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

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

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New post 23 Jan 2017, 02:40
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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
[Reveal] Spoiler: OA

Kudos [?]: 36 [0], given: 40

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Math Expert
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D
Joined: 02 Aug 2009
Posts: 5349

Kudos [?]: 6128 [3], given: 121

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

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New post 23 Jan 2017, 03:32
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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



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

Only B is correct
_________________

Absolute modulus :http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html

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

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New post 23 Jan 2017, 15:27
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



I got to B easily through few steps...
100Q-RQ = RP - 50P
Q(100-R) = P(R-50)
from the above, we can get:
50<R<100
A and E are right away out.

if R =90, then 100-R=10 and 100-50 = 50, but this doesn't satisfy the condition P>Q, but rather vice versa Q>P - D is out.

at least we got now a 50% chance to get the right answer, m? :D

let's try C:
100-75=25
75-50=25

25Q=25P -> P=Q -> doesn't satisfy the condition P>Q

the only option left is B.

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

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New post 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.

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

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New post 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).

Thanks
Rohit

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

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New post 08 Jun 2017, 10:25
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ANS: B
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Re: If P,Q and R are positive integers such that 50P + 100Q = R(P + Q) and [#permalink]

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New post 08 Jun 2017, 10:56
Leo8 wrote:
Attachment:
FullSizeRender (4).jpg


ANS: B


One doubt...

P(50-R) = Q(R-100)
P/Q=(R-100)/(50-R)------(1)
Also, P>Q
IMPLIES P/Q>1
BUT FROM eq (1), this means
(R-100)/(50-R) > 1
R-100>50-R
2R>150
R>75

Which is opposite of what you have shown.

Can you spot the issue with my approach??

Kudos [?]: 20 [0], given: 284

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

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New post 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.

Please advise.

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

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New post 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.

Kudos [?]: 5 [0], given: 150

Re: If P,Q and R are positive integers such that 50P + 100Q = R(P + Q) and   [#permalink] 13 Oct 2017, 06:20
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If P,Q and R are positive integers such that 50P + 100Q = R(P + Q) and

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