word problem for mathematics

label Mathematics
account_circle Unassigned
schedule 1 Day
account_balance_wallet $5

how much of a(n) 60% orange juice drink must be mixed with a 16 gallons of a 40% orange juice drink to obtain a mixture that is 50% orange juice 

May 18th, 2015

Let the amount of the 60% needed be x (number of gallons)

The total pure juice in the mixture is 60%*x+40%*16 = 0.6x+0.4*16=0.6x+6.4

The number of gallons in the mixtures: x+16

We want it to be 50%, that is: (0.6x+6.4)/(x+16)=50%=0.5

Multiply by x+16

0.6x+6.4=0.5x+8

0.6x-0.5x=8-6.4

0.1x=1.6

x=1.6/0.1 = 16

Check: We add 16 gallons of 60%, that is we add 0.6*16=9.6 juice. The total juice is now 9.6+6.4=16

The total amount of liquid in the micture: 16+16 = 32 gallons

So indeed we got a 50% mixture, as required

May 18th, 2015

Did you know? You can earn $20 for every friend you invite to Studypool!
Click here to
Refer a Friend
...
May 18th, 2015
...
May 18th, 2015
Sep 20th, 2017
check_circle
Mark as Final Answer
check_circle
Unmark as Final Answer
check_circle
Final Answer

Secure Information

Content will be erased after question is completed.

check_circle
Final Answer