If p, q, and r are positive integers such that q is a multiple of p

Question

If p, q, and r are positive integers such that q is a multiple of p, and r is a factor of p, which of the following is  NOT  necessarily an integer?

A) (q-p)/p
B) (q-p)/r
C) (q-r)/p
D) (p^2)q/r
E) (q+pr)/p

sachin91 · Answer

Since,q is multiple of p
We can write as q = py where y is any postive integer

Similarly,p = rx

q in terms of r 
q = rxy

Now evaluate each option

A) (q-p)/p = q/p - 1 = y -1

Can be integer
B) (q-p)/r = (rxy - rx)/r
 = xy - x 
Can be integer
C) (q-r)/p = (rxy - r)/rx
 = y- 1/x

Clearly 1/x will always be fraction.
So,this expression can never
be integer.

You can check for other two options as well.

Hence, OA - C

Posted from my mobile device

IanStewart · Answer

This question is a close copy (with some different answer choices) of a question in several editions of the OG. The above solution is right except in one detail - it says answer C is never an integer, but it is in one case, the case where r = p.

Here, q is a multiple of p, and p is a multiple of r. You could use algebra now (q = mp, and p = kr, so q = mkr) and substitute into each answer choice to see which cancel down to integers. You could also pick numbers, say q = 8, p = 4 and r = 2. Or you can just notice that r will always cancel out if it's the denominator of any fraction like those in the answer choices, and p will cancel with anything containing p or q, so the only answer that doesn't cancel down is C, because r/p is not an integer unless r and p are equal.

(edited to fix a typo where I flipped around p and q in the 'pick numbers' sentence)

Satishqwerty · Answer

As the question stem and the options both contain only variables, we can always assume some values. 
Let's take p = 8, q = 16, r = 4

Option A :

\frac{(q-p)}{p} = \frac{(16-8)}{8} = 1

Option B :

\frac{(q-p)}{r} = \frac{(16-8)}{4} = 2

Option C :

\frac{(q-r)}{p} = \frac{(16-4)}{8} = \frac{12}{8}

Option D :

\frac{(p^2)q}{r} = \frac{(8^2)16}{4} = 8^2 	imes 4

Option E :

\frac{(q+pr)}{p} = \frac{(16 + 8 	imes 4)}{8} = \frac{16 + 32}{8} = 6

OA, C

mSKR · Answer

If p, q, and r are positive integers such that q is a multiple of p, and r is a factor of p, which of the following is NOT necessarily an integer?
q= pk1
p= rk2
==> q= rk1k2

A) (q-p)/p=> p common factor- integer
B) (q-p)/r= r can be common factor
C) (q-r)/p = rk1k2-r/( rk2)= not necessarily integer
D) (p^2)q/r= r common factor
E) (q+pr)/p= p common factor

C

Invincible49 · Answer

What if q=5, p=1, r=1.

C stands if this is the scenario.

Please help.

Parnika182 · Answer

Sachin19  , the question says not necessarily an integer. Rest all options will be integers

Invincible49 · Answer

Parnika182  Got it! Thanks!

bumpbot · Answer

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If p, q, and r are positive integers such that q is a multiple of p

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If p, q, and r are positive integers such that q is a multiple of p

Prep Toolkit

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