How much pure acid should be mixed with 4 gallons of a 50% acid solution in order to get a 60% acid solution?

In this case you have 50% solution, so it is 1/2 pure stuff and 1/2 other stuff.

You have 4 gallons, so the include 2 gallons of pure stuff and 2 gallons of other stuff (perhaps water).

You need 60% mixture.

And you have pure stuff to add.

So,

x = amount of pure stuff to add

50%(4) + 100%(x) = 60%(4+x)

Notice the final volume is 4+x.

.5(4) + 1(x) = .6(4) + .6(x)

multiply by 10 to eliminate decimals

5(4) + 10x = 6(4) + 6(x)

20 + 10x = 24 + 6x

4x = 4

x = 1

This suggests you add 1 gallon of pure stuff to 4 gallons of 50% mixture to end up with 5 gallons of 60% mixture.

Always check your solution.

5 * 60% = 3 gallons of pure stuff and 2 gallons of water.

4 * 50% = 2 gallons of pure stuff

1 * 100% = 1 gallon of pure stuff.

Correct!

Add 1 gallon of pure stuff.

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Step-by-step explanation:

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well there are two answers for that.

1: you need 120 ones then that will equal up to 120.

2: you could also do one 100 and 2 tens.

those are the answers that i have thought of.