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by Rastis » Mon Dec 08, 2014 12:41 pm
An investment a 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 over a 3-year period?

a) 2d/3

b) 3d/4

c) 4d/3

d) 3d/2

e) 8d/3
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by GMATGuruNY » Mon Dec 08, 2014 12:49 pm
An investment of d dollars at k percent simple 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
The interest rate is irrelevant.
This problem is really about a PROPORTION.
$d invested over 2 years is equivalent to $600 in interest.
$x invested over 3 years must be equivalent to $2400 in interest.
Thus:
(d * 2)/600 = (x * 3)/2400
d/300 = x/800
d/3 = x/8
x = (8d)/3.

The correct answer is E.
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by GMATGuruNY » Mon Dec 08, 2014 12:49 pm
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

2d/3
3d/4
4d/3
3d/2
8d/3
An alternate approach is to PLUG AND CHUG:
1. Choose a good value for k.
2. Use the value of k to determine the value of d.
3. Answer the question and get a target.
4. Determine which answer choice yields the target.

Step 1:
Let the interest rate = 100%:
k=100.

Step 2:
Since $600 is earned over 2 years, $300 is earned each year.
To earn $300 at an interest rate of 100%, the amount invested = $300:
d=300.

Step 3:
To earn $2400 over 3 years, $800 must be earned each year.
To earn $800 at an interest rate of 100%, the amount invested = 800.
This is the target.

Step 4:
Now plug d=300 into the answers to see which yields the target of 800.
A quick scan of the answers reveals that only E works:
8d/3 = 8(300)/3 = 800.

The correct answer is E.
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by Rastis » Mon Dec 08, 2014 1:00 pm
That makes more sense. I was trying to use the interest rate formula and ended up guessing and moving on. Thanks!
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by Brent@GMATPrepNow » Mon Dec 08, 2014 5:01 pm
"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

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
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by Mathsbuddy » Fri Dec 19, 2014 6:42 am
d x k/100 x 2 = 600
So k/100 =300/d

e x k/100 x 3 = 2400
e x 300/d x 3 = 2400
e x 3/d = 8
e = 8d/3
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by Matt@VeritasPrep » Mon Dec 22, 2014 12:47 pm
Here's another pretty straightforward approach. $d at k% is currently giving us $300 a year. To get $800 a year, we'd need to multiply our principle by 8/3, since 300 * (8/3) = 800.

Hence we multiply d by (8/3), giving us (8/3)d, and we're done!
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