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Is (a-k)/(b-k)>(a+k)/(b+k) ?

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by RBBmba@2014 » Thu Feb 05, 2015 10:50 pm
Is (a-k)/(b-k)>(a+k)/(b+k) ?

(1) a>b>k

(2) k>0

P.S: What is the best way to tackle this sort of problem - plugging in numbers or just going by solving inequality algebraic way ?

Experts(Mitch/Rich/Brent and others) - can you please share your analysis & explanation ?

OA:C
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Source: — Data Sufficiency |
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by DavidG@VeritasPrep » Fri Feb 06, 2015 2:44 am
There's not typically a "best" way to approach questions. Often these issues are a matter of taste and comfort. Some might prefer algebra. Some might prefer picking numbers.

That said, what makes this one challenging to do algebraically is that if we were to, say, multiply both sides of the inequality by (b - k) we'd have to consider a scenario in which (b-k) >0, in which case the sign of the original inequality would not flip, and we'd get "Is (a - k) > (a +k)(b-k)/(b+k)?" And we'd have to consider a case when (b - k) < 0, in which case, we'd get "Is (a - k) > (a +k)(b-k)/(b+k)?" You could also cross-multiply, and knock out scenarios for when (b-k) and (b+k) are both positive or negative, or one of them is positive and one of them is negative. It'll work. But ugh.

So I'd pick numbers. The key is to pick easy numbers and choose them strategically to minimize calculations.

For Statement 1
a = 3
b = 2
k = 1

Is (3-1)/(2-1) > (3+1)/(2+1) ? 2>4/3, so YES

a = 3
b = 2
k = 0
Is (3-0)/(2-0) > (3+0)/(2+0) ? NO, they're the same

S1: Not Suff

Statement 2
We can reuse our first scenario from S1:
a = 3
b = 2
k = 1

Is (3-1)/(2-1) > (3+1)/(2+1) ? YES 2>4/3

Or
a = 2
b = 3
k = 1
Is (2-1)/(3-1) > (2+1)/(3+1) Is 1/2 > 3/4? NO

Statement 2: NOT SUFF


Together:

No matter what we pick, the answer will always be YES. The reason is that if you take any positive fraction in which the numerator is larger than the denominator (because a > b) and subtract a positive constant from both numerator and denominator, that fraction must grow larger. Start with 4/3. Subtract 1 from both numerator and denominator, and we get 3/2. 3/2 is bigger. Start with 10/3, subtract 2 from numerator and denominator, we get 8/1. 8/1 is bigger. (Logically, if we were to add the same positive term to both numerator and denominator for this scenario, the fraction would get progressively smaller and smaller.)

Together they are sufficient, so the answer is C.
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by GMATGuruNY » Fri Feb 06, 2015 2:51 am
RBBmba@2014 wrote:Is (a-k)/(b-k)>(a+k)/(b+k) ?

(1) a>b>k

(2) k>0
To prevent division by 0, the prompt should state that b≠k and b≠-k.

Statement 1: a>b>k
Test one case that also satisfies statement 2.
Case 1: a=3, b=2 and k=1.
Plugging these values into (a-k)/(b-k)>(a+k)/(b+k), we get:
(3-1)/(2-1) > (3+1)/(2+1)
2 > 4/3
YES.

Test one case that does NOT also satisfy statement 2.
Case 2: a=3, b=2, k=-1
Plugging these values into (a-k)/(b-k)>(a+k)/(b+k), we get:
(3+1)/(2+1) > (3-1)/(2-1)
4/3 > 2
NO.

Since the answer is YES in Case 1 but NO in Case 2, INSUFFICIENT.

Statement 2: k>0
Case 1 also satisfies statement 2.
In Case 1, the answer to the question stem is YES.

Test one case that does NOT also satisfy statement 1.
Case 3: a=2, b=3, k=1
Plugging these values into (a-k)/(b-k)>(a+k)/(b+k), we get:
(2-1)/(3-1) > (2+1)/(3+1)
1/2 > 3/4
NO.

Since the answer is YES in Case 1 but NO in Case 3, INSUFFICIENT.

Statements combined:
Case 1 satisfies both statements.

Test one more random case that satisfies both statements.
Case 4: a=20, b=10, k=5
Plugging these values into (a-k)/(b-k)>(a+k)/(b+k), we get:
(20-5)/(10-5) > (20+5)/(10+5)
15/5 > 25/15
3 > 5/3.
YES.

Cases 1 and 4 illustrate that -- when the statements are combined -- (a-k)/(b-k)>(a+k)/(b+k).
SUFFICIENT.

The correct answer is C.
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by GMATGuruNY » Fri Feb 06, 2015 7:39 am
RBBmba@2014 wrote:Is (a-k)/(b-k)>(a+k)/(b+k) ?

(1) a>b>k

(2) k>0
Algebraic approach:
Is (a-k)/(b-k) > (a+k)/(b+k)?

Move everything to the left side:
(a-k)/(b-k) - (a+k)/(b+k) > 0

Put the left side over a common denominator:
(a-k)(b+k) - (a+k)(b-k) / (b-k)(b+k) > 0

Distribute:
(ab + ak - bk - k²) - (ab - ak + bk - k²) / (b-k)(b+k) > 0

Cancel out the terms in red and combine like terms:
2ak - 2bk / (b+k)(b-k) > 0

Simply the numerator by factoring out 2k:
2k(a-b) / (b+k)(b-k) > 0.

Divide each side by 2:
k(a-b) / (b+k)(b-k) > 0

Question stem, rephrased:
Is k(a-b) / (b+k)(b-k) > 0?

To answer the rephrased question stem, we need to know the signs of the following factors:
k, a-b, b+k and b-k.

Statement 1: a>b>k
Thus, a-b>0 and b-k>0.
No information about k or b+k.
INSUFFICIENT.

Statement 2: k>0
No information about a-b, b+k or b-k.
INSUFFICIENT.

Statements combined: a>b>k>0
Thus, k>0, a-b>0, b-k>0 and b+k>0.
SUFFICIENT.

The correct answer is C.

Algebra requires more work upfront but almost no work when evaluating the statements.
Testing cases requires no work upfront but more work when evaluating the statements.
While I suspect that most test-takers would find it easier to test cases, each test-taker should proceed according to his or her personal strengths.
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by RBBmba@2014 » Fri Feb 06, 2015 8:58 am
Thanks Dave and Mitch. Much appreciate your reply.

@Mitch- a quick question for you Sir.

What is the best way to tackle this sort of problem - plugging in numbers or just going by solving inequality algebraic way ? Is there any way to understand this quickly or it's just one's *6th sense*?
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by [email protected] » Fri Feb 06, 2015 11:06 am
Hi RBBmba@2014,

When choosing among the various ways to approach a question, there are often "clues" in how the question is written that will help you to quickly choose.

Here, the prompt is just one big question, with no information. In this context, A, B and K can be ANYTHING. As such, we're expected to consider ALL of the possibilities - TESTing VALUES is a perfect approach here.

Another detail worth noticing is that the K is the 'hinge point' of this question (in some spots it's subtracted, in others it's added). Making the K positive, 0 or negative will effect the answer to the question, so we have to make sure to "play around" with that variable a bit.

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by GMATGuruNY » Sat Feb 07, 2015 4:20 am
RBBmba@2014 wrote:Thanks Dave and Mitch. Much appreciate your reply.

@Mitch- a quick question for you Sir.

What is the best way to tackle this sort of problem - plugging in numbers or just going by solving inequality algebraic way ? Is there any way to understand this quickly or it's just one's *6th sense*?
There is no such thing as a "best way."
The best way for one test-taker might not be the best way for another.
The key is to discover what is the best way for YOU.
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by Matt@VeritasPrep » Mon Feb 09, 2015 12:44 am
One other thing: the best way doesn't always come to you, especially on test day. It's better to try something and see what ideas it sparks - they may not be great ones, but they'll often be enough for you to gut out the problem.

Thinking of a great solution is a lot like thinking of a witty retort: you can often come up with something decent on the spot, but you usually think of something even better a few minutes after the moment has passed!
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by Matt@VeritasPrep » Mon Feb 09, 2015 1:09 am
By the way, let me add one more approach to this.
Is (a-k)/(b-k) > (a+k)/(b+k) ?


Notice that (a + k) = (a - k + 2k), so the question is REALLY asking
Is (a - k)/(b - k) > (a - k + 2k)/(b - k + 2k) ?


If we let (a - k) = x and (b - k) = y, it might be easier to see what's happening here.
Is x/y > (x + 2k)/(y + 2k) ?
At this point, let's consider two cases. If y and k are positive, cross-multiplication gives x(y + 2k) > y(x + 2k). But if y is positive and (y + k) is negative (or vice versa), cross-multiplication gives x(y + 2k) < y(x + 2k). These are two different questions to answer, obviously, so we need to know about the signs of these variables to properly approach the problem.

From S1, we get a - k > b - k > 0. (Subtract k from the inequality.) Since x = a - k and y = b - k, we know that x > y > 0. This tells us y > 0, but we still don't know about k; INSUFFICIENT.

From S2, we know k > 0, but not about y; INSUFFICIENT.

Together, we know that y (i.e. (b - k)) and k are positive. Returning to our work from S1, our question is really
Is x(y + 2k) > y(x + 2k) ?
This reduces to
Is kx > ky ?
Since k is positive, we can safely divide both sides and reduce the problem to
Is x > y ?


S1 gave us the answer to this (x > y > 0), so we can finally answer the question affirmatively, and the two statements together are SUFFICIENT.

Note: in my first version of this post, I made a real gaffe en route to the solution. (This is why I shouldn't post at 1:22 AM :D) Caught it immediately after pushing submit, though, so I fixed it ... but if you saw that, one step was wrong. (I got cute trying to circumvent the algebra, and made a foolish deduction - whoops!)
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by RBBmba@2014 » Wed Feb 11, 2015 10:41 pm
Much appreciate replies from all the experts!
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