If the average (arithmetic mean) of the assessed values of x houses is

\(SumX/X = 212\) ==> \(SumX = 212X\)
\(SumY/Y = 194\) ==> \(SumY = 194Y\)

Question:\((\frac{SumX + SumY}{X+Y})\)?
Question:\((\frac{212X + 194Y}{X+Y})\)?

(1) \(X+Y=36\)

Substitute into question:
Question:\((\frac{212X + 194Y}{36})\)?

Insuficient

(2) X=2Y

Substitute into question
Question:\((\frac{212(2Y) + 194Y}{2Y+Y)})\)?
Question:\((\frac{618Y}{3Y})\)?
Question:\((\frac{618}{3})\)?
Question:\((206)\)?

Wait a second-- do you know what \(206\) is? Of course we do... Sufficient!

Final answer, \(B\).

The answer should certainly be B here; it's a weighted average question, and if we know the average of the two groups, to find the overall average we only need the ratio of the sizes of the two groups. Statement 2 tells us the ratio of x to y is 2 to 1, so the combined average will be "twice as close" to the average of the x houses - that is, the combined average will be $206,000.

Who is claiming that the answer is C?

Corrected the OA it should be B, not C.

If the average (arithmetic mean) of the assessed values of x houses is $212,000 and the average of the assessed values of y other houses is $194,000, what is the average of the assessed values of the x+y houses?

This is a simple weighted average question, and if we knew the ratio of x to y it would be sufficient to answer the question

(1) x+y=36. Not sufficient.
(2) x=2y. Sufficient.

Answer: B.

P.S. General formula for weighted average: \(weighted \ average=\frac{total \ weight}{sum \ of \ values}\) --> \(weighted \ average=\frac{weight_1*value_1+weight_2*value_2}{value_1+value_2}\).

So, as you can see the ratio of values (the ratio of x to y) would give us the answer for the given question.

Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.

If the average (arithmetic mean) of the assessed values of x houses is $212,000 and the average of the assessed values of y other houses is $194,000, what is the average of the assessed values of the x+y houses?

(1) x+y=36
(2) x=2y

This is a '2by2'question, one of the most common type of questions in GMAT math


We have 2 variables (x,y) and 2 equations are given by the 2 conditions, which gives (C) high chance of being the answer
Looking at the conditions together, x=24, y=12 so this seems sufficient, but this is a commonly made mistake; if we look at the conditions separately,
condition 2 contains ratios and when one condition contains numbers and another contains ratios, there is high chance the one with ratio is the answer, so if condition 2 is examined,

average assessed value= (212,000x+194,000y)/(x+y)=(212,000*2y+194,000y)/(2y+y).
y can be canceled out, so the condition is sufficient and the answer becomes (B).

For cases where we need 2 more equations, such as original conditions with “2 variables”, or “3 variables and 1 equation”, or “4 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 70% chance that C is the answer, while E has 25% chance. These two are the majority. In case of common mistake type 3,4, the answer may be from A, B or D but there is only 5% chance. Since C is most likely to be the answer using 1) and 2) separately according to DS definition (It saves us time). Obviously there may be cases where the answer is A, B, D or E.

average of x houses = $212,000
average of y = $194,000

What is the average of the x + y houses?

(1) x + y = 36
We can't determine x or y. INSUFFICIENT.

(2) x = 2y

x is weighted 2/3 and y is weighted 1/3.

212,000 - 194,000 = 18,000

18,000 / 3 = 6,000

6,000 * 2 = 12,000

194,000 + 12,000 = 206,000

SUFFICIENT.

Answer is B.

If the average (arithmetic mean) of the assessed values of x houses is $212,000 and the average of the assessed values of y other houses is $194,000, what is the average of the assessed values of the x+y houses?

212000x + 194000y / x + y = ?

(1) x+y=36
Insufficient b/c there is still 1 unknown

(2) x=2y

212000(2y) + 194000y / 2y + y

At this point, I want to make sure the y's cancel. This next step isn't necessary, but I took any random number (e.g. 5 and replaced 212000 and 194000 with them) to see if they do cancel (that way I can avoid computing 212000 + 194000

5(2y) + 5y / 3y = 15y/3y = 5

A constant remains...so sufficient

B

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