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# GRE Math Challenge #69- Quantity A: p^4 – p^6

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GRE Math Challenge #69- Quantity A: p^4 – p^6 [#permalink]  03 May 2015, 00:59
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Question Stats:

52% (00:39) correct 48% (01:18) wrong based on 25 sessions
$$0 < p <1$$

Quantity A: $$p^4$$ – $$p^6$$
Quantity B: $$p^3$$– $$p^5$$

• Quantity A is greater.
• Quantity B is greater.
• Both Quantities are Equal
• Cannot be determined
[Reveal] Spoiler: OA

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Re: GRE Math Challenge #69 [#permalink]  18 May 2015, 11:32
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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.
[Reveal] Spoiler:
B

Cheers,
Brent
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Re: GRE Math Challenge #69- Quantity A: p^4 – p^6 [#permalink]  22 Jan 2017, 21:57
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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
Re: GRE Math Challenge #69- Quantity A: p^4 – p^6   [#permalink] 22 Jan 2017, 21:57
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