A wholesaler makes up a @a pound mixture of two kinds of coffee. Brand A costs $@b per pound and Brand B cost $@c per pound. How many pounds of each kind of coffee must be used if the mixture is to cost $@d per pound?
Let x=pounds of Brand A, and y=pounds of Brand B. {grocery} Since it is to be a @a pound mixture, {x+y=@a} Since the cost of the mixture is to be $@d per pound, and the mixture will weigh @a pounds, the total cost of the mixture is {(@d)(@a)} From the given costs of each mixture we can compute the total cost from {@bx+@cy=(@d)(@a)} Solve equations 2 and 4 for x and y.
A @a pound mixture of walnuts and almonds costs $@b. If the walnuts cost $@c per pound, and the almosts cost $@d per pound, how many pounds of each kind are there in the mixture?
Let w=pounds of walnuts and a=pounds of almonds. {grocery} Since the total weight is to be @a pounds, we can write {a+w=@a} From the two given costs-per-pound numbers, we can write that {@cw+@da=@b} Solve equations 2 and 3 for a and w to get your answers.
A person needs to mix @a pounds of macadamia nuts that cost $@b per pound with pecans that cost $@c per pound. How many pounds of pecans should be used if the mixture is to cost $@d per pound?
Let p=the number of pounds of pecans that should be used. {grocery} The total weight of the mixture will be 20+p pounds, and the total cost of the final mixture will be {($@d)(p)} (remember, $@d is the desired cost per pound). In this mixture, the total cost of the macadamia nuts will be {(@a)(@b)} and the total cost of the pecans will be {(@c)(p)} Now, the total cost of the nuts will be the sum of the costs of the macadamic and pecans, or {(@a)(@b)+@cp=@dp} Solve this for p to get your answer.
How many {volume} of pure alcohol are there in @a {volume} of a @b% solution of alcohol?
Let x=the amount of alcohol. {perconvert} Your answer is just @b% of @a or {x=(@b/100)(@a)} Crunch all of these numbers to find x, the amount of pure alcohol.
How many {volume} of a @a% alcohol solution must be added to @b {volume} of a @c% alcohol solution to make a @d% solution?
Let x={volume} of @a% alcohol solution, y={volume} of @c% solution (=@b {volume}). {perconvert} The total {volume} in the final solution will be x+y or x+@b. The amount of @a% alcohol will be {(@a/100)x} and the amount of @c% alcohol will be {(@c/100)y or (@c/100)(@b)} The amount of alcohol in the final mixture will the sum of these two, which must equal {(@d/100)(x+y) or (@d/100)(x+@b)} In equation form, this all gives {(@a/100)x+(@c/100)(@b)=(@d/100)(x+@b)} Solve this equation for x to get your answer.
How much water should be added to a @a% solution of alcohol to obtain @b {volume} of a @c% solution?
Let x={volume} of water, and y={volume} of alcohol. {perconvert} Since water and alcohol add in volume, we can write {x+y=@b} Now, pure water doesn't contain ANY alcohol, so it doesn't contribute alcohol to the @c% solution. This means the amount of alcohol in the final solution comes solely from the @a% solution and is {(@a/100)y=(@c/100)@b} Solve equations 1 and 2 for x and y. x will be the {volume} of water required.
A pharmacist wishes to make @a {volume} of a @b% solution of acid by mxing @c% and @d% solutions. How much of each type should be used?
Let x={volume} of @c% solution and y={volume} of @d% solution. {perconvert} Since the volumes of each solution add together, we can write that {x+y=@a} The amount of acid in the @c% solution is (@c/100)x and the amount of acid in the @d% solution is (@d/100)y. The amount of acid in the final solution is the sum of these two, or {(@c/100)x+(@d/100)y=(@b/100)@a} Solve equations 1 and 2 for x and y to find your answers.
