If a

Question

If a<0 and b<0 , is \frac{(a-2p)}{(b-2p)} > \frac{a}{b} ? (1) p>0 (2) a

D3N0 · Answer

Ans: C (IMO)
here we one relation is given and also it is given that a and b both are negative.
so we can actually cross multiply this and solve the equation for clarity. after solving we get 2p (a-b)> 0 is to be prove.
for this we need to know two things (1) sign of p and (2) relation between a and b. remember we already know sign of a and b
from State1) we know the sign of p but relation between a and b is not given (Not suff)
fromState2) we know the realtion but not the sign of the p

together: we know so Ans is (C) together they are sufficient.

conniezh · Answer

The question can be simplified to is b>a? 
Here's how: from the equation given, we can simplify it to b(a-2p)>a(b-2p) 
If you further simplify it, it'd be ab-2bp>ab-2ap -->  b>a

Now looking at 1 and 2, 2 answers the question directly.

Answer:  B

push12345 · Answer

Here is my take on it

I) p>0 
But a/b is ratio
And we dont know whether a-2p/b-2p would be greater than a/b
Depends on a>b then no 
a<b then yes

II) no info about p so insufficient

Together we can answer question

C is answer

Eg ldt p=1 a=-3 b=-2

a-2p/b-2p= 5/4

a/b=3/2
 3/2 is greater than 5/4

Give kudos if it helps

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chetan2u · Answer

\frac{(a-2p)}{(b-2p)} > \frac{a}{b} ...... here the answer would depend on if A) a/b<1 or otherwise B) -2p is positive or negative 1) x and y are positive numbers or both are <0 and x\frac{x}{y} b) If you subtract same positive number from x and y, the fraction \frac{x-a}{y-a}<\frac{x}{y} 2) x and y are positive numbers or both are <0 and x>y or fraction \frac{x}{y}>1 a) If you ADD same positive number to x and y, the fraction \frac{x+a}{y+a}<\frac{x}{y} b) If you subtract same positive number ( same as adding a negative number) from x and y, the fraction \frac{x-a}{y-a}>\frac{x}{y} you can check other cases too https://gmatclub.com/forum/posting.php?mode=edit&f=7&p=2086955 (1) p>0 so -2p is negative but nothing about x/y say \frac{x}{y}<1.....\frac{3}{4} and p =1.....\frac{3-2}{4-2}=\frac{1}{2}<\frac{3}{4} .. \frac{x}{y}>1.....\frac{4}{3} and p =1.....\frac{4-2}{3-2}=\frac{2}{1}>\frac{4}{3} insuff (2) a1 but nothing about p say \frac{x}{y}>1.....\frac{4}{3} and p =-1...-2p=2......\frac{4+2}{3+2}=\frac{6}{5}<\frac{4}{3} .. \frac{x}{y}>1.....\frac{4}{3} and p =1....-2p=2.....\frac{4-2}{3-2}=\frac{2}{1}>\frac{4}{3} insuff combined.. case Ib \frac{x}{y}>1.....\frac{4}{3} and p =1.....\frac{4-2}{3-2}=\frac{2}{1}>\frac{4}{3} suff C

amanvermagmat · Answer

We can slightly modify (a-2p)/(b-2p) > a/b by bringing a/b to the left side. Then we get: (a-2p)/(b-2p) - a/b > 0 OR {ab - 2pb - ab + 2pa}/{b*(b-2p)} > 0 OR {2p*(a-b)}/{b*(b-2p)} > 0 Now since b is negative, we can multiply both sides by b and change the sign of inequality. So now the question is basically asking whether: {2p*(a-b)}/{b-2p} < 0 or not. (1) p > 0 So 2p is positive, and denominator (b-2p) is negative (since b is already negative). But we dont know whether (a-b) in the numerator is negative or positive. Not sufficient. (2) a < b So (a-b) in the numerator is negative. But we dont know whether '2p' and 'b-2p' are negative or positive. Not sufficient. Combining the statements, p is positive. This makes '2p' as positive, 'b-2p' as negative. And since a < b this makes 'a-b' as negative. So overall {2p*(a-b)}/{b-2p} becomes positive. And this gives NO as an answer to the question asked. Sufficient. Hence C answer

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If a < 0 and b < 0 , is (a - 2p)/(b - 2p) > a/b ?

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