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Re: 0 < a < b/2 < 9 [#permalink]
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- He actually simplified the inequality and for each term subtracted a.


\(0 < a < \frac{b}{2} < 9\)

using -a

\(-a < 0 < \frac{b}{2} - a < 9 - a\)

As you can see inside the inequality \(\frac{b}{2} - a < 9 - a\)
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Re: 0 < a < b/2 < 9 [#permalink]
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Carcass wrote:
\(0 < a < \frac{b}{2} < 9\)

Quantity A
Quantity B
\(9-a\)
\(\frac{b}{2} - a\)



GIVEN: 0 < a < b/2 < 9
Subtract a from all sides to get: 0 - a < a - a < b/2 - a < 9 - a
Simplify: -a < 0 < b/2 - a < 9 - a

If we ignore the first two values, we can see that: b/2 - a < 9 - a
Since 9 - a is GREATER THAN b/2 - a, the correct answer is A

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Re: 0 < a < b/2 < 9 [#permalink]
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We can easily eliminate -a from the inequality since it's on both the sides. After that, only 9 and b/2 are left, and 9 is bigger is given in the question.
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Re: 0 < a < b/2 < 9 [#permalink]
I got this correct! I am improving my quant score.! Gotta top GRE
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Re: 0 < a < b/2 < 9 [#permalink]
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Adding +a in both the quantities,
Quantity A = 9
Quantity B= b/2
Given that,
b/2 <9, hence Quantity A > Quantity B
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Re: 0 < a < b/2 < 9 [#permalink]
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0<a<b/2<9 which is equal to 0<2a<b<18

QA -> 9 - a
QB -> b/2 - a

Add a to both sides
QA -> 9
QB -> b/2

Multiply both sides by 2

QA - > 18
QB -> b

since b is less than 2 as given in the 0<2a<b<18

QA > QB
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Re: 0 < a < b/2 < 9 [#permalink]
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