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Let P = (x + y)k. If P = 10 and k = 3

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by DavidG@VeritasPrep » Tue Dec 05, 2017 7:48 am
lheiannie07 wrote:Let P = (x + y)k. If P = 10 and k = 3, what is the average of x and y?

(A) 0
(B) 1/2
(C) 5/3
(D) 10/3
(E) 7/2

Can some experts help me find the best solution in this problem?

OA C
Ultimately, we want to solve for the average of x and y, or (x + y)/2. Initially, we're given 10 = (x + y) * 3
10/3 = x + y
1/2 * (10/3) = (x + y)/2
5/3 = (x + y)/2

The answer is C
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by Brent@GMATPrepNow » Tue Dec 05, 2017 7:50 am
lheiannie07 wrote:Let P = (x + y)k. If P = 10 and k = 3, what is the average of x and y?

(A) 0
(B) 1/2
(C) 5/3
(D) 10/3
(E) 7/2
Given: P = (x + y)k, P = 10, and k = 3
Plug in P = 10 and k = 3, to get: 10 = (x + y)(3)
Divide both sides by 3 to get: 10/3 = (x + y)

What is the average of x and y?
The average of x and y = (x + y)/2
= (10/3)/2
= 10/6
= 5/3
= C

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Brent
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by EconomistGMATTutor » Wed Dec 06, 2017 9:59 am
Let P = (x + y)k. If P = 10 and k = 3, what is the average of x and y?

(A) 0
(B) 1/2
(C) 5/3
(D) 10/3
(E) 7/2
Hi lheiannie07,
Let's take a look at your question.

$$P\ =\left(x+y\right)k$$
$$10\ =\left(x+y\right)3$$
$$x+y=\frac{10}{3} ... (i)$$

We know that average of x and y can be calculated as,
$$Average\ =\ \frac{x+y}{2}$$

Plugin the value of x+y from eq(i), we get:
$$Average\ =\ \frac{10}{3\times2}$$
$$Average\ =\ \frac{5}{3}$$

Therefore, option C is correct.

Hope it helps.
I am available if you'd like any follow up.
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by Scott@TargetTestPrep » Thu Jan 11, 2018 2:10 pm
lheiannie07 wrote:Let P = (x + y)k. If P = 10 and k = 3, what is the average of x and y?

(A) 0
(B) 1/2
(C) 5/3
(D) 10/3
(E) 7/2
We are given that P = (x + y)k, P = 10, and k = 3. Thus:

10 = (x + y)3

10/3 = x + y

Since x + y = 10/3, the average of x and y is (x +y)/2 = (10/3)/2 = 10/6 = 5/3.

Answer: C

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