School-Safe Puzzle Games

## A Unique Blend

A Cafe owner purchases 2 kinds of Kopi Luwak coffee beans-one at \$320 per pound, and the other,  an even better quality, at \$400 per pound. He mixes together some of each, and he plans on selling this mix at \$430 per pound. In doing so, he plans to make a profit of 25%  on the cost. How many pounds of each kind of coffee must he use to make a mixture of a hundred pounds weight?

Answers to this challenge can be entered into the section below; submissions will automatically be revealed when time is up!

### 9 Comments to “A Unique Blend”

1. Shawn | PUZZLE GRANDMASTER | Profile

70# of the \$320/# beans & 30# of the \$400/# beans

If he buys X pounds of \$320/# and Y pounds of \$400/#, then:

X + Y = 100#

100# * \$430/# = 1.25 * [(\$320/# * X) + (\$400/# * Y)]

2. Bobo The Bear | PUZZLE MASTER | Profile

First, if \$430 reflects cost plus profit (C+P), and his profit margin is 25% (P = .25C), then
430 = (C + .25C)
C = 344 = average cost per pound

Now we use an equation for the combined costs
320(x) + 400(100-x) = 344(100)
x = 70

He should use 70 lbs of the lesser quality coffee mixed with 30 lbs of the better quality coffee.

3. John Lewis Highsmith | Profile

This description may be a bit lengthy. I get into my teaching mode on this type problem and I make no apologies for that.

We can set up 2 equations (these are known as simultaneous equations, since they deal with the same conditions.)

Using the usual suspects, x and y, where x = the weight of the lower quality beans and y = the weight of the better quality, we have:

x + y = 100 This is the weight relationship.

320x + 400y = ? This is the cost relationship.

The reason for the ? is that the cost is yet to be determined.

A third equation is needed: C(ost) + P(rofit) = S(elling price), or, in this case C + 0.25C = 430 and C = 344/pound, for a total cost of 34400. Remember, the profit was set as 25% of the cost.

This gives us the second equation as 320x + 400y = 34400.

From here the procedure is:

Multiply the first equation by -320 (or -400, if you wish), giving -320x – 320y = -32000. Needless to say, (but I will anyway, for the heedless) the equation will be different if you use -400, but the final results will be the same either way.

Add this result to the second equation.

320x + 400y = 34400
-320x – 320y = -32000

The x coefficient is 0 and we are left with the equation 80y = 2400, or y = 30.

The result is 30 pounds of the better quality beans and 70 pounds of the “inferior” (if that word could be applied to this type coffee) beans.

70*320 + 30*400 = 34400, 0.25*34400 = 8600, 34400 + 8600 = 43000, or 430/#

And, as the Latins would have it “QED”.

4. suineg | PUZZLE MASTER | Profile

Ok, I think this could be the answer:
X= Number of pounds of the 320\$ coffee beans
Y= Number of pounds of the 400\$ coffee beans
C= Cost of the mix
Equations:
1)320X + 400Y = C
2)X + Y = 100

You know that you have to make 25% of profit and you also know the total sale of the mix is 100* 430= 43000 because its 100 pounds of the mix sell at 430 per pound, cool.

Now, the profit in this example is 25% of the cost is the (total sale) = 1,25C, but total sale is 43000 so: C= 43000/1.25 or C= 34400

Now substitute C in (1) and solve the 2 equations:
-80X = -5600; X= 70 so Y= 30

Now lets check this: 70*320 + 30*400= 34400;
43000-34400= 8600; 0.25*34400= 8600 , cool its verified so:
the Mix should have: 70 pounds of the 320\$ coffee bean and 30 pounds of the 400\$ coffee bean

5. Mashplum | PUZZLE MASTER | Profile

70 pounds of the \$320 coffee would cost \$22,400, and 30 pounds at \$400 would be \$12,000. The 100 pounds of blend would cost the owner \$34,000. \$34,400 plus 25% (\$8,600) is \$43,000 or \$430 per pound.

6. Hex | PUZZLE MASTER | Profile

70 and 30 pounds

7. Mashplum | PUZZLE MASTER | Profile