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GRE Prep Club Legend  Joined: 07 Jun 2014
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GRE Math Challenge #107- p + q = 1 and 0 < p < q [#permalink]
Expert's post 00:00

Question Stats: 63% (00:38) correct 36% (01:19) wrong based on 11 sessions
$$p + q = 1$$
$$0 < p < q$$

Quantity A: $$1/pq$$
Quantity B: 1
• Quantity A is greater.
• Quantity B is greater.
• Both Quantities are Equal
• Cannot be determined
[Reveal] Spoiler: OA

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Sandy
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Re: GRE Math Challenge #107 [#permalink]
Expert's post
sandy wrote:
$$p + q = 1$$
$$0 < p < q$$

Quantity A: $$1/pq$$
Quantity B: 1

Two important facts:
1) since p and q are both positive, the product pq is POSITIVE
2) since p and q are both positive, AND p + q = 1, we can conclude that p and q are each less than 1

The fraction in quantity A might cause issues for us, so let's eliminate that fraction.
Since fact #1 tells us that the product pq is POSITIVE, let's multiply both quantities by pg to get:
Quantity A: 1
Quantity B: pq

Next, since p and q are each less than 1, the product pq must be less than 1.
So, we get:
Quantity A: 1
Quantity B: some number less than 1

[Reveal] Spoiler:
A

Cheers,
Brent
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Re: GRE Math Challenge #107- p + q = 1 and 0 < p < q [#permalink]
sandy wrote:
$$p + q = 1$$
$$0 < p < q$$

Quantity A: $$1/pq$$
Quantity B: 1
• Quantity A is greater.
• Quantity B is greater.
• Both Quantities are Equal
• Cannot be determined

p + q = 1

and 0<p<q.

From the given data , it's clear that both p ans q are positive fraction.

When we divide an integer by a fraction , the integer becomes much bigger. So, 1 will be greater than before. Re: GRE Math Challenge #107- p + q = 1 and 0 < p < q   [#permalink] 11 Aug 2018, 13:01
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