A mechanic needs a radiator to have 40% antifreeze solution. The radiator currently is filled with 5 gallons of 15% antifreeze solution.

How much antifreeze mixture should be drained from the car if the mechanic replaces it with pure antifreeze.

Drain__ gallons from the radiator.

let the amount drained be x gallons

These x gallons will contain .4x gallons of antifreeze

amount left of the old stuff = 5-x gallons
so ....

.15(5-x) + (1)x = .4(5)

solve for x

Let me know what you got, so I can check your answer.

x=2

To find the amount of antifreeze mixture that needs to be drained from the car, we can set up an equation.

Let x represent the number of gallons that need to be drained.

The total amount of antifreeze in the radiator before draining can be calculated by multiplying the volume (5 gallons) by the percentage of antifreeze in the mixture (15%).
So, the amount of antifreeze before draining is 5 gallons * 15% = 0.15 gallons.

After draining the x gallons of the antifreeze mixture, the remaining volume in the radiator will be (5 - x) gallons.

The desired concentration of the antifreeze solution after replacement is 40%. So, the amount of antifreeze after replacing with pure antifreeze is (5 - x) gallons * 40% = 0.4 * (5 - x) gallons.

Since we are replacing the drained amount with pure antifreeze, the amount of antifreeze after replacing is also equal to the amount of pure antifreeze drained, which is x gallons.

Setting up the equation:
0.4 * (5 - x) = x

Now, let's solve for x.

0.4 * 5 - 0.4 * x = x
2 - 0.4x = x
2 = 1.4x
x = 2 / 1.4
x ≈ 1.43

Therefore, the mechanic should drain approximately 1.43 gallons from the radiator.

To find out how much antifreeze mixture should be drained from the car, we need to set up an equation using the given information.

Let's assume the amount of antifreeze mixture to be drained from the car is x gallons.

The initial amount of antifreeze in the radiator is 5 gallons * 15% (0.15) = 0.75 gallons of antifreeze.
The remaining amount of water in the radiator is 5 gallons - 0.75 gallons = 4.25 gallons.

After draining x gallons of the antifreeze mixture, the total amount of liquid in the radiator will be 5 gallons - x gallons.

According to the requirements, the antifreeze solution should be 40% of the total amount of liquid in the radiator. So,

0.40 * (5 - x) = x

Now we can solve this equation to find the value of x.

0.40 * 5 - 0.40 * x = x

2 - 0.40x = x

2 = 1.40x

x = 2 / 1.40

x ≈ 1.43

Therefore, approximately 1.43 gallons of the antifreeze mixture should be drained from the car.