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If a, b, c and d are positive integers and a/b < c/d, which

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If a, b, c and d are positive integers and a/b < c/d, which  [#permalink]

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New post 27 May 2013, 14:36
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If a, b, c and d are positive integers 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

Doh! I've seen my mistake now... :)
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Re: If a, b, c and d are positive integers and a/b < c/d, which  [#permalink]

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New post 27 May 2013, 15:02
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If a, b, c and d are positive integers 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

Since the numbers are positive we can safely cross-multiply. So, we are given that \(ad<bc\).

I. \(\frac{a+c}{b+d} < \frac{c}{d}\) --> \(ad+cd<bc+cd\)--> \(cd\) cancels out: \(ad<bc\). This is given to be true.

II. \(\frac{a+c}{b+d} < \frac{a}{b}\) --> \(ab+bc<ab+ad\) --> \(ab\) cancels out: \(bc<ad\). Opposite what is given, thus this option is not true.

III. \(\frac{a+c}{b+d} = \frac{a}{b} + \frac{c}{d}\). From I we already know that \(\frac{a+c}{b+d} < \frac{c}{d}\), thus when we add a positive value (\(\frac{a}{b}\)) to \(\frac{c}{d}\) we make the right hand side even bigger. Thus \(\frac{a+c}{b+d} = \frac{a}{b} + \frac{c}{d}\) cannot be true.

Answer: B.
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Re: If a, b, c and d are positive integers and a/b < c/d, which  [#permalink]

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New post 17 Jan 2015, 22:25
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stormbind wrote:
If a, b, c and d are positive integers 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

Doh! I've seen my mistake now... :)


a/b < c/d ; since all are positive ; a/c< b/d ; add 1 to both sides ; (a+c)/c < (b+d)/d; so we have (a+c)/(b+d) < c/d;

I. (a+c)/(b+d) < c/d -YES , we proved it above.
II. (a+c)/(b+d) < a/b - we know a/b < c/d and we also know (a+c)/(b+d) < c/d; does that mean it is always the case that (a+c)/(b+d) < a/b < c/d. cannot say with certainty .
III. (a+c)/(b+d) = a/b + c/d ; Given (a+c)/(b+d) < c/d ; adding a/b which is +ive will make the RHS more bigger . so YES.
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Re: If a, b, c and d are positive integers and a/b < c/d, which  [#permalink]

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New post 18 Jan 2015, 00:18
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Hi All,

This Roman Numeral question can be solved with a mix of Algebra and TESTing VALUES. Also notice that the three Roman Numerals have the SAME fraction on the "left side" - this means that you don't have to do calculations 3 times, you just have to do them ONCE and put them in all 3 Roman Numerals.

We're told that A, B, C and D are all POSITIVE and that A/B < C/D. We're asked which to the Roman Numerators MUST be true (which is the same as saying "which of these is ALWAYS TRUE no matter how many different examples you try?").

Since the variables are all POSITIVE, we can cross-multiply the inequality...

A/B < C/D
AD < BC

It's also worth noting that this inequality has a couple of great "weak spots": You can make the A, B and D the same value and make the C really BIG. You can also make the A, C and D the same value and make the B really BIG.

With those ideas in mind, I'm going to TEST VALUES.

IF....
A = C = D = 1
B = 100

(A+C)/(B+D) = 2/101

I. Is 2/101 < C/D = 1/1?
The answer is YES
This is NOT enough proof to say that this is ALWAYS TRUE though....

II. Is 2/101 < A/B = 1/100?
The answer is NO
ELIMINATE Roman Numeral II. Eliminate Answers C and D.

III. Is 2/101 = A/B + C/D = 1/100 + 1/1?
The answer is NO
ELIMINATE Roman Numeral III. Eliminate Answer E.

We're now down to just Roman Numeral I. Notice how it has the came fraction (C/D) as the inequality from the prompt. That is INTERESTING. Let's see what happens when we cross-multiply the inequality in Roman Numeral I.....

(A+C)/(B+D) < C/D

D(A+C) < C(B+D)
AD + CD < BC + CD

The "CD"s cancel...

AD < BC

This is what we PROVED early on in the work. Roman Numeral I MUST be TRUE.

Final Answer:

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Re: If a, b, c and d are positive integers and a/b < c/d, which  [#permalink]

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New post 18 Jan 2015, 00:29
We have to keep in mind the time constraint we have .

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Re: If a, b, c and d are positive integers and a/b < c/d, which  [#permalink]

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New post 24 Aug 2015, 04:08
can please somebody explain where in m wrong here

from stem a/b <c/d = d/b < c/a
add 1 on both sides
d+b /b < c+a /a
d+b/c+a < b/a

a+c/b+d > a/b form here we can deduce that 1 is true
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Re: If a, b, c and d are positive integers and a/b < c/d, which  [#permalink]

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New post 24 Aug 2015, 08:34
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If a, b, c and d are positive integers and a/b < c/d, which  [#permalink]

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New post 25 Aug 2015, 01:20
Bunuel wrote:
vipulgoel wrote:
can please somebody explain where in m wrong here

from stem a/b <c/d = d/b < c/a
add 1 on both sides
d+b /b < c+a /a
d+b/c+a < b/a

a+c/b+d > a/b form here we can deduce that 1 is true


Please check the options. Expression in red proves that option II ((a+c)/(b+d) < a/b) is NOT true.


That I understood , I proceeded as given above way, which i think mathematically correct, what i am asking , from there how we can deduce that "1" is correct OR I am wrong somewhere in that approach ( this approach is somewhat same to lucky's approach) ???
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If a, b, c and d are positive integers and a/b < c/d, which  [#permalink]

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New post 04 Jan 2019, 22:36
My method was to pick easy numbers and substitute them in, since we're given all numbers are positive integers (phew):
a=1, b=2, c=3, d=4 because we need to make sure a/b < c/d (i.e. we're saying 1/2 < 3/4).

I. (a+c)/(b+d) < c/d ---> 4/6 < 3/4 (TRUE)
II. (a+c)/(b+d) < a/b ---> 4/6 < 1/2 (FALSE)
III. (a+c)/(b+d) = a/b + c/d ---> 4/6 = 1/2 + 3/4 = 5/6 (FALSE)

Answer: B (Only statement I must be true).
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If a, b, c and d are positive integers and a/b < c/d, which   [#permalink] 04 Jan 2019, 22:36
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