SOLUTION: Please Help! Solve the problem. Use = 3.14 when necessary. At $3.40 per in3, how much will it cost to fill an aquarium with dimensions of 6 1/3 inch x 3 4/5 inch x 4 1/2 inch?

Algebra ->  Volume -> SOLUTION: Please Help! Solve the problem. Use = 3.14 when necessary. At $3.40 per in3, how much will it cost to fill an aquarium with dimensions of 6 1/3 inch x 3 4/5 inch x 4 1/2 inch?      Log On


   

Question 468433: Please Help!
Solve the problem. Use = 3.14 when necessary.
At $3.40 per in3, how much will it cost to fill an aquarium with dimensions of 6 1/3 inch x 3 4/5 inch x 4 1/2 inch?

A. $613.70
B. $261.63
C. $368.22
D. $310.94
The answer is C. What are the steps to figure this out. I converted each faction: 6 1/3= 19/3=190/30; 3 4/5= 19/5=114/30; 4 1/2= 9/2=135/30.
Total: 439/30. I am confusing myself.
Answer by rfer(16322) About Me  (Show Source):
You can put this solution on YOUR website!
19/3*19/5*9/2=3249/30 cu in
=108 3/10 cu in total
3249/30*3.40=$368.22
Your working to hard, you need improper fractions to multiply, but no need for common denominators in multiplcaton.
Also to find cubic measurments, multiply not add.
V=L*W*H