Given that A > 0 and C > 0, is (A + B)/(C + B) > A/C?

Since we are told that A and C are both greater than zero, that means we can freely multiply and divide these two variables in the inequality but not B since we aren't told if B is positive or negative.

With that in mind, lets play around with the question first.
(A+B) A
------- > ---
(C+B) C

Since we can multiply/divide A and C around the inequality lets cross multiply

AC + AB > AC + BC

Subtract AC on both sides and we're left with "is AB > BC"

Now on to the choices

1) B>0
Insufficient since this doesn't tell us anything about C and A.

2) A<C
Again, insufficient since if you consider this statement alone, we aren't told anything about the value of B. If B is positive then BC will be be greater than AB if C > A. However, if B is a negative value then BC will be less than AB if C > A

1) and 2)
Sufficient. We know B is positive. This fills the gap in the second statement. If B is positive then having C > A ensure that BC > AB

So answer is C

Additions of B in numerator and denominator can be tricky,

I guess answer should be E

Given that A > 0 and C > 0, is (A + B) / (C + B) > A/C?

(1) B > 0

(2) A < C

st1 -- B is positive given
also A and C are positive

hence (A + B) / (C + B) > A/C

We can cross multiply

AC + BC > AC + AB
BC > AB
BC - AB > 0
B( C-A) >0

already that B is >0 hence --- C >A

Therefore ---- We got the relation between c and a .

Now

(A + B) / (C + B) > A/C Use number any positve number that satisfy C>A and B is also positive .We will get a definitive ans .

ST B . Not sufficient . NO clues given for B

Hence Ans is A .

VERITAS PREP OFFICIAL SOLUTION:

For (1), we are only given that b is a positive number. Picking values is a good strategy here. If we can choose values such that we get a “yes” and choose values so that we get a “no” then we can quickly eliminate. If a = 1 and c = 1 and b = 1, then the inequality is NOT correct. However, if a = 1 and c = 2, and b = 3, we get 4/5 > 1/2, which IS correct. (2) can be proved insufficient similarly. Combined, if b is positive and a < c, then it will continue to INCREASE the left-hand side of the inequality no matter what values we pick.

The answer is (C).

Remember, even the most seemingly complex Data Sufficiency questions can be overcome with solid strategy!

I solved it by remembering that if 0 < x/y < 1, then adding same positive number to the numerator and denominator, the value will increase.

1. B>0
just listed 2 possible options
A=C
B=2
A+2/C+2 = A/C so equal

A=3
C=5
B=1
3+1/5+1 = 4/6 = 2/3 > 3/5

2 outcomes, A alone is not sufficient.

2. A<C
ok, so 0<A/C<1 but what about B? if B is negative, then adding it to numerator and denominator would decrease the new fraction.
clearly 2 alone is not sufficient.

1+2
A<C
and B>0
sufficient.

whenever we add something to numerator and denominator, percentages come into play! like if the percentage increase in Numerator is more than the percentage increase in denominator, ratio increases and vice versa.

So here we are adding a number B to numerator (A) and denominator (C) but we don't know which one is larger so as know whose percentage increase is more.

St1: B>0 now we know that we are adding a number and not subtracting since B is positive but we don't know whether A >C or C>B so we can't comment on the change in ratio.

St2: A<C but this does not tell about B( whether we are subtracting a number or adding a number)

1 & 2 combined: now B>0 so we are adding something. and A<C so percentage change in numerator(A) would be more as compared to percentage change in denominator(C) . So the ratio will increase.

Hence C.

The first temptation is to cross-multiply the expressions, but we can't do it because we only know that A, C are positive, but we don't know the sign of B.
So for start let the question stem as it is.

from 1) \(b>0\) Now we know all the signs and we can cross multiply. We get:

\(CA+CB>CA+AB\)
CA+CB>CA+AB
\(CB>AB\)
\(CB-AB>0\)
\(B(C-A)>0\)
Now we have that B multiplied by (C-A) is positive. How can that be? only if B>0 and (C-A)>0 or B<0 and (C-A)<0

From the stem we are told that B>0, so the question becomes "Is (C-A) positive?" From 1) we can't extract any info on C-A so it's insufficient.

from 2) A<C, so 0<C-A We don't know the sign of B, so we can't cross-multiply and that impairs us from working on the question. Insufficient.

1) + 2) from 1 we got our rephrased question, "Is (C-A) positive?" and from 2) we get that A<C, so 0<C-A and we now have enough info to answer.

The analysis of your Question Stem says we have to find whether B<0, statement A says b>0, then how is that insufficient?