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2 ≤ x ≤ 10 y > 3
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27 Mar 2020, 02:40
27 Mar 2020, 02:40
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\(2 ≤ x ≤ 10\)
\(y > 3\)
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.
Kudos for the right answer and explanation
\(y > 3\)
Quantity A |
Quantity B |
\(x-y\) |
\(xy\) |
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.
Kudos for the right answer and explanation
Re: 2 ≤ x ≤ 10 y > 3
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27 Mar 2020, 03:41
27 Mar 2020, 03:41
1
Carcass wrote:
\(2 ≤ x ≤ 10\)
\(y > 3\)
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.
Kudos for the right answer and explanation
\(y > 3\)
Quantity A |
Quantity B |
\(x-y\) |
\(xy\) |
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.
Kudos for the right answer and explanation
Here,
By looking at the equation, we can find out
x and y are positive numbers
and product of two positive numbers will always be greater than the substraction of the same positive numbers
i.e. (x-y) < xy
Option B
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2 ≤ x ≤ 10 y > 3
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30 Mar 2020, 06:30
30 Mar 2020, 06:30
1
Carcass wrote:
\(2 ≤ x ≤ 10\)
\(y > 3\)
\(y > 3\)
Quantity A |
Quantity B |
\(x-y\) |
\(xy\) |
We can solve this question using matching operations (my favorite Quantitative Comparison strategy)
Given:
Quantity A: \(x-y\)
Quantity B: \(xy\)
Add \(y\) to both quantities to get:
Quantity A: \(x\)
Quantity B: \(xy+y\)
Factor Quantity B as follows:
Quantity A: \(x\)
Quantity B: \(y(x+1)\)
Since it is given that \(x\) is positive, we can also be certain that \(x+1\) is positive, which means we can safely divide both quantities by \(x+1\) to get:
Quantity A: \(\frac{x}{x+1}\)
Quantity B: \(y\)
Since \(x\) is positive, and since \(x<x+1\), we can be certain that \(\frac{x}{x+1}<1\)
We get:
Quantity A: Some number that's less than 1
Quantity B: \(y\) (which is some number greater than 3)
Answer: B
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