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Linear agebra problem

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Linear agebra problem

by Mogorosi » Wed Dec 01, 2010 2:14 pm
the problem below keeps killing me; i want a methodical solution to it.

A company pays project contractors a rate of "a" dollars for the first hour and "b" dollars for each additional hour after the first, where a > b.

In a given month, a contractor worked on two different projects that lasted 2 and 4 hours, respectively. The company has the option to pay for each project individually or for all the projects at the end of the month. Which arrangement would be cheaper for the company and how much would the company save?

A. Per month, with savings of $(a + b)
B. Per month, with savings of $(a - b
C. The two options would cost an equal amount.
D. Per project, with savings of $(a + b
E. Per project, with savings of $(a - b)
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by trangle » Wed Dec 01, 2010 3:11 pm
We can pick numbers to solve this problem. Let's assume a=$2 and b=$1

If the company pays per month, it pays 6 hours total. We have (1hr*$2)+(5hr*$1)=$7

If the company pays each project individually:
The first project: (1hr*$2)+(1hr*$1)=$3
The second project: (1hr*$2)+(3hr*$1)=$5
So the company pays $3+$5=$8

We can clearly see that the company will save $1 (a-b) for paying per month. The correct answer is B.

What you do think?
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by N:Dure » Wed Dec 01, 2010 4:11 pm
trangle wrote:
We can pick numbers to solve this problem. Let's assume a=$2 and b=$1

If the company pays per month, it pays 6 hours total. We have (1hr*$2)+(5hr*$1)=$7
Notice here that there are two distinct projects, one lasting for 2 hrs and the other 4. So by plugging in the numbers, it should be:
1st project + 2nd project together = a (1st hour) +b (extra hour) + a +3b = 2a + 4b= 2($2) + 4($1) = $8

1st project alone: a+b = 3
2nd project: a+3b = 5

Total $8

So I think the answer is C. The two options would cost an equal amount.
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by Rahul@gurome » Thu Dec 02, 2010 12:20 am
Mogorosi wrote:A company pays project contractors a rate of "a" dollars for the first hour and "b" dollars for each additional hour after the first, where a > b. In a given month, a contractor worked on two different projects that lasted 2 and 4 hours, respectively. The company has the option to pay for each project individually or for all the projects at the end of the month. Which arrangement would be cheaper for the company and how much would the company save?

A. Per month, with savings of $(a + b)
B. Per month, with savings of $(a - b)
C. The two options would cost an equal amount.
D. Per project, with savings of $(a + b)
E. Per project, with savings of $(a - b)
If the company pays in per month basis,
  • Total number of work hours = (2 + 4) = 6
    Thus the company pays $(a + 5b)
If the company pays in per project basis,
  • For 1st project it pays $(a + b)
    For 2nd project it pays $(a + 3b)
    Total payment = $(a + b) + $(a + 3b) = $(2a + 4b)
Now, per project payment is greater than per project payment, because [(2a + 4b) - (a + 5b)] = (a - b), which is a positive quantity as a > b.

Thus per month payment is less and cheaper for the company with a savings equal to the difference of those payments, i.e. $(a - b)

The correct answer is B.
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by Rahul@gurome » Thu Dec 02, 2010 12:23 am
Mogorosi wrote:A company pays project contractors a rate of "a" dollars for the first hour and "b" dollars for each additional hour after the first, where a > b. In a given month, a contractor worked on two different projects that lasted 2 and 4 hours, respectively. The company has the option to pay for each project individually or for all the projects at the end of the month. Which arrangement would be cheaper for the company and how much would the company save?

A. Per month, with savings of $(a + b)
B. Per month, with savings of $(a - b)
C. The two options would cost an equal amount.
D. Per project, with savings of $(a + b)
E. Per project, with savings of $(a - b)
If the company pays in per month basis,
  • Total number of work hours = (2 + 4) = 6
    Thus the company pays $(a + 5b)
If the company pays in per project basis,
  • For 1st project it pays $(a + b)
    For 2nd project it pays $(a + 3b)
    Total payment = $(a + b) + $(a + 3b) = $(2a + 4b)
Now, per project payment is greater than per project payment, because [(2a + 4b) - (a + 5b)] = (a - b), which is a positive quantity as a > b.

Thus per month payment is less and cheaper for the company with a savings equal to the difference of those payments, i.e. $(a - b)

The correct answer is B.
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by Rahul@gurome » Thu Dec 02, 2010 5:49 am
Mogorosi wrote:Hold on a second, sir, I am still lost.

the formula here is a+b(x-1) where x is the total number of hours worked, right?

Project 1, with 2 hours yields a payment of a+b; and

Project 2, with 4 hours yields a payment of a+3b.

Please take it from here and show me how you treat individual payment vs combined payment at the end of the month. Thanks in advance.
Yes, that's correct.

Now if the company pay on per project basis, they pays $(a + b) for first project and then $(a + 3b) for the second project, a total of $(2a + 4b). Hope this is clear to you.


But when the company pays per month basis, it will just see how many hours a contractor has worked in that month. In this case which is (2 + 4) = 6 hours. Thus for the month the company will pay $a for the first hour and $b for the next 5 hour. In your words, for the month with 6 hours yields a payment of (a + 5b)

Thus two payments are different. Is that clear to you?
Which of them is smaller and smaller by what amount is already explained in the previous post.
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