AAPL wrote: ↑

Tue May 12, 2020 7:48 am

GMAT Paper Tests

If w, x, y, and z are non-negative integers, each less than 3, and w(3^3)+x(3^2)+y(3)+z=34, then w+z=

A. 0
B. 1
C. 2
D. 3
E. 4

OA C

Given that w(3^3)+x(3^2)+y(3)+z=34, we have

27w + 9x + 3y + z = 34

Since it is given that w, x, y, and z are non-negative integers, each less than 3, w, x, y and z can only be one among (0, 1, 2).

Looking at 27w + 9x + 3y + z = 34, we see that w ≠ 2 as at w = 2, we have 27*2 = 54 > 34.

Also, w ≠ 0, as at w = 0, we have 9x + 3y + z = 34. Even if we take the maximum value of x, y, and z, each equal to 2, we have 9x + 3y + z = 26 < 34.

So, w = 1.

Thus, from 27*1 + 9x + 3y + z = 34 => 9x + 3y + z = 7.

It's evident that x must be 0. Thus, we have 3y + z = 7. The only solution is y = 2 and z = 1.

Thus, w + z = 1 + 1 = 2.

The correct answer: C

Hope this helps!

-Jay
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AAPL wrote: ↑

Tue May 12, 2020 7:48 am

GMAT Paper Tests

If w, x, y, and z are non-negative integers, each less than 3, and w(3^3)+x(3^2)+y(3)+z=34, then w+z=

A. 0
B. 1
C. 2
D. 3
E. 4

OA C

Solution:

The given equation can be rewritten as 27w + 9x + 3y + z = 34. If w = 0, then either x, y or z must be greater than or equal to 3 so that the sum can equal 34. If w = 2, then 27w is already greater than 34. We see that the only possible value for w is 1.

Since w = 1, we have:

27 + 9x + 3y + z = 34

9x + 3y + z = 7

We see that x must be 0, y must be 2 and z must be 1. Therefore, w + z = 1 + 1 = 2.

Answer: C