SOLUTION: A shop that sells nuts has one mixture that is 30% pecans and 70% peanuts. A second mixture is 40% pecans and 60% peanuts. The manager wants to make a mixture that is 38% pecan

Algebra.Com
Question 1069813: A shop that sells nuts has one mixture that is 30% pecans and
70% peanuts. A second mixture 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 the
first. How many ounces of each mixture should she use?
Thanks!
Found 3 solutions by Boreal, addingup, Edwin McCravy:
Answer by Boreal(15235)   (Show Source): You can put this solution on YOUR website!
x of the first with .30pk+.70p
x+25 of the second with .40pk+.70p.
Total is 2x+25(.38)=.76x+9.50
.30x+.40(x+25)=.76x+9.50
.30x+.40x+10=.76x+9.50
0.5=0.06x
50=6x
x=8 1/3 oz of the first--.30(8.33)=2.50 oz pecans
x+25=33 1/3 oz of the second=.40(33 1/3)=13.33 oz pecans
That would be 41 2/3 oz with 15.83 oz pecans, and that is 38%.
Answer by addingup(3677)   (Show Source): You can put this solution on YOUR website!
Let's call the weight of the first mixture x. Thus, the weight of the second mixture will be x+25, since you want to use 25oz. more of the second mixture than the first.
So, the total weight for our final mixture will be:
x+x+25 = 2x+25
:
Now for the weight of the pecans:
first mixture: 30% pecans = 0.3x.
second mixture:40% pecans = 0.4(x+25)
Lastly, we are looking for a mix with 38% pecans = 0.38(x+x+25)= 0.38(2x+25)
:
Put it all together:
0.3x+0.4(x+25) = 0.38(2x+25)
0.3x+0.4x+10 = 0.76x+9.5
0.7x = 0.76x-0.5 subtract 7x and add 0.5 to both sides
0.5 = 0.06x flip the equation, it looks better with the variable x on the left
0.06x = 0.5 divide both sides by 0.06
x = 8.33 So this is the weight of the first mixture. The second mixture, the problem says, will be 25 oz. more: 8.33+25 = 33.33oz.
--------------------------------
Additional info:
What will be the total weight of the final mixture? 8.33+33.33 = 41.66 ounces.
:
Happy learning,
John
Answer by Edwin McCravy(20056)   (Show Source): You can put this solution on YOUR website!
The other tutors are correct, but I'll bet your 
teacher wants you to make a chart for mixture
problems. 

So begin by making this chart:


         | ounces |  percent   | ounces of | 
         |of nuts |(as decimal)|  PECANS  |
---------|--------|------------------------|
 1st mix |        |            |           |
 2nd mix |        |            |           | 
---------|--------|------------|-----------|
final mix|        |            |           |

She wants to use 25 ounces more of the second mixture
than the first.
Let x = the number of ounces of the 1st mix.
Then x+25 = the number of ounces of the 2nd mix.
Fill those in

         | ounces |  percent   | ounces of | 
         |of nuts |(as decimal)|  PECANS  |
---------|--------|------------------------|
 1st mix |   x    |            |           |
 2nd mix |  x+25  |            |           | 
---------|--------|------------|-----------|
final mix|        |            |           |

Fill in the three percents expressed as decimals (hundredths)

         | ounces |  percent   | ounces of | 
         |of nuts |(as decimal)|  PECANS  |
---------|--------|------------------------|
 1st mix |   x    |    0.30    |           |
 2nd mix |  x+25  |    0.40    |           | 
---------|--------|------------|-----------|
final mix|        |    0.38    |           |

Add the number of ounces of 1st and 2nd mixes to get the
total number of ounces of final mix.
 x + x+25 = 2x+25.  So fill that in for the ounces of
final mix:

         | ounces |  percent   | ounces of | 
         |of nuts |(as decimal)|  PECANS  |
---------|--------|------------------------|
 1st mix |   x    |    0.30    |           |
 2nd mix |  x+25  |    0.40    |           | 
---------|--------|------------|-----------|
final mix| 2x+25  |    0.38    |           |

Next we fill in the last column with PECANs 
by taking the percentages of ounces of each of 
the three mixes.  So we just multiply the two 
columns:

         | ounces |  percent   | ounces of | 
         |of nuts |(as decimal)|  PECANS  |
---------|--------|------------------------|
 1st mix |   x    |    0.30    |0.30x      |
 2nd mix |  x+25  |    0.40    |0.40(x+25) | 
---------|--------|------------|-----------|
final mix| 2x+25  |    0.38    |0.38(2x+25)|           

The equation comes from adding up the ounces of PECANS
in 1st and 2nd mixes and setting it equal the the number
of ounces of PECANS in the final mix:



0.30x + 0.40(x+25) = 0.38(2x+25)

Clear of decimals by moving them 2 places right:

    30x + 40(x+25) = 38(2x+25)

Solve that and get x =  or  

Since we let x = the number of ounces of the 1st mix,
the number of ounces of the 30% mixture is  

Since then x+25 = the number of ounces of the 2nd mix,
the number of ounces of the 40% mixture is .

Edwin
RELATED QUESTIONS

the manager of nutt's nuts regularly sells cashews for $6.50 per pound, pecans for $7.50... (answered by josgarithmetic)
A bulk store has two types of mixed nuts: One is 60% peanuts, and the other is 35%... (answered by stanbon)
A company sells a container of mixed nuts that is 50% peanuts and the other container... (answered by josgarithmetic,ikleyn)
9. A company sells a container of mixed nuts that is 12% peanuts and another container... (answered by jorel1380,Boreal)
Whats an example of a mixture problem that could be represented by the expression 8 + x (answered by jorel555)
emily is employed by sunset foods. she was instructed to create 30 pund mixture of... (answered by josmiceli)
A machine is supposed to mix peanuts, hazel-nuts, cashews, and pecans in the ratio... (answered by LChester89)
Hi A store owner has a pound of a mixture of nuts that is 90% peanuts and 10% Brazil... (answered by josgarithmetic,greenestamps)