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It is given that 0<p<1
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20 Apr 2021, 23:31
20 Apr 2021, 23:31
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It is given that \(0<p<1\)
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
Question part of the project GRE Quantitative Reasoning Daily Challenge - (2021) EDITION
GRE - Math Book
Quantity A |
Quantity B |
\(\frac{1}{p^2}\) |
\(\frac{1}{(p+1)^2}\) |
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
Question part of the project GRE Quantitative Reasoning Daily Challenge - (2021) EDITION
GRE - Math Book
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Re: It is given that 0<p<1
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21 Apr 2021, 11:18
21 Apr 2021, 11:18
1
Carcass wrote:
It is given that \(0<p<1\)
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 |
\(\frac{1}{p^2}\) |
\(\frac{1}{(p+1)^2}\) |
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
\(0<p<1\) mean \(p\) is a positive number
Col. A: \(\frac{1}{p^2}\)
Col. B: \(\frac{1}{(p+1)^2}\)
Cross-multiplying the denominators (Since \(p > 0\));
Col. A: \((p+1)^2\)
Col. B: \(p^2\)
Col. A: \(p^2 + 2p + 1\)
Col. B: \(p^2\)
Subtracting \(p^2\) from both sides;
Col. A: \(2p + 1\)
Col. B: \(0\)
Subtracting \(1\) from both sides and dividing by \(2\);
Col. A: \(p\)
Col. B: \(\frac{-1}{2}\)
Hence, option A
Re: It is given that 0<p<1
[#permalink]
05 May 2021, 22:49
05 May 2021, 22:49
2
If you know the fact that squaring a number between 0 and 1 results in a number smaller than the original number, this question can be solved quickly without any calculations.
We can easily realize that the denominator for :
option A = 1 / (number between 0 and 1)^2;
option B = 1 / (number NOT between 0 and 1)^2;...since you are adding +1 to p, the denominator is greater than 1;
From the fact that squaring a number between 0 and 1 results in a number smaller than the original number, it can be understood that the denominator of option A will be much lesser than option B. Since it is denominator of a fraction, option A will be bigger in value than option B.
Hence, answer is option A.
Hit Kudos if this reply helped!:)
We can easily realize that the denominator for :
option A = 1 / (number between 0 and 1)^2;
option B = 1 / (number NOT between 0 and 1)^2;...since you are adding +1 to p, the denominator is greater than 1;
From the fact that squaring a number between 0 and 1 results in a number smaller than the original number, it can be understood that the denominator of option A will be much lesser than option B. Since it is denominator of a fraction, option A will be bigger in value than option B.
Hence, answer is option A.
Hit Kudos if this reply helped!:)
