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dddanny2006: Master | Next Rank: 500 Posts; Posts: 209; Joined: Thu Jan 12, 2012 12:59 pm

by dddanny2006 » Tue Nov 05, 2013 2:38 am

If b is not equal to Zero,and a>b ,is a>c?
(1)(a/b) >(c/b)

(2) 5ab > 6bc

I know why statements 1 and 2 alone cant be sufficient.

Now combining statements 1 and 2

From statement 1,we have a > c----------(1) ,a<c--------------(2) ,we have 2 equations because we dont know whether b is negative or positive

From statement 2,we have 5a>6c----------(3) ,5a<6c-------------(4) ,again we have 2 equations.

If we combine 1 and 3, then a>c.Also combining equations 1 and 4 then indeed a>c

Why cant we combine equations this way?
It gives a whole range of possibilities.

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Source: Beat The GMAT — Data Sufficiency | Tue Nov 05, 2013 2:38 am

theCodeToGMAT: Legendary Member; Posts: 1556; Joined: Tue Aug 14, 2012 11:18 pmFollowed by: 34 members; Thanked: 448 times; GMAT Score:650

by theCodeToGMAT » Tue Nov 05, 2013 3:32 am

Is it {E}

b != 0
a>b

To find: a > c?

Statement 1:
a/b > c/b
if b = +ve
then, a>c
if b = -ve
then,a<c
INSUFFICIENT

Statement 2:
5ab > 6bc
if b = +ve
then, 5a > 6c ==> a > 6/5 (c) ==> means a>c
if b = -ve
then, 5a < 6c ==> a < 6/5 (c) ==> we cannot say
INSUFFICIENT

Combining...
We still dont know whether "b" is +ve or -ve
Answer [spoiler]{E}[/spoiler]

R A H U L

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theCodeToGMAT: Legendary Member; Posts: 1556; Joined: Tue Aug 14, 2012 11:18 pmFollowed by: 34 members; Thanked: 448 times; GMAT Score:650

by theCodeToGMAT » Tue Nov 05, 2013 3:36 am

The equations (1) & (3) are for the cases when "b" is "+ve"

&

The equations (2) & (4) are for the cases when "b" is "-ve"

When we do combining of equations.. we do combination taking care of our assumption. i.e. whether "b" is "+ve" or "-ve" .. cross linking is incorrect.

R A H U L

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GMATGuruNY: GMAT Instructor; Posts: 15539; Joined: Tue May 25, 2010 12:04 pmFollowed by: 1906 members; Location: New York, NY; Thanked: 13060 times; GMAT Score:790

by GMATGuruNY » Tue Nov 05, 2013 5:48 am

dddanny2006 wrote:If b is not equal to Zero,and a>b ,is a>c?
(1)(a/b) >(c/b)

(2) 5ab > 6bc

Since bâ‰ 0, we can rephrase statement 1 by multiplying each side by bÂ², which must be a POSITIVE value:
bÂ² * (a/b) > bÂ² * (c/b)
ab > bc.

Statement 2 can be rephrased as follows:
ab > (6/5)bc.

The result is that the two statements are almost IDENTICAL.

Case 1: Plug b=1 into both statements
Statement 1: a > c.
Statement 2: a > (6/5)c.
Here, it's possible that c=1 and a=2, in which case a>c.

Case 2: Plug b=-1 into both statements
Statement 1: -a > -c, implying that a < c.
Statement 2: -a > -(6/5)c, implying that a < (6/5)c.
Here, it's possible that c=1 and a=0, in which case a<c.

Thus, the two statements combined are INSUFFICIENT.

The correct answer is E.

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by [email protected] » Tue Nov 05, 2013 3:01 pm

Hi dddanny2006,

This type of DS question can be dealt with using algebra skills, Number Properties or TESTing Values. Here's how you can TEST Values...

We're told B not= 0 and A>B. We're asked IF A > C? This is a YES/NO question.

Fact 1: A/B > C/B

First, an easy set of POSITIVE values:
A=2
B=1
C=1
Is 2>1? YES

Next, something different; NEGATIVE values:
A= -2
B= -3
C= -1
Is -2 > -1? NO

Fact 1 is INSUFFICIENT

Fact 2: 5AB > 6BC

The great thing about this information is that the TEST CASES that I used in Fact 1 will fit perfectly here too:

A=2
B=1
C=1
YES

A= -2
B= -3
C= -1
NO

Fact 2 is INSUFFICIENT

Combined, I don't need to do any more work. Both sets of values fit both Facts, so I still have a YES and a NO answer.
Combined, INSUFFICIENT

Final Answer: E

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Rich

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