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Re: P, Q, R are positive numbers such that [#permalink]
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.

Re: If P, Q, R are positive numbers such that R = 100P + 20Q [#permalink]
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 stmnt1) no info about q 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)

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