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If P,Q and R are positive integers such that 50P + 100Q = R(P + Q) and if P>Q, which of the following could be the value of R?

A. 50 B. 65 C. 75 D. 90 E. 100

Hi, Easiest way would be to substitute choices in given equation 50p+100q=r(p+q) and check for P>Q..

A. 50.....50p+100q=50(p+q)....50p+100q=50p+50q....100q=50q... Wrong B. 65.....50p+100q=65(p+q)....50p+100q=65p+65q....35q=15p... p=35p/15...P>Q.. Correct C. 75.....50p+100q=75(p+q)....50p+100q=75p+75q....25q=25p.. p=Q.... Wrong D. 90.....50p+100q=90(p+q)....50p+100q=90p+90q....10q=40p.. ..P<q..Wrong E. 100.....50p+100q=100(p+q)....50p+100q=100p+100q....100p=50p... Wrong

Re: If P,Q and R are positive integers such that 50P + 100Q = R(P + Q) and [#permalink]

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31 May 2017, 09:30

siddharthsinha123 wrote:

If P,Q and R are positive integers such that 50P + 100Q = R(P + Q) and if P>Q, which of the following could be the value of R?

A. 50 B. 65 C. 75 D. 90 E. 100

From the equation above we see that P(50-R)+Q(100-R)=0. Therefore R<>50 or 100, so A and E are out. Also, P<>Q so C is out. Now we see that P will be with negative sign and Q - with Positive. Taking into account that P>Q |50-R| should be <|100-R|. Thus B is the answer |50-65|=15<|100-65|=35 (with D it would be |50-90|=40>|100-90|=10.

If P,Q and R are positive integers such that 50P + 100Q = R(P + Q) and [#permalink]

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08 Jun 2017, 10:15

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Hi

Simplest method would be to re-arrange the given statement as :

(50P + 100Q)/(P+Q) = R

From this you know that this is nothing but weighted average. It is given that P > Q, hence the average would be closer to 50 than to 100 and not in the middle value (75). You are left with only 65 (B).

Re: If P,Q and R are positive integers such that 50P + 100Q = R(P + Q) and [#permalink]

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10 Jul 2017, 07:02

On substituting the equation R=65 then the we have 35q=15p... or p=35q/15...P>Q.. But this value can be non integer if q=1,2.... etc then doesn't it violate the statement that P,Q and R are positive integers.

Re: If P,Q and R are positive integers such that 50P + 100Q = R(P + Q) and [#permalink]

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13 Oct 2017, 06:20

\(50P + 100Q = R(P + Q)\)

\(50(P + 2Q) = R(P + Q)\)

\(50[(P + Q) + Q] = R(P + Q)\)

\(50Q = (R - 50)(P + Q)\)

\(\frac{Q}{(P + Q)}= \frac{(R - 50)}{50}\)

Let's try with the answer in the middle (C) \(R=75 => \frac{(75 - 50)}{50}= \frac{25}{50} = \frac{1}{2} => \frac{Q}{(P + Q)}= \frac{1}{2}\) => Q=1 and P=1, which cannot be (P must be greater than Q). With this, the bigger answers D and E (90 and 100) cannot be either.

Let's with B: \(R=65 => \frac{(65 - 50)}{50}= \frac{15}{50} = \frac{3}{10} => \frac{Q}{(P + Q)}= \frac{3}{10}\) => Q=3 and P=7, and this matches with the statement which says that P > Q. => Correct Answer: B.