# Math Help - Word problem

1. ## Word problem

Hello all,

Experiencing a brain freeze here. There are two related word problems in the book I am reading. The first one I solved, the second one is giving me fits. The first problem reads:

How much water must be evaporated from 500 grams of a 10% salt solution in order to obtain a solution containing 15% salt?

My solution was to make x represent the amount of water to be evaporated. My equation was:

90% (500) - x = 85% (500 - x)

Solving: 450 - x = 425 - .85x
25 = .15x
x = 166 2/3 grams

which is correct.

The second question states:

Generalize Exercise 57 as follows: Let a, b, and d be positive real numbers with a < b < 100. Given a solution of d grams of salt water that is a% salt, how much water must be evaporated in order that the resulting mixture will be b% salt? Express the answer in terms of a, b, and d.

Here I've tried expressing the equation as

Solving:
ad - x = bd -bx
ad - bd = x - bx
d(a - b) = x (1 - b)
(d(a - b))/(1 - b) = x

However, this gives an incorrect answer with the original numbers. The solution given in the book is

x = (d(b - a))/b

I can't for the life of me figure out how they got to this equation. Please help!

2. Hello, earachefl!

In the generalization, your set-up is wrong . . .

How much water must be evaporated from 500 grams of a 10% salt solution
in order to obtain a solution containing 15% salt?

My solution was to make x represent the amount of water to be evaporated.

My equation was: . $90\%(500) - x \;= \;85\%(500 - x)$ . Right!
Look at what we've got . . .

$\text{We have: }\;\underbrace{90\%}_{1 - 10\%}(500) - x \;=\;\underbrace{85\%}_{1 - 15\%}(500 - x)$

We have $d$ grams of solution which is $a\%$ salt.
We evaporate $x$ grams of water to get a solution which is $b\%$ salt.

The general equation is: . $(1 - a)d - x \;=\;(1 - b)(d - x)$

3. D'oh!!

thank you!