Question

You have one type of nut that sells for $2.80/lb and another type of nut that sells for $5.30/lb. You would like to have 12.5 lbs of a nut mixture that sells for $3.80/lb. How much of each nut will you need to obtain the desired mixture?

Answer 1

Answer:

$7.5\text{ lbs and }5\text{ lbs}$

Step-by-step explanation:

GIVEN: You have one type of nut that sells for $\ 2.80\text{/lb}$ and another type of nut that sells for $\ 5.30\text{/lb}$. You would like to have $12.5\text{ lbs}$ of a nut mixture that sells for $\ 3.80\text{/lb}$  

TO FIND: How much of each nut will you need to obtain the desired mixture.

SOLUTION:

Rate of first type of nut $=\ 2.80\text{/lb}$  

Rate of second type of nuts  $=\ 5.30\text{/lb}$

Total amount of nut mixture $=12.5\text{ lbs}$

let the total quantity of first type of nut be $=x\text{ lbs}$

quantity of second type of nuts $=12.5-x \text{ lbs}$

Now,

rate of new nut mixture $=\ 3.80\text{/lb}$

rate of new mixture $=\frac{\text{rate of first type of nuts}\times\text{quantity of first type nuts}+\text{quantity of second type of nuts}\times\text{rate of second type of nuts}}{\text{total quantity}}$putting values

            $\frac{2.80\times x+5.3\times(12.5-x)}{12.5}=3.8$

           $2.8x+5.3\times12.5-5.3x=12.5\times3.8$

           $2.5x=18.75$

          $x=7.5\text{ lbs}$

quantity of first type of nuts $=7.5\text{ lbs}$

quantity of second type of nuts $=12.5-7.5=5\text{ lbs}$

Hence $7.5\text{ lbs}$ of first type of nuts and $5\text{ lbs}$ of second type of nuts are required to obtain the desired mixture.