­A 10-kilogram soil mixture containing 30 percent sand and 70 percent

­A 10-kilogram soil mixture containing 30 percent sand and 70 percent humus, by weight, is thorougly mixed together. In order to create a 10-kilogram mixture that contains 50 percent of each component, approximately how many kilograms of the soil mixture should be removed and replaced by pure sand?

It's easy to get the impression that we should focus on the sand since the question involves replacing some of the mixture with pure sand. However, it's much easier to find the answer by focusing on the humus since we need to handle only removal of humus whereas, if we focus on the sand, we'll have to handle first subtracting and then adding sand.

We need to create a 10-kilogram mixture such that each component makes up 50 percent of the mixture. So, the final mixture will contain 10 × 0.5 = 5 kilograms of humus. If it contains 5 kilograms of humus, it will contain 5 kilograms of sand as well.

So, since the mixture currently has 10 × 0.7 = 7 kilograms of humus, we need to remove enough of the current mixture to remove 2 kilograms of humus.

The soil mixture is "thoroughly mixed." So, the humus is evenly distributed throughout the mixture. So, to remove 2 kilograms of humus, we remove 2/7 of the entire mixture.

1/7 is a little over 0.14. So, 2/7 is about 0.29.

0.29 × 10 = 2.9

(In case you didn't know the decimal equivalent of 1/7, you could quickly see that 2/7 is going to be a little less 0.3 because 21/7 = 3.)

(A) 2.0

(B) 2.9

(C) 3.4

(D) 4.0

(E) 5.0

Correct answer:­­­­­­