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0 < xy
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07 Jun 2017, 01:18
07 Jun 2017, 01:18
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Question Stats:
72% (01:48) correct
27% (01:45) wrong based on 247 sessions
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\(0 < xy\)
Quantity A |
Quantity B |
\(\frac{5}{2x} - \frac{2}{2y}\) |
\(\frac{5y-2x}{2x-2y}\) |
A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.
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Re: 0 < xy
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09 Jun 2017, 10:37
09 Jun 2017, 10:37
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Carcass wrote:
\(0 < xy\)
Quantity A |
Quantity B |
\(\frac{5}{2x} - \frac{2}{2y}\) |
\(\frac{5y-2x}{2x-2y}\) |
Given:
Quantity A: 5/2x - 2/2y
Quantity B: (5y-2x)/(2x-2y)
Let's see what happens if we rewrite Quantity A as one big fraction.
First get common denominators...
Quantity A: 5y/2xy - 2x/2yx
Quantity B: (5y-2x)/(2x-2y)
Combine to get:
Quantity A: (5y-2x)/2yx
Quantity B: (5y-2x)/(2x-2y)
NICE!! This numerators are the same for both fractions.
This gives me the opportunity to use a strategy I call "Looking for Equality" (see our video below for more on this)
I can see that if I choose an x- and y-value that makes the numerators both zero, this will speed up my solution. You'll see why shortly.
I can see that, if x = 5 and y = 2, then we can plug these values into the quantities to get:
Quantity A: [5(2)-2(5)]/(2)(2)(5)
Quantity B: [5(2)-2(5)]/[2(5)-2(2)]
Evaluate:
Quantity A: 0
Quantity B: 0
So, when x = 5 and y = 2 the quantities are EQUAL.
The correct answer is EITHER C or D
At this point, every pair of x- and y-values will either yield equal quantities or not.
Let's plug in super-easy numbers.
We'll plug in x = 1 and y = 1
We get:
Quantity A: [5(1)-2(1)]/(2)(1)(1)
Quantity B: [5(1)-2(1)]/[2(1)-2(1)]
Evaluate:
Quantity A: 3/2
Quantity B: 3/0
These quantities are definitely NOT EQUAL.
Answer:
Show: ::
D
RELATED VIDEOS
General Discussion
Re: 0 < xy
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30 Jan 2018, 10:47
30 Jan 2018, 10:47
2
From the given condition, x and y have to both positive or negative.
Now putting value x=1, y=1
Q.A 3/2 Q.B undefined
Now putting x=1,y=2
Q.A 2 Q.B -4 , Q.A>Q.B
Now putting x=-1,y=-1
Q.A -3/2 Q.B undefined
So the relation can't be determined.
Now putting value x=1, y=1
Q.A 3/2 Q.B undefined
Now putting x=1,y=2
Q.A 2 Q.B -4 , Q.A>Q.B
Now putting x=-1,y=-1
Q.A -3/2 Q.B undefined
So the relation can't be determined.
