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

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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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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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