Problem Solving

Two liquids are mixed in the ratio 3:2 and the vendor gains 10% by selling the mixture at $11/liter. If the first liquid costs $2 more than the second, find the cost price of first liquid.

A. $8
B. $10.8
C. $6
D. $8.8
E. $4

The OA is B.

Vendor gains 10% by selling the mixture at $11/liter i.e he sells the mixture at 110% cost,
so Cost of mixture = 11/(110%) = $10 liter.

Let x the cost of the first liquid, so the cost of second liquid = x - 2
so, cost of mixture = x*60%+(x-2)*40% = 10
0.6x+0.4x-0.8 = 10
x = 10.8. Option B.

Has anyone another strategic approach to solve this PS question? Regards!

11/1.1 = $10 cost of liter
3:2 = 6:4

Let 6x = cost of the first liquid
6x+4(x-2) = 10
x = 1.8

6x = 10.8. Option B.

Regards!

GMATGuruNY wrote:

AAPL wrote:Two liquids are mixed in the ratio 3:2 and the vendor gains 10% by selling the mixture at $11/liter. If the first liquid costs $2 more than the second, find the cost price of first liquid.

A. $8
B. $10.8
C. $6
D. $8.8
E. $4

Since the selling price of $11 per liter represents a profit of 10%, the cost per liter = $10.
The cost of the first liquid is $2 more than the cost of the second liquid.
For the average cost per liter to be $10, the cost of the first liquid must be MORE THAN $10, while the cost of the second liquid must be LESS THAN $10.
Thus, the correct answer -- which represents the price of the first liquid -- must be more than $10.

The correct answer is B.

Dear Mitch,
While your solution is very logic, I tried to use the alligation which led me to different answer.
The cost is $10.
I used the alligation method as follows:

A.........2.......M..........3......B
First liquid/ Second liquid = 3/2 = 6/4 ( the difference is 2)

Where did I go wrong with using alligation? Can you help please.

Thanks