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Coffe Blen Question

by tamiri » Sat Jun 01, 2013 11:19 am
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 =>7.3

Thanks
Tamir
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by fcabanski » Sat Jun 01, 2013 12:06 pm
This is data sufficiency. It's not problem solving.

This is a yes/no question. Is x < .8? If the statement(s) tells you yes or no, then it or they are sufficient.

For data sufficiency try to anticipate an answer. Then examine the statements one at a time.

The problem gives two equations.

EQ 1: Two types of coffee (x and y) make up 1 kilogram of blended coffee.

x + y = 1

EQ 2:

C= 6.5x + 8.5y

Is x < .8

It's easy to anticipate.

EQ 1: Plug .8 into the equation. .8 + y = 1 so y = 1-.8 = .2. But the question is if x < .8 When x gets smaller, y gets bigger, so y > .2

EQ 2: For x > .8, C must be larger than the cost of a blend with the max amount of x (.8) and min amount of y (.2). That's because x is the cheaper of the types. Using the max amount of x makes the coffee as cheap as possible. If there's less x, the coffee blend becomes more expensive. Use > instead of >= because x is never .8. x is less than .8.
C > 6.5(.8) + 8.5(.2)
C > 5.2 + 1.7 = 6.9
If C > 6.9, then x < .8.

Any statement that tells either

y > .2 ( If it is, then x < .8 so the answer is yes. If it isn't, then x = .8 or x > .8 so the answer is no.)
or
C > 6.9 (If it is, then x < .8 (yes). If it isn't, then x >= .8 (no).

is sufficient. Knowing either answers the yes/no question.

Examine the statements.

(1) y > 0.15 It's not sufficient. We need y > .2. Eliminate A and D. Possible answers are now B, C, E.

If you're not convinced.

y could be .16, which makes x = .84. Is x < .8? No. y could be 1, which makes x = 0. Is x < .8? Yes. Unless you're the Zodiac, who answered every question with "yes, no, no, yes", the two different answers aren't a good thing.

(2) C =>7.3 This is sufficient. As long as C > 6.9, there must be less than .8 of x. The answer is B.

If you're not convinced.

Imagine x = .79 (that's less than .8)

x+y = 1
.79 + y = 1
y = 1-.79 = .21

C = .79(6.5) + .21(8.5) = 5.135 + 1.785 = 6.92 As there is less of the cheaper (x) type there is more of the more expensive (y) type, and the cost increases. That means when C >= 7.3, there's even less of x than .79 and more of y than .21. x is certainly < .8 because it's even less than .79. Yes.

B 2 alone is sufficient, but 1 alone is not sufficient.
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by tamiri » Sun Jun 02, 2013 7:21 pm
Thank you.
Great answer !
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by faraz_jeddah » Mon Jun 03, 2013 1:19 am
fcabanski wrote:This is data sufficiency. It's not problem solving.

This is a yes/no question. Is x < .8? If the statement(s) tells you yes or no, then it or they are sufficient.

For data sufficiency try to anticipate an answer. Then examine the statements one at a time.

The problem gives two equations.

EQ 1: Two types of coffee (x and y) make up 1 kilogram of blended coffee.

x + y = 1

EQ 2:

C= 6.5x + 8.5y

Is x < .8

It's easy to anticipate.

EQ 1: Plug .8 into the equation. .8 + y = 1 so y = 1-.8 = .2. But the question is if x < .8 When x gets smaller, y gets bigger, so y > .2

EQ 2: For x > .8, C must be larger than the cost of a blend with the max amount of x (.8) and min amount of y (.2). That's because x is the cheaper of the types. Using the max amount of x makes the coffee as cheap as possible. If there's less x, the coffee blend becomes more expensive. Use > instead of >= because x is never .8. x is less than .8.
C > 6.5(.8) + 8.5(.2)
C > 5.2 + 1.7 = 6.9
If C > 6.9, then x < .8.

Any statement that tells either

y > .2 ( If it is, then x < .8 so the answer is yes. If it isn't, then x = .8 or x > .8 so the answer is no.)
or
C > 6.9 (If it is, then x < .8 (yes). If it isn't, then x >= .8 (no).

is sufficient. Knowing either answers the yes/no question.

Examine the statements.

(1) y > 0.15 It's not sufficient. We need y > .2. Eliminate A and D. Possible answers are now B, C, E.

If you're not convinced.

y could be .16, which makes x = .84. Is x < .8? No. y could be 1, which makes x = 0. Is x < .8? Yes. Unless you're the Zodiac, who answered every question with "yes, no, no, yes", the two different answers aren't a good thing.

(2) C =>7.3 This is sufficient. As long as C > 6.9, there must be less than .8 of x. The answer is B.

If you're not convinced.

Imagine x = .79 (that's less than .8)

x+y = 1
.79 + y = 1
y = 1-.79 = .21

C = .79(6.5) + .21(8.5) = 5.135 + 1.785 = 6.92 As there is less of the cheaper (x) type there is more of the more expensive (y) type, and the cost increases. That means when C >= 7.3, there's even less of x than .79 and more of y than .21. x is certainly < .8 because it's even less than .79. Yes.

B 2 alone is sufficient, but 1 alone is not sufficient.
Brilliant and detailed solution.

I am wondering if the allegation method can be used here? And if so, would it be shorter?
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by Matt@VeritasPrep » Mon Jun 03, 2013 10:06 am
tamiri 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 =>7.3
Just to add a direct solution:

From the stem,

x + y = 1
and
C = 6.5x + 8.5y

Which really gives us

y = 1 - x, so
C = 6.5x + 8.5(1-x), or
C = 6.5x + 8.5 - 8.5x, or
C = 8.5 - 2x

Since we want to know if x is less than .8, we can rewrite that statement as

2x = 8.5 - C, or
x = (8.5-C)/2

Since x and (8.5-C)/2 are thus the same thing, the question is really

"Is (8.5 - C)/2 less than .8?"

or

"Is (8.5 - C) less than 1.6?"

Doing this much work BEFORE evaluating the statements is often really helpful. Once we've done this, we can see immediately that S2 is insufficient. As for S1, since y = (1 - x) and y > .15, we know (1 - x) > .15, or that x is less than .85. Since we want to know if x is less than .8, this is clearly insufficient.

Clarifying and simplifying the initial statement in DS goes a long way, and often prevents us from having to test lots of of numbers and hope that they don't contradict one another.
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by GMATGuruNY » Mon Jun 03, 2013 4:37 pm
tamiri 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 =>7.3

Thanks
Tamir
Since C is equal to the cost of one kilogram of the mixture of X and Y, we get:
X+Y = 1.

Since C = 6.5X + 8.5Y:
The cost per kilogram of X = 6.5.
The cost per kilgram of Y = 8.5.
The cost per kilogram of the MIXTURE of X and Y = C.

Statement 1: y > .15
If Y = .16, then X = .84, in which case X > .8.
If Y = .3, than X = .7, in which case X < .8.
INSUFFICIENT.

Statement 2: C ≥ 7.3
Use ALLIGATION -- a great way to handle mixture problems.
Let C = 7.3.

Step 1: Plot the 3 percentages on a number line, with the costs of X and Y (6.5 and 8.5) on the ends and the cost of the mixture (7.3) in the middle.
X 6.5------------7.3--------------8.5 Y

Step 2: Calculate the distances between the percentages.
X 6.5-----.8-----7.3-----1.2-----8.5 Y

Step 3: Determine the ratio in the mixture.
The ratio of X to Y in the mixture is the RECIPROCAL of the distances in red.
X:Y = 1.2 : .8 = .6 : .4.
Here, each kilogram of the mixture will be composed of X = .6 and Y = .4.

If C = 8, we get:
X 6.5-----1.5-----8-----.5-----8.5 Y
X:Y = .5 : 1.5 = .25 : .75.
Here, each kilogram of the mixture will be composed of X = .25 and Y = .75.

As the value of C INCREASES (from 7.3 to 8), the value of X DECREASES (from .6 to .25).
Thus, given the constraint that C ≥ 7.3, the greatest possible value of X occurs when C = 7.3 and X = .6.
Thus, X < .8.
SUFFICIENT.

The correct answer is B.
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