How much pure anti freeze must be added to 10 gal of 10% antifreeze solution to increase its concentration to 60 % anti freeze? If X represents the number of pure antifreeze solution, what is the correct statement of the problem?
Add up the amounts of anti-freeze in the parts and the mixture. They must be the same. If x gallons of 100% anti-freeze are added, then
.10*10 + 1.00x = .60(10+x)
original 10 gal is
1 gal af, 9 gal water
so
1 gal + x gal = 0.60 ( 10 + x)
1 + x = 6 + .6 x
.4 x = 5
x = 12.5
thanks,Steve and Anonymous,but i do not understand.please,i will need you to elucidate a little bit more.if you could give me the steps explaining them,that would not only provide the answer but also teach me how to do it and the logic behind it.
thanks anyway.i have an idea,but it is not gelling for me
so far this what i clearly understand
10 gallons of 10% antifreezee solution
so 1 gallon has 10% antifreeze and 90%water.
so it's 1 gallon of antifreeze and 9 gallons of water which make up the 10 gallons total
but i do not know how to proceed from here
The correct statement of the problem is:
How much pure antifreeze solution, represented by X, needs to be added to 10 gallons of a 10% antifreeze solution in order to increase the concentration of antifreeze to 60%?
To solve this problem, we can use the concept of concentrations and mixtures. We know that the original solution is 10% antifreeze, which means that for every 100 parts of the solution, 10 parts are antifreeze.
Since 10 gallons of this solution are given, we can determine the amount of pure antifreeze present:
Amount of antifreeze in 10 gallons = 10 gallons * 10% = 1 gallon
Now, let's assume that X gallons of pure antifreeze solution is added to the existing 10 gallons of 10% antifreeze solution. We want the final concentration to be 60%, which means that for every 100 parts of the final solution, 60 parts should be antifreeze.
To find the amount of pure antifreeze in the final solution, we need to consider the total volume. The total volume of the final solution is 10 gallons + X gallons. So, the amount of antifreeze in the final solution is:
Amount of antifreeze in final solution = (10 gallons * 10%) + (X gallons * 100%)
We can set up an equation based on the concentration of antifreeze:
Amount of antifreeze in final solution = (10 gallons + X gallons) * 60%
Finally, by equating these two expressions for the amount of antifreeze in the final solution, we can solve for the value of X, which represents the number of gallons of pure antifreeze that needs to be added.