Good one

[ontopofit]

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

[mals24]

Given: x+y = 1
C = 6.5X+8.5Y

To Find: Is x<0.8

St1: y<0.15

y = 1-x

1-x < 0.15
x<0.85---Insuff

St2: C>7.30

7.3 < 6.5X+8.5(1-X)

Solving you get: X< 0.6 Hence X<0.8 Suff

Answer: B

[sanju09]

Given: x+y = 1
C = 6.5X+8.5Y

To Find: Is x<0.8

St1: y<0.15

y = 1-x

1-x < 0.15
x<0.85---Insuff

St2: C>7.30

7.3 < 6.5X+8.5(1-X)

Solving you get: X< 0.6 Hence X<0.8 Suff

Answer: B

IMO D

This is the only error in your work that revented you say st 1 sufficient.
(1-x < 0.15
x<0.85---Insuff)

[gmatmachoman]

IMO B

[ajith]

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

x+y =1

(1)
y>0.15
x<0.85

Not sufficient

C = 6.5x+8.5y = 6.5x+6.5y +2 y >= 7..30
6.5 +2y>= 7.3
2y>= 0.8
y>=0.4

x<0.6 Sufficient


Hence B

[sanju09]

Given: x+y = 1
C = 6.5X+8.5Y

To Find: Is x<0.8

St1: y<0.15

y = 1-x

1-x < 0.15
x<0.85---Insuff

St2: C>7.30

7.3 < 6.5X+8.5(1-X)

Solving you get: X< 0.6 Hence X<0.8 Suff

Answer: B

IMO D

This is the only error in your work that prevented you say st 1 sufficient.
(1-x < 0.15
x<0.85---Insuff)

With apologies...

Our question is about x, so let's deal in x only, as much as it is possible. We can read it as "blend consists of x kilogram of type I coffee and 1 - x kilogram of type II coffee" and that "C = 6.5 x + 8.5 (1 - x)" or "C = 8.5 - 2 x".

(1) says that 1 - x > 0.15, or x < 0.85. Now, what is less than 0.85 is not necessarily less than 0.8. So, (1) is not sufficient.

(2) says that 8.5 - 2 x ≥ 7.30 or x ≤ 0.6. Now, what is less than or equal to 0.6 is necessarily less than 0.8. So, (1) is sufficient.

OK...[spoiler]B[/spoiler]