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Grid with sums

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Source: — Problem Solving |
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by parallel_chase » Sun Dec 28, 2008 12:11 pm
2w + y + x = 6-------I
w + y + 2x = 3-------II

Subtract the 2 equations, we get w-x =3 or w = 3+x

we have to find

3x + w = n

Substitute the value of w

3x + x + 3 = n

4x+3 = n

plug the answer choices.


4x + 3 = 6, x = 3/4 not an integer - eliminate

4x + 3 = 7 , x = 1 integer--answer

4x + 3 = 8, x = 5/4 not an integer - eliminate

4x + 3 = 9, x = x=6/4 or 3/2 not an integer - eliminate

4x + 3 = 10, x = 7/4 not an integer - eliminate

Hence B.


If B is the answer, my solution goes against 1 condition in the question stem.

Z<W<X<Y

if W is less than X, then how come i am getting W=X+3 or is it that W is negative. If W is negative then X also has to to be negative.

Is there something that i am missing.
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by cramya » Sun Dec 28, 2008 12:25 pm
Can someone please hlep me underatand this?

w = x+3 like PC said

I am working wiht 2 other equations and getting x as a non intger(I must be missing something also wiht the problem constraints and whats given)


x+w+2z = -2 I

x+2y+w = 5 II
w+2y+z = 7 III

Subtract III from II
x-z = -2

z = x+2


Sub z=x+2 in I

x+w+2(x+2) = -2

3x+w=-6 IV
w-x = 3 V

Subtract V FROM IV

4X =-9
X = -9/4???


May be I messed up the grid....
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by Brent@GMATPrepNow » Sun Dec 28, 2008 12:40 pm
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Wow, am I ever sloppy.
The letter in the top right corner is wrong. It should have been a Z.
I have corrected it here.

Having said that, I recommend that you do not solve this system for W, X, Y, and Z. Solving 4 equations with 4 unknowns is beyond the scope of the GMAT.

As you probably guessed, there is a quick (10-second) way to solve this question.

A thousand apologies for posting the wrong question.
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by rajataga » Sun Dec 28, 2008 12:51 pm
yup, right now...

solve first row and first column,

u get z = x - 5

substitute in 1st row,

u get 3x + w = 8

therefore, n = 3x + w = 8
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by parallel_chase » Sun Dec 28, 2008 1:05 pm
x+w+2z = -2----------I
2z+w+y = 0 or 2z = -w -y

substitute the value in the first equation
x + w - w -y = -2 or x-y = -2-------II

2x + w + z = 3
2x + y + z = 9

Subtract both the equations

y-w = 6 or y = w+6-------III

substitute III in II

x - w -6 = -2

x = 4+w

n=3x + w

n = 12 + 3w +w = 12+ 4w

12 + 4w = n

plug in the answer options

12 + 4w = 6, w= -3/2 not an integer

12 + 4w = 7, w= -5/4, not an integer

12 + 4w = 8, w= -1, integer

12 + 4w = 9, w= -3/4, not an integer

12 + 4w = 10, w=-1/2, not an integer

Hence Answer is C.
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by Brent@GMATPrepNow » Sun Dec 28, 2008 1:45 pm
This one is quite evil.
The sums of the 4 rows are given, so if we add all 4 of those sums (3+6+9+0=18) we will get the sum of all 16 numbers in the grid.

If we add the 4 column sums, we better get the same sum since we are adding the same 16 numbers.

So, we can conclude that (-2)+7+n+5 = 18
Solve for n to get n=8

Told you it was evil :D
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by parallel_chase » Sun Dec 28, 2008 1:47 pm
Brent Hanneson wrote:This one is quite evil.
The sums of the 4 rows are given, so if we add all 4 of those sums (3+6+9+0=18) we will get the sum of all 16 numbers in the grid.

If we add the 4 column sums, we better get the same sum since we are adding the same 16 numbers.

So, we can conclude that (-2)+7+n+5 = 18
Solve for n to get n=8

Told you it was evil :D
Nice question, but Awesome solution.

Thanks.
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by rajataga » Sun Dec 28, 2008 10:48 pm
Damn cool solution.......
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by pbanavara » Tue Dec 30, 2008 4:03 pm
The solution is way too cool. Thanks Brent.
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by logitech » Tue Dec 30, 2008 4:07 pm
0+3+6+9 = n + 5 - 2 + 7

n = 8
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