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GRE Math Challenge #107- p + q = 1 and 0 < p < q : Quantitative Comparison Questions

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Re: GRE Math Challenge #107- p + q = 1 and 0 < p < q [#permalink]
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Re: GRE Math Challenge #107- p + q = 1 and 0 < p < q [#permalink]
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I'm making this question complicated just to add another discussion point.

0<p<q
p+q = 1

So, p = 1- q

Now 0< 1-q < q (replaced p with 1-q)

0<1-q or 1-q<q
So q<1 or q>1/2

Therefore q can be 3/4 and p can be 1/4

3/4 times 1/4 = 3/16

1/(3/16) = 16/3 or 5.33 . Therefore A is greater than B
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Re: GRE Math Challenge #107- p + q = 1 and 0 < p < q [#permalink]
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Re: GRE Math Challenge #107- p + q = 1 and 0 < p < q [#permalink]
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Solution:

Rather than solving each and every case, we can solve this question in under 20 sec, if we know that if the numerator is bigger than the denominator it will yield a value greater than 1.

In the above case just because p+q=1, we know p & q are either fractions or decimal values. And when we multiply decimal values it further decreases.

IMO A

Hope this helps!
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Re: GRE Math Challenge #107- p + q = 1 and 0 < p < q [#permalink]
If p+q = 1 and both are bigger than 0, we could say for instance p=0.1 and q=0.9 so that 0.1+0.9=1.

Then to compute A: 1/0.1*0.9= 11.11 which is bigger than 1, so A is bigger than B.
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Re: GRE Math Challenge #107- p + q = 1 and 0 < p < q [#permalink]
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Re: GRE Math Challenge #107- p + q = 1 and 0 < p < q [#permalink]
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