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?
A. 3lbs peanuts, 6 lbs almonds, 2 lbs raisins
B. 8 lbs peanuts, 1 lb almonds, 2 lbs raisins
C. 6 lbs peanuts, 3 lbs almonds, 2 lbs raisins
D. 8 lbs peanuts, 2 lbs almonds, 1 lbs raisins

Answer 1

Given: 11-pound mixture of peanuts, almonds, and raisins
Cost:
peanuts - 1.5 per pound
almonds - 3 per pound
raisins - 1.5 per pound

mixture:
twice as many peanuts as almond; total cost of mixture is 21.

a + p + r = 11 lbs
a + 2a + r = 11 lbs
3a + r = 11
r = 11 - 3a

1.5(2a) + 3a + 1.5r = 21
3a + 3a + 1.5r = 21
6a + 1.5r = 21
6a + 1.5(11-3a) = 21
6a + 16.5 - 4.5a = 21
6a - 4.5a = 21 - 16.5
1.5a = 4.5
1.5a/1.5 = 4.5/1.5
a = 3

almonds = 3 lbs
peanuts = 2a = 2(3) = 6lbs
raisins = 11 - 3a = 11 - 3(3) = 11 - 9 = 2 lbs

My answer is: C. 6 lbs peanuts, 3 lbs almonds, 2 lbs raisins 

Answer 2

Let

x--------> pounds of peanuts

y--------> pounds of almonds

z--------> pounds of raisins

we know that

$x+y+z=11$ -----> equation $1$

$1.5x+3y+1.5z=21$ -----> equation $2$

$x=2y$ -----> equation $3$

Substitute the equation $3$ in equation $1$ and equation $2$

So

$[2y]+y+z=11$

$3y+z=11$ -------> equation $4$

$1.5[2y]+3y+1.5z=21$

$3y+3y+1.5z=21$

$6y+1.5z=21$ -----> equation $5$

Solve the system

$3y+z=11$ -----> equation $4$

$6y+1.5z=21$ -----> equation $5$

Multiply by -2 equation $4$

$-6y-2z=-22$

Adds equation $4$ and equation $5$

$-6y-2z=-22\\ 6y+1.5z=21\\ --------\\ -0.5z=-1\\ z=2lb$

find the value of y

$6y+1.5z=21\\ 6y+1.5*2=21\\ 6y=21-3\\ y=18/6\\ y=3lb$

find the value of x

$x=2y\\ x=2*3\\ x=6lb$

therefore

the answer is the option

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