If x, y and k are positive numbers such that [x/(x+y)][10] + [y/(x+y)][20] = k and if x<y, which of the following could be the value of k?
A. 10
B. 12
C. 15
D. 18
E. 30
A. 10
B. 12
C. 15
D. 18
E. 30
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OG 12 PS 148
Source: Beat The GMAT — Problem Solving |
k=10(x+2y)/(x+y)
now plug in values :
x.....y....2y.....x+2y......x+y......k
1.....2.....4.......5............3.......
1.....4.....8.......9............5........18(OA)
now plug in values :
x.....y....2y.....x+2y......x+y......k
1.....2.....4.......5............3.......
1.....4.....8.......9............5........18(OA)
let me attempt this..
lets try to simplify this expression.
10x/(x+y) + 20y/(x+y) = k
= (10x +20y)/ (x+y) = k
= (10x +10y +10y)/ (x+y) = k
= 10 + [10y/(x+y)] = k
Now we try back solving by plugging values of k.
The goal is to get an expression/relation such that x < y.
Start from D.
k= 18.
ie. 10y/(x+y) = 8
10y = 8x +8y
=>2y = 8x
=> y = 4x.
This implies that y > x OR x < y.
let me try what nitya34 's answer C
ie. let K = 15
ie 10y = 5x + 5y
=> 5y = 5x
=> y = x
This is not possible coz we know that x < y.
Hope I havent missed anything...
My answer is D
lets try to simplify this expression.
10x/(x+y) + 20y/(x+y) = k
= (10x +20y)/ (x+y) = k
= (10x +10y +10y)/ (x+y) = k
= 10 + [10y/(x+y)] = k
Now we try back solving by plugging values of k.
The goal is to get an expression/relation such that x < y.
Start from D.
k= 18.
ie. 10y/(x+y) = 8
10y = 8x +8y
=>2y = 8x
=> y = 4x.
This implies that y > x OR x < y.
let me try what nitya34 's answer C
ie. let K = 15
ie 10y = 5x + 5y
=> 5y = 5x
=> y = x
This is not possible coz we know that x < y.
Hope I havent missed anything...
My answer is D
10x+20y=kx+ky
(10-k)x+(20-k)y=0
with the conditions 0<x<y and k>0
x/y=(k-20)/(10-k)
by given condition
0<x/y<1
x/y>0
(k-20)/(10-k)>0
=>k-20>0 and 10-k>0=> k>20 and k<10 not possible
or
k-20<0 and 10-k<0
=>k<20 and k>10
=> 10<k<20---(A)
using x/y<1
=>(k-20)/(10-k)<1
=>(2k-30)/(10-k)<0
=> either 2k-30<0 and 10-k >0
or 2k-30 >0 and 10-k<0
taking first set we get
k<15 and k<10=> k<10
or k>15 and k>10 =>k>15
=>k<10 cannot be a solution from (A) => k>15 will satisfy
only k=18 for k>15 therefore k=18 is a possible value.
(10-k)x+(20-k)y=0
with the conditions 0<x<y and k>0
x/y=(k-20)/(10-k)
by given condition
0<x/y<1
x/y>0
(k-20)/(10-k)>0
=>k-20>0 and 10-k>0=> k>20 and k<10 not possible
or
k-20<0 and 10-k<0
=>k<20 and k>10
=> 10<k<20---(A)
using x/y<1
=>(k-20)/(10-k)<1
=>(2k-30)/(10-k)<0
=> either 2k-30<0 and 10-k >0
or 2k-30 >0 and 10-k<0
taking first set we get
k<15 and k<10=> k<10
or k>15 and k>10 =>k>15
=>k<10 cannot be a solution from (A) => k>15 will satisfy
only k=18 for k>15 therefore k=18 is a possible value.
