Mixture

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bia: Senior | Next Rank: 100 Posts; Posts: 30; Joined: Wed May 07, 2008 6:16 am

by bia » Sun May 18, 2008 5:09 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 >= 7.30

Bia

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Source: Beat The GMAT — Data Sufficiency | Sun May 18, 2008 5:09 am

amitansu: Master | Next Rank: 500 Posts; Posts: 294; Joined: Tue Feb 26, 2008 9:05 pmFollowed by: 1 member; Thanked: 13 times

by amitansu » Sun May 18, 2008 9:15 am

I think ans is E here.

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wawatan: Master | Next Rank: 500 Posts; Posts: 171; Joined: Sat May 03, 2008 3:51 pm; Thanked: 8 times

by wawatan » Sun May 18, 2008 2:33 pm

I believe the answer is also E. In order to find out if x<8>.15, well x could equal .5 which will satisfy the question but x could also equal 200, and this will not satisfy the equation. ok, the second statement says C>=7.3, well this just means that cost has to be greater or equal to 7.3. So cost could equal 7.3. example, x=0, and y=7.3/8.5. (this will satisfy the question =yes) another example is C could equal 10000000 and that means that x>.8 this could be true, so this would not satisfy the equation. So even if you put both statements together, it will still not answer the question. The best way to solve this problem is to plug in values.

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Stuart@KaplanGMAT: GMAT Instructor; Posts: 3225; Joined: Tue Jan 08, 2008 2:40 pmFollowed by: 614 members; Location: Toronto; Thanked: 1710 times; GMAT Score:800

Re: Mixture

by Stuart@KaplanGMAT » Sun May 18, 2008 8:20 pm

bia 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

Well, we know the answer isn't a, b or d, since each statement by itself is missing an entire variable.

So, let's try combining.

Let's make C and y as small as we possibly can and see what happens to x.

c = 7.30
y = tiny bit more than .15, let's say .15 for now.

So:

7.3 = 6.5x + 8.5(.15)
7.3 = 6.5x + 1.275
7.300 - 1.275 = 6.5x
6.025 = 6.5x
6.025/6.5 = x

(By the way, where is this question from? Way too nasty numbers to show up on the real GMAT - we can't even estimate, because it's just too close.)

approx 0.93 = x

So, it's possible to get a value of x greater than 0.8. If we increase y some more, the value of x goes down, so it's also possible to get a vlue of x less than 0.8: choose (e).

Stuart Kovinsky | Kaplan GMAT Faculty | Toronto

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comrade: Newbie | Next Rank: 10 Posts; Posts: 7; Joined: Fri Jul 11, 2008 1:30 pm

by comrade » Sat Aug 02, 2008 3:27 pm

ok sorry for bumping an old thread (if thats against the rules) but I have a different solution:

We have a kilogram of coffee here, composed of x kilograms of type I coffee and y kilograms of type II. Since there's only 1 kg of coffee altogether, x & y must be less than 1.

Because we have 1 kg of the mixture, x=1-y and y=1-x

C=6.5x+8.5y, therefore by a simple substitution:

(equation 1) C=-2x+8.5
(equation 2) C=6.5+2y

now lets consider the choices:

1) y > 0.15
therefore C > 6.5+2*0.15, C > 6.8

thus from equation 1 above, -2x+8.5>6.8, x<0.85, INSUFFICIENT

2) C >= 7.30

-2x+8.5>=7.30, x =< 0.6, SUFFICIENT

hence answer is B

any thoughts?

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