algebra word problem

Question

A chemist has 1000 g of a solution that is 40% acid. How many grams of water must be added to reduce the acidity to 25%?

Answer 1

Let the answer be x

1000(40%) +x(0%) = (1000+x)(25%)

water has zero percent acid

25% = .25 = 1/4

400 = 250 + .25x

.25x = 150

x = 150/.25 = 150/1/4 = 4(150 )= 600 grams

Answer 2

A 1000gm solution that is 40% acidic means that 40% of the solution is acid. In other words, the solution is made up "x" grams of acid + "y" grams of non-acid (such as water). Therefore, the original mixture contains 40% X 1000grms, or 400gms, of acid: (400gms acid/1000gms solution) X 100% = 40%. If we add water to get a 25% acid solution, that means we need to solve this: (400gms/"x" gms solution) = 25%, where the "x" gms of solution will be 1000 (original amount) + the new amount of water added, call it "y".

400/(1000 + y) = .25

400 = .25(1000 + y)

400/.25 = 1000 + y

1600 - 1000 = y

y = 600

Therefore, the amount of water needed to be added to the original mixture to get acidity to 25% is 600gms.

Hope this helps!

Answer 3

Suppose x grams of water should be added.

Then 25% = 0.25 = (1000 * 0.40 ) / (1000+x).

The numerator is the amount of salt and the denominator is the amount of acid.

Multiple 1000 + x on both sides,

0.25 (1000 + x) = 1000 * 0.4 = 400

250 + 0.25 x = 400

0.25x = 400 -250 = 150.

x = 150/0.25 = 600.

The answer is 600g.

Answer 4