Mixture - Amount of Each / Partial Solution - Explanation

[Guest]

Another mixture:

A store has 2 varieties of candy, one at 80 cents per lb. and another at 60 cents per lb. They want to combine both for a mix of 50 lbs. to sell at 72 cents per lb. Determine amt. of each.

Partial solution with problem:

Let n = lbs. of 80 cent candy.

50 - n = lbs. of 60 cent candy.

Total of 80 cent: 80n

Total of 60 cent: 60(50 - n)

Value of all: 80n + 60(50 - n)

Value of 1 lb. when mixed: 80n + 60(50 - n) / 50

Value of 1 lb. of mix: 80n + 60(50 - n) / 50 = 72

Questions:

1) Is the 60 cent subtracted from the 80 cent due to both amounts are unknown ? In the other problems one value is known.

2) Is the value divided by 50 to obtain 1 lb. ?

[Guest]

Your Partial solution extended................

Partial solution with problem:

Let n = lbs. of 80 cent candy.

50 - n = lbs. of 60 cent candy.

Total of 80 cent: 80n ...........cents

Total of 60 cent: 60(50 - n) ..........cents

Value of all orig: 80n + 60(50 - n) ....original cost

new mix = 72 x 50 ..........newmix selling price

80n + 60(50 - n) = 72 x 50 .....setup equation
80n + 3000 - 60n = 3600
20n = 600
n = 30

So 30 lbs of 80cent candy = 2400 cents
and 20 lbs of 60cent candy = 1200 cents
thats total of............... 3600 cents
same as.....
gives 50 lbs of 72cent candy = 3600 cents

[Guest]

Thanks. Did you forget about my questions ?

[Guest]

Your Questions:......................I didn't think you still needed these answered

1) Is the 60 cent subtracted from the 80 cent due to both amounts are unknown ? In the other problems one value is known.

I assume you mean...."Is the (80 cent amount) subtracted from the (50 lbs mix amount) due to both amounts are unknown....Yes, n and (50 - n)

The problem question asks.....Determine amount. of each. ..........
There is only one unknown in this problem....The amount.?..... You let the amount of 80cent = n
It doesn't matter which candy....If we know any one we can find the other as you have shown, because we know the total amount is 50 lbs.

2) Is the value divided by 50 to obtain 1 lb. ?
If you know the value of 50 lbs, you can divide by 50 to obtain the value of 1 lb..... The question did not ask for this.

In the equation the value of each candy (cents per lb) x (number of lbs) was added together on LHS and put equal to newmix selling price of (72 cents per lb) x (50 lbs) on the RHS.

[Guest]

" I assume you mean...."Is the (80 cent amount) subtracted from the (50 lbs mix amount) due to both amounts are unknown....Yes, n and (50 - n) " Yes, that is what I meant.

" If you know the value of 50 lbs, you can divide by 50 to obtain the value of 1 lb..... The question did not ask for this. "

Value of 1 lb. when mixed: 80n + 60(50 - n) / 50 : This is why I was asking the question.

Thanks again.

[Guest]

Value of 1 lb. when mixed: 80n + 60(50 - n) / 50 : This is why I was asking the question.....?????

The batch of mixed candy is 50 lb and you let n = amount in lbs of 80cent candy in the batch, so the 60cent amount was (50-n) lbs.
The total value of the batch of 50 lbs is: ... {80n + 60(50 - n)} cents.

If you divide this by 50 you will get the value of 1 lb.
So..... {80n + 60(50 - n)} / 50 is value of 1 lb, and we are told in the question this is 72 cents ....the selling price of 72 cents per lb.

So ..... {80n + 60(50 - n)} / 50 = 72

This is just a re-arrangement of the equation we had earlier.

[Guest]

Thanks. I understand now.