If a, b, c, and d are positive numbers and a/b < c/d , which of the fo

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) We can't prove equality from inequality. Hence can't be 'must be true'. Bunuel, is my logic correct in here?

Hence correct answer is B

Your method is correct as well as all the 4 numbers are positive.

This question can be tackled easily by the method below:

You are given a/b < c/d ---> as all the numbers are positive you can cross multiply to get , a/c < b/d ---> (a/c) + 1 < (b/d) + 1 ---> (a+c)/c < (b+d)/d ---> (a+c) / (b+d) < c/ d which is statement I and is hence true.

Now, again consider a/b < c/d ---> d/b < c/a ---> (d/b) + 1 < (c/a) + 1 ---> (d+b)/b < (c+a)/a ---> a/b < (c+a)/(b+d) but this is not what statement II mentions, making statement II false.

Finally, for (a+c) / (b+d) = a/b + c/d , you can clearly prove this wrong by taking a,b,c,d as 1,2,3,4. So this is also not true.

For a MUST BE TRUE question, you need to get a yes ALWAYS.

Thus, only statement I is true , making B the correct answer.

Hope this helps.
_________________

Thanks for your reply.

But can you also comment on the following-
St3) We can't prove equality from inequality. Hence st3 can't be 'must be true'. Is my logic correct specifically for st3?

No this is a very broad statement that you are mentioning. Take a simpler example: If I say x<3 and then ask whether x =2 is true ? You will say "yes" but here you used the inequality to arrive at your answer for an equality.

I have shown 1 way of using particular values to see that III is NOT must be true.

The other way is to assume that (a+c) / (b+d) = a/b + c/d is indeed true.

Then you end up getting \(b^2*c+a*d^2=0\) ---> \(-(a/c)= b^2/d^2\) but this can not possible as a and c are given to be of the same sign (both are positive) and as such this will make -(a/c) a negative quantity.

Additionally, you also know that \(b^2/d^2\) can never be <0 . Thus you have this contradiction, making this statement not a must be true.

Hope this helps.
_________________

Bunuel I am very confused why 3rd statement is not correct. Here is my take on it -

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

Taking LCM -
so, (a+c)/(b+d)= (ad+cb)/(bd)

cross multiplying -
abd + bcd =abd +bcd, which is true. So the answer should be E. Can you please let me know why this approach is wrong. Thanks for all your help !

I used real values and came to the answer. Used 1/2 and 3/4.

Hi vtomar20

I see a problem in your algebra.

Even i did the same mistake and eventually found what i was doing wrong.

When you simplify the expression:

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

you will get

abd + bcd =abd +bcd + ad^2 + cb^2

and this however is not true, since in the above expression we have an extra ad^2 + cb^2 on the right side which will make the right side bigger (and we know this because all the a, b, c, and d, are positive numbers ).

so E is out and B is the answer.

I hope its clear

Thankyou
Arunabh saxena

a/b<c/d
we multiple two side by bd. remember all number are positive.
ac<cb.

this simple multiply make the fraction from into factors form. this is key to do inequality problem.

A good idea to eliminate III is if you look at option 1 it tells (a+c)/(b+d)<c/d, which we have already proved to be true.

Now go to option III (a+c) / (b+d) = a/b + c/d , which denotes (a+c) / (b+d) > c/d hence it is false as we have already proved I to be true !!

Pick numbers... noticing the LHS is the same for all 3 roman numerals.
a=1/b=1 < c=2/d=1

I. 1+2/1+1 < 2/1
3/2 < 2, TRUE

II. 3/2 < 1, FALSE

III. 3/2 = 3, FALSE

Pick some other numbers to retest I.
a=1/b=3 < c=2/d=3
1+2/3+3 < 2/3
3/6 < 2/3
1/2 < 2/3, TRUE
Good enough, B it is.

I solved it by taking convenient values.
Lets take a= 4 ; b=2 ; c=12 ; d = 3

c/d = 4 which is greater than a/b=2

I must be true as ,
(a+c) /( b+d) = 16 / 5 = 3.2
c/d = 4
So (a+c) /( b+d) < c/d

II may not be true ,

(a+c) /( b+d) = 16 / 5 = 3.2
a/b = 2
2 is not greater than 3.2

III may not be true ,

(a+c) /( b+d) = 16 / 5 = 3.2

a/b + c/d = 2 +4 =6

Only I must be true.
B is the answer.

Please give me Kudos if you liked my explanation.

As each letter represents a positive number, we can cross multiply to get \(ad < cb\).

I. Cross multiply: \(ad + cd < cb + cd\)
\(ad < cb\)

Statement I must be true.

II. Cross multiply: \(ab + bc < ab + ad\)
\(bc < ad\)

We can't say for certain that this is true.

III. \(\frac{(a+c)}{(b+d)} = \frac{a}{b} + \frac{c}{d}\)

let \(a = 1, b = 2, c = 2, d = 3\)

\(\frac{3}{5} = \frac{1}{2 }+ \frac{2}{3}\\
\)
Not true.

Only statement that must be true is statement I.
_________________

GIVEN:

• a, b, c, and d are positive integers.
• \(\frac{a}{b< c/d }\)
  
  • In another form, we can say that ad < cb …. (1) [Multiplying bd to both sides of the inequality]
  • Note: Since b and d are positive, bd > 0. So, multiplying both sides by bd doesn’t change the inequality sign.

TO FIND:

Which of the following must be true:

1. \(\frac{(a+c)}{(b+d)}\) < \(\frac{c}{d}\)
2. \(\frac{(a+c)}{(b+d)}\) < \(\frac{a}{b}\)
3. \(\frac{(a+c)}{(b+d)}\) = \(\frac{a}{b}\) + \(\frac{c}{d}\)

APPROACH:

We will take each statement one-by-one and start with considering it true. We will then simplify it to find the conditions under which it will be true. We will finally check these conditions with the given information.
Don’t worry if this feels confusing right now; you’ll get it once you see this in action!

WORKING OUT:

• \(\frac{(a+c)}{(b+d)}\) < \(\frac{c}{d}\)
  
  Suppose \(\frac{(a+c)}{(b+d)}\) < \(\frac{c}{d}\) is true. Let’s simplify this now:
  
  • Cross-multiplying gives (a + c)d < (b + d)c.
    
    • That is, ad + cd < bc + dc.
      
    • Or ad < bc. (Since cd = dc, we can cancel it out from both sides)
  
  This means that statement I is true if ad < bc.
  But that is precisely what we are given! Hence, based on our given information, Statement (I) is always true!
  Conclusion: Statement (I) must be true.
  
  
• \(\frac{(a+c)}{(b+d)}\) < \(\frac{a}{b}\)
  
  Suppose \(\frac{(a+c)}{(b+d)}\) < \(\frac{a}{b}\) is true. Let’s simplify this now:
  
  • Cross-multiplying gives (a + c)b < (b + d)a.
    
    • That is ab + cb < ba + da.
      
    • Or cb < da (since ab = ba, we can cancel it out from both sides)
  
  This means that statement (II) is true if ad > bc.
  
  BUT from (1), we know that ad < bc. Thus, the necessary condition for Statement (II) is NEVER possible – it is the exact opposite of the condition we are given.
  
  Hence, statement (II) must NOT be true. (In fact, it must be false!)
  
  
• \(\frac{(a+c)}{(b+d)}\) = \(\frac{a}{b}\) + \(\frac{c}{d} \)
  
  Suppose \(\frac{(a+c)}{(b+d)}\) = \(\frac{a}{b}\) + \(\frac{c}{d}\) is true. Let us again simplify it:
  
  • Taking LCM on the right side gives: \(\frac{(a+c)}{(b+d)}\) = \(\frac{(ad+bc)}{bd }\)
    
  • Cross-multiplying, we get: (a + c)bd = (b + d)(ad + bc).
    
    • abd + cbd = bad + \(b^2\)c + a\(d^2\) + bcd.
      
    • 0 = \(b^2\)c + a\(d^2\) (since abd = bad and cbd = bcd, we can cancel them out from both sides)
      
    • \(b^2\)c = -a\(d^2\).
  This means that statement (III) is true IF \(b^2\)c = -a\(d^2\).
  
  But this is NOT POSSIBLE because a, b, c and d are positive integers, making -a\(d^2\) negative and \(b^2\)c positive. We know that a positive and a negative number can never be equal!
  Thus, the necessary condition for Statement (III) cannot be true. And hence, statement (III) must NOT be true.

Conclusion: Out of the 3 only statement (I) must be true.

Therefore, Correct choice: Option B.

Hope this helps!

Best Regards,
Ashish Arora
Quant Expert, e-GMAT
_________________