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# If P, Q, R are positive numbers such that R = 100P + 20Q

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Manager
Joined: 04 Mar 2012
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If P, Q, R are positive numbers such that R = 100P + 20Q [#permalink]

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13 May 2012, 07:11
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Question Stats:

57% (01:44) correct 43% (01:20) wrong based on 93 sessions

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If P, Q, R are positive numbers such that R = 100P + 20Q, and P + Q = 2, is R < 120?

(1) P > 1.1
(2) P > Q
[Reveal] Spoiler: OA
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Re: P, Q, R are positive numbers such that [#permalink]

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13 May 2012, 22:57
R = 100P + 20Q, and P + Q = 2

Using statement (1), if P>1.1, Q<0.9
As R = 100P + 20Q, R is smallest when P is at its smallest and Q at its largest. This happens when P=1.11 and Q=0.89
R(min)= 100(1.11) + 20(0.89) = 111 + 17.8 which is greater than 120
Therefore R>120. Sufficient.

Using statement (2),
P>Q
R(min) = 100 (1.01) + 20(0.99) = 101 + 19.8 = which is more than 120. Sufficient.

D it is.
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Senior Manager
Joined: 18 Sep 2009
Posts: 349
Re: If P, Q, R are positive numbers such that R = 100P + 20Q [#permalink]

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14 May 2012, 06:52
R=100(p) + 20(q), p+q=2,is R<120
after rephrasing:
R= 100(2-Q) +20(q)
R= 200-100(q)+20(q)
R= 200-80(q)
200-80(q)<120
200-120<80(q)
80<80(q)
Is 1<q or q>1
stmnt2) not sufficient

(1+2) p>1.1,p>q
minimum value of p= 1.2 then q maybe 1.1 ( q>1 yielding yes)
q may be .75 ( q<1, yielding no)

whats wrong with my answer , plz explain
Math Expert
Joined: 02 Sep 2009
Posts: 43898
Re: If P, Q, R are positive numbers such that R = 100P + 20Q [#permalink]

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14 May 2012, 07:31
Expert's post
2
This post was
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TomB wrote:
R=100(p) + 20(q), p+q=2,is R<120
after rephrasing:
R= 100(2-Q) +20(q)
R= 200-100(q)+20(q)
R= 200-80(q)
200-80(q)<120
200-120<80(q)
80<80(q)
Is 1<q or q>1

stmnt2) not sufficient

(1+2) p>1.1,p>q
minimum value of p= 1.2 then q maybe 1.1 ( q>1 yielding yes)
q may be .75 ( q<1, yielding no)

whats wrong with my answer , plz explain

(1) P > 1.1 --> since P=2-Q then 2-Q>1.1 --> Q<0.9. Sufficient.
(2) P>Q --> 2-Q>Q --> Q<1. Sufficient.

If P, Q, R are positive numbers such that R = 100P + 20Q, and P + Q = 2, is R < 120?

$$R=100P+20Q=80P+20(P+Q)=80P+40$$. The question asks: is $$80P+40<120$$ --> is $$P<1$$?

(1) P > 1.1. Sufficient.
(2) P > Q --> P>2-P --> P>1. Sufficient.

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Joined: 07 Apr 2012
Posts: 441
Re: If P, Q, R are positive numbers such that R = 100P + 20Q [#permalink]

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16 May 2012, 06:05
Could you explain why you tried to solve for Q instead of P?
I did it for P and worked out well, but is there a reason?
Math Expert
Joined: 02 Sep 2009
Posts: 43898
Re: If P, Q, R are positive numbers such that R = 100P + 20Q [#permalink]

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16 May 2012, 07:00
ronr34 wrote:
Could you explain why you tried to solve for Q instead of P?
I did it for P and worked out well, but is there a reason?

You can do either way.
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Intern
Joined: 24 Feb 2017
Posts: 39
Re: If P, Q, R are positive numbers such that R = 100P + 20Q [#permalink]

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11 May 2017, 10:07
is its definite NO answers with option A ?
Math Expert
Joined: 02 Sep 2009
Posts: 43898
Re: If P, Q, R are positive numbers such that R = 100P + 20Q [#permalink]

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11 May 2017, 10:15
mkumar26 wrote:
is its definite NO answers with option A ?

Yes, both statements give definite NO answer to the question.
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Re: If P, Q, R are positive numbers such that R = 100P + 20Q [#permalink]

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11 May 2017, 10:27
mkumar26 wrote:
is its definite NO answers with option A ?

Simplify the question stem first. the question is R<120
or 100P+20Q<120. this implies - 5P+Q<6----(1)

Now P+Q=2 or Q=2-P. Putting the value of Q in equation (1) above, we get
5P+2-P<6 OR P<1 -----(2)

So the question reduces to is P<1

Statement 1: Directly provides the answer as P>1.1. Hence sufficient

Statement 2: P>Q, putting the value of Q we get

P>2-P OR P>1. Hence Sufficient.

So Option D
Re: If P, Q, R are positive numbers such that R = 100P + 20Q   [#permalink] 11 May 2017, 10:27
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