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? Model the scenario then solve it. Then, in two or more sentences explain whether your solution is or is not reasonable. write out every step.
Answers
Answer:
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Step-by-step 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
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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")
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Check:
(35xx0.2)+(85xx0.68)=$64.8
120xx0.54 = $64.8
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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
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"
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→ 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? Model the scenario then solve it. Then, in two or more sentences explain whether your solution is or is not reasonable. write out every step.
→ 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.
→ 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
→ 0.20 x + ( 1 - x ) 0.68 = 1 x 0.54
→ 0.20 x - 0.68 x + 0.68 = 0.54
→ -0.48 x + 0.68 . 0.54
→ Lets get rid of the decimals for now and multiply everything by 100
→ -48 x + 68 = 54
→ -48 x = 68 - 54
→ -48 x = 14
→ x = 14/48
→ x = 7/28 of the blend at $0.20
→ So the proportion of the $0.68 is 1 - 7/24
→ = 17/24