Data Sufficiency

Can we find better ways to handle this problem.

[spoiler]OA - E; Source: Grockit[/spoiler]

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.

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)

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.

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

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

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

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.

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