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x and y are positive integers and x > y.
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20 Jul 2019, 02:34
20 Jul 2019, 02:34
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Question Stats:
64% (00:48) correct
35% (00:44) wrong based on 191 sessions
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x and y are positive integers and \(x > y\).
A)The quantity in Column A is greater.
B)The quantity in Column B is greater.
C)The two quantities are equal.
D)The relationship cannot be determined from the information given.
Quantity A |
Quantity B |
\(x^2-y^2\) |
\(x+y\) |
A)The quantity in Column A is greater.
B)The quantity in Column B is greater.
C)The two quantities are equal.
D)The relationship cannot be determined from the information given.
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Re: x and y are positive integers and x > y.
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20 Jul 2019, 07:48
20 Jul 2019, 07:48
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Carcass wrote:
x and y are positive integers and \(x > y\).
Quantity A |
Quantity B |
\(x^2-y^2\) |
\(x+y\) |
There are two methods for solving this Quantitative Comparison - Plugging in Values or Algebraically. Let's first try plugging in.
If we were to select the easy values of x = 2 and y = 1, then Quantity A = 2² - 1² = 3 and Quantity B = 2 + 1 = 3.
In this scenario the quantities would be equal, so eliminate Choice A and Choice B since neither quantity is greater than the other.
Now, plug in a different value for only one of the variables such as x = 3 and y = 1.
Then, Quantity A = 3² - 1² = 8 and Quantity B = 3 + 1 = 4.
In this scenario Quantity A is greater, so eliminate Choice C and select Choice D since the relationship between the quantities is inconsistent.
Now, let's consider the algebraic approach.
The quadratic equation x² - y² in Quantity A can be factored to (x + y)(x - y).
Since x - y can = 1, it should be apparent that (x + y)(1) could be equal to x + y in Quantity B, but doesn't have to be equal, so you could also select Choice D immediately based on this deduction.
x and y are positive integers and x > y.
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07 Jan 2024, 21:31
07 Jan 2024, 21:31
1
Proving this numerically
If \(x = 3,4,5 \text{ and } y = 2,3,5\) both sides are equal. In general if \(x = y+1\), then both columns have equal value. Thus C is the choice.
But when \(x-y>1\), then Quantity A is always greater.
Therefore, no relationship exists between the two columns and therefore the answer is D.
We can prove this algebraically as well.
Quantity A = \(x^2 - y^2\) can be written as \((x+y)(x-y)\). Quantity B is \(x+y\)
Cancelling \(x+y\) on both sides as \(x\) is always positive, we get
Quantity A = \(x - y\)
Quantity B = \(1\)
Now since \(x>y\), Quantity A can be \(1\) or greater than \(1\). Thus no relationship exists between the two quantities. Answer is D.
If \(x = 3,4,5 \text{ and } y = 2,3,5\) both sides are equal. In general if \(x = y+1\), then both columns have equal value. Thus C is the choice.
But when \(x-y>1\), then Quantity A is always greater.
Therefore, no relationship exists between the two columns and therefore the answer is D.
We can prove this algebraically as well.
Quantity A = \(x^2 - y^2\) can be written as \((x+y)(x-y)\). Quantity B is \(x+y\)
Cancelling \(x+y\) on both sides as \(x\) is always positive, we get
Quantity A = \(x - y\)
Quantity B = \(1\)
Now since \(x>y\), Quantity A can be \(1\) or greater than \(1\). Thus no relationship exists between the two quantities. Answer is D.
