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07 Jul 2022, 01:30
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A
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:
59% (01:46) correct
41%
(01:48)
wrong
based on 41
sessions
History
Date
Time
Result
Not Attempted Yet
If \(x ≠ –1,\) then \(\frac{2cx^2 + 4cx + cy}{c (x + 1)^2} =\)
(1) \(y = 2\)
(2) \(c = 3\)
Are You Up For the Challenge: 700 Level Questions
(1) \(y = 2\)
(2) \(c = 3\)
Are You Up For the Challenge: 700 Level Questions
07 Jul 2022, 01:54
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This question is a bit tricky because it only sates that x<>-1.
It is silent about the value of c. If c is 0, the value is undefined.
We have to proceed with the understanding that this is not an oversight, but a trap for unsuspecting test takers.
(1) Only
If c is 0, the value is not defined.
If c is not 0, we can simplify the value to be
(2x^2+4x+1)/(x+1)^2=2
Not sufficient
(2)Only
Without knowing the value of y, we cannot have one specific value.
Not sufficient
(1)(2) Together
The value is 2.
Sufficient.
The answer is thus (C).
It is silent about the value of c. If c is 0, the value is undefined.
We have to proceed with the understanding that this is not an oversight, but a trap for unsuspecting test takers.
(1) Only
If c is 0, the value is not defined.
If c is not 0, we can simplify the value to be
(2x^2+4x+1)/(x+1)^2=2
Not sufficient
(2)Only
Without knowing the value of y, we cannot have one specific value.
Not sufficient
(1)(2) Together
The value is 2.
Sufficient.
The answer is thus (C).
Sushil117
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07 Jul 2022, 10:30
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If x≠–1,x≠–1, then 2cx^2+4cx+cy/c(x+1)^2=
(1) y=2
by taking C common and cancelling , (2x^2+ 4x+y)/(x+1)^2
putting value of y
(2x^2+ 4x+2)/(x+1)^2
taking two 2 common
2(x^2+2x+1)/(x+1)^2
2(x+1)^2/(x+1)^2
=2 sufficient
(2) c=3
by putting value of x
(18+12+y)/16
insufficient
IMO A
(1) y=2
by taking C common and cancelling , (2x^2+ 4x+y)/(x+1)^2
putting value of y
(2x^2+ 4x+2)/(x+1)^2
taking two 2 common
2(x^2+2x+1)/(x+1)^2
2(x+1)^2/(x+1)^2
=2 sufficient
(2) c=3
by putting value of x
(18+12+y)/16
insufficient
IMO A
