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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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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
General Discussion
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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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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??
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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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Can be easily solved using weight average concept.

Given: 50 ----------------R------------------100
weight:P--------------------------------------Q
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 tests

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

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

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

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

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

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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Re: If P,Q and R are positive integers such that 50P + 100Q = R(P + Q) and [#permalink]
Leo8 wrote:
Attachment:
FullSizeRender (4).jpg


ANS: B


can u explain how you got 50-r as greater
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Re: If P,Q and R are positive integers such that 50P + 100Q = R(P + Q) and [#permalink]
rahulkashyap wrote:
Leo8 wrote:
Attachment:
FullSizeRender (4).jpg


ANS: B


can u explain how you got 50-r as greater




from our equation :

P(50 -R) = Q( R - 100)

But Its given P > Q so for the equality to be true the co-effficent of P must be greater than co-efficient of Q

50 - R > R - 100

also nitesh181989 :

P/q > 1 will give you equation : 50 - R > R - 100

you have put co-effiecient 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]
Leo8 wrote:
rahulkashyap wrote:
Leo8 wrote:
Attachment:
FullSizeRender (4).jpg


ANS: B


can u explain how you got 50-r as greater




from our equation :

P(50 -R) = Q( R - 100)

But Its given P > Q so for the equality to be true the co-effficent of P must be greater than co-efficient of Q

50 - R > R - 100

also nitesh181989 :

P/q > 1 will give you equation : 50 - R > R - 100

you have put co-effiecient 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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Re: If P,Q and R are positive integers such that 50P + 100Q = R(P + Q) and [#permalink]
chetan2u wrote:
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


Hi chetan2u, Might be silly but would this approach work as well?

=50p+100q=r(p+q)
=50p + 100q/(p+q) = r
=50p + 50q + 50q/(p+q) = r
=50 + 50q/(p+q) = r ......(The answer will be > 50 but always less than 75 because 50q/(p+q) can never be >=25 since p>q )
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Re: If P,Q and R are positive integers such that 50P + 100Q = R(P + Q) and [#permalink]
Expert Reply
avirupm97 wrote:
chetan2u wrote:
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


Hi chetan2u, Might be silly but would this approach work as well?

=50p+100q=r(p+q)
=50p + 100q/(p+q) = r
=50p + 50q + 50q/(p+q) = r
=50 + 50q/(p+q) = r ......(The answer will be > 50 but always less than 75 because 50q/(p+q) can never be >=25 since p>q )


That’s a perfectly fine solution. No error at all.
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Re: If P,Q and R are positive integers such that 50P + 100Q = R(P + Q) and [#permalink]
Asked: 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?

50P + 100Q = R (P + Q)
(50-R)P + (100-R)Q = 0
P = (100-R)Q/(R-50)
Since P > Q
(100-R)Q/(R-50) > Q
(100-R) > R-50
R < 75

Since P > 0
(100-R)Q/(R-50) > 0
R < 100
R > 50

IMO 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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