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One kilogram of a certain coffee blend consists of x kilogram of type

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

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New post 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
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New post 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 --> \(1-x>0.15\) --> \(x<0.85\). 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.
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Re: One kilogram of a certain coffee blend consists of x kilogram of type  [#permalink]

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New post 05 Feb 2013, 02:17
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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.
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Re: One kilogram of a certain coffee blend consists of x kilogram of type  [#permalink]

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New post 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.

Hope that helps
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Re: One kilogram of a certain coffee blend consists of x kilogram of type  [#permalink]

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New post 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??
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Re: One kilogram of a certain coffee blend consists of x kilogram of type  [#permalink]

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New post 13 Jun 2013, 02:15
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Re: One kilogram of a certain coffee blend consists of x kilogram of type  [#permalink]

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New post 16 Sep 2013, 03:15
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imadkho 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



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

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New post 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. :)
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Re: One kilogram of a certain coffee blend consists of x kilogram of type  [#permalink]

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New post 18 Sep 2013, 12:15
one kilogram of a certain coffee blend consists of X kilogram of type I and Y kilogram of type II" means that X+Y=1
Combined C=6.5X+8.5Y,

we get:
X=(8.5-C)/2, Y=(C-6.5)/2
Combined C>=7.3, X=(8.5-C)/2<1.2/2=0.6

Answer is B
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Re: One kilogram of a certain coffee blend consists of x kilogram of type  [#permalink]

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New post 28 Nov 2013, 09:10
@Bunuel,
You are just great!!!
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Re: One kilogram of a certain coffee blend consists of x kilogram of type  [#permalink]

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New post 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.
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Re: One kilogram of a certain coffee blend consists of x kilogram of type  [#permalink]

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New post 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.
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Re: One kilogram of a certain coffee blend consists of x kilogram of type  [#permalink]

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New post 19 Jul 2017, 06:24
This question is amazing and i guess dont need any hard calculations.
Statement 1 is clearly insufficient
Statement 2 sufficient...question asks is x < 0.8
Statement 2 tells us that C > or equal to 7.3...so we can write
6.5X + 8.5 y = 7.3 (remember C equals to 7.3 or greater than but not less than that)
Now see that 7.3 is between 6.5 and 8.5
7.3 -6.5 = 0.8
8.5-7.3 = 1.2
6.5--------7.3-------------8.5
0.8 1.2
One can see that 7.3 is more close to 6.5 it means the ratio of X is more than the ratio of y type of coffee...the ratio will be equal to 0.8/1.2 = x/y or x/y = 2/3
we can find the ratio of two coffee blends and thus we an find the ratio of x which is 3/5 =0.6
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Re: One kilogram of a certain coffee blend consists of x kilogram of type  [#permalink]

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