Lesson Word problems on mixtures for dry substances like coffee beans, nuts, cashew and peanuts

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Word problems on mixtures for dry substances like coffee beans, nuts, cashew and peanuts


Problem 1

A coffee shop has coffee which worth  $6  per pound.  They want to mix it with  20  pounds of coffee which worth  $9  per pound to get
a mixture that can be sold for  $8  per pound. How many pounds of the cheaper coffee should they use?

Solution

Let  x  be the number of pounds of coffee worth  $6  per pound what should be mixed.  It costs  6*x  dollars.
20 pounds of  $9  coffee costs  9*20  dollars.  In total,  the mix costs  6x + 9*20  dollars.
Hence,  one pound of the mix costs  %286x+%2B+9%2A20%29%2F%28x%2B20%29.  Thus you have an equation to determine  x

%286x+%2B+9%2A20%29%2F%28x%2B20%29 = 7.

Simplify and solve it:

6x + 180 = 7*(x+20),     (after multiplying both sides by (x+20) )

6x + 180 = 7x + 140,

180 - 140 = 7x - 6x,

40 = x.

Answer.  x = 40.   40  pounds of the  $6  coffee should be mixed with  20  pounds of  $9  coffee to get the mix at the price of  $7  per pound.


Problem 2

Chocolate coffee beans sell for  $7.00  per pound and hazelnut coffee beans sell for  $6.10  per pound.  One customer wants
a 6 pound mixture of both types of coffee.  How many pounds of each should be used if the mixture is to cost  $6.40  per pound?

Solution

Let  c  be the number of pounds of the chocolate coffee and  h  be the number of pounds of the hazelnut coffee to be mixed.
So,  your first equation is:

c + h = 6

The cost of the mix should be  6*$6.40 = $38.40.  From the other side,  it is 7*c + 6.10*h.  Thus you have the second equation

7*c + 6.10*h = 38.4.

Now,  you can solve this system of two equations by substitution.

c = 6 - h,

7(6-h) + 6.1h = 38.4,

42 - 7h + 6.1h = 38.4,

42 - 0.9h = 38.4,

-0.9h = -3.6,

h = 4.

c + 4 = 6.

c = 2.

Answer.  2  pounds of chocolate coffee and  4  pounds of hazelnut coffee should be used.


Problem 3

A customer has asked a caterer for  60 LB of nuts,  60% of which are to be cashews.
The caterer has available mixtures of  70% cashews and  45% cashews.  How many pounds of each mixture should be used?

Solution

The caterer needs to mix  x  lbs of the  70% cashew mixture with  (60-x)  lbs of the  45% cashew mixture to obtain the  60 lbs of  60% cashew mixture.
Change the percentages to their decimal equivalents and write the algebraic equation to solve for  x.

0.7*x + 0.45*(60-x) = 0.6*60.

Simplify and solve for  x.

0.7x + 27 - 0.45x = 36.

Combine like-terms.

0.25x = 36-27,

0.25x = 9.

Divide both sides by  0.25.

x = 36.

Answer.  The caterer should mix  36 lbs of the  70% cashew mixture with  (60-36 = 24) lbs of the  45% cashew mixture to obtain  60 lbs of  60% cashew mixture.


Problem 4

Peanuts sell for  $3.00  per pound.  Cashews sell for  $6.00  per pound.  How many pounds of cashews should be mixed
with  12  pounds of peanuts to obtain a mixture that sells for  $4.20  per pound?

Solution

Let  x  be the number of pounds of cashews to be mixed.
So,  the weight of the mix is  12 + x  pounds.

The cost of the mix should be  12*$4.20 = $50.40.  From the other side,  it is 12*3 + 6.00*x  dollars.  Thus you have an equation

12*3 + 6.00*x = 50.40.

Simplify and solve the equation:

6x =50.40 - 36,

6x = 14.40,

x = 14.40%2F6 = 2.4.

Answer.  2.4  pounds of cashews should be used.


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