Data Sufficiency Question

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Data Sufficiency Question

by Rospino » Thu May 24, 2012 5:20 am
Is the integer y a multiple of 4?

(1) 3(y^2) is a multiple of 18.
(2) y = p/q, where p is a multiple of 12 and q is a multiple of 3.

I. DVY NO SINFO (Acronym I use for Data Sufficiency. DVY NO = Determine Value, Yes/No. S = Simplify. Info = What information is needed?)
II. Solve for Statements 1 and Statements 2.

I. Yes/No. Simplify: Is y/4 an integer? Info: y value.
II.
- St. 1: 3(y^2) is a multiple of 18.
y^2 = multiple of 6.
if y^2 = 6, then y = 6^(1/2) ---> y/4 is not an integer.
if y^2 = 12, then y = 12^(1/2)---> y/4 is not an integer.
if y^2 = 18, then y = 18^(1/2)---> y/4 is not an integer.

Last step: Use multiples of both 6 & 4 and work backwards.
if y = 12, then y^2 = 144 ---> y^2 is a multiple of 6 and y/4 is an integer.
if y = 24, then y^2 = 576 ---> y^2 is a multiple of 6 and y/4 is an integer.

St. 1: Insufficient. Choices A & D are eliminated. Options B, C, and E remain.

- St. 2: y = p/q, where p is a multiple of 12 and q is a multiple of 3.
B = any integer
p = B * 3 * 4 ---> p is a multiple of 12

C = any integer
q = C * 3 ---> q is a multiple of 3

Let's say B = 2, C = 1:
y = (2 * 3 * 4)/(1 * 3 ) = 2 * 4. Divisible by 4.

Let's say B = 2, C = 3:
y = (2 * 3 * 4)/(3 * 3 ) = (2 * 4)/3. Not divisible by 4.

St. 2: Insufficient. B is eliminated. Options C & E remain.

- St. 1 & St. 2 Together:
Here I was going with the guess that the answer is E because both statements have numbers that are and are not divisible by 4. Is this a correct assumption? Is this assumption valid for other Yes/No Data Sufficiency problems as well (i.e., if St. 1 has yes & no answers and if St. 2 has yes & no answers, I can assume that St. 1 & St. 2 together will also have yes & no answers therefore the answer is E.)?

Also, is there a faster way to solve this problem than plugging in numbers?

I appreciate any help provided.

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by [email protected] » Thu May 24, 2012 10:46 pm
Yes even I am getting the answer as E.

Stimulus: Is Y perfectly divisible by 4?


Statement 1: 3y^2/18 is perfectly divisible, hence y^2 is a multiple of 6.

Firstly, if y^2 is a multiple of 6, it should be perfect square as well only then

'y' can atleast be perfectly divisible by 4.

Therefore, y ^2 = 36 ; y^2 = 144 ; y^2 = 324 thats what.

so y = 6; y = 12 ; y = 18 Out of the 3, 12 is divisible by 4 but the other 2 options are not divisible by 4.

Hence Insufficient.


Statement 2: y = (p/q) hence p = numerator and q = denominator and also y is an integer.

Case (i): 12 X 4 = 48 = p and 3 X 2 = 6 = q and y = 8 which is divisible by 4.

Case (ii): 12 X 1 = 12 = p and 3 X 2 = 6 = q and y = 2 which is not divisible by 4.

Case (iii): 12 X 6 = 72 = p and 3 X 8 = 24 = q and y = 3 which is also not divisible by 4.

Hence the statement 2 alone is also not sufficient.


Both statements combined:

Consider y = 6 or y = 12 or y = 18 as the quotient for the statement 2 and see if y is divisible by 4.

The thing is that you can get all the three answers as quotients but 1 is divisible and the other 2 are not divisible by 4. That is for sure.

Hence the correct answer is E.

This is my explanation.

Also just check the C option better, have I left something there...

Hope this helps...
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