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Manager  Joined: 20 Mar 2015
Posts: 56
Re: What is the value of y?  [#permalink]

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Bunuel wrote:
What is the value of y?

(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.

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

Hope it's clear.

Why is Y>=0? Can't we write 'Y>0'? If LHS is positive and RHS must also be positive then, RHS has to be greater than 0. As 0 is neither positive nor negative, how can we write Y>=0?

|some value|= some value. Since LHS is an absolute value, RHS has to be positive as well. Why is RHS>=0 and not RHS>0?
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Re: What is the value of y?  [#permalink]

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shekhar4847 wrote:
Bunuel wrote:
What is the value of y?

(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.

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

Hope it's clear.

what if y is between 2 and 3?
In that case we will have two values of y.

Hi shekhar4847,

I gives you $$y\geq{2}$$...
II gives you two cases..
a) $$y<3$$ --> $$3-y=11$$ --> $$y=-8$$;
this means if y<3, it can have ONLY -8 as the value,
so when we combine this with $$y\geq{2}$$,which does not fit between 2 and 3.. hence WRONG

b)$$y\geq{3}$$ --> $$-3+y=11$$ --> $$y=14$$
this means if $$y\geq{3}$$, it can have ONLY 14 as the value,
so when we combine this with $$y\geq{2}$$, 14 fits in both the ranges.. hence CORRECT
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Re: What is the value of y?  [#permalink]

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bimalr9 wrote:
Bunuel wrote:
What is the value of y?

(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.

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

Hope it's clear.

Why is Y>=0? Can't we write 'Y>0'? If LHS is positive and RHS must also be positive then, RHS has to be greater than 0. As 0 is neither positive nor negative, how can we write Y>=0?

|some value|= some value. Since LHS is an absolute value, RHS has to be positive as well. Why is RHS>=0 and not RHS>0?

Hi,
an ABSOLUTE value is NOT necessarily positive, it is NON-NEGATIVE, so it can include both 0 and +ive value..
if in statement I x is 2 or -2, LHS will be 0 and hence y=2..
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Re: What is the value of y?  [#permalink]

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dvinoth86 wrote:
What is the value of y?

(1) 3|x^2 – 4| = y – 2
(2) |3 – y| = 11

Y = ?

stmt:1 - it has two variables X AND Y. we need value of Y. insuff.

stmt:2 - Y=-8 or Y=11 insuff.

combined: -

frm stmt-1 3|x^2 – 4| = y – 2

X^2 is positive. value of mod is positive. so left hand side is POSITIVE.

frm stmt-2 Y=-8 or Y=11

so we can only pick 11 as the value to keep right hand side POSITIVE.

Option C is right.
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What is the value of y?  [#permalink]

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Hi Bunuel,

Please correct me if i am wrong.

i). 3|x^2 - 4| =y-2

3x^2 = y+ 10 OR 3x^2 = -y +14
equating the 2 values of 3x^2 Cant we equate these 2 values?
y=10 = -y +14
y=2

ii). gives 2 values of y -8 & 14

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Re: What is the value of y?  [#permalink]

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shrashtisinghal wrote:
Hi Bunuel,

Please correct me if i am wrong.

i). 3|x^2 - 4| =y-2

3x^2 = y+ 10 OR 3x^2 = -y +14
equating the 2 values of 3x^2 Cant we equate these 2 values?
y=10 = -y +14
y=2

ii). gives 2 values of y -8 & 14

How do you know that we have

3x^2 - 12 =y-2

or

12 - 3x^2 =y-2

Since, we donot know which of the above equations is valid. We cannot deduce a single value.

As per Modulus property,

Mod(x) = x if x>0
= -x if x<0.

Hence A is not sufficient.

In B also, we are getting two values of y, so not sufficient.

But, when you combine the statements, you will see the equation is satisfied only for y =14. hence, C.
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Re: What is the value of y?  [#permalink]

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dvinoth86 wrote:
What is the value of y?

(1) 3|x^2 – 4| = y – 2
(2) |3 – y| = 11

FROM 1
(y-2)/3 >= 0 , thus y>= 2 ... ... insuff

from 2

3-y = 11 or 3-y = -11 , thus Y could be either -8 or 14.. insuff

both together

since from 1 y>= 2 and from 2 y = -8 or 14 then y = 4 ( common domain) ...C
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Re: What is the value of y?  [#permalink]

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Bunuel wrote:
What is the value of y?

(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.

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

Hope it's clear.

Brilliant approach! I'd never thought it could be this easy.
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Re: What is the value of y?  [#permalink]

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Bunuel wrote:
What is the value of y?

(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.

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

Hope it's clear.

Since we have the absolute value of (x^2-4), couldn't it be considered as the absolute value of (x+2)(x-2)???
This way, you'd have two possible roots: -2 and 2. Then, since it's an absolute value, the LHS of the equation would give you always 6, making y=8.
What did I miss? Re: What is the value of y?   [#permalink] 01 Mar 2019, 14:08

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