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

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What is the value of y? (1) 3|x^2 4| = y 2 (2) |3 y| = 11 [#permalink]

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29 Sep 2010, 08:33
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What is the value of y?

(1) 3|x^2 – 4| = y – 2

(2) |3 – y| = 11
[Reveal] Spoiler: OA

Last edited by prashantbacchewar on 29 Sep 2010, 08:38, edited 1 time in total.

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29 Sep 2010, 08:37
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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.

(2) $$|3 - y| = 11$$:

$$y<3$$ --> $$3-y=11$$ --> $$y=-8$$;
$$y\geq{3}$$ --> $$-3+y=11$$ --> $$y=14$$.

Two values for $$y$$. Not sufficient.

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

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Re: DS problem 3|x^2 – 4| = y – 2 [#permalink]

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29 Sep 2010, 10:14
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prashantbacchewar wrote:
What is the value of y?

(1) 3|x^2 – 4| = y – 2

(2) |3 – y| = 11

from 1

equation with 2 unknown we can only deduct that y-2> 0 ie: y>2

from 2

y has to have 2 values +ve and -ve..........insuff

both

y>2 , then from 2 y has the +ve value.

C

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03 Oct 2010, 16:29
Bunuel wrote:
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.

(2) $$|3 - y| = 11$$:

$$y<3$$ --> $$3-y=11$$ --> $$y=-8$$;
$$y\geq{3}$$ --> $$-3+y=11$$ --> $$y=14$$.

Two values for $$y$$. Not sufficient.

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

Hey, pls explain the reasoning of LHS

LHS is absolute value which is never negative, hence RHS als can not be negative.???
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03 Oct 2010, 16:57
hirendhanak wrote:
Bunuel wrote:
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.

(2) $$|3 - y| = 11$$:

$$y<3$$ --> $$3-y=11$$ --> $$y=-8$$;
$$y\geq{3}$$ --> $$-3+y=11$$ --> $$y=14$$.

Two values for $$y$$. Not sufficient.

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

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.

Hope this helps.
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05 Apr 2011, 22:36
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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

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Re: DS problem 3|x^2 – 4| = y – 2 [#permalink]

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06 Apr 2011, 04:53
The answer is C. (1) says y is +ve, and (2) gives 2 values of y, one +ve and one -ve. So combining both we get y = 14.
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Re: DS problem 3|x^2 – 4| = y – 2 [#permalink]

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16 Apr 2011, 17:17
1. Insufficient

because all we know here is y>2

2. Insufficient.

because y can be -8 or 14

together . Sufficient.

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

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29 Aug 2013, 09:39
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.

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08 Sep 2013, 00:55
[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.

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08 Sep 2013, 05:01
ygdrasil24 wrote:
Bunuel wrote:
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.

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.
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08 Sep 2013, 05:23
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.[/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 ?

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08 Sep 2013, 05:33
ygdrasil24 wrote:
Bunuel wrote:
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.

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 --> $$|x|=-x=-negative=positive$$.

Theory on Abolute Values: math-absolute-value-modulus-86462.html

DS Abolute Values Questions to practice: search.php?search_id=tag&tag_id=37
PS Abolute Values Questions to practice: search.php?search_id=tag&tag_id=58

Hard set on Abolute Values: inequality-and-absolute-value-questions-from-my-collection-86939.html

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08 Sep 2013, 05:38
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.

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08 Sep 2013, 05:40
ygdrasil24 wrote:
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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Re: DS problem   [#permalink] 08 Sep 2013, 05:40
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