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x different from y
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20 Oct 2018, 09:24
20 Oct 2018, 09:24
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Question Stats:
53% (01:22) correct
46% (01:18) wrong based on 147 sessions
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For \(x \neq y\), \(x \Phi y = \frac{x+y}{x-y}\)
\(p > 0 > q\)
Quantity A is greater.
Quantity B is greater.
The two quantities are equal.
The relationship cannot be determined from the information given.
\(p > 0 > q\)
Quantity A |
Quantity B |
\(p \Phi q\) |
\(q \Phi p\) |
Quantity A is greater.
Quantity B is greater.
The two quantities are equal.
The relationship cannot be determined from the information given.
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Re: x different from y
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27 Nov 2018, 06:30
27 Nov 2018, 06:30
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Carcass wrote:
For \(x \neq y\), \(x \Phi y = \frac{x+y}{x-y}\)
\(p > 0 > q\)
\(p > 0 > q\)
Quantity A |
Quantity B |
\(p \Phi q\) |
\(q \Phi p\) |
Apply the rule to get:
Quantity A: (p + q)/(p - q)
Quantity B: (q + p)/(q - p)
Let's solve this question by testing values
case i: p = 1 and q = -1
Plug in values to get:
Quantity A: (1 + -1)/(1 - -1) = 0/2 = 0
Quantity B: (-1 + 1)/(-1 - 1) = 0/(-2) = 0
In this case, the two quantities are EQUAL
case ii: p = 2 and q = -1
Plug in values to get:
Quantity A: (2 + -1)/(2 - -1) = 1/3
Quantity B: (-1 + 2)/(-1 - 2) = 1/(-3)
In this case, Quantity A is GREATER
Answer: D
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Re: x different from y
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25 Nov 2018, 14:25
25 Nov 2018, 14:25
2
Answer: A
For x != y:
xΦy = (x+y) / (x-y)
A= pΦq = (p+q) / (p-q)
B=qΦp= (q+p)/(q-p)
We have q < 0 < p, so p-q>0 and q-p<0 thus, A is positive and B is negative. A is bigger than B.
For x != y:
xΦy = (x+y) / (x-y)
A= pΦq = (p+q) / (p-q)
B=qΦp= (q+p)/(q-p)
We have q < 0 < p, so p-q>0 and q-p<0 thus, A is positive and B is negative. A is bigger than B.
