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25 Mar 2012, 11:23
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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
(1) y > 0.15
(2) C >= 7.30
25 Mar 2012, 11:28
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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?
"One kilogram of a certain coffee blend consists of x kilogram of type I coffee and y kilogram of type II coffee"
So, \(x+y=1\). Question: is \(x<0.8\).
(1) y > 0.15
Not sufficient.
(2) C >=7.30
Sufficient.
Answer: B.
"One kilogram of a certain coffee blend consists of x kilogram of type I coffee and y kilogram of type II coffee"
So, \(x+y=1\). Question: is \(x<0.8\).
(1) y > 0.15
- \(1-x>0.15\);
\(x<0.85\).
Not sufficient.
(2) C >=7.30
- \(C=6.5x + 8.5y\geq{7.3}\);
\(C=6.5x + 8.5(1-x)\geq{7.3}\);
\(2x\leq{1.2}\);
\(x\leq{0.6}\).
Sufficient.
Answer: B.
16 Sep 2013, 03:15
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imadkho
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
(1) y > 0.15
(2) C >= 7.30
Responding to a pm:
Quote:
I could solve most of the problems using the scale method, though I was finding it difficult to solve this particular question using it.
C = Avg cost of the mix = Cost of the blend = 6.5x + 8.5y
If you remember your weighted average formula, Avg = (C1*w1 + C2*w2)/(w1 + w2) =
Avg Cost = (C1*x + C2*y)/(x + y)
(x+y) is 1 kg since the weight of the blend is 1 kg.
Avg Cost = C = (C1*x + C2*y) = (6.5x + 8.5y)
So cost of type I coffee is $6.5/kg and cost of type II coffee is $8.5/kg.
So everything is set for you. All you have to do is plug in the values of x and y now.
Question: Is x > 0.8?
(1) y > 0.15
If y is 0.16, x is 0.84
If y is 0.5, x is 0.5
This statement alone is not sufficient.
(2) C >= 7.30
Let's say C = 7.3
x/y = (8.5 - 7.3)/(7.3 - 6.5) = 12/8 = 3/2 (The Scale Method)
x = 3/5 = 0.6
If C is greater than 7.3, x/y will get smaller and hence the value of x will keep getting smaller too. Say if C = 7.5,
x/y = 1/1
x = 0.5
and so on...
So x can be at most 0.6. Hence it must be less than 0.8.
Statement II alone is sufficient.
Answer (B)
