SOLUTION: My friend sent me a mixture problem, asking for help, and this one stumps me.
"The cooling system of a batmobile has a capacity of 15 liters. The system is currently filled wit
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Question 761012: My friend sent me a mixture problem, asking for help, and this one stumps me.
"The cooling system of a batmobile has a capacity of 15 liters. The system is currently filled with a mixture that is 40% antifreeze. How much of this mixture should be drained and replaced with pure antifreeze so that the system is filled with a solution that is 50% antifreeze?"
I suppose what is tripping me up is the "drained and replaced" bit. I could work up an equation for diluting or strengthening the mixture, but not both. (Does that even make sense?) Word problems are generally not my strength.
Found 2 solutions by josgarithmetic, MathTherapy:
Answer by josgarithmetic(39617) (Show Source): You can put this solution on YOUR website!
The value for the quantity removed is equal to the value for the quantity added. Call this quantity, v, the liters both of the 40% antifreeze removed and of the pure antifreeze added.
represents the amount of pure antifreeze in the resulting adjusted 15 liter capacity cooling system. The volume of the mixture will be the same as it was before the adjustment. The pure antifreeze material is 100% antifreeze.
Solve for v.
Answer by MathTherapy(10552) (Show Source): You can put this solution on YOUR website!
My friend sent me a mixture problem, asking for help, and this one stumps me.
"The cooling system of a batmobile has a capacity of 15 liters. The system is currently filled with a mixture that is 40% antifreeze. How much of this mixture should be drained and replaced with pure antifreeze so that the system is filled with a solution that is 50% antifreeze?"
I suppose what is tripping me up is the "drained and replaced" bit. I could work up an equation for diluting or strengthening the mixture, but not both. (Does that even make sense?) Word problems are generally not my strength.
Let amount to be drained and replaced be D
Then: .4(15 D) + D = .5(15 D + D)
6 - .4D + D = 7.5
- .4D + D = 7.5 6
.6D = 1.5
D, or amount to be drained and replaced = , or L
You can do the check!!
Further help is available, online or in-person, for a fee, obviously. Send comments, thank-yous, and inquiries to D at MathMadEzy@aol.com
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