A coffee mixture has beans that sell for $0.20 a pound and beans that sell for $0.68. If 120 pounds of beans create a mixture worth $0.54 a pound, how much of each bean is used?

2 Answers
Jul 20, 2017

Answer:

#$0.68# type #-> 85^("lb")#

#$0.20# type #-> 35^("lb")#

Explanation:

One of the better ways to tackle this problem type is to consider them as a set of fraction that sum to the value of 1. Where 1 represents the whole of the final blend.

#color(blue)("Determine the fractional proportion of each type of bean")#

Let the proportion of the $0.20 be #x#

Then the proportion of the $0.68 is #1-x#

Set the target at $0.54

Dropping the unit of measurement ($ ) for now we have

#0.20x+(1-x)0.68=1xx0.54#

#0.20x-0.68x+0.68=0.54#

#-0.48x+0.68=0.54#

Lets get rid of the decimals for now and multiply everything by 100

#-48x+68=54#

#48x=68-54#

#x=14/48#

#x=7/24# of the blend at $0.20

So the proportion of the $0.68 is #1-7/24=17/24#
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Determine the weight proportion of each bean type")#

#$0.68# type #->17/24xx120^("lb") = 85^("lb")#

#$0.20# type #->7/24xx120^("lb")= 35^("lb")#
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Check:

#(35xx0.2)+(85xx0.68)=$64.8#

#120xx0.54 = $64.8#

May 3, 2018

Answer:

Method 2 of 2

An amazing approach that is not commonly used. Very fast once used to it.

#85^("lb")" of $0.68 beans"#
#35^("lb")" of $0.20 beans"#

Explanation:

This uses the principle of a straight line graph.

Let the weight of the $0.20 bean be #w_(0.2)#
Let the weight of the $0.68 bean be #w_(0.68)#

Then by considering only say #w_0.68# the value of #w_(0.2)# is directly inferred by #w_(0.2)=120-w_(0.68)#

So it is possible to determine the part of the answer by considering just #w_(0.68)#

If all #w_(0.68)# total value is #................" "120xx$0.68 = $ 81.60#
If all #w_(0.2)# total value is #................." "120xx$0.20=$24.00 #
If all the target blend total value is #." "120xx$0.54 = $64.80 #

Plotting this on the graph we have:
Tony B

The slope of all is the same as the slope of a part of it

#color(green)("Slope of all "->("change in y")/("change in x") ->(81.60-24.00)/(120)=57.6/120)#

#color(red)("Slope of part: "(64.8-24.00)/x=40.8/xcolor(green)(=57.6/120))#

#x=40.8xx120/57.6 = 85^("lb") = w_(0.68)#

So we have:
#85^("lb")" of $0.68 beans"#
#120-85=35^("lb")" of $0.20 beans"#