Antifreeze and Engine Coolant

How many gallons of antifreeze do i need for a 1992 k1500 305?

**We need you to answer this question!**

###### If you know the answer to this question, please register to join our limited beta program and start the conversation right now!

## Related Questions

###### Asked in Math and Arithmetic

### How many gallons of a 80 percent antifreeze solution must be mixed with 70 gallons of 20 percent antifreeze to get a mixture that is 70 percent antifreeze?

70 gallons of 20% solution contains 70*0.2 = 14 gallons of
antifreeze.
Suppose you need G gallons of the 80% antifreeze solution. This
will contain 0.8*G gallons of antifreeze.
Total volume of solution = G + 70 gallons
Volume of antifreeze required in this solutions to make it a 70%
solution is
0.7*(G + 70) = 0.7G + 49 gallons.
Volume of antifreeze = 14 + 0.8G gallons
So 0.7G + 49 = 14 + 0.8G
0.7G + 35 = 0.8G
35 = 0.1G
350 = G
Answer: 350 gallons.

###### Asked in Antifreeze and Engine Coolant, Math and Arithmetic, Algebra

### How many gallons of a 80 percent antifreeze solution must be mixed with 60 gallon of 15 percent antifreeze to get a mixture that's 70 percent antifreeze?

330 gallons of 80% antifreeze mixed with 60 gallons of 15%
antifreeze will provide 390 gallons of 70% antifreeze.
Let the multiplier of 60 gallons be x, then considering the
percentages of antifreeze in the solutions:
(15 + 80x) ÷ (1 + x) = 70
⇒ 15 + 80x = 70 + 70x
⇒ 10x = 55
⇒ x = 5.5
Therefore mix 60 x 5.5 = 330 gallons of 80% antifreeze.

###### Asked in Antifreeze and Engine Coolant, Math and Arithmetic, Algebra

### How many gallons of antifreeze must be mixed with 70 gallons of 25 percent antifreeze to get a mixture that is 80 percent antifreeze?

Find the overall balance of antifreeze: amount of antifreeze to
start with + the amount of pure antifreeze to add = amount of
antifreeze in the final solution
Let X be the amount of pure antifreeze to add in gallons.
(70 * 0.25) + X = 0.80 * (70+X)
17.5 + X = 56 + 0.8X
0.2X = 56 - 17.5
0.2X = 38.5
X = 38.5/0.2 = 192.5 gallons

###### Asked in Math and Arithmetic, Algebra, Percentages, Fractions, and Decimal Values

### How many gallons of a 70 percent antifreeze solution must be mixed with 70 gallons of 25 percent antifreeze to get a mixture that is 60 percent antifreeze?

Suppose G gallons are required. At 70% concentration, this will
contain 0.7*G gallons of antifreeze.
The 70 gallons of 25% contains 17.5 gallons of the active
ingredient.
In total, then, there are (G + 70) gallons containing 0.7G +
17.5 gallons of antifreeze and this represents 60%.
So (0.7G + 17.5) / (G + 70) = 60/100
10*(0.7G + 17.5) = 6*(G + 70)
7G + 175 = 6G + 420
G = 420 - 175 = 245.
Answer: 245 gallons.

###### Asked in Antifreeze and Engine Coolant, Math and Arithmetic, Algebra

### How many gallons of a 60 percent antifreeze solution must be mixed with 70 gallons of 10 percent antifreeze to get a mixture that is 50 percent antifreeze?

Suppose G gallons of 60% antifreeze are required.
G gallons of 60% contain 0.6G gallons of the active
ingredient.
70 gallons of 10% contain 7 gallons of the active
ingredient.
So, in the mixture, there are
G + 70 gallons containing 0.6G + 7 gallons of the active
ingredient. This is to represent 50%.
So 0.6G + 7 = 0.5*(G + 70)
= 0.5G + 35
0.1 G = 35 - 7 = 28
G = 280 gallons.

###### Asked in Antifreeze and Engine Coolant, Math and Arithmetic, Algebra

### How many gallons of 80 percent antifreeze solution must be mixed with 100 gallons of 10 percent antifreeze to get a mixture that is 70 percent antifreeze?

600 gallons.
To solve this think about the amount of antifreeze in the
solution.
When 10% antifreeze is added to x times as much 80% antifreeze,
the resultant percentage antifreeze will be:
(10 + 80x)/(1 + x)
So to obtain a 70% solution, x will need to solve:
(10 + 80x)/(1 + x) = 70
⇒ 10 + 80x = 70 + 70x
⇒ 10x = 60
⇒ x = 6
So to 100 gallons of 10% antifreeze solution, 100 x 6 = 600
gallons of 80% antifreeze solution will be needed to make it a 70%
solution.
Consider adding 100 gallons of 10% antifreeze and 100 gallons of
80% antifreeze together and then taking half the resultant
solution, that is so that you have 100 gallons of the mixture; the
amount of antifreeze in this 100 gallons is (10 gallons + 80
gallons) ÷ 2 = 45 gallons, that is a 45% mixture.
Now consider adding 100 gallons of 10% antifreeze and 200
gallons of 80% antifreeze together and then taking a third of the
resultant solution, that is so that you have 100 gallons of the
mixture; the amount of antifreeze in this 100 gallons is (10
gallons + 2 x 80 gallons) ÷ 3 = 562/3 gallons, that is a 562/3%
mixture.
If x times as much of the 80% antifreeze is added to the
If 1 gallon of 10% antifreeze was added together with 80%
antifreeze, the resultant mixture will still be (10+90)/2 =
45%.
So if x times as much of the 80%
Consider the amount of antifreeze in
the solution.
When two solutions of the same volume is added, the amount of
antifreeze in the same volume is half the new amount.
Consider the ratio of anti-freeze to water in each of the
solutions in fraction form of antifreeze/water:
10% is 10/90
70% is 70/30
80% is 80/20
When two ratios are added together, the amount of antifreeze in
the solution is added together and the amount of water is added
together to give some odd fraction maths:
10% + 80% = (10+80)/(90+20) = 90/110 =