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

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

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03 Jun 2018, 11:10
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75% (hard)

Question Stats:

48% (01:48) correct 52% (01:32) wrong based on 54 sessions

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

(1) $$p>0$$

(2) $$a<b$$

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Re: If a < 0 and b < 0 , is (a - 2p)/(b - 2p) > a/b ?  [#permalink]

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03 Jun 2018, 12:05
1
Bunuel wrote:
If $$a<0$$ and $$b<0$$ , is $$\frac{(a-2p)}{(b-2p)}$$ $$>$$ $$\frac{a}{b}$$ ?

(1) $$p>0$$

(2) $$a<b$$

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.
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Re: If a < 0 and b < 0 , is (a - 2p)/(b - 2p) > a/b ?  [#permalink]

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03 Jun 2018, 12:26
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.

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Re: If a < 0 and b < 0 , is (a - 2p)/(b - 2p) > a/b ?  [#permalink]

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05 Jul 2018, 19:53
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

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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Re: If a < 0 and b < 0 , is (a - 2p)/(b - 2p) > a/b ?  [#permalink]

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05 Jul 2018, 20:23
Bunuel wrote:
If $$a<0$$ and $$b<0$$ , is $$\frac{(a-2p)}{(b-2p)}$$ $$>$$ $$\frac{a}{b}$$ ?

(1) $$p>0$$

(2) $$a<b$$

$$\frac{(a-2p)}{(b-2p)}$$ $$>$$ $$\frac{a}{b}$$ ......
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<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 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
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) $$a<b$$
so a/b>1 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
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1) Absolute modulus : http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
2)Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html
3) effects of arithmetic operations : https://gmatclub.com/forum/effects-of-arithmetic-operations-on-fractions-269413.html

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Re: If a < 0 and b < 0 , is (a - 2p)/(b - 2p) > a/b ?  [#permalink]

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05 Jul 2018, 20:36
Bunuel wrote:
If $$a<0$$ and $$b<0$$ , is $$\frac{(a-2p)}{(b-2p)}$$ $$>$$ $$\frac{a}{b}$$ ?

(1) $$p>0$$

(2) $$a<b$$

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.

Re: If a < 0 and b < 0 , is (a - 2p)/(b - 2p) > a/b ? &nbs [#permalink] 05 Jul 2018, 20:36
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