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What is the value of y? (1) 3|x^2 -4| = y - 2 (2) |3 - y| = 11

(1) As 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.

What is the value of y? (1) 3|x^2 -4| = y - 2 (2) |3 - y| = 11

(1) As 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.

What is the value of y? (1) 3|x^2 -4| = y - 2 (2) |3 - y| = 11

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

Answer: C.

Hey, pls explain the reasoning of LHS

LHS is absolute value which is never negative, hence RHS als can not be negative.???

what values you can put for x? if you put x = 2, left hand side becomes zero, making y=2. if you put x as any other value, it will make 3*|x^2 - 4| positive, and add 2 to that, making y > 2. so, y is either 2 or greater.

1st equation: x^2 - 4 can be positive, or negative ( say x = 1, then x^2 - 4 is -3, and say x = 3, then x^2-4 is 5. However since its absolute value is taken, it will always be a positive number, which when multiplied by 3 will give either zero (if x=2) or a positive number. This added to 2 brought pver fom RHS of the equation will have to be a positive number still. So while Statement 1 is insufficient in telling us the value of y, it does tell us that y is positive. Equation 2: by telling us that absolute value of 3-y is 11 implies that the vlue inside the absolute value lines is either 11 or -11, which means that y is either 14 or -8. since it gives two values of y it is insufficient by itself. However since statement tells us for certain that y i s positive and statement two give sus two values of y one positive and one negative, both combined we can pick the positive value and be sure that that is the correct value of y hence both together ae sufficient

Re: What is the value of y? (1) 3|x^2 4| = y 2 (2) |3 y| = 11 [#permalink]
29 Aug 2013, 08:39

What is the value of y? (1) 3|x^2 – 4| = y – 2 (2) |3 – y| = 11

Stmt 1: Y=3|x^2-4|+2. The value of y will change with the value of x. insufficient. Stmt2: 3-y=11 OR 3-y=-11. We get y=-8 in first case and y=14 in second. Insufficient. Together: Anything inside a Modulus is ALWAYS positive. Therefore, |x^2-4| will always positive. And when we add 2 to it, the result still remains positive. That means, y cannot be -8 from stmt2. Sufficient. Answer C.

[quote="Bunuel"]What is the value of y? (1) 3|x^2 -4| = y - 2 (2) |3 - y| = 11

(1) As 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."

Bunuel,

In this case we said since the LHS has a mod so its value is positive always. But we also know mod function is negative for values of x < 0 , or we dont always assume for a given equation say |x-2| =2 , that LHS is positive always ? we say x-2 = x-2 , when x> 2 and 2-x when x < 2. So this looks a contradiction, can you please explain ?

What is the value of y? (1) 3|x^2 -4| = y - 2 (2) |3 - y| = 11

(1) As 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."

Bunuel,

In this case we said since the LHS has a mod so its value is positive always. But we also know mod function is negative for values of x < 0 , or we dont always assume for a given equation say |x-2| =2 , that LHS is positive always ? we say x-2 = x-2 , when x> 2 and 2-x when x < 2. So this looks a contradiction, can you please explain ?

In this case for statement 1,

as it is in modulus so LHS will always be positive, its x whose value can be -ve or +ve, but LHS will always be +ve. Therefore we conclude that y-2>=0 or y>=2

From statement 2 we get y=-8,14

Combining 1 and 2, we get y=14

Hope it is clear.
_________________

--It's one thing to get defeated, but another to accept it.

as it is in modulus so LHS will always be positive, its x whose value can be -ve or +ve, but LHS will always be +ve. Therefore we conclude that y-2>=0 or y>=2

From statement 2 we get y=-8,14

Combining 1 and 2, we get y=14

Hope it is clear.[/quote]

Let us take an example. y =|X-2| So y = x-2 , when x-2>0 or x-2=0 and y = 2-x, when x-2<0 ro x <2

If we take x = 3, then y = -1, how ? I am doing something silly here for sure! If you see graph of mod function Y is always positive, so how is Y = -1 above at x =3 ?

What is the value of y? (1) 3|x^2 -4| = y - 2 (2) |3 - y| = 11

(1) As 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."

Bunuel,

In this case we said since the LHS has a mod so its value is positive always. But we also know mod function is negative for values of x < 0 , or we dont always assume for a given equation say |x-2| =2 , that LHS is positive always ? we say x-2 = x-2 , when x> 2 and 2-x when x < 2. So this looks a contradiction, can you please explain ?

It seems that you should brush up your fundamentals.

First of all: absolute value is always greater than or equal to zero (\geq{0}), not >0.

Next, |x|=-x, when x<0 --> |x|=-x=-negative=positive.

Ok Thanks. I was missing this - of -x = +x, So basically in mod function x can range from - to +, but y will always be >0 or = 0. Correct me if I understood it wrongly.

Ok Thanks. I was missing this - of -x = +x, So basically in mod function x can range from - to +, but y will always be >0 or = 0. Correct me if I understood it wrongly.

Apart from the red part all is correct. The red part does not make sense.
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