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x>0
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27 Apr 2021, 19:39
27 Apr 2021, 19:39
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Question Stats:
57% (01:07) correct
42% (00:44) wrong based on 40 sessions
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\(x > 0\)
Quantity A is greater.
Quantity B is greater.
The two quantities are equal.
The relationship cannot be determined from the information given.
Quantity A |
Quantity B |
\(\frac{1}{x}\) |
\(\frac{x+1}{x^2}\) |
Quantity A is greater.
Quantity B is greater.
The two quantities are equal.
The relationship cannot be determined from the information given.
Re: x>0
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29 Apr 2021, 11:13
29 Apr 2021, 11:13
1
x>0
QA: 1/x and QB: (x+1)/x^2
multiplying x on both side,
QA: 1 and QB:(x+1)/x , now
QB: (x/x)+(1/x)=1+1/x,
subtracting 1 from both side,
QA: 0 and QB: 1/x hence, option B is correct!
QA: 1/x and QB: (x+1)/x^2
multiplying x on both side,
QA: 1 and QB:(x+1)/x , now
QB: (x/x)+(1/x)=1+1/x,
subtracting 1 from both side,
QA: 0 and QB: 1/x hence, option B is correct!
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x>0
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29 Nov 2021, 13:21
29 Nov 2021, 13:21
Melat wrote:
\(x > 0\)
Quantity A is greater.
Quantity B is greater.
The two quantities are equal.
The relationship cannot be determined from the information given.
Quantity A |
Quantity B |
\(\frac{1}{x}\) |
\(\frac{x+1}{x^2}\) |
Quantity A is greater.
Quantity B is greater.
The two quantities are equal.
The relationship cannot be determined from the information given.
We can solve this question using matching operations (my favorite Quantitative Comparison strategy)
Given:
Quantity A: \(\frac{1}{x}\)
Quantity B: \(\frac{x+1}{x^2}\)
Since we're told \(x\) is positive, we can be certain that \(x^2\) is also positive, which means we can safely multiply both quantities by \(x^2\) to get:
Quantity A: \(x\)
Quantity B: \(x+1\)
At this point, we can see that quantity B is greater. However, we can perform one extra step to be absolutely sure of this.
Subtract \(x\) from both quantities:
Quantity A: \(0\)
Quantity B: \(1\)
Answer: B
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