if a, b, c, and d are positive

[ting.cui10]

if a, b, c, and d are positive numbers and a/b < c/d , which of the following must be true?

I. (a+c) / (b+d) < c/ d

II. (a+c) / (b+d) < a/b

III. (a+c) / (b+d) < a/b + c/d

a) none
b) I only
c) II only
d) I and II
e) I and III

OA: B


how do you solve this problem?

[krishnan.anju1987]

B looks about right. This is how I tried it
Given

a/b<c/d implies

ad<bc

Consider 1) a+c/b+d < c/d is true
then, cross multiplying
ad+dc<bc+cd
thus, ad<bc which is true and hence 1) is true.

Consider 2) a+c/b+d < a/b
implies

ab+bc<ba+ad
bc<ad
which we know to be false

3) a+c/b+d < a/b+ c/d

we know 2 is not true so 3) is also not true as the same expression is a part of 3).


Hence correct answer is B

[ting.cui10]

could you explain your logic for the last part?

this part: we know 2 is not true so 3) is also not true as the same expression is a part of 3).

i dont see how you arrived at that conclusion. take for instance a hypothetical example, if (a+c) / (b+d) = 5 and a/b = 4 and c/d = 2 then (a+c) / (b+d) < a/b is false but (a+c) / (b+d) < a/b + c/d is true.

[krishnan.anju1987]

Guess I have explained better

(a+c)/(b+d) >a/b

(a+c)/(b+d) +c/d > a/b +c/d

Cross multiply and cancel and you get

Bc>ad

Which its false, hence c is wrong.

Also, the example you have taken agrees with only 2nd statement but I got your point.

Hope this

[tim]

looks like you've reversed the inequality, so your approach won't work on 3. the idea of cross multiplying is a good one though..

[krishnan.anju1987]

Hi Tim,

Please help me out if I am doing something wrong.

After I cross multiplied and got the answer bc>ad, comparing it to the inequality given in the question (ad>bc) would let me know that the inference from the third statement is wrong and hence the statement is insufficient.

I went over my solution once more and came up with the same answer. Is there something wrong with this approach that I maybe missing.

Thanks for your help in advance.

[desiwolverine]

I think it should be E meaning I and III are correct.

You can cross multiply and then cancel out terms.

ad < bc is what we get from the given data.

1st case cross multiplication lines up to that.
2nd case does not.
3rd case after cross multiplication and cancelling out gives cb^2 + ad^2 > 0.

Which should be true since c,b,a and d are positive numbers.

[tim]

Krishnan, looks like you've done it right. Wolverine, I just ran through the calculations and got the same result as you. Unless we both made the same mistake it seems E should be the answer. Can someone post a screenshot from GMAT Prep to let us know for sure what they say the correct answer is?

[krishnan.anju1987]

Hi,

Could someone please post the answer to this question. I went over C once more and came up with another solution which makes E the answer.

We know 1) is true and hence (a+c)/(b+d)< (c/d)

now, (a+c)/(b+d) < a/b+ c/d

can be written as

(a+c)/(b+d) < a/b+ (a+c)/(b+d) + y

since c/d = (a+c)/(b+d) + y where y is some fraction or integer which when added to the fraction would give c/d

hence from above

0< a/b +y

since and b is positive and y has to be positive, this is true and hence iii) has to be true.

Does this look good?

[tim]

still waiting for a screenshot to confirm the answer. i got E as well..

[Ankit Gupta]

After solving the equation 3, I am getting the following inequality.

bc+ad >0

Please confirm.

My answer is also coming out to be E !!

[crissro]

(a+c)\ (b+d)=a/(b+d) + c/(b+d)< a/b + c/d when b,d#0

[jnelson0612]

Hi,

Could someone please post the answer to this question. I went over C once more and came up with another solution which makes E the answer.

We know 1) is true and hence (a+c)/(b+d)< (c/d)

now, (a+c)/(b+d) < a/b+ c/d

can be written as

(a+c)/(b+d) < a/b+ (a+c)/(b+d) + y

since c/d = (a+c)/(b+d) + y where y is some fraction or integer which when added to the fraction would give c/d

hence from above

0< a/b +y

since and b is positive and y has to be positive, this is true and hence iii) has to be true.

Does this look good?

Looks good to me! That is creative.

[jnelson0612]

After solving the equation 3, I am getting the following inequality.

bc+ad >0

Please confirm.

My answer is also coming out to be E !!

My answer is also E. For III, I got:

cb^2 + ad^2 > 0.

[krishnan.anju1987]

Thanks Jamie :)