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

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:

GMAT assassins aren't born, they're made,
Rich

We have to keep in mind the time constraint we have .

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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) ???

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).

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

Everytime you see must be true you can choose easy value and plug in.
Must be true means that if does hol for a certain set of numbers will hold for every set of number

\(\frac{a}{b}<\frac{c}{d} \to \frac{2}{1}<\frac{8}{2}=2<4\)
\(\frac{(a+c)}{(b+d)}=\frac{10}{3}\)

I. (a+c)/(b+d) < c/d
\(\frac{10}{3}<4\) I ok
II. (a+c)/(b+d) < a/b
\(\frac{10}{3}<4\) II
III. (a+c)/(b+d) = a/b + c/d
\(\frac{10}{3}=2+4\) III

I only will always hold
B

We know that when you take a positive fraction x/y and add a to the numerator, b to the denominator then the resulting fraction moves closer to a/b. So then in this case you will clearly see that only I can be correct or always true as we are adding c to the numerator and d to the denominator for the +ve fraction a/b and we know that this will move close to c/d or in other words the fraction a+c/b+d is in between a/b and c/d on the number line.