Question

A nursery is selling one kind of grass mixture for $0.75/lbs while another kind is going for $1.10/lbs. How much of each kind would need to be mixed to produce 50 lbs mixture of seeds that will sell for 0.90/lbs. Thanks.

Answer 1

seed @ .75 = 50(.90 - 1.10)/(.75 - 1.10), i believe

Answer 2

200/7 perhaps at .75 ; 28' 4/7

Answer 3

we can either move the numbers about or simply determine whats left:

+ 3/7 and add 21 more to get 50; 21' 3/7

28' 4/7
21' 3/7
-------
49' 7/7

Answer 4

You must express what you know in terms of equations. Let \[x _{1}\] be the amount of seed 1 in pounds and \[x _{2}\] be the amount of seed 2 in pounds.

Then the total cost of the mixture will be:
\[0.75x _{1}+1.10x _{2} = 0.9\left( x_{1} + x_{2} \right)\]
And the total weight of the mixture will be:
\[ x_{1} + x_{2} = 50\]
Substitute equation 2 in equation 1 to get (remember, this equation is about the cost of the mix):
\[0.75x _{1}+1.10x _{2} = 0.9*50 = 45\]
Solve equation 1 for one of the variables
\[x_{1} = \frac{45 - 1.10x_{2}}{0.75} = 60 - 1.466^{-} x_{2} \]

Substitute this proportion back in equation 2:
\[ 60 - 1.466^{-} x_{2} + x_2 = 50 \]
\[ 0.466^{-} x_{2} = 10 \]
\[ x_{2} = \frac{10}{0.466^{-}} = 21.428571428571428571428571428571 \]

Substitute this value back in equation 2 to find the missing value:
\[x_{1} + 21.428571428571428571428571428571 = 50 \]
\[x_{1} = 28.571428571428571428571428571429 \]

Answer 5

a +b = t ; total the amount needed
ax +by = tz ; amounts * prices equals total amount * needed price

then eliminate the one your not looking for.


a +b = t <-- * -x
ax +by = tz


-ax -bx = -tx
ax +by = tz
------------
b(y-x) = t(z-x) ; and solve for "b"

b@y = t(z-x)/(y-x)

Answer 6

seed@.75 = 50 (.90 - 1.10) / (.75 - 1.10)

seed@ 1.10 = 50 (.90 - .75) / (1.10 - .75)