Question 65087This question is from textbook An Incremental Development
: How much pure antifreeze should be added to 300 gallons of 30 percent antifreeze solution to raise the antifreeze solution to 40 percent? This question is from textbook An Incremental Development
You can put this solution on YOUR website! How much pure antifreeze should be added to 300 gallons of 30 percent antifreeze solution to raise the antifreeze solution to 40 percent?
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Amt of 100% afreeze = x gallons ; amt of active ingrediant=1.00x gallons
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Amt of 30% afreeze = 300 gallons; amt of active ingrediant=0.30(300)=90 gals
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Amt of 40$ afreeze = x+300; amt of active ingrediant = 0.4(x+300) gals
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EQUATIOn:
active + active = active
x+90=0.4x+120
0.6x=30
x=50 gallons of pure antifreeze must be added
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Cheers,
Stan H.
You can put this solution on YOUR website! How much pure antifreeze should be added to 300 gallons of 30 percent antifreeze solution to raise the antifreeze solution to 40 percent?
Let the amount of pure antifreeze to be added be: x
Then the amount of 100% (pure) antifreeze added is 1.00x
The amount of 30% antifreeze is: .30(300)
Then the total of 40% antifreeze would be:.40(300+x)
Equation to solve is:
1.00x+.30(300)=.40(300+x)
x+90=120+.4x
x-.4x+90=120+.4x-.4x
.6x+90=120
.6x+90-90=120-90
.6x=30
.6x/.6=30/.6
x=50
You need 40 gallons of 100% (pure) antifreeze to change 300 gallons of a 30% into a 40% solution.
Note: If decimals freek you out more than big numbers you can get rid of your decimals by multiplying both sides by 100, then your equation would be:
100x+30(300)=40(300+x)
100x+9000=12000+40x
100x-40x+9000=12000
60x+9000=12000
60x+9000-9000=12000-9000
60x=3000
60x/60=3000/60
x=50 You get the same answer---it's up to you.
Happy Calculating!!!
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