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Elis [28]
3 years ago
9

A metal company's old machine makes bolts at a constant rate of 100 bolts per hour.The company's new machine makes bolts at a co

nstant rate of 150 bolts per hour. If both machines start at the same time and continue making bolts simultaneously, how many minutes will it take the two machines to make a total of 300 bolts?
A. 36
B. 72
C. 120
D. 144
E. 180
Mathematics
2 answers:
zloy xaker [14]3 years ago
7 0

Answer:72minutes B

Step-by-step explanation:

100t + 150t =300

250t =300

t=300/250

t=6/5 hours

Converting to minutes

6/5×60=72 minutes

weqwewe [10]3 years ago
4 0

Answer:

B

Step-by-step explanation:

Let’s label the old machine as A while the new machine be labeled B

Firstly we establish the individual rates which have been given in the question:

A = 100 bolts/hour

B = 150 bolts/hour

Now, number of bolts made is rate in bolts/hour * time taken

Since we do not know the time taken, let the time taken be represented by x.

Since they are working together, the number of bolts produced by A in x hours would be 100x while the number of bolts produced by B in x hours is 150x.

We add both together to give the total number of bolts since they are working together.

The total number of bolts is thus: 150x + 100x. We have been given this total value to be 300. Hence:

300 = 150x + 100x

300 = 250x

x = 300/250

x = 1.2 hours

Since 1 hour is 60 minutes, 1.2 hours is same as 1.2 * 60 = 72 minutes

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Answer:

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Step-by-step explanation:


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Step-by-step explanation:

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Savatey [412]

Maximizing profit, is a way of getting the highest possible profit, from a function.

The bakery should make 45 loaves of A and 0 loaves of B, to maximize profit

To do this, we make use of the following representations.

x represents source A, and y represents source B

So, we have:

<u>Constraint 1: </u>

A uses 5 pounds, and B uses 2 pounds of oats.

Available: 180

The above condition is represented as;

\mathbf{5x + 2y \le 180}

<u>Constraint 2: </u>

A and B use 3 pounds of flour each.

Available: 135

The above condition is represented as;

\mathbf{3x + 3y \le 135}

<u>Objective function</u>

A yields $40, while B yields $30

So, the objective function is:

\mathbf{Maximize\ Z = 40x + 30y}

So, we have:

\mathbf{Maximize\ Z = 40x + 30y}

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\mathbf{5x + 2y \le 180}

\mathbf{3x + 3y \le 135}

\mathbf{x,y \ge 0}

See attachment for the graph of the subjects

From the graph, we have the corner points to be:

\mathbf{(x,y) = \{(0,45),(30,15),(45,0)\}}

Substitute these values in the objective function

\mathbf{Z = 40(0) +30(45) = 1350}

\mathbf{Z = 40(30) +30(15) = 1650}

\mathbf{Z = 40(45) +30(0) = 1800}

The maximum value of Z is at: (45,0)

This means that: the bakery should make 45 loaves of A and 0 loaves of B, to maximize profit

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3.485

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3.485

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