Word problem

Question

How many pounds of coffee costing $3.20 per pound must be mixed with 18 pounds of coffee costing $2.80 per pound to produce a mixture that costs $2.96 per pound ?

Answer 1

Realize that price per pound of coffee means the amount of money needed for one pound of that coffee. This can be written mathematically as:

 

[Price Per Pound] = [Total Price for Coffee] ÷ [Total Weight of Coffee]  (Equation 1)

 

So this problem basically wants you to to use two coffees of known price per pound to create a coffee mixture whose average price per pound is $2.96. In equation form this can be written as:

 

[Average Price Per Pound of Mixture] = $2.96   (Equation 2)

 

What we need to do is express the [Average Price Per Pound of Mixture] in terms of:

• the weight of Coffee 1                   (unknown)
• the price per pound of Coffee 1      ($3.20)
• the weight of Coffee 2                   (18 pounds)
• the price per pound of Coffee 2      ($2.80)

These are quantities that we know or are looking for.

 

Using Equation 1 we can first write that:

 

[Average Price Per Pound of Mixture] = [Total Price for Coffee Mixture] ÷ [Total Weight of Coffee Mixture]   (Equation 3)

 

However the [Total Price for Coffee Mixture] is equivalent to the sum of the total prices of each coffee alone:

 

[Total Price for Coffee Mixture] = [Total Price for Coffee 1] + [Total Price for Coffee 1]      (Equation 4)

 

From Equation 1, by multiplying by [Total weight of Coffee] on both sides it can be shown that:

 

[Total Price of Coffee] = [Price Per Pound of Coffee] × [Total Weight of Coffee]        (Equation 5)

 

Using Equation 5, we can rewrite Equation 4 in terms of the quantities that we know or are looking for.

 

 

[Total Price for Coffee Mixture] = ( [Price Per Pound of Coffee 1] × [Total Weight of Coffee 1] ) + ( [Price Per Pound of Coffee 2] × [Total Weight of Coffee 2] )       (Equation 6)

Additionally, the [Total Weight of Coffee Mixture] is equivalent to the sum of the weights of each coffee alone:

 

[Total Weight of Coffee Mixture] = [Total Weight of Coffee 1] + [Total Weight of Coffee 2]       (Equation 7)

 

From here on, for simplicity's sake the quantities will be represented as variables:

• [Price Per Pound of Coffee 1] ⇒ R₁
• [Price Per Pound of Coffee 2] ⇒ R₂
  
• [Total Weight of Coffee 1] ⇒ W₁
• [Total Weight of Coffee 2] ⇒ W₂
• [Total Price of Coffee 1] ⇒ P₁
• [Total Price of Coffee 2] ⇒ P₂
• [Average Price per Pound of Coffee Mixture] ⇒ R_mix

 

By substituting Equation 6 and Equation 7 into Equation 3, it can be stated that:

 

            (R₁ × W₁) + (R₂ × W₂)
Rmix = ----------------------------       (Equation 8)
                      (W₁ + W₂)

 

 

All quantities accept for W₁ are known, thus we can begin to rearange the equation to solve for W₁. First by substituting in all known values:

 

              ($3.20 × W₁) + ($2.80 × 18)           ($3.20 × W₁) + ($50.40) 
$2.96 = ---------------------------------------- = ----------------------------------    (Equation 9)
                             (W₁ + 18)                                     (W₁ + 18)

 

Then by multiplying by (W₁ + 18) on both sides:

 

$2.96 × (W₁ +18) = ($3.20 × W₁) + $50.40

 

 

($2.96 × W₁) + $53.28 = ($3.20 × W₁) + $50.40         (Equation 10)

 

 

Grouping like terms:

 

$53.28 – $50.40 = ($3.20 × W₁)  – ($2.96 × W₁)

 

$2.88 = $0.24 × W₁     (Equation 11)

 

 

And finally dividing by $0.24 on both sides:

 

$2.88 / $0.24 =  W₁  

 

W₁ = $2.88 / $0.24 = 12 

 

Meaning that 12 pounds  of the coffee costing $3.20 must be added to 18 pounds of coffee costing $2.80 to produce a mixture with a unit price of $2.96 per pound.