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GMATPrepNow
If p and q are positive integers, and p < q, then which of the following MUST be true?

I) \(\frac{p}{q} < \frac{p+1}{q+1}\)

II) \(\frac{p–1}{q} < \frac{p+1}{q}\)

III) \(\frac{p–1}{q} < \frac{p–1}{p+1}\)

A) I only
B) I and II only
C) I and III only
D) II and III only
E) I, II and III

*kudos for all correct solutions

Equation 1 can be re written as:
pq + p < q + pq
p<q -----------true.(given in question stem)

Equation 2
qp - q < pq + q
-q < q
true as q is positive-------------2

Equation 3
(p-1)(p+1) < q(p-1)
(p-1)(p+1-q) < 0
No ..take p=1 and q=2
it will violate the equation .
so , this is not true..

Answer is B
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If we substitute each equation with p=4 and q=8
the outcome is
1. 1/2<5/9=-1<-4
2. 3/8<5/8 = -5<-3
3. 3/8<1/3 = -5<-2
So conclusion should be options 2 & 3.

Please advise why do we get different values on plugging values and directly solving the algebraic expression.

These (-)ve values were assumed as they were considered as integers.
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If we substitute each equation with p=4 and q=8
the outcome is
1. 1/2<5/9=-1<-4
2. 3/8<5/8 = -5<-3
3. 3/8<1/3 = -5<-2
So conclusion should be options 2 & 3.

Please advise why do we get different values on plugging values and directly solving the algebraic expression.

These (-)ve values were assumed as they were considered as integers.

I'm confused with the steps you took to make the conclusions in blue above

For example, how does 1/2<5/9 turn into -1<-4?
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GMATPrepNow
If p and q are positive integers, and p < q, then which of the following MUST be true?

I) \(\frac{p}{q} < \frac{p+1}{q+1}\)

II) \(\frac{p–1}{q} < \frac{p+1}{q}\)

III) \(\frac{p–1}{q} < \frac{p–1}{p+1}\)

A) I only
B) I and II only
C) I and III only
D) II and III only
E) I, II and III

*kudos for all correct solutions

Given : p<q
-> p/q<1

I. p/q . When a constant is added to nr and dr, the fraction tends to move towards 1.
Since p/q<1 so (p+1)/(q+1) will move towards 1.
Hence p/q < (p+1)/(q+1)


II. (p-1)/q < (p+1)/q.. Since dr. is same and Nr. on right side is greater than left side.

III. (p–1)/q < (p–1)/p+1 .. NOT TRUE . Since p<q -> p+1<=q,
So, (p–1)/q <= (p–1)/p+1

Hence only I & II are correct
Answer B.
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shashankism
GMATPrepNow
If p and q are positive integers, and p < q, then which of the following MUST be true?

I) \(\frac{p}{q} < \frac{p+1}{q+1}\)

II) \(\frac{p–1}{q} < \frac{p+1}{q}\)

III) \(\frac{p–1}{q} < \frac{p–1}{p+1}\)

A) I only
B) I and II only
C) I and III only
D) II and III only
E) I, II and III

*kudos for all correct solutions

Given : p<q
-> p/q<1

I. p/q . When a constant is added to nr and dr, the fraction tends to move towards 1.
Since p/q<1 so (p+1)/(q+1) will move towards 1.
Hence p/q < (p+1)/(q+1)


II. (p-1)/q < (p+1)/q.. Since dr. is same and Nr. on right side is greater than left side.

III. (p–1)/q < (p–1)/p+1 .. NOT TRUE . Since p>q -> p+1>q,
So, p–1)/q > (p–1)/p+1


Hence only I & II are correct
Answer B.

shashankism , I can't follow the bolded part. p isn't greater than q, and in analysis of I you note that, so I think something went wrong here ...?
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shashankism
GMATPrepNow
If p and q are positive integers, and p < q, then which of the following MUST be true?

I) \(\frac{p}{q} < \frac{p+1}{q+1}\)

II) \(\frac{p–1}{q} < \frac{p+1}{q}\)

III) \(\frac{p–1}{q} < \frac{p–1}{p+1}\)

A) I only
B) I and II only
C) I and III only
D) II and III only
E) I, II and III

*kudos for all correct solutions

Given : p<q
-> p/q<1

I. p/q . When a constant is added to nr and dr, the fraction tends to move towards 1.
Since p/q<1 so (p+1)/(q+1) will move towards 1.
Hence p/q < (p+1)/(q+1)


II. (p-1)/q < (p+1)/q.. Since dr. is same and Nr. on right side is greater than left side.

III. (p–1)/q < (p–1)/p+1 .. NOT TRUE . Since p>q -> p+1>q,
So, (p–1)/q > (p–1)/p+1


Hence only I & II are correct
Answer B.

shashankism , I can't follow the bolded part. p isn't greater than q, and in analysis of I you note that, so I think something went wrong here ...?

Yes i didi a mistake over there. I have corrected my solution
Basically p < q
so p+1<=q
and hence
(p-1)/q <= (p-1)/(p+1)
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We can take p=1 and q=2 ( positive integers where p<q)
Substitute these values one by one in all 3 options
Hence 1 and 2 holds the condition.
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