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# Problems of Mixture Ratios

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Problems of Mixture Ratios [#permalink]  13 Oct 2011, 18:53
I'm having problems with these math questions:

a) To paint a bathroom a painter needs 4 gallons of light blue paint mixed in a proportion of 5 parts whit to 3 parts blue. From a previous job, she has 1.5 gallons of another shade of blue paint that was mixed in the proportion of 1 part white to 3 parts blue. Exactly how much white paint and how much blue paint must the painter buy to be able to mix the old and the new paint together to achieve the desired shade?

b) I have two gallons of grape drink that is one-third water, and I mix it with three gallons of lemonade that is three-fourths water. After mixing, I spill one gallon of the mixture. How much water is ther in the remaining four gallons of the mixture?
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Re: Problems of Mixture Ratios [#permalink]  14 Oct 2011, 13:28

2.125 Gallon White Paint
0.375 Gallon Blue Paint

B) Remaining 4 gallons of mixture has 2.33 Gallon water.

Explanation:

A) Req. mix composition is: 5unit White + 3unit Blue = 8 units of Req mix.
We require 4 Gallon Mix = 8 units, hence each unit here is 0.5 gallon.

Therefore required final mix is :
5 x 0.5 = 2.5 Gallon White ...... (Ref i)
3 x 0.5 = 1.5 Gallon Blue. ....... (Ref ii)

We currently have 1.5 Gallon of 1part White + 3part blue. So of the 4 parts of mixture we have 1 part white & 3 part blue. 1/4 White + 3/4 Blue.

Therefore white in leftover mixture = (1/4) x 1.5 gallon = 0.375 gallon
Additional white required = 2.5 - 0.375 = 2.125 gallon white .... Using 2.5 gallon from (Ref i) above

Blue in leftover mixture = (3/4) x 1.5 = 1.125 gallon
Additional white required = 1.5 -1.125 =0.375 gallon of blue paint. ... Using 1.5 gallon from (Ref ii) above.

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P.S - We know we need another 2.5 gallon paint to increase the current mix of 1.5 gallon to 4 gallon. So to speed up my calculations, I only calculated white req. and then reduced it from total of 2.5 to arrive at blue required. Hence calculations in "blue font" can be avoided. Pun intended.
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(B)

Mix 1 : 2 gallon Grape juice with 1/3 water = 2 x 1/3 = 2/3 gallon water.
Mix 2 : 3 gallon Lemonade with 3/4 water = 3 x 3/4 = 9/4 gallon water.

total water in 5 gallon of final mix (mix 1 + 2) = 2/3 + 9/4 = 35/12 gallon water.
Fruit Juice in 5 gallon of final mix = 5 - (35/12) = 25/12 gallon Fruit juice
Final ratio = Fruit Juice : Water = 25/12 : 35/12 = 25 : 35 = 5:7

Now this ratio will be preserved even if some of the final mix is spilled. So after 1 gallon spill, 4 gallons of the new mix will have the same ratio; Fruit Juice : Water = 5:7 i.e. if 12 units of this mix is taken it will have 5 part Fruit juice and 7 part water. So water is 7/12 of the final mix.

7/12 of 4 gallon = (7/12) x 4 = 7/3 gallon = 2.33 gallon.

Cheers !
Re: Problems of Mixture Ratios   [#permalink] 14 Oct 2011, 13:28
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