One kilogram of a certain coffee blend consists of x kilogram of type I coffee and y kilogram of type II coffee. The cost of the blend is C dollars per kilogram, where C = 6.5x + 8.5y. Is x < 0.8 ?
(1) y > 0.15
(2) C >= 7.30
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nice question
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atulmangal
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Anurag@Gurome
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x + y = 1atulmangal wrote:One kilogram of a certain coffee blend consists of x kilogram of type I coffee and y kilogram of type II coffee. The cost of the blend is C dollars per kilogram, where C = 6.5x + 8.5y. Is x < 0.8 ?
(1) y > 0.15
(2) C >= 7.30
C = 6.5x + 8.5y
(1) y > 0.15 does not give a definite answer if x < 0.8.
So, (1) is NOT SUFFICIENT.
(2) C ≥ 7.30 implies C = 6.5x + 8.5(1 - x) = 6.5x + 8.5 - 8.5x ≥ 7.30
8.5 - 2x ≥ 7.3
2x ≤ 1.2
x ≤ 0.6
So, x ≤ 0.6, which implies x < 0.8
So, (2) is SUFFICIENT.
The correct answer is B.
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vineeshp
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1) if y = 0.15 x=0.85
if y = 0.3 x=0.7
Not sufficient.
2) C >= 7.30
At x =0.8 y =0.2
C = 6.9. If you increase x from this point, the value of C will only decrease as the coeff of x is smaller than coeff of y. Hence x has to be less than this.
hence sufficient.
To ilustrate
6.5x + 8.5(1-x) = 7.3
1.2 = 2x
x = 0.6
If you increase x from this point, value of C will decrease. So for it to stay above, x has to be <= 0.6.
hence B is sufficient.
if y = 0.3 x=0.7
Not sufficient.
2) C >= 7.30
At x =0.8 y =0.2
C = 6.9. If you increase x from this point, the value of C will only decrease as the coeff of x is smaller than coeff of y. Hence x has to be less than this.
hence sufficient.
To ilustrate
6.5x + 8.5(1-x) = 7.3
1.2 = 2x
x = 0.6
If you increase x from this point, value of C will decrease. So for it to stay above, x has to be <= 0.6.
hence B is sufficient.
Vineesh,
Just telling you what I know and think. I am not the expert.
Just telling you what I know and think. I am not the expert.
