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GRE Math Challenge #69- Quantity A: p^4 – p^6
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Updated on: 19 Feb 2020, 11:55
Updated on: 19 Feb 2020, 11:55
1
Expert Reply
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Question Stats:
71% (00:45) correct
29% (01:09) wrong based on 100 sessions
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\(0 < p <1\)
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
Quantity A |
Quantity B |
\(p^4\) – \(p^6\) |
\(p^3\)– \(p^5\) |
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
Retired Moderator
Joined: 10 Apr 2015
Posts: 6218
Given Kudos: 136
Re: GRE Math Challenge #69
[#permalink]
18 May 2015, 11:32
18 May 2015, 11:32
1
sandy wrote:
\(0 < p < 1\)
Quantity A: \(p^4\) – \(p^6\)
Quantity B: \(p^3\)– \(p^5\)
Given: 0 < p < 1
If we factor each quantity, something nice happens:
Quantity A: p^4(1 – p^2)
Quantity B: p^3(1 – p^2)
NOTE: If 0 < p < 1, then p^2 is LESS THAN 1, which means 1 - p^2 must be POSITIVE.
As such, we can divide both quantities by (1 - p^2) to get:
Quantity A: p^4
Quantity B: p^3
NOTE: If 0 < p < 1, then p^3 is POSITIVE.
As such, we can divide both quantities by p^3 to get:
Quantity A: p
Quantity B: 1
Since 0 < p < 1, we can see that quantity B must be greater.
Answer:
Show: ::
B
Cheers,
Brent
Re: GRE Math Challenge #69- Quantity A: p^4 – p^6
[#permalink]
22 Jan 2017, 21:57
22 Jan 2017, 21:57
2
A: P^4 - P^6
B: P^3 - P^5
Expand---
A: P^3 ( P^1 - P^3)
B: P^3 (1 - P^2)
Divide both by P^3
A: P^1 - P^3
B: 1 - P^2
Now, Expand A
A: P (1 - P^2)
B: 1 - P^2
Divide Both by (1 - P^2)
A: P
B: 1
Since P is Always less than 1, Answer is B
B: P^3 - P^5
Expand---
A: P^3 ( P^1 - P^3)
B: P^3 (1 - P^2)
Divide both by P^3
A: P^1 - P^3
B: 1 - P^2
Now, Expand A
A: P (1 - P^2)
B: 1 - P^2
Divide Both by (1 - P^2)
A: P
B: 1
Since P is Always less than 1, Answer is B
