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there are 16 workers employed on a highway project, some at $200 per day and some at $165 per day. The daily payroll is $2745. Find the number of workers employed at each wage. could u tell me how u did it too if u just happen to figure this out. Thanx so much!

2007-01-09 09:18:31 · 8 answers · asked by ♥kimbo141232 2 in Science & Mathematics Mathematics

8 answers

Let x be the number making $200 per day and y be the number making $165 per day.

The total number of workers is 16:

x + y = 16

Solve for x:

x = 16 - y

Then, $200 times the number of workers making $200, plus $165 times the number of workers making $165, equals the total daily payroll of $2745:

200x + 165y = 2745

To make the numbers smaller, divide both sides of the equation by 5:

40x + 33y = 549

Then plug in the value for x:

40(16 - y) + 33y = 549

Distribute the 40 through (16 - y):

640 - 40y + 33y = 549

Combine the y's and subtract 640 from both sides:

-7y = -91

Divide by 7 to get y:

y = 13

Then plug that value of y into the x = 16 - y equation to get x:

x = 16 - 13 = 3

So there are 3 workers who make $200 per day, and 13 who make $165 per day.

Double check: 13 + 3 = 16, duh. :c)

13(165) + 3(200) = 2145 + 600 = 2745, check!

floodtl's method is really interesting though...

2007-01-09 09:25:37 · answer #1 · answered by Jim Burnell 6 · 1 0

Well, let's say x equals the number of workers employed for $200/day and y equals the number of workers employed for $165/day. If the company's daily payroll is $2745, that means the company spends that much to pay their laborers' wages every day.

200x + 165y = 2745

Now, to solve an equation with two variables, you need two equations. Since the total number of workers is given to you, you can make this other equation:

x + y = 16

Now, you've got...

200x + 165y = 2745
x + y = 16

You need to get "rid of" a variable. Multiply the entire bottom equation by NEGATIVE 200.

-200(x + y = 16)
-200x - 200y = -3200

Now, you've got...
200x + 165y = 2745
-200x - 200y = -3200
---------------------------
-35y = -455
y = 13

Now, plug this y value into the easiest equation to find x.

x + y = 16
x + 13 = 16
x = 3

That means your answer is...

13 workers employed at $165/day
3 workers employed at $200/day

2007-01-09 17:29:16 · answer #2 · answered by Anonymous · 1 0

You kids have it easy.....imagine if you didn't know how many workers there were? Might make it harder, huh? That kind of problem even has its own name: Diophantine, after Diophantus, the Greek guy who first proposed the problem.

Anyway, there is a really easy way to solve this problem: You know that at least 16*165 = $2640 is spent, so just see how many times you need to give someone a "raise" of 200-165=$35 to get up to $2745.

2007-01-09 20:26:56 · answer #3 · answered by aristotle2600 3 · 0 0

Let a be the number of workers making 200 and b the number of workers making 165.

You have two equations:

200a + 165b = 2745
a + b = 16

You want to modify the second equation so that when it's added to the first, one of the variables disappears. You can't change the value of the second equation, so whatever you do to the left side, you have to do to the right.

In other words, multiply boths sides of the second equation by -200:

200a + 165b = 2745
-200a - 200b = -3200

Add the two equations together:

0 - 35b = -455

Divide both sides by -35

b = 13

Plug 13 into either equation for b and solve for a. The second equation is easiest to work with:

a + 13 = 16
a = 3

2007-01-09 17:31:58 · answer #4 · answered by Bob G 6 · 0 0

it is not impossible. Here is how you would do it.
Let x be the number of workers who make $200 and y be the number of workers who make 165/day.
x + y = 16 (the number of workers)
200x + 165y = 2745 (dollar amount times the number of workers added together gives you the daily total)
Solve for x and y

x+y = 16 'x= 16-y
200x+165y = 2745 'substitute the value for x from above
200 (16-y) +165y = 2745
3200 - 200y + 165y = 2745
-35y = -455
y = 13
Now solve for x
X+ y = 16
x + 13 = 16, therefore, x = 3

2007-01-09 17:27:31 · answer #5 · answered by QnA 2 · 1 0

Algebra will handle this; there are two unknowns, High-paid workers, and Low-paid workers. We are given:
H + L = 16;
200L + 165H = 2745.
The usual algebraic substitution will solve this easily.

2007-01-09 17:31:37 · answer #6 · answered by Anonymous · 0 0

x is the number of workers getting $200
y is the number of workers getting $165

x+y=16
so
x=16-y

200x+165y=2745

substitute value for x

200(16-y)+165y=2745
solve:
3200-200y+165y=2745
3200-35y=2745
455-35y=0
13-y=0
y=13

x=16-y
x=16-13
x=3

check:
200x+165y=2745
600+2145=2745

hooray!

2007-01-09 17:33:07 · answer #7 · answered by sudonym x 6 · 0 0

200x+165y=2745
x+y=16
y=16-x
substitute- 200x+2640-165x=2745
35x+2640=2745
35x=105
x=3
3+y=16
y=13
x=3
3 at $200
13 at $165

2007-01-09 17:29:40 · answer #8 · answered by Anonymous · 1 0

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