Question

a distributor wants to make 1000 pounds of trail mix that is 8% raisins. He will use a mix that is 5% raisins and a mix that is 20% raisins. How many pounds of each should he use?

Answer 1

Answer:

200 pounds 20% raisins

800 pounds 5% raisins

Step-by-step explanation:

Let number of 5% be x

and number of 20% be y

x + y = 1000

Also;

5% of x + 20-% of y = 8% of 1000 pounds

= 0.05x + 0.2y = 80

From i , x = 1000-y

0.05(1000-y) + 0.2y = 80

50-0.05y + 0.2y = 80

0.15y = 30

y = 200

x = 1000 - y

= 1000-200 = 800

Answer 2

He will use 800 pounds for a mix that is 5% raisins and 200 pounds for a mix that is 20% raisins.

Given that,

A distributor wants to make 1000 pounds of trail mix that is 8% raisins.

He will use a mix that is 5% raisins and a mix that is 20% raisins.

We have to determine,

How many pounds of each should he use.

According to the question,

Let number of 5% raisins be x,

And number of 20% raisins be y.

Then,

He will use a mix that is 5% raisins + A mix that is 20% raisins = Distributor trail mix.

$x + y = 1000$

And,

He will use a mix that is 5% raisins of x + and a mix that is 20% raisins of y =  1000 pounds of trail mix that is 8% raisins.

5% of x + 20% of y = 1000 of 8% raisins

$0.05 x+ 0.20y = 80$

On solving both the equation,

$x + y =1000\\\\x = 1000-y$

Substitute the value of x in the equation 2,

$0.05 (1000-y) + 0.20y = 80\\\\0.05 \times 1000 - 0.05y + 0.20y = 80\\\\50 + 0.15y = 80\\\\0.15y = 80-50 \\\\0.15y = 30\\\\y = \dfrac{30}{0.15}\\\\y = 200$

Substitute the value of y in the equation 1,

$x + y = 1000\\\\x + 200 = 1000\\\\x = 1000-200\\\\x = 800$

He will use x = 800 pounds a mix that is 5% raisins and y = 200 pounds for a mix that is 20% raisins.

Hence, He will use 800 pounds for a mix that is 5% raisins and 200 pounds for a mix that is 20% raisins.

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