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Two liquids are mixed in the ratio 3:2 and the vendor gains

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by AAPL » Sat May 19, 2018 4:17 am

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Answers

A

B

C

D

E

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Difficulty—

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!
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by GMATGuruNY » Sat May 19, 2018 4:32 am
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.
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by swerve » Sat May 19, 2018 9:53 am
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.

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by Scott@TargetTestPrep » Tue May 22, 2018 5:14 pm
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
We can let 3n = the number of liters of the first liquid and 2n = the number of liters of the second liquid. Also, we can let x = the cost of the second liquid per liter and thus x + 2 = the cost of the first liquid per liter. We can create the following equation:

1.1[3n(x+2) + 2nx]/(3n + 2n) = 11

[3nx + 6n + 2nx]/5n = 10

5nx + 6n = 50n

5x + 6 = 50

5x = 44

x = 8.8

Thus the first liquid costs 8.8 + 2 = $10.80 per liter.

Answer: B

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by Mo2men » Wed Sep 26, 2018 1:28 am
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
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