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Coffee A and coffee B are mixed in the ratio x:y by weight. A costs $50/kg and B...

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christiano-cr7 | (Level 1) Salutatorian

Posted September 20, 2013 at 5:49 AM via web

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Coffee A and coffee B are mixed in the ratio xy by weight. A costs $50/kg and B costs $40/kg. If the cost of A is increased by 10% while that of B is decreased by 15%, the cost of the mixture per kg remains unchanged.

Find x :

A. 2 : 3

B. 5 : 6

C. 6 : 5

D. 3 : 2

E. 55 : 34

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jeew-m | College Teacher | (Level 1) Educator Emeritus

Posted September 20, 2013 at 5:57 AM (Answer #1)

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According to the given data we can derive the following equations.

Cost of coffee mixture `= 50x+40y ----(1)`

It is given that the cost of A is increased by 10% while that of B is decreased by 15%.

Cost of Coffee A `= 50+50/100xx10 = 55`

Cost of coffee B `= 40-40/100xx15 = 34`

Since the cost of coffee remains same;

Cost of coffee mixture `= 55x+34y ----(2)`

(1) = (2)

`50x+40y= 55x+34y`

`6y = 5x`

`6/5 = x/y`

`x:y = 6:5`

So the ratio of x:y = 6:5 and the correct answer is at C)

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tonys538 | TA , Undergraduate | (Level 1) Valedictorian

Posted September 30, 2014 at 12:49 PM (Answer #2)

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Coffee A and coffee B are mixed in the ratio x:y by weight. A costs $50/kg and B costs $40/kg.

When x kg of A and y kg of B are mixed, the cost of the resulting mixture is (50x+40y)/(x+y).

If the cost of A is increased by 10% and the cost of B is decreased by 15%, the cost of the mixture is (x*50*1.1 + y*40*0.85)/(x+y)

As the two mixtures cost the same:

(50x+40y)/(x+y) = (x*50*1.1 + y*40*0.85)/(x+y)

(50x+40y)/(x+y) = (55*x + 34*y)

50x + 40y = 55x + 34y

5x = 6y

x:y = 6:5

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