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## What is the average (arithmetic mean) of A, B, and 4C?

tagged by: M7MBA

This topic has 1 expert reply and 2 member replies

### Top Member

M7MBA Master | Next Rank: 500 Posts
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#### What is the average (arithmetic mean) of A, B, and 4C?

Thu Mar 01, 2018 6:52 am
What is the average (arithmetic mean) of A, B, and 4C?

(1) A + B = 17

(2) C^2 = 49

The OA is the option E.

Using both statements together is not sufficient to get an answer? Experts, can you show me what is the best way to solve this PS question?

### Top Member

Vincen Legendary Member
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Tue Mar 06, 2018 3:45 am
Hello Gmat_mission.

We are asked for $$\frac{A+B+4C}{3}=?$$

(1) A+B=17.

This implies that $$\frac{A+B+4C}{3}=\frac{17+4C}{3}=?$$ Hence, we cannot determine the average.

$$\left(2\right)\ \ C^2=49.$$

This implies that C=7 or C=-7. Hence we get $$\frac{A+B+4C}{3}=\frac{A+B+4\left(\pm7\right)}{3}=\frac{A+B\pm28}{3}=?$$ Hence, we cannot determine the average.

Using both statements together we get $$\frac{A+B+4C}{3}=\frac{17\pm28}{3}.$$ We have two options $$\frac{A+B+4C}{3}=\frac{17+28}{3}=\frac{45}{3}=15$$ or $$\frac{A+B+4C}{3}=\frac{17-28}{3}=-\frac{11}{3}.$$ Since the options are different, hence this option is NOT sufficient.

Therefore, the correct answer is E.

### GMAT/MBA Expert

Jay@ManhattanReview GMAT Instructor
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Wed Mar 07, 2018 9:09 pm
M7MBA wrote:
What is the average (arithmetic mean) of A, B, and 4C?

(1) A + B = 17

(2) C^2 = 49

The OA is the option E.

Using both statements together is not sufficient to get an answer? Experts, can you show me what is the best way to solve this PS question?
You are supposed to calculate the value of (A + B + 4C)/3.

Certainly (1) and (2) alone are not sufficient. While (1) does not have the value of C, (2) does not have the value of A and B.

Even combining the two statements will not help as C^2 = 49 gives two values of C = 4 and -4. Thus, with each value of C, you get different values of (A + B + 4C)/3. In DS, we want a unique value, which is not the case here. Insufficient.

Hope this helps!

-Jay
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### Top Member

deloitte247 Senior | Next Rank: 100 Posts
Joined
02 Mar 2018
Posted:
35 messages
Sat Mar 03, 2018 12:33 pm
$$mean\ of\ A,\ B\ and\ 4C\ is\ =\frac{A+B+4C}{3}$$
$$If\ c^2=49$$
$$then\ c=\sqrt{49}=\left(\frac{+}{-}7\right)$$
$$i.e\ c=+7\ and\ c=-7$$
[A+B=17], when c=+7
$$mean=\ \frac{17+4\left(7\right)}{3}=15$$
when c=-7
$$mean=\ \frac{17+4\left(-7\right)}{3}=\frac{11}{3}$$

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