If 0

Question

If 0 < q ≤ p, then for all possible values of p and q, p must be greater than which of the following?

A.  \frac{1}{p+q}

B.  \frac{1}{p-q}

C.  \frac{p-q}{p+q}

D.  \frac{p^2-q^2}{p+q}

E.  \frac{pq+q^2}{p}

chetan2u · Answer

If 0 < q ≤ p, then for all possible values of p and q, p must be greater than which of the following?

A. \(p>\frac{1}{p+q}.........p(p+q)>1.......0.1(0.1+0.1)>1....NO\)

B. \(p>\frac{1}{p-q}.........p(p-q)>1.......0.1(0.1-0.1)>1....NO\)

C. \(p>\frac{p-q}{p+q}.......0.2>\frac{0.2-0.1}{0.2+0.1}......0.2>\frac{0.1}{0.3}......…0.1*0.3>0.1………0.03>0.1……..NO\)
If x is an integer, the expression will always be true. Only when p and q are <1 or rather closer to 0, when you multiply p+q with a fraction that is very close to 0, it will make p+q very small.

D. \(p>\frac{p^2-q^2}{p+q}...........p>\frac{(p-q)(p+q)}{p+q}............p>p-q..........q>0.......YES\)

E. \(p>\frac{pq+q^2}{p}\)............p^2>pq+q^2[/quote]. When p=q \(p^2>p^2+p^2.........NO\)

D

IanStewart · Answer

If q is positive, then p-q < p for sure (if you subtract a positive value from something, it gets smaller), and answer D is equal to p-q, so D is correct.

adijoshi21 · Answer

. When p=q  p^2>p^2+p^2.........NO

D

How do you rule out C?

IanStewart · Answer

Yes, there does seem to be a mistake in chetan's solution for C, unless I've misunderstood what he's doing.

If q is very close to zero, then p-q and p+q will both roughly be equal to p, and (p-q)/(p + q) will be roughly equal to p/p = 1. So if q is tiny and p is less than 1, p can be smaller than answer choice C. For example, if q = 0.01 and p = 0.09, you'll find that p is smaller than answer C.

chetan2u · Answer

Hi

It seems while solving the question, I must have misread and must have been in a hurry. 
 IanStewart  is correct.

The reason why C is incorrect. 
Whenever we have p<1 and q very close to 0, both p-q and p+q will be nearly equal to p. 
So we will have p>~(p/p) or p>~1…..

Example 
Let p=0.9 and q=0.01
 \frac{p-q}{(p+q)}=\frac{(0.9-0.01)}{(0.9+0.01)}=\frac{0.89}{0.91}

So  0.9>\frac{0.89}{0.91} 
 0.9*0.91>0.89 
 0.819>0.89 …….NO

Kinshook · Answer

Asked: If 0 < q ≤ p, then for all possible values of p and q, p must be greater than which of the following?

A. \(\frac{1}{p+q}\)
If p = .5 ; q = .5
p = .5 ; 1/p+q = 1/1 ; p<1/p+q : Incorrect

B. \(\frac{1}{p-q}\)
If p = .6 ; q = .5
\(p = .5 ; \frac{1}{p-q} = 1/.1 = 10 ; p<\frac{1}{p-q} \): Incorrect

C. \(\frac{p-q}{p+q}\)
If p = .6 ; q = .1
\(p = .6 ;\frac{ p-q}{p+q} = .5/.7= .71 ; p<\frac{p-q}{p+q}\) : Incorrect

D. \(\frac{p^2-q^2}{p+q}\)
If p=q; p>0
If p>q: p> p - q; q>0
Always Correct

E. \(\frac{pq+q^2}{p}\)
If p=2; q=2
\(\frac{pq + q^2}{p} = [m]8/2\) = 4; p < \frac{pq + q^2}{p}[/m]

IMO D

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If 0 < q ≤ p, then for all possible values of p and q, p must be great

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