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If (x^2 + 1)y=3, which of the following is not a possible value for y?

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If (x^2 + 1)y=3, which of the following is not a possible value for y?  [#permalink]

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New post 17 Oct 2015, 15:37
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If (x^2 + 1)y=3, which of the following is not a possible value for y?

A. 3/2
B. 2
C. 8/3
D. 3
E. 7/2

This is from a Kaplan CAT test.

Kaplan states that "The variable x is being squared, and we know that whenever any value is squared, the result is always a positive number"
So when I do the match with a, b and c I understand how those can be values for y given that x^2 = a positive number.
For D however, you will get x^2=0, now here is where I have my question. According to the web, zero is neither positive nor negative, according to kaplan's explanation, any squared number must be a positive number, then how can x^2=0 be correct in saying that x^2= a positive number if 0 is nor positive nor negative.
For e, I understand that x^2 = (a negative number) hence this cannot be a value so this is not a possible value of y.


I understand why e is correct, but I do not understand why D is technically wrong!, can someone please help?
best
Oloman
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Re: If (x^2 + 1)y=3, which of the following is not a possible value for y?  [#permalink]

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New post 17 Oct 2015, 17:40
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Hi oloman,

To start, NOT every number that's squared is positive; squaring the number 0 will get you 0. Here though, since we're dealing with (X^2 + 1), that sum will always be AT LEAST 1. That piece of information is the 'key' to this question. Here's why:

We're given (X^+1)(Y) = 3

As (X^2+1) increases in value, Y will DECREASE in value.

So, when (X^2+1) = 1, we have...

(1)(Y) = 3 and Y = 3

But if X=1, then (X^2+1) = 2....

(2)(Y) = 3 and Y = 3/2

You'll find that no matter what value you plug in for X, Y can NEVER be greater than 3.

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If (x^2 + 1)y=3, which of the following is not a possible value for y?  [#permalink]

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New post 18 Oct 2015, 12:55
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oloman wrote:
If (x^2 + 1)y=3, which of the following is not a possible value for y?

A. 3/2
B. 2
C. 8/3
D. 3
E. 7/2

This is from a Kaplan CAT test.

Kaplan states that "The variable x is being squared, and we know that whenever any value is squared, the result is always a positive number"
So when I do the match with a, b and c I understand how those can be values for y given that x^2 = a positive number.
For D however, you will get x^2=0, now here is where I have my question. According to the web, zero is neither positive nor negative, according to kaplan's explanation, any squared number must be a positive number, then how can x^2=0 be correct in saying that x^2= a positive number if 0 is nor positive nor negative.
For e, I understand that x^2 = (a negative number) hence this cannot be a value so this is not a possible value of y.


I understand why e is correct, but I do not understand why D is technically wrong!, can someone please help?
best
Oloman


Its simple
equation is (x^2 +1)y = 3 ===> x^2 = (3/y) - 1 . As , in LHS, it is x^2 the value has to be positive or zero in RHS.
Hence, (3/y) - 1>= 0 ===> y =< 3. Now only option E has greater value of 'y' than 3.
Thus, correct ans is E

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Re: If (x^2 + 1)y=3, which of the following is not a possible value for y?  [#permalink]

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New post 18 Oct 2015, 12:30
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thanks! makes sense,
So, can I just say that x^2 +1 >= 1 , so, 3/y>=1, then plug in answer choices and see which one wont satisfy?? The underlining lesson here is that x^2 can be 0 as well?
best

EMPOWERgmatRichC wrote:
Hi oloman,

To start, NOT every number that's squared is positive; squaring the number 0 will get you 0. Here though, since we're dealing with (X^2 + 1), that sum will always be AT LEAST 1. That piece of information is the 'key' to this question. Here's why:

We're given (X^+1)(Y) = 3

As (X^2+1) increases in value, Y will DECREASE in value.

So, when (X^2+1) = 1, we have...

(1)(Y) = 3 and Y = 3

But if X=1, then (X^2+1) = 2....

(2)(Y) = 3 and Y = 3/2

You'll find that no matter what value you plug in for X, Y can NEVER be greater than 3.

Final Answer:

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Rich
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Re: If (x^2 + 1)y=3, which of the following is not a possible value for y?  [#permalink]

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New post 18 Oct 2015, 18:16
Hi oloman,

Yes, based on the concepts involved, your approach would absolutely work.

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Re: If (x^2 + 1)y=3, which of the following is not a possible value for y?  [#permalink]

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New post 18 Oct 2015, 19:04
VikashAlex wrote:

Its simple
equation is (x^2 +1)y = 3 ===> x^2 = (3/y) - 1 . As , in LHS, it is x^2 the value has to be positive or zero in RHS.
Hence, (3/y) - 1>= 0 ===> y =< 3. Now only option E has greater value of 'y' than 3.
Thus, correct ans is E

+1 kudos if it helped you


A bit of correction in your solution. The text in red above is not the complete range. The complete range of values of y from 3/y-1 \(\geq\) 0 will be 0 \(\leq\) y \(\leq\) 3
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Re: If (x^2 + 1)y=3, which of the following is not a possible value for y?  [#permalink]

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New post 19 Oct 2015, 00:20
Engr2012 wrote:
VikashAlex wrote:

Its simple
equation is (x^2 +1)y = 3 ===> x^2 = (3/y) - 1 . As , in LHS, it is x^2 the value has to be positive or zero in RHS.
Hence, (3/y) - 1>= 0 ===> y =< 3. Now only option E has greater value of 'y' than 3.
Thus, correct ans is E

+1 kudos if it helped you


A bit of correction in your solution. The text in red above is not the complete range. The complete range of values of y from 3/y-1 \(\geq\) 0 will be 0 \(\leq\) y \(\leq\) 3


Hi,
Yes, its true the y range will be 0 ≤ y ≤ 3. Missed it 0 part. But in any case the option E does not fall in between this reason.
Thank you for pointing out. This is the silly thing that I miss during DS ques.
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Re: If (x^2 + 1)y=3, which of the following is not a possible value for y?  [#permalink]

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New post 22 Oct 2015, 11:39
y=3/(x^2+1) => y<=3
for all answer choices y<=3 except E
ans E
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Re: If (x^2 + 1)y=3, which of the following is not a possible value for y?  [#permalink]

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New post 05 Jun 2017, 05:29
IMO E
3\(x^2+1)=Y
From the above equation we can infer that the value of y <=3
We have X^2 in denominator which is always positive.
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Re: If (x^2 + 1)y=3, which of the following is not a possible value for y?  [#permalink]

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New post 05 Jun 2017, 06:36
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oloman wrote:
If (x^2 + 1)y=3, which of the following is not a possible value for y?

A. 3/2
B. 2
C. 8/3
D. 3
E. 7/2

This is from a Kaplan CAT test.

Kaplan states that "The variable x is being squared, and we know that whenever any value is squared, the result is always a positive number"
So when I do the match with a, b and c I understand how those can be values for y given that x^2 = a positive number.
For D however, you will get x^2=0, now here is where I have my question. According to the web, zero is neither positive nor negative, according to kaplan's explanation, any squared number must be a positive number, then how can x^2=0 be correct in saying that x^2= a positive number if 0 is nor positive nor negative.
For e, I understand that x^2 = (a negative number) hence this cannot be a value so this is not a possible value of y.


I understand why e is correct, but I do not understand why D is technically wrong!, can someone please help?
best
Oloman


The only constraint of \((x^2 + 1)*y=3\) is that \(x^2\) cannot be negative.

So, \(x^2 = 3/y - 1\)
\(3/y - 1\) cannot be negative.
Plug in the values to see which one gives a negative value.

\(3/y - 1\)

Only option (E) gives you a negative value.
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Re: If (x^2 + 1)y=3, which of the following is not a possible value for y?  [#permalink]

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New post 06 Jun 2017, 16:36
oloman wrote:
If (x^2 + 1)y=3, which of the following is not a possible value for y?

A. 3/2
B. 2
C. 8/3
D. 3
E. 7/2


We are given that (x^2 + 1)y = 3.

Since x^2 is always nonnegative, we see that x^2 + 1 must always be greater than or equal to 1. Thus, in order for the product of a number greater than or equal to 1 and y to be 3, y has to be less than or equal to 3. Thus, y cannot equal 7/2.

Answer: E
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Re: If (x^2 + 1)y=3, which of the following is not a possible value for y?  [#permalink]

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