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Simple interest

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Simple interest

by gmattesttaker2 » Thu Feb 13, 2014 10:36 pm
Hello,

Can you please tell me if my solution is correct here:

If $24,000 invested at x percent simple annual interest yields an interest of z dollars
at the end of one year. How much is it necessary to invest during one year at y
percent simple annual interest in order to get an interest of z dollars?

1) x = 4

2) x/y = 2/3


Sorry, I don't have the OA.


My approach:

100 - x
24,000 - z

=> 24,000 (x) = 100 (z)

=> z = 240/x

To find:

100 - y
? - z

i.e.

100 - y
? - 240x

=> (100)(240x) = ?(y)
=> ? = 24000x/y

Hence we need to find x/y

1) x = 4

In-suff.

2) x/y = 2/3

Suff.

Hence, ans is B
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by akash.delsaria » Fri Feb 14, 2014 1:54 am
Hi

According to me, Option 1 alone is in-sufficient as it does not give us a value for y. This one was correctly pointed out by you.

As for option 2, it gives us a value of x/y = 2/3. However, this is the reduced fraction value and does not necessarily mean that x=2 and y=3. It could also mean x=4, y=6 and so on. Therefore, even Option B alone does not give us 1 single solution to the problem.

If we are to combine the two options, we get x=4 and y=6. Using this value of x, we can obtain the value of z ($960).

And using this value of z, and the value of y=6, we can obtain the amount ($16000) to be invested to obtain an interest of $960 at a 6% simple interest rate.

So, in my opinion, the correct answer will be both the options combined together.
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by Brent@GMATPrepNow » Fri Feb 14, 2014 8:53 am
gmattesttaker2 wrote:
If $24,000 invested at x percent simple annual interest yields an interest of z dollars
at the end of one year. How much is it necessary to invest during one year at y
percent simple annual interest in order to get an interest of z dollars?

1) x = 4
2) x/y = 2/3

Target question: How much must we invest?

Given: $24,000 invested at x percent simple annual interest yields an interest of z dollars at the end of one year.

IMPORTANT: x percent = x/100

So, the given information tells us that "x percent of $24000 = z dollars"
In other words (x/100)(24,000) = z

Let's turn the target question into a similar equation.
Let I = the investment necessary to get an interest of z dollars
So, we want "y percent of I to equal z dollars"
In other words (y/100)(I) = z

Now that we have two equations that are set equal to z, we can write...
(x/100)(24,000) = (y/100)(I)
Multiply both sides by 100 to get: 24000x = yI
Divide both sides by y to get: I = 24000x/y

NOTE: Our target question asks us to find the value of I. Now that we know that I = 24000x/y, we can rephrase the target question as...

REPHRASED target question: What is the value of 24000x/y?

When we plug x = 4 into 24000x/y, we get (24000)(4)/y
Since we don't know the value of y, we cannot determine the value of 24000x/y.
So, statement 1 is NOT SUFFICIENT

Statement 2: x/y = 2/3
Take the expression 24000x/y and rewrite it as (24000)(x/y)
Now replace x/y with 2/3 to get (24000)(2/3) = 16000
Perfect. We're able to evaluate 24000x/y
So, statement 2 is SUFFICIENT

Answer = B

Aside: We have a free video with tips on rephrasing the target question: https://www.gmatprepnow.com/module/gmat- ... cy?id=1100

Cheers,
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by GMATGuruNY » Fri Feb 14, 2014 9:09 am
An interest rate is just that: a RATE.
The problem above is no different from the following:
gmattesttaker2 wrote:
If 24000 seconds are required to travel z miles at a speed of x miles per hour, how many seconds are required to travel z miles at a speed of y miles per hour?

1) x = 4

2) x/y = 2/3
Statement 1: x = 4
No information about y.
INSUFFICIENT.

Statement 2: x/y = 2/3
y = (3/2)x.
Rate and time are RECIPROCALS.
Since the rate for y is equal to 3/2 the rate for x, the time required at y miles per hour is equal to 2/3 the time required at x miles per hour.
Thus, the time required at y miles per hour = (2/3)(24000) = 16000 seconds.
SUFFICIENT.

The correct answer is B.

To illustrate that statement 2 in the problem as posted is sufficient:

Case 1: x=20% and y=30%
Here, the amount of interest yielded at x% = 20% of 24,000 = 4800.
If the amount invested at y% is equal to 2/3 of $24000 -- $16000 -- then the interest yielded at y% = 30% of 16000 = 4800.
The same amount of interest -- $z -- is earned in each case.

Case 2: x=40% and y=60%.
Here, the amount of interest yielded at x% = 40% of 24,000 = 9600.
If the amount invested at y% is equal to 2/3 of $24000 -- $16000 -- then the interest yielded at y% = 60% of 16000 = 9600.
The same amount of interest -- $z -- is earned in each case.

Thus, to earn the same amount of interest in each case, the amount invested at y% must be 2/3 the amount invested at x%:
(2/3)(24000) = 16000.
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