Question

A restaurant has one type of lemonade that is 12% sugar and another that is 7% sugar. How many gallons of each type does the restaurant need to make 20 gallons of a lemonade mixture that is 10% sugar?

A) 12 gallons of the 12%, and 8 gallons of the 7%
B) 10 gallons of the 12% and 10 gallons of the 7%
C) 8 gallons of the 12% and 12 gallons of the 7%
D) 2 gallons of the 12% ad 18 gallons of the 7%

Answer 1

10% of 20 is 2
12 × 0.12 (this is 12%) = 1.44
8 × 0.07 (this is 7%) = 0.56
1.44 + 0.56 = 2.00

Therefore, the correct answer is A.

Answer 2

Answer: 12 gallon of 12% sugar and 8 gallon of 7% sugar.

Explanation:

Let we use x gallon of 12% of sugar mixture and y gallon of 7% of sugar mixture.

Thus, According to the question,

The resultant mixture having quantity 20 gallon,

Therefore, x + y = 20 ------(1)

Also, Sugar in x gallon mixture + sugar in y gallon mixture = sugar in resultant mixture.

Again according to the question,

Resultant mixture has 10% of sugar.

Therefore, 12% of sugar in x gallon + 7% sugar in y gallon = 10% sugar in 20 gallon

0.12 x + 0.07 y = 0.2

12 x + 7 y = 200 --------(2)

7× equation (1),

We get, 7x + 7y = 140 -----(3)

Equation(1) - equation (3)

5 x = 60

x = 12

By putting the value of x in equation (1),

12 + y = 20

⇒ y = 8

Thus, He will add 12 gallon of 12% sugar mixture and 8 gallon of 7% sugar mixture for obtaining 20 gallon of 10% sugar mixture.

Therefore, First Option is correct.