or can refer your solution lol. These kinda things happen because of time constraint; you can't write everything brother. I don't understand why you're even bothering about if you understood the solution. Anyways, i'm outta this thread.

nick1816,

I thought we explain every question properly for the other members of GMAT Club Community.
That is the only reason for posting explained answers. I think i misunderstood the purpose of explanation.
Same here bro, outta from this thread. Regards!!

Another way..

What is compound interest --- \(P(1+\frac{r}{100})^t-P=I\)

The Compound Interest on an investment in 6 years is $400----- --
\(P(1+\frac{r}{100})^6-P=400........\)..
\(P(1+\frac{r}{100})^6=P+400\)......
\((1+\frac{r}{100})^6=\frac{P+400}{P}\)...(i)

Similarly, Compound Interest on the same investment in 12 years is $ 960--------
\(P(1+\frac{r}{100})^{12}-P=400........\)..
\(P((1+\frac{r}{100})^6)^2=P+960\)...Substitute (i) in it.
\(P(\frac{400+P}{P})^2=960+P.........(400+P)^2=P(960+P)...P=1000\)

C

chetan2u, Can you please expand (400+p)^2 = P(960+p), for some reason my solution isn't the same as yours.

Thanks.

(400+p)^2=p(960+p).....160000+800p+p^2=960p+p^2
160000=960p-800p=160p.....p=1000

This took me a little over 3 mins, but here is my approach:

Let P be the original amount and r = rate
There 400 + P = P\((1+r)^{6}\) ==> P(\((1+r)^{6}\) - 1) = 400 ==> Let's call this eq 1
Similarly, P(\((1+r)^{12}\) - 1) = 960 ==> Let's call this eq 2
Let x = \((1+r)^{6}\) ==> eq 1: P(x - 1) = 400 and eq 2: P(\(x^{2}\) - 1) = 960
Divide eq 1 by eq 2 (to cancel out P and x - 1) ==> \(\frac{1}{x+1}\) = \(\frac{5}{12}\) ==> x = \(\frac{7}{5}\)
Therefore, using eq 1, P = \(\frac{400}{x - 1}\) ==> P = 400/\(\frac{2}{5}\) ==> P = $1,000 (the answer is C)

A $100 investment earns 20% interest compounded every 10 years.
In this case:
Interest earned after the FIRST 10-year interval = 20% of 100 = 20
Resulting amount after the first 10-year interval = 100+20 = 120
Interest earned in the SECOND 10-year interval = 20% of 120 = 24
Notice that the interest earned in the second 10-year interval ($24) is 20% greater than the interest earned in the first 10-year interval ($20).
The underlined percentages in blue are the SAME.
Implication:
If the interest increases by x% from one interval to the next, then x% is the rate applied to the initial investment.

For the sake of clarity, I've revised the wording of the problem:



Interest earned in the second 6-year interval = (total earned after 12 years) - (interest earned in the first 6 years) = 960 - 400 = 560

The interest earned in the second 6-year interval ($560) is 40% greater than the interest earned in the first 6-year interval ($400).
Since the interest increases by 40% from one interval to the next, 40% is the rate applied to the initial investment.
Thus, the $400 earned in the first 6-year interval must be equal to 40% of the initial investment:
\(400 = (0.4)x\)
\(x = \frac{400}{0.4} = \frac{4000}{4} = 1000\)

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