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Investment of d dollars at k percent

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Investment of d dollars at k percent

by gander123 » Wed Jan 02, 2013 7:16 am
Hey guys,

I've been posting recently, and I hate it cause its a sign of my poor performance on lots of question types. However, I cant help it...

Here we go:

"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

Correct Answer: E"

I tried out this one using the common equation for simple interest C= P + Prt where I substituted k for r and 2 for t.

So I set up the following 2 equations:

d + 600 = d + 2kd
d + 2400 = d + 3kd

I never ended up with a really good solution..

I'd appreciate your advise.

Tobi
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by Brent@GMATPrepNow » Wed Jan 02, 2013 8:27 am
gander123 wrote: "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.

So, 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,
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by Param800 » Wed Jan 02, 2013 8:59 am
Hey Tobi,

I don't think that it is sign of poor performance, I think it is a sign of good performance because you are practicing questions and you are learning from your mistakes :)

Here's my solution to this problem.

We know this formula = SI = P * i * t

where SI is simple interest
i is interest rate
t is time

So..just substitute in this formula-

d*(k/100)*2 = 600

You will get.. d*k = 300 * 100 ------- (i)

Now, second statement ..

2400 = P * ( k/100 ) * 3

where P is the amount which we have to find

thus, P = (800 * 100 )/k ........ (ii)

Substituting the value of k from (i) to ( ii)

we will get...

P = 8d/3 (E)

Hope this helps
gander123 wrote:Hey guys,

I've been posting recently, and I hate it cause its a sign of my poor performance on lots of question types. However, I cant help it...

Here we go:

"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

Correct Answer: E"

I tried out this one using the common equation for simple interest C= P + Prt where I substituted k for r and 2 for t.

So I set up the following 2 equations:

d + 600 = d + 2kd
d + 2400 = d + 3kd

I never ended up with a really good solution..

I'd appreciate your advise.

Tobi
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by gander123 » Wed Jan 02, 2013 9:46 am
Hey Param, hey Brent,

two lovely approaches.. Thanks a lot!

Cheers,

Tobi
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by GMATGuruNY » Wed Jan 02, 2013 9:50 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

2d/3
3d/4
4d/3
3d/2
8d/3
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 gander123 » Wed Jan 02, 2013 10:02 am
Hey Mitch,

Thanks man... Pinned "PLUG and CHUG" to my door ;)

cya ,

Tobi
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by Scott@TargetTestPrep » Mon Mar 19, 2018 3:30 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
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, 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 deloitte247 » Thu Mar 29, 2018 2:10 pm
Currently , the investment pays $600 in 2 years. This implies that the pay is 1 year is $300
What is needed is $2400 in 3 years= $800 per year

Investment= 800/300 * d (current investment) = $8d/3
At the same rate, $8d/3 would yield $2400 interest over 3 years
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