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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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23 Jan 2017, 03:40
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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
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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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23 Jan 2017, 04:32
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.....\(50p+100q=65(p+q)....50p+100q=65p+65q....35q=15p... p=\frac{35q}{15}...P>Q\).. 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.....\(50p+100q=100(p+q)....50p+100q=100p+100q....100p=50p....P=0\)... Wrong Only B is correct
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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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23 Jan 2017, 16:27
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... 100QRQ = RP  50P Q(100R) = P(R50) from the above, we can get: 50<R<100 A and E are right away out. if R =90, then 100R=10 and 10050 = 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: 10075=25 7550=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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31 May 2017, 10:30
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(50R)+Q(100R)=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 50R should be <100R. Thus B is the answer 5065=15<10065=35 (with D it would be 5090=40>10090=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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08 Jun 2017, 11:15
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Hi
Simplest method would be to rearrange 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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08 Jun 2017, 11:25
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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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08 Jun 2017, 11:56
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Leo8 wrote: Attachment: FullSizeRender (4).jpg ANS: B One doubt... P(50R) = Q(R100) P/Q=(R100)/(50R)(1) Also, P>Q IMPLIES P/Q>1 BUT FROM eq (1), this means (R100)/(50R) > 1 R100>50R 2R>150 R>75 Which is opposite of what you have shown. Can you spot the issue with my approach??



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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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10 Jul 2017, 08:02
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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13 Oct 2017, 07:20
\(50P + 100Q = R(P + Q)\)
\(50(P + 2Q) = R(P + Q)\)
\(50[(P + Q) + Q] = R(P + Q)\)
\(50Q = (R  50)(P + Q)\)
\(\frac{Q}{(P + Q)}= \frac{(R  50)}{50}\)
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.



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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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06 Jan 2018, 06:34
Can be easily solved using weight average concept. Given: 50 R100 weight:PQ if P = Q, then R will be at centre i.e = 75 however given, P > Q, weighted average will be pulled towards 50 and will be less than 75 only choice B fits perfectly, as weighted average can't be equal to 50 (since P = 50) Kudos please, if you like my approach, need to unlock my gmat club testsThanks



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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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06 Jan 2018, 07:15
Leo8 wrote: Attachment: FullSizeRender (4).jpg ANS: B can u explain how you got 50r as greater



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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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06 Jan 2018, 10:13
rahulkashyap wrote: Leo8 wrote: Attachment: FullSizeRender (4).jpg ANS: B can u explain how you got 50r as greater from our equation : P(50 R) = Q( R  100) But Its given P > Q so for the equality to be true the coeffficent of P must be greater than coefficient of Q 50  R > R  100 also nitesh181989 : P/q > 1 will give you equation : 50  R > R  100 you have put coeffiecient of Q in the numerator and that of P in denominator
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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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06 Jan 2018, 10:37
Leo8 wrote: rahulkashyap wrote: Leo8 wrote: Attachment: FullSizeRender (4).jpg ANS: B can u explain how you got 50r as greater from our equation : P(50 R) = Q( R  100) But Its given P > Q so for the equality to be true the coeffficent of P must be greater than coefficient of Q 50  R > R  100 also nitesh181989 : P/q > 1 will give you equation : 50  R > R  100 you have put coeffiecient of Q in the numerator and that of P in denominator Well, first of all if p if greater than q, the inequality sign should be there other way. Not what you've pointed. (bigger) x (smaller)= (smaller) x(bigger) Even then, you don't know whether lhs and rhs are both positive / negative. So you can't say with certainty Posted from my mobile device




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