A shop that sells nuts is one mixture that is 30% pecans and 70% peanuts. The second picture is 40% pecans and 60% peanuts.FULL PROBLEM IS IN DESCRIPTION!thanks

Question

A shop that sells nuts is one mixture that is 30% pecans and 70% peanuts. The second picture is 40% pecans and 60% peanuts. The manager wants to make a mixture that is 38% pecans. She wants to use 25 ounces more of the second mixture than of the first. How many ounces of each mixture should she use?

Answer 1

We want to find the weight of each mixture which are unknown to us right now, therefore:

 

Let the weight of the first mixture be x. Then the weight of the second mixture would be (x+25) and the weight of final mixture would be (x + x + 25) = 2x+25.

 

The first mixture had 30% pecans, therefore the weight of pecans from the first mixture would be 0.3x.

 

The second mixture had 40% pecans, therefore the weight of pecans from the second mixture would be 0.4 * (x+25). 

 

We will have 38% pecans in the final mixture hence the weight of pecans is 0.38 of the total weight of pecans in the final mixture.

0.38 * (2x + 25)

 

Therefore 

 

0.38 * (2x + 25) = 0.3x + 0.4*(x + 25).

Now we have one equation one unknown and solve for x:

 

=> 0.76x + 9.5 = 0.7x + 10 

=> 0.76x - 0.7x = 10 - 9.5

=> 0.06x = 0.5
=> x = 50/0.06
=> x = 50/6 = 25/3
=> x = 8.33 oz weight of the first mixture therefore weight of the second mixture would be:

 

x+25= 8.33+25=33.33 oz