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(1) \(3|x^2-4|=y-2\). Now, since we are asked to find the value of y, from this statement we can conclude only that \(y\geq{2}\), as LHS is absolute value which is never negative, hence RHS als cannot be negative. Not sufficient.

(1) \(3|x^2-4|=y-2\). Now, since we are asked to find the value of y, from this statement we can conclude only that \(y\geq{2}\), as LHS is absolute value which is never negative, hence RHS als can not be negative. Not sufficient.

(1) \(3|x^2-4|=y-2\). Now, since we are asked to find the value of y, from this statement we can conclude only that \(y\geq{2}\), as LHS is absolute value which is never negative, hence RHS als can not be negative. Not sufficient.

(1)+(2) Since from (1) \(y\geq{2}\), then from (2) \(y=14\). Sufficient.

Answer: C.

Hope it's clear.

Sorry I can't to figure out why y >= 2........

1) 3x^2 - 4 = y-2 and -3x^2 + 4 = y-2 and then ??'

thanks

We are given that \(3|x^2-4|=y-2\). Now, the left hand side in this expression (\(3|x^2-4|\)) is an absolute value, so it cannot be negative, so the right hand side of the expression (\(y-2\)) must also be non-negative: \(y-2\geq{0}\) --> \(y\geq{2}\).

pls point out if i am wrong 1. 3(x^2- 4) = y- 2 and 3(x^2-4)= - (y-2) y=-4 y=-16

from 2 y = 8 y= 14

so e is the answer

from 2 y = 8 - should be y = -8; |3-8| = |-5| = 5 and not 11; |3-(-8)| = |3+8| = 11 y= 14

3(x^2- 4) = y-2 it should be \(3|x^2-4|=y-2\) if \(y-2\geq0\). and 3(x^2-4)= -(y-2) it should be \(3|x^2-4|=-(y-2)\) if \(y-2<0\) but we don't have any information about \(y\). y=-4 y=-16 No justification for these values, we don't know what is the value of \(x.\) Statement (1) is not sufficient.

Statement (2) provides two possible values for \(y\), not sufficient.

Taken together (1) and (2): since \(y-2\) equals an absolute value, it should be non-negative. Only \(y = 14\) is acceptable. Sufficient, therefore answer C. _________________

PhD in Applied Mathematics Love GMAT Quant questions and running.

Bunuel, I understand from statement 1 we get \(y>=2\) , but when we combine both statements together we get \(y>=2\) and \(y=14\). Now how can we just assume y to be 14, because y can also take the value of 2 right . I chose E on this basis .

Bunuel, I understand from statement 1 we get \(y>=2\) , but when we combine both statements together we get \(y>=2\) and \(y=14\). Now how can we just assume y to be 14, because y can also take the value of 2 right . I chose E on this basis .

Thanks TT

That's not correct. Y cannot be 2, otherwise the second equation would not be true. \(|3-2|=11\), \(1=11\) As you can see if we pick 2, the second equation is not verified. If take a look at the explanations above, you'll find out that statement 2 defines two possible vaules for y (-8,14)[ not enough to say the value of y]; and that statement 1 is true only for \(y>=2\), because \(3|x^2-4|=y-2\), the left part is always \(>=0\) (thanks to the abs value), and so the right part must be \(>=0\) too. If we merge those conditions:\(y=-8\) or \(y=14\) with \(y>=2\), the only value that y can have is 14

Hope it's clear now _________________

It is beyond a doubt that all our knowledge that begins with experience.

I know this is not needed for this problem but can someone show me how to solve 3|x^2 - 4| = y-2 from statement 1? Like what are the equations you could form if you tried to isolate y in this example?

I know this is not needed for this problem but can someone show me how to solve 3|x^2 - 4| = y-2 from statement 1? Like what are the equations you could form if you tried to isolate y in this example?

We have y = \(2+3|x^2-4|\)

Case I: x>2 -->

We can remove the modulus sign as it is and y = \(2+3(x^2-4) = 3x^2-10\)

(1) 3|x^2 – 4| = y – 2 It would take a long time to solve out for the positive and negative cases for x and even if we did I don't believe it would help us get an exact value for y because we would still have two variables to solve for (x, y) with no additional given information.

What we do know is that (y-2) must be ≥0 as it equals an absolute value, so y≥2 INSUFFICIENT

(2) |3 – y| = 11

Two cases:

Positive: y≤3: 3-y=11 ===> -y=8 ===> y=-8 Valid, as -8 falls within the range of ≤3 Negative: y>3: -3+y=11 ===> y=14 Valid, as 14 falls within the range of y>3 INSUFFICIENT - there are two valid solutions for y.

1+2)

What we know y≥2 y=-8 OR y=14

The only solution for y that also satisfies y≥2 is y=14 SUFFICIENT.

Again solving intuitively, and reaching the answer faster.

Picking the statement 2 first as it is the simpler of the two. (2) |3 – y| = 11

we get two values of Y; one +ve(14) and one -ve(-8). Not sufficient.

(1) 3|x^2 – 4| = y – 2 or, y= 3|x^2-4|+2 (RHS expression can never be -ve under any circumstance, since it involves a mod expression and '+2'). Therefore the value of Y is always +ve. But the statement in itself is not sufficient since we don't know the value of x.

Hence combining the two statements we can select the +ve value of Y from statement 2.

No way to determine exact values for x or y. INSUFFICIENT

(2) |3 – y| = 11 y<3 3 - y = 11 y = -8 Valid as -8 falls within the range of y<3 OR y>3 -(3 - y) = 11 -3 + y = 11 y = 14 Valid as y falls within the range or y > 3 Two valid solutions for y INSUFFICIENT

1+2) 3|x^2 – 4| = y – 2 and |3 – y| = 11 We could plug in values of y for the positive and negative cases of 3|x^2 - 4| or we could look at#1 and realize that because y-2 = an absolute value, it must be positive. Between the two solutions for y (-8 and 14) the only one that makes y - 2 positive is y = 14. SUFFICIENT

(1) \(3|x^2-4|=y-2\). Now, since we are asked to find the value of y, from this statement we can conclude only that \(y\geq{2}\), as LHS is absolute value which is never negative, hence RHS als cannot be negative. Not sufficient.

(1) \(3|x^2-4|=y-2\). Now, since we are asked to find the value of y, from this statement we can conclude only that \(y\geq{2}\), as LHS is absolute value which is never negative, hence RHS als cannot be negative. Not sufficient.

(1)+(2) Since from (1) \(y\geq{2}\), then from (2) \(y=14\). Sufficient.

Answer: C.

Hope it's clear.

Hi,

Can I solve statement 1 like this:

3|x^2-4|=y-2

Now since this is an absolute value

I would 1st solve for x

x^2-4=0 x2=4 and x=+/-2 now if I substituent the value of x in the above expression If x= +2 3|x^2-4|=y-2 3|(2)^2-4|=y-2 3|0|=y-2 therefore y=2

now if x=-2 3|x^2-4|=y-2 3|(-2)^2-4|=y-2 3|0|=y-2 and therefore y=2

In both the cases I will get the same value for Y.

Can someone please explain what is wrong with this approach.

We don't know whether x^2-4=0, thus all your further steps are based on that false assumption. If we knew that x^2-4=0, then x^2-4=0=y-2 --> y-2=0 --> y=2.

Also, you can notice that your approach is not correct from the fact that on the GMAT, two data sufficiency statements always provide TRUE information and these statements never contradict each other. From (2) we have that y is -8 or 14, and if from (1) you get that y is 2 it would mean that the statements clearly contradict.

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