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One kilogram of a certain coffee blend consists of x [#permalink]

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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?

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 --> \(1-x>0.15\) --> \(x<0.85\). Not sufficient.

I tried doing this via using the concept of weighted average. Its given in statement 2 that 6.5x + 8.5y>=7.3. Now, when x=.5 and y=.5, then the c would be 7.5. (GIVEN FACT IS THAT X+Y=1.) Total distance in terms of parts is 2 units and in terms of numerical distance is 1. For every 2 part change, there is a 0.1 deviation in the numerical separation. Since c>=7.3( note the 0.2 deviation), the numerical deviation is 0.1. Hence x must be <=0.6. Regards.
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Re: One kilogram of a certain coffee blend consists of x [#permalink]

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01 May 2013, 16:39

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x+y=1 c= 6.5x + 8.5y Question ---- is x<0.8 ?

As seen from above equations , for a given sum of x and y ( =1) , the value of C increases as x decreases and Y increases ( since y is multipled by greater number than x)

If x=0.8 , then y=0.2 and C= (6.5*0.8) + (8.5*0.2) = 6.9 .

Now if we increase value of x above 0.8 , the value of C decreases below 6.9 ( consider x=0.9 , then y would be 0.1 , which pulls down the value of C below 6.9--- as per the highlighted text above). And if we decrease the value of x below 0.8 , the value of c increases from 6.9 ( cos y is increased when x is reduced).

Statement 1 - y>0.15 , Y could be 0.2 , in which case x=0.8 ( No to the answer) or Y = 0.3 , in which case x=0.7 ( yes to the answer) Insufficient

Statement 2 C>=7.5 , as seen from above , for any c > 6.9 demands a x <0.8 . Hence sufficient.

Re: One kilogram of a certain coffee blend consists of x [#permalink]

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03 May 2013, 21:29

It is really nice to find this place as it is full of fun. Looking forward for new friends. Hope I'll definitely enjoy my stay here. ood question and quite a learning...

Out of curiousity, can this sum be solved in less than 2mins that too under "GMAT exam" conditions that too at first sight??

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

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.

Re: One kilogram of a certain coffee blend consists of x [#permalink]

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16 Sep 2013, 04:15

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Idk, in my opinion after you figure out that x+y=1 and that C=6.5x+8.5y (knowing that C can be 7.30), we have two equations and we can solve for X. Please correct me if I'm wrong but this was my line of reasoning and I've chosen B after 1min.

Re: One kilogram of a certain coffee blend consists of x [#permalink]

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21 Apr 2014, 01:09

Initially I got the answer wrong but there was a calculation mistake. I am posting my approach anyway :D,

Question Says, 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

Sol. x + y = 1 (from question statement) now, cost C = 6.5x + 8.5y = 6.5 (x + y) + 2y = 6.5 +2y------(eq. 1) we need to check whether x < 0.8 or y > 0.2 (since x + y = 1)

option 1). y > 0.15 clearly not sufficient alone.

option 2). C >= 7.30 from eq.1, 6.5 + 2y >= 7.30 => 2y => 0.80 => y>= 0.40 or x =< 0.60 Hence, sufficient.

Re: One kilogram of a certain coffee blend consists of x [#permalink]

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12 Jan 2015, 05:43

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With x+y = 1, we can simplify the equation for C:

C = 6.5 + 2y

Statement 1: Not sufficient. x can still be anywhere between 0 and 0.85

Statement 2: Sufficient: 7.3 = 6.5 + 2y. This leads to y = 0.4 and x smaller than 0.8. If C is greater than 7.3, this will only increase y and therefore decrease x, which will therefore be smaller than 0.8 in any case.

Answer B.
_________________

\(\sqrt{-1}\) \(2^3\) \(\Sigma\) \(\pi\) ... and it was delicious!

Re: One kilogram of a certain coffee blend consists of x [#permalink]

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04 Apr 2016, 01:53

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Re: One kilogram of a certain coffee blend consists of x [#permalink]

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12 Apr 2017, 09:40

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