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

00:00

A

B

C

D

E

Difficulty:

(N/A)

Question Stats:

0% (00:00) correct
0% (00:00) wrong based on 0 sessions

This topic is locked. If you want to discuss this question please re-post it in the respective forum.

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.

Re: DS -coffee blend [#permalink]
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

Re: DS -coffee blend [#permalink]
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.

Re: DS -coffee blend [#permalink]
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

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.

Re: DS -coffee blend [#permalink]
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

Re: DS -coffee blend [#permalink]
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

Re: DS -coffee blend [#permalink]
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.

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