If x ≠ 0 and x^2z – 4xy + y = 0, what is the value of z ?

Question

If  x ≠ 0  and  x^2z – 4xy + y = 0 , what is the value of z ?

(1)  z = 32x

(2)  z = 4y

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nick1816 · Accepted Answer

Statement 1-

x^2z – 4xy + y = 0

x^2*32x -4xy +y = 0

We can't find x or z using the above equation.

Insufficient

Statement 2-

x^2z – 4xy + y = 0

x^2*4y – 4xy + y = 0

y( 4x^2 - 4x+1) = 0

y(2x-1)^2 = 0

either y= 0 or x= 1/2

If y = 0, we can find z.
If x=1/2, we can't find z.

Insufficient

Combining both equations

1. If y=0, then x=0 ( reject this case )

2. If x = 1/2; x= 16

Sufficient

Kinshook · Answer

Asked: If  x ≠ 0  and  x^2z – 4xy + y = 0 , what is the value of z ?

(1)  z = 32x 
 x^2*32x - 4xy + y = 0 
 32x^3 - 4xy + y = 0 
Since y is unknown
NOT SUFFICIENT

(2)  z = 4y 
 4x^2y -4xy + y = 0 
 y(4x^2 - 4x + 1) = 0 
 y (2x - 1)^2 = 0 
x = 1/2; if y ≠ 0
Since y is unknown 
z is unknown
NOT SUFFICIENT

(1) + (2)
(1)  z = 32x 
(2)  z = 4y 
z = 32x = 4y; y = 8x
 32x^3 - 32x^2 + 8x = 0 
 8x(4x^2 - 4x + 1) = 0 
 x(2x-1)^2 = 0 
x =1/2 ; z = 16; Since  x 
eq 0  
SUFFICIENT

IMO C

Basshead · Answer

x^2z – 4xy + y = 0

Each statement is not sufficient alone so lets see if the two statements combined are sufficient:

x^2(32x) – xz + y = 0 
 32x^3 – 32x^2 + 8x = 0 
 8x(4x^2 – 32x + 1) = 0 
 x(2x^2 – 1)^2 = 0

x = \frac{1}{2}  and  z = 16 . SUFFICIENT.

Answer is C.

ubertu · Answer

Why when y=0 we cannot find z?

TestPrepUnlimited · Answer

ubertu

He wrote "we CAN find z"

TestPrepUnlimited · Answer

Note we have 3 variables and only one equation. The expectation is we need at least 3 equations at least to solve for z.

Statement 1:

We can kick z out of the equation and search for x instead now. The equation is  32x^3 - 4xy + y = 0  which still has 2 variables, insufficient.

Statement 2:

Again replace z and find y instead now. The equation is  4x^2y - 4xy + y = 0 . We are able to factor this into  y*(4x^2 - 4x + 1) = y*(2x - 1)^2 = 0 . We cannot tell what y is when  2x - 1 = 0  so insufficient.

Combined:

Combined we have  z =32x = 4y , so we have  y = 8x . Let us take the easier equation from statement 2, plug in  y = 8x  to get  8x*(2x - 1)^2 = 0 . We cannot have x = 0 as the question mentions x cannot be 0. Then we can only have  (2x - 1)^2 = 0  and  2x - 1 = 0 . Sufficient as finding x gives us y and z.

Ans: C

MathRevolution · Answer

Forget the conventional way to solve DS questions.

We will solve this DS question using the variable approach.

DS question with 3 variables and 1 Equation: Let the original condition in a DS question contain 3 variables and 1 Equation. Now, we know that each condition (1) and (2) would usually give us an equation each and we need 2 equations to match the numbers of variables and equations in the original condition, therefore the most likely answer is C.

To master the Variable Approach, visit https://www.mathrevolution.com and check our lessons and proven techniques to score high in DS questions.

Let’s apply the 3 steps suggested previously. [Watch lessons on our website to master these 3 steps]

Step 1 of the Variable Approach: Modifying and rechecking the original condition and the question.

We have to find value of 'z' .

=> $x^2z - 4xy + y = 0$

Second and the third step of Variable Approach: From the original condition, we have 3 variables (x, y, and z) + 1 equation (x^2z - 4xy + y = 0) .To match the number of variables with the number of equations, we need 2 equations. Since conditions (1) and (2) will provide 1 equation each, C would most likely be the answer.

Let's take both conditions together.

From condition(1), z = 32x and from the condition(2) z = 4y

=> z = 32x = 4y

=> 4y = 32x or y = 8x

=> $x^2 * 32x - 4x(8x) + (8x) = 0$

=> $32x^3 - 32x^2 + 8x = 0 $

=> $8x (4x^2 - 4x + 1) = 0$

=> $8x (2x - 1)^2 = 0$

=> 'x' is not equal to zero and hence, $(2x - 1)^2 = 0$

=> $(2x - 1)^2 = 0$ or x = $\frac{1}{2}$.

Therefore, z = 32x = 32 * $\frac{1}{2}$ = 16

Since the answer is unique, both conditions together are sufficient by CMT 2.

So, C is the correct answer.

Answer: C

Utkarsha20 · Answer

Clearly each statement alone is not sufficient.

On combining both statements we have

$$z = 32x = 4y$$

Therefore,

$$x = \frac{z}{32} \quad 	ext{and} \quad y = \frac{z}{4}$$

Putting these values into the equation we get

$$\left(\frac{z}{32}ight)^2 z - 4\left(\frac{z}{32}ight)\left(\frac{z}{4}ight) + y = 0$$

Simplifying we get

$$z\left(\frac{z^2}{32} - 4\left(\frac{z}{4}ight) + 8ight) = 0$$
$$z(z^2 - 32z + 8 \cdot 32) = 0$$
$$z(z^2 - 16z - 16z + 8 \cdot 32) = 0$$

Hence we obtain the following values
$$z = 0, 16, 16$$
We cannot take $$z = 0$$ since that would make $$x = 0$$.
Therefore,
$$z = 16$$

Thus both statements together are sufficient.

Answer: C

If x ≠ 0 and x^2z – 4xy + y = 0, what is the value of z ?

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GMAT Data Sufficiency (DS) Questions

If x ≠ 0 and x^2z – 4xy + y = 0, what is the value of z ?

Prep Toolkit

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My Rewards