Work Rate, word problem

Question

I have been working on this problem for almost a week now and I just cant seem to get it.

 

Math department development committee members Alan, Amy,Jessica and Julia were preparing for a conference. They had to pack 1,200 bags of conference material, and they expected to finish packing within the number of hours for which they had budgeted. Jessica was the fastest and could pack X in one hour. Amy could pack 5 bags less then Jessica could, Julia could pack 10 less then Jessica and Alan could pack 5 less then Julia. 

One hour into packing Alan got a call and had to leave. Jessica left when the budgeted time was up, because she had a class to teach. Amy and Julia stayed over 2 hours to finish the job. How many hours were budgeted for them to pack?

 

Answer 1

First, let’s draw a diagram (not to scale) of the labor hours on the project:

              Start           1 hr                                  Budgeted time (t)                    (t+2) hours
Alan          |=======|
Amy          |=============================================|
Jessica      |===========================|
Julia         |=============================================|

The problem asks you, “How many hours were budgeted for them to pack?” We called that value t.

Now, before we can find the time t, we must find how many bags each person packed (to total 1200). This is the usual “trick” of rate problems. The very important relationship to remember for this problem is:

                    (unit rate of bags per hour)*(number of hours) = (number of bags)

Put each person’s (unit rate)*(hours) into that formula:

                  Rate        Time                bags
Alan           (x-15)         1              (x-15)*1             (5 bag/hr less than Julia)
Amy           (x-5)       (t+2)           (x-5)(t+2)           (5 bag/hr less than Jessica)
Jessica          x             t                 x*t                    (x in 1 hr)

Jula           (x-10)     (t+2)            (x-10)(t+2)          (10 bag/hr less than Jessica)

The problem states: “pack 1,200 bags,” so:

   Alan’s bags +    Amy’s bags    + Jessica’s bags + Julia’s bags = 1200

    (x-15)*1   +    (x-5)(t+2)      +     xt             + (x-10)(t+2) = 1200      [eq.A]

There is also a very important piece of information somewhat hidden in the statement, “they expected to finish packing within the number of hours for which they had budgeted.” That means:

    (x-15)*t   +     (x-5)t    +   xt   +     (x-10)t   =  1200          [eq.B]
This equation represents everyone working for t hours (that is, the budgeted time) at each person’s own rate will get the 1200 bags packed.

Let’s solve for x in terms of t (to eliminate x by setting equations equal), then, finally, solve for t):
Using eq.A:

    x - 15   +   xt + 2x - 5t - 10   +   xt   +   xt + 2x - 10t - 20 =1200

          3xt - 15t + 5x – 45 = 1200

          3xt - 15t + 5x = 1245

          x(3t + 5) - 15t = 1245

          x(3t + 5) = 1245 + 15t

          x = (1245 + 15t)/(3t+5)         [eq.C]

Also (using eq.B)

       xt - 15t   +   xt - 5t   +   xt   +   xt - 10t = 1200

      4xt - 30t = 1200

      t(4x-30) = 1200

      4x-30 = 1200/t

      2x-15 = 600/t

      2x = 600/t + 15

      x = 300/t + 15/2            [eq.D]

Set them equal:

    x = (1245 + 15t)/(3t+5) = 300/t + 15/2

       2t(1245 + 15t) = (2)(3t+5)(300) + (t)(3t+5)(15)         [multiply by LCM=2t(3t+5)]

     2490t + 30t^2 = 1800t + 3000 + 45t^2 + 75t

     -15t^2 + 615t - 3000 = 0              [collect terms]

      15t^2 - 615t + 3000 = 0

      t^2 - 41t + 200 = 0                    [divide by 15] 

      (t-35.3)(t- 5.7) = 0                  [use quadratic formula]

       Either   t = 35 1/3   or    t = 5 3/4        [rounded]

Use eq.D (it’s easiest) to determine x:

    x = 300/(35 1/3) + 15/2

    x = 16 [rounded; Jessica can make 16 bags per hour]

In 5 3/4 hours [again, rounded], the number of bags packed is:

              Rate               Time              bags
Alan          1                   1                   1            (5 bag/hr less than Julia)
Amy         11                7 3/4             85 1/4      (5 bag/hr less than Jessica)

Jessica     16                5 3/4             92

Jula           6                7 3/4             46 1/2       (10 bags/hr less than Jessica)

                                      TOTAL      224 3/4 bags
Sorry, that’s not 1200 bags.

Use eq.D again to determine x:

    x = 300/(5 3/4) + 15/2            [make it 5 3/4 hours]

    x = 59.7                         [round; Jessica can make 59 3/4 bags per hour]

In 5 3/4 hours [again, rounded], the number of bags packed is:

                     Rate       Time            bags
Alan               44 3/4      1                44 3/4           [rounded]

Amy               54 3/4    7 3/4          424 1/4           [rounded]

Jessica           59 3/4    5 3/4           343 1/2           [rounded]

Julia              49 3/4     7 3/4          385 1/2           [rounded]

                                   TOTAL       1198 bags

1198 bags is close to 1200 bags, so there may be a calculation error (not just rounding) here, but this is the lengthy process.

 

The team planned to work 5 3/4 hours.

Answer 2

This is an involved problem with 2 sets of variables. First off:

 

Alan = x-15

Amy = x-5

Julia = x-10

Jessica = x

 

So in the first hour, the four of the made 4x-30 bags.

Alan dropped out for the remaining given time, they packed at a rate of 3x-15 per hour.

Then Amy and Julia worked for 2 extra hours which can be represented by 2(2x-15)

 

And then they packed all 1200 bags.

 

Now let T be the alloted time so that t-1 is the time the three worked after Alan left.

So you get:

 

(4x-30) + (t-1)(3x-15) + 2(2x-15) = 1200 

1st hour + middle time + last 2 hours

 

Simplifying, you get

8x-60 + 3tx-15t-3x+15 = 1200

5x-45 + 3tx-15t           = 1200

 

Now I solved for x

5x + 3tx                      = 1245 + 15t

x(5+3t)                       = 1245 + 15t

x                                 = (1245+15t)/(5+3t)

 

Since x was the number of bags they could do in an hour, I assumed that x was a whole number. In order for that expression to be a whole number, since the numerator is a multiple of 5, the denominator must also be a multiple of 5. In order for the denominator to be a multiple of 5, t must be a multiple of 5.

So  if t=5, x=66 I assume this is the answer but I tried other multiple of 5

t=10 leads to x=39.8

t=15 leads to x=29.4

t=20 leads to x=23.76

 

So I am fairly convinced that t=5 and x = 66 is the answer you are looking for. Let's check.

 

Alan=51

Amy = 61

Julia = 56

Jessica = 66

 

All worked for the first hour so they made ....................................234 bags.

Alan left and we have 4 hours left...so 4* (61+56+66)...................732 bags

2 hours overtime for amy and julia so 2(61+56)............................234 bags

 

For a total of 1200 bags.

 

Hope this helped