You are making 10 tons of concrete that is 37.9% cement by mixing a 25% cement mixture with a 68% cement mixture.
a) how much of the 68% cement mixture must you use?
b) how much of the 25% cement mixture must you use?

The 10 tons should have 37.9% cement.

Let the 25% mixture be x tons

Then the 68% mixture will be (10 -x) tons.

Using formula for mixtures:
Let p for percentage and w for weight in tons.

p₁w₁ + p₂w₂ = pw

25*x + 68*(10-x) = 37.9*10

25x + 680 - 68x = 379

680 - 43x = 379

680 - 379 = 43x

301 = 43x

43x = 301

x = 301/43

x = 7

Recall that x, was for the 25% mixture, 7 tons.

a) For the 68% mixture, = 10 - x = 10 -7 = 3 tons.

b) For the 25% mixture = x = 7 tons.  

Iagree that is the answer to the question

The answer to your question is d 68%