1. ## 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  Reply With Quote

2. If I understand the problem correctly, C.

Original solution was all 50&#37; 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.  Reply With Quote

3. 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  Reply With Quote

4. 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.  Reply With Quote

5. 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.

Code:
```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]  Reply With Quote

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