Is x=1 y=2 and z=3? Experts pls take a look

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mathewmithun: Senior | Next Rank: 100 Posts; Posts: 57; Joined: Thu Oct 08, 2009 12:02 am; Location: India

by mathewmithun » Tue May 29, 2012 1:46 am

is x=1, y=2 and z=3?
1: 5x+2z+3=3x+4y=y+2z+3
2: 5x+z+3=3x+4y and 3x+4y=y+2z+3

OA is E but I am getting the answer as D. Pls help

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Source: Beat The GMAT — Data Sufficiency | Tue May 29, 2012 1:46 am

mathewmithun: Senior | Next Rank: 100 Posts; Posts: 57; Joined: Thu Oct 08, 2009 12:02 am; Location: India

by mathewmithun » Tue May 29, 2012 9:39 am

mathewmithun wrote:is x=1, y=2 and z=3?
1: 5x+2z+3=3x+4y=y+2z+3
2: 5x+z+3=3x+4y and 3x+4y=y+2z+3

OA is E but I am getting the answer as D. Pls help

I don't know how Anurag's post is missing while I got mail intimation about the same. Anyways, I will explain the reason I went with A (i am correcting choice from D to A):

from statement1: 5x+2z+3=3x+4y ==> 2x-4y+2z=-3 substituting x=1,y=2 and z=3 on LHS: 2-8+6=0 while RHS is -3 and hence x y and z are not 1,2 and 3 respectively and hence I is sufficient to answer this question.

From statement 2: we have 3 variable and 2 equation and these equation are true when x=1, y=2 and z=3. But x y and z could take other values also and so we cannot for sure say statement 2 is sufficient. Hence A is the answer. Can pls someone check this question...

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Anurag@Gurome: GMAT Instructor; Posts: 3835; Joined: Fri Apr 02, 2010 10:00 pmFollowed by: 523 members; Location: Milpitas, CA; Thanked: 1854 times; GMAT Score:770

by Anurag@Gurome » Tue May 29, 2012 10:14 am

Algebraic Approach:

Statement 1: (5x + 2z + 3) = (3x + 4y) = (y + 2z + 3)
(3x + 4y) = (y + 2z + 3)

(5x + 2z + 3) = (3x + 4y) ---> (2x - 4y + 2z) = -3 ................. (1)
(3x + 4y) = (y + 2z + 3) ---> (3x + 3y - 2z) = -3 ................... (2)
(5x + 2z + 3) = (y + 2z + 3) ---> (5x - y) = 0 ................... (3)

From (3), y = 5x. Replacing this into (1) and (2) we have

(18x - 2z) = 3 ...................(4)
(18x - 2z) = -3 ...................... (5)

If we observe (4) and (5) carefully, we can see that no values of x and z can satisfy both of them simultaneously. Hence, there are no such x, y, and z that satisfies the given equation. Hence, answer to the question is NO.

Sufficient

Statement 2: (5x + z + 3) = (3x + 4y) and (3x + 4y) = (y + 2z + 3)

(5x + z + 3) = (3x + 4y) ---> (2x - 4y + z) = -3 .................. (A)
(3x + 4y) = (y + 2z + 3) ---> (3x + 3y - 2z) = -3 .................. (B)

We cannot solve for 3 unknowns from 2 equations. A infinite number of sets of values for x, y, and z will satisfy the given equation.

Not Sufficient

The correct answer is A.

Last edited by Anurag@Gurome on Tue May 29, 2012 10:18 am, edited 1 time in total.

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by Anurag@Gurome » Tue May 29, 2012 10:17 am

mathewmithun wrote:I don't know how Anurag's post is missing while I got mail intimation about the same.

Sorry, I was editing my post.

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