mixtures

[bmwhype]

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

[DWarrior]

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.

[coyote]

IMO the answer is 1/2.

Method 1: the ratio of the two parts will be (50-40) : (40-30)

Method 2: Let x be original quantity.
Let y be the replaced fraction .

0.5x - 0.5xy + 0.3xy = 0.4x

y=1/2

[DWarrior]

Originally Posted by coyote

Method 1: the ratio of the two parts will be (50-40) : (40-30)

Enlighten me please, this method seems much faster.

[coyote]

Originally Posted by DWarrior

Enlighten me please, this method seems much faster.

This is a standard problem applicable to mixtures as well as some problems on averages.

A                 :           B
a                             b
                        
                  c
    
(b~c)          :        (a~c)

What I've shown above is that if A and B are the items whose certain parameter is given as 'a' and 'b' respectively and same parameter of resultant mixture C is given to be 'c',
Then the ratio in which A and B have been mixed is (c~b) : (c~a)
['~' sign stands for difference not subtraction]

[DWarrior]

Nice, thanks

[narinder82]

answer is 1/2.

[bmwhype]

OA is C.