pappueshwar wrote:If x, y are positive numbers and x+y=1, which of the following can be the value of 100x+200y?
I. 80 II. 140 III. 199
A. I only
B. â…¡ only
C. â… ,â…¡ only
D. â…¡,â…¢ only
E. â… ,â…¡ and â…¢
OA IS C WHICH COMBINATION OF X AND Y SHOULD BE TAKEN ?
Put 100x+200y in terms of one of the variables.
Rephrasing x+y=1 as y=1-x and substituting for y in 100x+200y, we get:
100x + 200(1-x) = 100x + 200 - 200x = 200-100x.
Since x+y=1 and both x and y are positive, 0<x<1.
To determine the viable range here, convert 0<x<1 to 200-100x:
0 < x < 1
0 > -100x > -100
200 > 200-100x > 100.
Thus, 200-100x -- which is the equivalent of 100x+200y -- can be any value between 100 and 200.
The correct answer is
D.
II: 140
200-100x = 140
-100x = -60
x=.6, implying that y=.4.
100(.6) + 200(.4) = 60+80 = 140.
III: 199
200-100x = 199.
-100x = -1
x = .01, implying that y=.99.
100x + 200y = 100(.01) + 200(.99) = 1 + 198 = 199.
I: 80
200-100x = 80
-100x = -120
x = 1.2, implying that y=-.2.
Doesn't work, since y must be positive.
Only II and III are possible.
Last edited by
GMATGuruNY on Wed Mar 14, 2012 7:14 am, edited 1 time in total.
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