Mixture Problem

[pg]

Any help explain this one...Thanks

A certain shade of gray paint is obtained by mixing 3 parts of white paint with 5 parts of black paint. If 2 gallons of the mixture is needed and the individual colors can be purchased only in one-gallon or half- gallon cans, what is the least amount of paint, in gallons, that must be purchased in order to measure out the portions needed for the mixture?
(A) 2
(B) 21/2
(C) 3
(D) 31/2
(E) 4

[a user]

Originally posted by pg

Any help explain this one...Thanks

A certain shade of gray paint is obtained by mixing 3 parts of white paint with 5 parts of black paint. If 2 gallons of the mixture is needed and the individual colors can be purchased only in one-gallon or half- gallon cans, what is the least amount of paint, in gallons, that must be purchased in order to measure out the portions needed for the mixture?
(A) 2
(B) 21/2
(C) 3
(D) 31/2
(E) 4

In order to make a 2 gallon misture, we need (3/8) * 2 gallon of white paint and (5/8)*2 of blank paint.
(3/8)*2 = 3/4 gallon --> there are no 3/4 gallon for sale we have to buy 1 gallon instead.
(5/8)*2 = 5/4 gallon --> similarly, we have to buy 1 and half gallon of Blank to cover the black part of paint.

So, we need buy 1 gallon of while and 1 and half gallons of blank.
Total 2 and half gallons. therefore answer B is correct.

[pg]

Thanks a lot[^]

[pawanjsvn]

W+B = 2
W = 3/8 * (W+B)
= 3/4 (You need 1 gallon for white paint)
B = 2 - 3/4 = 5/4. (You need 1.5 gallons for Black)

Total = 2.5 gallons of paint