DS - interesting problem

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DS - interesting problem

by patanjali.purpose » Sat Jul 30, 2011 3:59 pm
Can we find better ways to handle this problem.

[spoiler]OA - E; Source: Grockit[/spoiler]
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by edge » Sat Jul 30, 2011 4:47 pm
If x Φ y = (2x - y)/(2y - x), where x ≠ 2y, then is (a Φ b) > (b Φ a)?

(1) a < b

(2) 2a < b
My answer is B.
St1 is insufficient (inequality will change sign based on different numbers chosen, for example (a, b) = (1, 3) and (a, b) = (2, 3)).
St2 is sufficient. Both (aΦb) and (bΦa) will be negative, and LHS > RHS.

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by patanjali.purpose » Sun Jul 31, 2011 2:12 am
edge wrote:
If x Φ y = (2x - y)/(2y - x), where x ≠ 2y, then is (a Φ b) > (b Φ a)?

(1) a < b

(2) 2a < b
My answer is B.
St1 is insufficient (inequality will change sign based on different numbers chosen, for example (a, b) = (1, 3) and (a, b) = (2, 3)).
St2 is sufficient. Both (aΦb) and (bΦa) will be negative, and LHS > RHS.
sT2 is also not sufficient check with (-5, -1)

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by GMATGuruNY » Sun Jul 31, 2011 2:56 am
If x Φ y = (2x - y)/(2y - x), where x ≠ 2y, then is (a Φ b) > (b Φ a)?

(1) a < b

(2) 2a < b
Question: Is (2a-b)/(2b-a) > (2b-a)/(2a-b)?

Look for values that satisfy both statements.

Statements 1 and 2: a < b and 2a < b.

Let a=1 and b=3.
2a-b = 2*1 - 3 = -1.
2b-a = 2*3 - 1 = 5.
Is (2a-b)/(2b-a) > (2b-a)/(2a-b)?
-1/5 > 5/-1
-1/5 > -5.
Yes.

Let a = -3 and b = -1.
2a-b = 2(-3) - (-1) = -5.
2b-a = 2(-1) - (-3) = 1.
Is (2a-b)/(2b-a) > (2b-a)/(2a-b)?
-5/1 > 1/-5
-5 > -1/5.
No.

Since in the first case the answer is Yes and in the second case the answer is No, both statements combined are insufficient.

The correct answer is E.
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by clock60 » Sun Jul 31, 2011 3:16 am
hi all. it seems that it is kind of problems in which plugging and obtaining yes and no answer works better than any other ways, i honestly tried to simplify but looks like it does not work here
at first we need to determine
does ((2a-b)/(2b-a))>((2b-a)/2a-b))
after simplify i got
does (a^2-b^2)/((2b-a)(2a-b)) not great relieve....
(1)b=3, a=1. provided that b>a (3>1)
(2-3)/(6-1)=-1/5
(6-1)/2-3)=-5 as a result -1/5>-5 and the answer is yes

but b=1, a=-3 (b>a)
(-6-1)/(2+3)=-7/5
and the right part of the inequality=-5/7, here the answer is no as -7/5<-5/7 so 1 st insuff
(2) the same values for a and b works here as for st1
b=3, a=1.3>2*1, and the answer is yes
b=1, a=-3, 1>2(-3) the answer is no
both: the same values work for combined st
3a<2b
a=1,b=3 3*1<2*3 the answer is yes
a=-3 b=1, 3*(-3)<2*1 the answer is no
so E looks valid answer

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by patanjali.purpose » Sun Jul 31, 2011 5:03 am

Thanks Mitch. I always enjoy the simplicity of all your posts. Thanks.

Could you suggest how shall we pick nos to solve DS problems, esp when the problem involves 2 variables. Do you recommend some standard approach for the same.


thanks.
Patanjali

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by sl750 » Sun Jul 31, 2011 5:12 am
The question asks whether 3(a-b)(a+b)/(2b-a)(2a-b) > 0 ?

For statement 1 a<b,
a=1,b=3 (For b=2, we get a divide by zero)
3(-2)(4)/(5)(-1) > 0
a=-3,b=-1
3(-2)(-4)/(1)(-5) < 0. eliminate A and D

For statement 2 2a<b,
a=1,b=3, we get a yes,
a=-3,b=-1, we get a no. eliminate B

Combining both, we don't get any new information. So answer is E

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by GMATGuruNY » Tue Aug 02, 2011 7:11 am
patanjali.purpose wrote:
Thanks Mitch. I always enjoy the simplicity of all your posts. Thanks.

Could you suggest how shall we pick nos to solve DS problems, esp when the problem involves 2 variables. Do you recommend some standard approach for the same.


thanks.
Patanjali
Try to determine what's being tested.

The DS question above is about inequalities: whether one value is greater than another.
Both a and b are being subtracted.
Subtracting a positive number yields a decrease; subtracting a negative number yields an increase.
Hence the approach that I used above: trying both positive and negative values for a and b that satisfy the two statements.
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by pemdas » Sun Feb 17, 2013 3:55 pm
IF a(b)=(2a-b)/(2b-a) THEN b(a)=(2a- (2a-b)/(2b-a)) / (2*(2a-b)/(2b-a) - a)
Simplifying, b(a)=b/(2b-a) / (2a-b)/(b-a) = b(b-a) / (2b-a)(2a-b)
We compare a(b) and b(a) OR (2a-b)/(2b-a) and b(b-a) / (2b-a)(2a-b)
We may cancel out (2b-a) in the denominators to get (2a-b) and b(b-a)/(2a-b)
Further we may notice that (2a-b) is repeated in the numerator and denominator, so we handle it as (2a-b)^2 and b(b-a)

The main question is whether (2a-b)^2 > b(b-a), we can immediately check that if b(b-a)<0 then the answer is Yes, because the LHS will be squared and positive.

St(1) a<b implies that a is less than b merely, and we don't know either the value nor the sign of b. Hence Insuff

St(2) 2a<b the same as statement 1

Combining both we get no additional info, hence answer is E
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