Interest Rate Growth

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Interest Rate Growth

by vrn2vw » Sun Nov 01, 2015 9:36 am
An investment of d dollars at k percent simple annual interest yields $600 interest over a 2-year period. In terms of d, what dollar amount invested at the same rate will yield $2,400 interest over a 3 year period?

A)2d/3
B)3d/4
C)4d/3
D)3d/2
E)8d/3

Can someone explain this one? OA is E
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by theCEO » Sun Nov 01, 2015 10:06 am
vrn2vw wrote:An investment of d dollars at k percent simple annual interest yields $600 interest over a 2-year period. In terms of d, what dollar amount invested at the same rate will yield $2,400 interest over a 3 year period?

A)2d/3
B)3d/4
C)4d/3
D)3d/2
E)8d/3

Can someone explain this one? OA is E
Interest = principal x rate x time
600 = d * (k/100) * 2
k = 60000/2d = 30000/d

what dollar amount invested at the same rate will yield $2,400 interest over a 3 year period?

Principal = Interest / (rate x time)
Principal = 2400 / (30000/100d x 3) = 2400 / (900/1d) = 2400 * d/900
Principal = 24d/9 = 8d/3

ans = e

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by [email protected] » Sun Nov 01, 2015 10:53 am
Hi vrn2vw,

To start, we can simplify the equations by focusing on the INTEREST ONLY.

D dollars at K percent simple interest for 2 years = $600

D(K/100)(2) = 600

This becomes....

DK = 30,000

Now we have to look at....

E dollars at K percent simple interest for 3 years = $2400

E(K/100)(3) = 2400

EK = 80,000

So, we need to go from 30,000 to 80,000. What do we need to multiply by? 8/3.

Final Answer: E

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by Brent@GMATPrepNow » Sun Nov 01, 2015 11:22 am
"An investment of d dollars at k percent simple annual interest yields $600 over a 2-year period. In terms of d, what dollar amount invested at the same rate will yield $2,400 interest over a 3-year period?

(A) 2d/3
(B) 3d/4
(C) 4d/3
(D) 3d/2
(E) 8d/3
An investment of d dollars at k percent simple annual interest yields $600 over a 2-year period.
In other words, an investment of d dollars yields $300 in interest each year.

What dollar amount invested at the same rate will yield $2,400 interest over a 3-year period?
In other words, how much money must we invest to earn $800 in interest EACH YEAR?

If a d dollar investment yields $300 in interest EACH YEAR, then:
- a 2d dollar investment would yield $600 (2 times $300) in interest EACH YEAR
- a 3d dollar investment would yield $900 (3 times $300) in interest EACH YEAR
- a 4d dollar investment would yield $1200 (4 times $300) in interest EACH YEAR
- etc.

From here there are two approaches.

APPROACH #1
We want the ANNUAL interest to be $800.
This means we must invest an amount that is BETWEEN 2d dollars and 3d dollars [since $800 is BETWEEN $600 and $900].
When we check the answer choices, only [spoiler]E, which can be written as (8/3)d,[/spoiler] is BETWEEN 2d dollars and 3d
So, the correct answer must be E


APPROACH #2
To increase the ANNUAL interest from $300 to $800, we must invest 800/300 TIMES as much money.
800/300 = 8/3, so we must invest (8/3)d dollars [aka 8d/3 dollars]
Answer = E

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by vinoo81 » Sun Dec 31, 2017 10:40 pm
Hi,

I approached this problem as following:

600 = d x 2 x k/100 (Given and rephrasing in pnr)

Let's say, to get an interest of 2400 after 3 years, the principal amount invested is P

2400 = P x 3 x k/100 Dividing this by 4(as our goal is equate to above equation)

600 = P/4 x 3 x k/100 = d x 2 x k/100

So "P" in terms of d => P = (dx2x4)/3 => 8d/3 which is option E. Is this approach correct?

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by GMATWisdom » Mon Jan 01, 2018 7:59 am
vrn2vw wrote:An investment of d dollars at k percent simple annual interest yields $600 interest over a 2-year period. In terms of d, what dollar amount invested at the same rate will yield $2,400 interest over a 3 year period?

A)2d/3
B)3d/4
C)4d/3
D)3d/2
E)8d/3

Can someone explain this one? OA is E
If A is the desired amount then applying the simple interest formula we get
d*k*2/100 =600....................(1)
A*k*3/100 =2400....................(2)
dividind (2) by (1) we get
3A/2d=4
Or A=8d/3
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by Scott@TargetTestPrep » Sat Aug 24, 2019 4:22 am
vrn2vw wrote:An investment of d dollars at k percent simple annual interest yields $600 interest over a 2-year period. In terms of d, what dollar amount invested at the same rate will yield $2,400 interest over a 3 year period?

A)2d/3
B)3d/4
C)4d/3
D)3d/2
E)8d/3

Can someone explain this one? OA is E

Recall that the amount of simple interest, I, is equal to the principal, P, times the annual interest rate, r, times the number of years, t. That is, I = Prt.

We are given that I = 600, P = d, r = k/100 and t = 2, so we have:

600 = d(k/100)(2)

300 = dk/100

dk = 30,000

However, we are being asked for the dollar amount invested, in terms of d, at the annual interest rate of k percent that will yield $2,400 interest over a 3-year period. If we let this dollar amount be x, then

2,400 = x(k/100)(3)

800 = xk/100

xk = 80,000

Since dk = 30,000 and xk = 80,000,we see that k = 30,000/d and k = 80,000/x. Thus, we can equate 30,000/d and 80,000/x and isolate x.

30,000/d = 80,000/x

30,000x = 80,000d

3x = 8d

x = 8d/3

Answer: E

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by Brent@GMATPrepNow » Sat Aug 24, 2019 5:31 am
vrn2vw wrote:An investment of d dollars at k percent simple annual interest yields $600 interest over a 2-year period. In terms of d, what dollar amount invested at the same rate will yield $2,400 interest over a 3 year period?

A)2d/3
B)3d/4
C)4d/3
D)3d/2
E)8d/3

Can someone explain this one? OA is E
An investment of d dollars at k percent simple annual interest yields $600 over a 2-year period.
In other words, an investment of d dollars yields $300 in interest each year.

What dollar amount invested at the same rate will yield $2,400 interest over a 3-year period?
In other words, how much money must we invest to earn $800 in interest EACH YEAR?

If a d dollar investment yields $300 in interest EACH YEAR, then:
- a 2d dollar investment would yield $600 (2 times $300) in interest EACH YEAR
- a 3d dollar investment would yield $900 (3 times $300) in interest EACH YEAR
- a 4d dollar investment would yield $1200 (4 times $300) in interest EACH YEAR
- etc.

From here there are two approaches.

APPROACH #1
We want the ANNUAL interest to be $800.
This means we must invest an amount that is BETWEEN 2d dollars and 3d dollars [since $800 is BETWEEN $600 and $900].
When we check the answer choices, only E, which can be written as (8/3)d, is BETWEEN 2d dollars and 3d
So, the correct answer must be E


APPROACH #2
To increase the ANNUAL interest from $300 to $800, we must invest 800/300 TIMES as much money.
800/300 = 8/3, so we must invest (8/3)d dollars [aka 8d/3 dollars]
Answer = E

Cheers,
Brent
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