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One kilogram of a certain coffee blend consists of x

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One kilogram of a certain coffee blend consists of x [#permalink] New post 24 May 2007, 05:36
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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

A. Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.

B. Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.

C. BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.

D. EACH statement ALONE is sufficient.

E. Statements (1) and (2) TOGETHER are NOT sufficient.
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Re: DS -coffee blend [#permalink] New post 24 May 2007, 06:13
Balvinder 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> 0.15
(2) C >=7.30

A. Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.

B. Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.

C. BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.

D. EACH statement ALONE is sufficient.

E. Statements (1) and (2) TOGETHER are NOT sufficient.


E is the answer for this one.
(1) insufficient because y can be anything and x can be either <or> than 0.8
(2) insufficient not enough data

(1)+(2) insufficient because y can be anything and x can be either <or> that 0.8
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Re: DS -coffee blend [#permalink] New post 24 May 2007, 06:32
Balvinder 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> 0.15
(2) C >=7.30

A. Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.

B. Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.

C. BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.

D. EACH statement ALONE is sufficient.

E. Statements (1) and (2) TOGETHER are NOT sufficient.



Its B..
Here, X+Y = 1kg

From 1) If y>0.15 then X can be anything

From 2) if c>=7.30 then X must be lower than 0.8

So sufficient.
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Re: DS -coffee blend [#permalink] New post 24 May 2007, 08:38
johnycute wrote:
Balvinder 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> 0.15
(2) C >=7.30

A. Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.

B. Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.

C. BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.

D. EACH statement ALONE is sufficient.

E. Statements (1) and (2) TOGETHER are NOT sufficient.



Its B..
Here, X+Y = 1kg

From 1) If y>0.15 then X can be anything

From 2) if c>=7.30 then X must be lower than 0.8

So sufficient.


y>15 so let's take y=0.16 => 7.3= 6.5x + 8.5*0.16 x=0.91
y also can be let's say 0.7 => 7.3=6.5x+5.95 x=0.2

guys it's a trick question and GMAT tricked you.

the answer is E
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 [#permalink] New post 24 May 2007, 09:31
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> 0.15

C = 6.5x + 1.36

x = (c - 1.36)/6.5

Not sufficient !!

(2) C >=7.30

7.30 = 6.5x + 8.5y

Not sufficient

Together

x = ( 7.30 - 1.36)/6.5 => 5.94/6.5 => 0.91

hence x is not < 0.8 - we get an answer
If C increase so will X

My Answer :
C. BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
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Re: DS -coffee blend [#permalink] New post 24 May 2007, 09:56
Balvinder 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> 0.15
(2) C >=7.30


St. (1) y > 0.15 is clearly nsf.
St. (2) C >=7.30

since we know that:
x + y = 1
y = 1 - x

if so,

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

if C >= 7.3

8.5 - 2x >= 7.3
x =< 0.6

so suff.

B.
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Re: DS -coffee blend [#permalink] New post 24 May 2007, 11:14
Himalayan wrote:
Balvinder 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> 0.15
(2) C >=7.30


St. (1) y > 0.15 is clearly nsf.
St. (2) C >=7.30

since we know that:
x + y = 1
y = 1 - x

if so,

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

if C >= 7.3

8.5 - 2x >= 7.3
x =< 0.6

so suff.

B.


thanks, this question tricked all of us, except you :)
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Re: DS -coffee blend [#permalink] New post 24 May 2007, 11:32
Balvinder 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> 0.15
(2) C >=7.30



I vote for B.


(1) If y=0.16, x=0.84. But if y=0.21, x=0.79. INSUFF.
(2) Let's use x=0.8 as a benchmark, then C=6.9. Increasing the x (which we can look at as a percentage) would only lower C because x costs lower than y. So in order to get C=7.30, you must increase y and lower x from 0.8. Therefore x must be lower than 0.8. SUFF.
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 [#permalink] New post 24 May 2007, 15:54
thanks everybody

OA is B
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 [#permalink] New post 25 May 2007, 15:53
probably didn't use the best method but i just plugged in

at .9x it would cost $4.90

at .8x it would cost $6.90

at .7 x it would cost $7.10

so i figured x<.7 so sufficient
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Simple Method!!! [#permalink] New post 15 Jun 2007, 19:55
given x+y=1-----------------------Equation1
--stmt1 is clearly insuff as X can satisfy equation with more then 1 value
--Stmt 2 saying C is equal or > 7.3 so lets take C=7.30 so equation becomes 6.5x+8.5y=7.30 --------------Eq2

Solve Eq1 & Eq2 you get X always < 0.8 so 'B' is sufficient to answer the question

same goes for any value higher then 7.30

So Stmt2 is enough to Anser the Question asked
Simple Method!!!   [#permalink] 15 Jun 2007, 19:55
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