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Re: One kilogram of a certain coffee blend consists of x kilogram of type [#permalink]
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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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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]
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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Bumping for review and further discussion*. Get a kudos point for an alternative solution!

*New project from GMAT Club!!! Check HERE

All DS Mixture Problems to practice: search.php?search_id=tag&tag_id=43
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Re: One kilogram of a certain coffee blend consists of x kilogram of type [#permalink]
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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]
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]
@Bunuel,
You are just great!!!
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Re: One kilogram of a certain coffee blend consists of x kilogram of type [#permalink]
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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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]
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]
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



This is essentially a weighted average problem. I tend to go with the visual graph over the formulas.

1.) Statement 1 says that y is greater than 0.15 but there's no info about x, so off the bat it's insufficient!

2.) C ≥ 7.30
I like to use the "see-saw method" so if we draw the graph out (Refer to attached picture):
Statement 2 is sufficient

Answer is B!!

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Screenshot_4.png
Screenshot_4.png [ 23.87 KiB | Viewed 20059 times ]

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Re: One kilogram of a certain coffee blend consists of x kilogram of type [#permalink]
The real deal here is x+y = 1
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One kilogram of a certain coffee blend consists of x kilogram of type [#permalink]
Bunuel 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?

"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.


Hi brunel, what's the reason behind substituting y with 8.5(1−x) here? How will I know to substitue here?Thanks
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Re: One kilogram of a certain coffee blend consists of x kilogram of type [#permalink]
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Re: One kilogram of a certain coffee blend consists of x kilogram of type [#permalink]
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