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Elodia [21]
3 years ago
6

You have one type of candy that sells for $2.50/lb and another type of candy that sells for $8.20/lb. You would like to have 17.

1 lbs of a candy mixture that sells for $3.50/lb. How much of each candy will you need to obtain the desired mixture?
you will need
___lbs of the cheaper candy
___lbs of the expensive candy
Mathematics
1 answer:
Ilya [14]3 years ago
8 0

Answer: you will need 14.1 lbs of the cheaper candy 3 lbs of the expensive candy.

Step-by-step explanation:

Let x represent the number of pounds of the cheaper candy that you will need.

Let y represent the number of pounds of the expensive candy that you will need.

You would like to have 17.1 lbs of the candy mixture. It means that

x + y = 17.1- - - - - - - - -1

The mixture would sell for $3.50/lb. It means that the total cos of the mixture would be

17.1 × 3.5 = $59.85

If the cheaper candy sells for $2.50/lb and the expensive candy sells for $8.20/lb, it means that

2.5x + 8.2y = 59.85- - - - - - - - - 1

Substituting x = 17.1 - y into equation 1, it becomes

2.5(17.1 - y) + 8.2y = 59.85

42.75 - 2.5y + 8.2y = 59.85

- 2.5y + 8.2y = 59.85 - 42.75

5.7y = 17.1

y = 17.1/5.7

y = 3

x = 17.1 - y = 17.1 - 3

x = 14.1

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