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marusya05 [52]
3 years ago
14

The average of three different positive integers is 8. What is the largest integer that could be one of them.

Mathematics
1 answer:
Trava [24]3 years ago
7 0
Supposing that the integers are x, y and z, and that their average is 8, then

x+y+z
-------- = 8, or x + y + z = 24.
    3

Let's experiment.    Making up a table, we get:

x   y   z   sum (sum MUST be 24)
5  9  10    24
4  9   11    24
3  10  11   24
2   11 11    24
1   11  12   24
1   10  13   24
1    1   22   24

Can't choose integers smaller than 1, so it appears that 22 is the largest possible integer that could be one of x, y and z.
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Kyle works at a donut​ factory, where a​ 10-oz cup of coffee costs 95¢​, a​ 14-oz cup costs​ $1.15, and a​ 20-oz cup costs​ $1.5
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kyle works at a donut shop, where a 10oz. cup of coffee costs 95c, a 14 oz. cup of coffee costs $1.15, and a 20oz. cup costs $1.50. during one busy period Kyle served 24 cups of coffee, using 384 ounces of coffee, while collecting a total of $30.60. How many cups of each size did kyle fill.

Let x = no. of 10 oz cups sold

Let y = no. of 14 oz cups sold

Let z = no. of 20 oz cups sold

Equation 1: total number of cups sold:

x + y + z = 24

Equation 2: amt of coffee consumed:

10x + 14y + 20z = 384

Equation 3: total revenue from cups sold

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Mult the 1st equation by 20 and subtract the 2nd equation from it:

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------------------------ subtracting eliminates z

10x + 6y = 96; (eq 4)

Mult the 1st equation by 1.5 and subtract the 3rd equation from it:

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.55x + .35y = 5.40; (eq 5)

Multiply eq 4 by .055 and subtract from eq 5:

.55x + .35y = 5.40

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0x + .02y = .12

y = .12/.02

y = 6 ea 14 oz cups sold

Substitute 6 for y for in eq 4

10x + 6(6) = 96

10x = 96 - 36

x = 60/10

x = 6 ea 10 oz cups

That would leave 12 ea 20 oz cups (24 - 6 - 6 = 12)

Check our solutions in eq 2:

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