DS(Statistics)

[rintoo22]

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

The way I solved it.
Total cost of the x houses = 212000x
Total cost of the y houses = 194000y
So the average of the assessed values of the x+y houses is

(212000x + 194000y)/(x+y) .......(1)

Statement 1 : Not Sufficient as it only gives the value of x+y=36
Statement 2 : In my opinion Sufficient

(212000(2y) + 194000(y))/(2y+y)
(212000(2y) + 194000(y))/(3y)
618000y/3y = 206000
Hence the answer is B.

However the QA is given a C.

[srcc25anu]

I would also think the answer should be B
If we know the ratio of X and Y, that should be sufficient to know the weighted average of those two.

Since we know x = 2y:
212,000 *2y + 194,000*y / (3Y)
=> 618,000y/3y (Y cancels out)
Thus average of X+Y houses = 206,000
hence SUFFICIENT

A is clearly NOT sufficient. we do not know X and Y individually. Insufficient.

Ans B

[Anju@Gurome]

Statement 2 : In my opinion Sufficient

(212000(2y) + 194000(y))/(2y+y)
(212000(2y) + 194000(y))/(3y)
618000y/3y = 206000
Hence the answer is B.

You are correct. The answer should be B.
However, I'd like to point out that the two green lines are unnecessary time wasting for DS problem. From the first line itself we can see that all the y's will cancel out from the numerator and denominator, i.e. we have a definite number which doesn't include any unknown variable. Hence, the statement is sufficient.

For DS problems if we can definitely solve or eliminate the unknowns, we are done. We don't have to solve them.

[bharat.bondalapati]

@rintoo22
The way you solved it is completely correct. The answer has to be B.