48.6k views
3 votes
Sandy mixed together 9 gallons of one brand of juice and 8 gallons of a second brand of juice which contains 48% real fruit juice. Find the percent of real fruit juice in the first brand if the mixture contained 30% real fruit juice.

A) 18%

B) 140%

C) 180%

D) 14%

User Novitchi S
by
5.3k points

2 Answers

3 votes

Answer:D

Step-by-step explanation:jus took test

User Itsathere
by
5.4k points
2 votes

Answer:

D) 14%

Explanation:

Let the percent of the real fruit in the first brand = x

Given that number of gallons of real fruit in the first brand = 9 gallons

Given that percent of the real fruit in the second brand = 48%

Given that number of gallons of real fruit in the second brand = 8 gallons

Given that percent of the real fruit in the mixture = 30%

then number of gallons of real fruit in the mixture = (9+8) = 17 gallons

Then we get equation:

[x% of 9 gallons] + [48% of 8 gallons] = [30% of 17 gallons]

9x+48(8)=30(17)

9x+384=510

9x=510-384

9x=126

x=126/9

x=14

Hence final answer is D) 14%

User Adam Hunyadi
by
4.7k points