Math word problems

Question

1. The manager of a store selling tea plans to mix a more expensive tea that costs $10 per pound with a less expensive tea that costs $5 per pound to create a 110-pound blend that will sell for $8.00 per pound. How many pounds of each type of tea are required?

2.  How much of a(n) 60% orange juice drink must be mixed with 19gallons of a 30% orange juice drink to obtain a mixture that is 50% orange juice? 

The amount of 60% orange juice drink is how many gallons? 

Answer 1

 

1) Let x = more expensive, y = less expensive.

 

System of equations:

x + y =110

10x + 5y = 110 * 8 = 880

 

x = 110-y

 

10(110-y) + 5y = 880

1100 - 10y + 5y =880

220 = 5y

44 = y

 

x = 110-44 = 66

 

So they require 66 and 44 pounds.

 

2) Let x = the amount of the second juice added to the mix. Set up an equation for the total amount of juice mixed into the formula = to the total amount of liquid added to the formula.

 

(19*.3) + .6x = .5(19+x)

5.7 + .6x = .5(19+x)
5.7 + .6x = 9.5 + .5x
.1x = 3.8
x=38

Double checking the work.

19 * .3 + 38 * .6 = .5(19+38)
5.7 + 22.8 = 28.5
28.5=28.5

So it's 38 gallons.