SOLUTION: One grade of oil has 0.50% of an additive, and a higher grade has 0.75% of the additive. How many liters of each must be used to have 1000 L of a mixture with 0.65% of the additive
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Question 159806: One grade of oil has 0.50% of an additive, and a higher grade has 0.75% of the additive. How many liters of each must be used to have 1000 L of a mixture with 0.65% of the additive. Found 2 solutions by checkley77, KnightOwlTutor:Answer by checkley77(12844) (Show Source):
You can put this solution on YOUR website! .75x+.5(1000-x)=.65*1000
.75x+500-.5x=650
.25x=650-500
.25x=150
x=150/.25
x=600 liters of .75% additive.
1000-600=400 liters of 50% additive.
Proof:
.75*600+.5*400=650
450+200=650
650=650
You can put this solution on YOUR website! Convert % in decimal format
.5%=.5/100=0.005
.75%=.75/100=.0075
.65%=.65/100=.0065
X= the higher grade of oil that has .75% of additive
1000-X= volume of grade of oil that has .5% additive
The total volume on both sides of the equation =1000L
.0075X+.005(1000-X)=.0065(1000)
Use the distributive property
.0075X+5-.005X=6.5
.0025X+5=6.5
subtract 5 from both sides
.0025X=1.5
divide both sides by .0025
X=600L of .75% additive oil
1000-600=400L of .50% additive
Plug in to check answer
600(.0075) + 400(.005)=.0065(1000)
4.5 + 2=6.5
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