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17 Oct 2015, 15:37
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E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
55%
(hard)
Question Stats:
66% (01:57) correct
34%
(01:53)
wrong
based on 519
sessions
History
Date
Time
Result
Not Attempted Yet
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
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
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
18 Oct 2015, 12:55
Kudos
Bookmarks
oloman
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
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
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
+1 kudos if it helped you
17 Oct 2015, 17:40
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Bookmarks
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:
GMAT assassins aren't born, they're made,
Rich
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:
GMAT assassins aren't born, they're made,
Rich
