Posted by **coried** on Wednesday, August 5, 2009 at 3:27pm.

Word problem:

Radiator in a certain make needs to contain 50 liters of 40% antifreeze. The radiator now contains 50 liters of 20% antifreeze. How many liters of this solution must be drained and replaced with 100% antifreeze to get the desired strength?

- Algebra -
**drwls**, Wednesday, August 5, 2009 at 4:16pm
Let X be the number of liters that must be drained and replaced with 100% antifreeze.

[0.2(50-X)+X]/50 = 0.4

Solve that for X.

- Algebra -
**bobpursley**, Wednesday, August 5, 2009 at 4:17pm
You want to double the concentration?

If you remove X liters, you remove .2X antifreeze, and .8X water.

You have left in the radiator after draining X, 10-.2X liters of pure antifreeze. You need a total of 20 liters to get 40percent antifreeze.

So you add 10+.2X of antifreeze.

You drain out 10-2X of solution.

X=10+.2X

.8X=10

x=12.5 liters

## Answer This Question

## Related Questions

- college - The radiator in a certain make of car needs to contain 20 liters of 40...
- Algrbra I - Help!! I do not understand how to work these types of problems. My ...
- Word Problem - It is necessary to have a 40% antifreeze solution in the radiator...
- algebra - A mechanic is working on a car that has a 10 Liter radiator that is ...
- Math algebra - The radiator in a car is filled with a solution of 60% antifreeze...
- Math - The radiator of a car contains 20 liters of water. Five liters are ...
- algebra - A radiator contains 6 liters of a 25% antifreeze solution. How much ...
- Algebra - The radiator in a car is filled with a solution of 75 per cent ...
- math - The radiator in a car is filled with a solution of 70 per cent ...
- math - The radiator in a car is filled with a solution of 70 per cent ...

More Related Questions