A food store makes a 11–pound mixture of peanuts, almonds, and raisins. The cost of peanuts is $1.50 per pound, almonds cost $3.00 per pound, and raisins cost $1.50 per pound. The mixture calls for twice as many peanuts as almonds. The total cost of the mixture is $21.00. How much of each ingredient did the store use?

Question

A food store makes a 11–pound mixture of peanuts, almonds, and raisins. The cost of peanuts is $1.50 per pound, almonds cost $3.00 per pound, and raisins cost $1.50 per pound. The mixture calls for twice as many peanuts as almonds. The total cost of the mixture is $21.00. How much of each ingredient did the store use?

3 lbs peanuts, 6 lbs almonds, 2 lbs raisins
8 lbs peanuts, 1 lbs almonds, 2 lbs raisins
6 lbs peanuts, 3 lbs almonds, 2 lbs raisins
8 lbs peanuts, 2 lbs almonds, 1 lbs raisins

Answer 1

Let the amount of peanuts be P, the amount of almonds A, and the amount of raisins R.

From the given information, we have the following equations:
P + A + R = 11 (equation 1)
P = 2A (equation 2)
1.5P + 3A + 1.5R = 21 (equation 3)

Substituting equation 2 into equation 1:
2A + A + R = 11
3A + R = 11 (equation 4)

Multiplying equation 4 by 1.5 to match the coefficients in equation 3:
4.5A + 1.5R = 16.5 (equation 5)

Subtracting equation 5 from equation 3:
1.5P + 3A + 1.5R - (4.5A + 1.5R) = 21 - 16.5
1.5P - 1.5A = 4.5
P - A = 3 (equation 6)

Adding equation 6 to equation 5:
4A = 19.5
A = 4.875 (approximately 4.9 lbs)

Substituting A back into equation 6:
P - 4.9 = 3
P = 7.9 (approximately 7.9 lbs)

Substituting A = 4.9 and P = 7.9 into equation 4:
3(4.9) + R = 11
R = 11 - 14.7
R = -3.7

Since R cannot be negative, this solution is not valid.

Therefore, the correct answer is: 8 lbs peanuts, 1 lb almonds, and 2 lbs raisins.

Answer 2

To find out how much of each ingredient the store used, we can set up a system of equations based on the given information.

Let's assume the store used x pounds of almonds.

The mixture requires twice as many peanuts as almonds, so the store used 2x pounds of peanuts.

The total weight of the mixture is 11 pounds, so the store used 11 - (2x + x) = 11 - 3x pounds of raisins.

The cost of peanuts per pound is $1.50, so the cost of peanuts used is 2x * $1.50 = $3x.
The cost of almonds per pound is $3.00, so the cost of almonds used is x * $3.00 = $3x.
The cost of raisins per pound is $1.50, so the cost of raisins used is (11 - 3x) * $1.50 = $16.50 - $4.5x.

The total cost of the mixture is $21.00, so we have the equation:
$3x + $3x + $16.50 - $4.5x = $21.00.

Simplifying the equation, we have:
$1.5x + $16.50 - $4.5x = $21.00.
-$3x + $16.50 = $21.00.
-$3x = $4.50
x = -$4.50 / -$3
x = 1.5.

So, the store used 1.5 pounds of almonds.
The store used twice as many peanuts as almonds, so they used 2 * 1.5 = 3 pounds of peanuts.
The total weight of the mixture is 11 pounds, so they used 11 - (1.5 + 3) = 11 - 4.5 = 6.5 pounds of raisins.

Therefore, the store used 3 lbs of peanuts, 1.5 lbs of almonds, and 6.5 lbs of raisins.

Answer 3

To solve this problem, let's assign variables to the quantities of peanuts, almonds, and raisins used in the mixture.

Let's say:
P = pounds of peanuts
A = pounds of almonds
R = pounds of raisins

We are given that the total weight of the mixture is 11 pounds, so we can write the equation:
P + A + R = 11 (equation 1)

We are also given that the cost of peanuts is $1.50 per pound, almonds are $3.00 per pound, and raisins are $1.50 per pound. The total cost of the mixture is $21.00, so we can write the equation:
1.50P + 3.00A + 1.50R = 21.00 (equation 2)

Additionally, the mixture calls for twice as many peanuts as almonds, so we can write the equation:
P = 2A (equation 3)

Now we have a system of three equations with three variables. We can solve it using any method we prefer (substitution, elimination, or matrices). Let's solve it using the substitution method:

First, let's solve equation 3 for P in terms of A:
P = 2A

Next, substitute this expression for P in equations 1 and 2:
2A + A + R = 11 (substituting P = 2A into equation 1)
1.50(2A) + 3.00A + 1.50R = 21.00 (substituting P = 2A into equation 2)

Simplifying these equations, we have:
3A + R = 11 (equation 4)
3.00A + 3.00A + 1.50R = 21.00
6A + 1.50R = 21.00 (equation 5)

Now, we have a system of two equations with two variables (A and R). We can solve this system using the elimination or substitution method. Let's solve it using the elimination method:

Multiply equation 4 by -1.50 and equation 5 by 3. This allows us to eliminate the R variable:
-4.50A - 1.50R = -16.50 (equation 6)
18.00A + 4.50R = 63.00 (equation 7)

Adding equations 6 and 7, we have:
13.50A = 46.50

Dividing both sides by 13.50, we find that:
A = 3.44

Now substitute this value for A in equation 4:
3(3.44) + R = 11
10.32 + R = 11

Subtracting 10.32 from both sides, we find that:
R = 0.68

Finally, we can substitute the values of A and R into equation 3 to find the value of P:
P = 2A
P = 2(3.44)
P = 6.88

Therefore, the store used 6.88 pounds of peanuts, 3.44 pounds of almonds, and 0.68 pounds of raisins in the mixture.