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.

Respuesta :

beans that sell for 0.2 aer t
beans that sell for 0.68 are s

total is 120 pounds
s+t=120

cost is
0.2t+0.68s=0.54*120

solve

0.2t+0.68s=0.54*120
0.2t+0.68s=64.8
multilpy boht sides by 10
2t+6.8s=648
divide boht sides by 2
t+3.4s=324
other equaiton
t+s=120
multily this equaiton by -1 and add to other equaiation

-t-s=-120
t+3.4s=324 +
0t+2.4s=204

2.4s=204
divide both sides by 2.4
s=85
sub back

s+t=85
85+t=120
minus both sides 85
t=35


35 pounds of the $0.20 variety
85 pounds of the 0.68 variety

check
35*0.20+85*0.68=
7+57.8=
64.8

120*0.54=64.8

64.8=64.8

my answer is reasonable


35lb of $0.20 beans
85lb of $0.68 beans

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.

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 × 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

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

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Determine the weight proportion of each bean type

$

0.68

type  

17

24

×

120

lb

=

85

lb

$

0.20

type  

7

24

×

120

lb

=

35

lb

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Check:

(

35

×

0.2

)

+

(

85

×

0.68

)

=

$

64.8

120

×

0.54

=

$

64.8

Step-by-step explanation:

Q&A Education