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#1 (permalink) |
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TestMagic Guru-in-Training
![]() ![]() ![]() Join Date: Dec 2006
Posts: 938
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mixtures
Some part of the 50% solution of acid was replaced with the equal amount of 30% solution of acid. As a result, 40% solution f acid was obtained. what part of the original solution was replaced?
1/5 1/4 1/2 3/4 4/5 |
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#2 (permalink) |
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TestMagic Guru-in-Training
![]() ![]() ![]() Join Date: Oct 2007
Location: New Jersey
Posts: 642
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If I understand the problem correctly, C.
Original solution was all 50% acid, then x % of that was removed and x % of 30% acid added. So: (1-x)50+x30=40 solve for x Or: the average of 50 and 30 is 40, so you need a 50/50 split. |
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#5 (permalink) |
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TestMagic Fan!
![]() ![]() Join Date: Aug 2007
Location: India
Posts: 347
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This is a standard problem applicable to mixtures as well as some problems on averages.
Code:
A : B
a b
c
(b~c) : (a~c)
Then the ratio in which A and B have been mixed is (c~b) : (c~a) ['~' sign stands for difference not subtraction] |
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