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's assume the amount of almonds in the mixture is x pounds.

Since the mixture calls for twice as many peanuts as almonds, the amount of peanuts would be 2x pounds.
The total weight of the mixture is 11 pounds, so we can write the equation:
x + 2x + 1.5x = 11
Combining like terms, we get:
4.5x = 11
Dividing both sides by 4.5, we get:
x = 11/4.5
x ≈ 2.44
To get whole numbers, we can round x up to 3.
So, the amount of almonds is 3 pounds and the amount of peanuts is 2(3) = 6 pounds.
The total cost of peanuts would be 6 * 1.5 = $<<6*1.5=9.00>>9.00
The total cost of almonds would be 3 * 3 = $<<3*3=9.00>>9.00
The total cost of raisins would be 11 - 9 - 9 = $<<11-9-9=-7.00>>-7.00 (which doesn't make sense since it can't be negative)

So, it seems there is an error in the given information. The correct proportions of the ingredients cannot be determined.

Answer 2

To solve this problem, we can set up a system of equations:

Let the number of pounds of peanuts be represented by 'p',
the number of pounds of almonds be represented by 'a',
and the number of pounds of raisins be represented by 'r'.

We know that the mixture calls for twice as many peanuts as almonds, so we can write the equation: p = 2a

The cost of peanuts is $1.50 per pound, almonds cost $3.00 per pound, and raisins cost $1.50 per pound. The total cost of the mixture is $21.00, so we can write the equation: 1.5p + 3a + 1.5r = 21

Now we have two equations with two unknowns, so we can solve the system of equations to find the values of p, a, and r.

Substituting p = 2a into the second equation, we get: 1.5(2a) + 3a + 1.5r = 21
Simplifying, we get: 3a + 3a + 1.5r = 21
Combining like terms, we get: 6a + 1.5r = 21

Now we have a system of equations:
p = 2a
6a + 1.5r = 21

From the first equation, we can substitute 2a in place of p in the second equation:
6a + 1.5r = 21
6a + 1.5r = 21

Simplifying the equations, we get:
6a + 1.5r = 21

Now we can solve for the values of a, r, and p by substituting values:

Let's try the answer choice: 8 lbs peanuts, 1 lbs almonds, 2 lbs raisins
Substituting these values into the first equation, we get:
8 = 2(1)
8 = 2

This answer choice is not valid because it does not satisfy the first equation.

Let's try the answer choice: 6 lbs peanuts, 3 lbs almonds, 2 lbs raisins
Substituting these values into the first equation, we get:
6 = 2(3)
6 = 6

This answer choice is valid because it satisfies the first equation.

Substituting the values into the second equation, we get:
6(3) + 1.5(2) = 21
18 + 3 = 21

This answer choice is also valid because it satisfies the second equation.

Therefore, the correct answer is: 6 lbs peanuts, 3 lbs almonds, and 2 lbs raisins.

Answer 3

To solve this problem, we need to set up a system of equations based on the given information.

Let's represent the number of pounds of peanuts, almonds, and raisins in the mixture as follows:
P = number of pounds of peanuts
A = number of pounds of almonds
R = number of pounds of raisins

According to the problem, the mixture calls for twice as many peanuts as almonds, so we can write the equation:
P = 2A

The cost of peanuts is $1.50 per pound, almonds cost $3.00 per pound, and raisins cost $1.50 per pound. The total cost of the mixture is $21.00, so we can write another equation based on the cost:
(1.50P) + (3.00A) + (1.50R) = 21.00

Now we have two equations with two variables, P and A. We can solve this system of equations to find the values of P and A.

Let's substitute the first equation (P = 2A) into the second equation to eliminate the variable P:
(1.50(2A)) + (3.00A) + (1.50R) = 21.00
3A + 3A + 1.50R = 21.00
6A + 1.50R = 21.00

Now we have one equation with one variable. We can solve for A:
6A + 1.50R = 21.00
6A = 21.00 - 1.50R
A = (21.00 - 1.50R) / 6

To find the value of A, we need to substitute R = 2 into the equation above:
A = (21.00 - 1.50(2)) / 6
A = (21.00 - 3.00) / 6
A = 18 / 6
A = 3

Now that we have the value of A, we can substitute it back into the first equation (P = 2A) to find the value of P:
P = 2(3)
P = 6

We also need to find the value of R. We can substitute the values of P and A into the second equation (1.50P + 3.00A + 1.50R = 21.00) to find R:
1.50(6) + 3.00(3) + 1.50R = 21.00
9 + 9 + 1.50R = 21.00
1.50R = 21.00 - 18.00
1.50R = 3.00
R = 3.00 / 1.50
R = 2

Therefore, the food store used 6 pounds of peanuts, 3 pounds of almonds, and 2 pounds of raisins in the mixture. So the correct answer is:

6 lbs peanuts, 3 lbs almonds, 2 lbs raisins.