# Rule of three (?) problem

#### ss_0001

##### New member
Hi,

The airport workers Jurgen, Manfred and Phillip are assigned to load luggage onto the aircraft. If Jurgen does the job alone he would need 4hrs 20mins. Manfred would need 5hrs 40mins to do it on his own, and Phillip 6hrs. How long does it take if they all load the aircraft together?

Is there a simple way to do this? (I have to work without a calculator) I tried setting Jurgen to 1.0, and then worked out the others relative to him, but it got quite messy.

The answer shown was 1hrs 45mins.

#### Subhotosh Khan

##### Super Moderator
Staff member
ss_0001 said:
Hi,

The airport workers Jurgen, Manfred and Phillip are assigned to load luggage onto the aircraft. If Jurgen does the job alone he would need 4hrs 20mins. Manfred would need 5hrs 40mins to do it on his own, and Phillip 6hrs. How long does it take if they all load the aircraft together?

Is there a simple way to do this? (I have to work without a calculator) I tried setting Jurgen to 1.0, and then worked out the others relative to him, but it got quite messy.

The answer shown was 1hrs 45mins.
First get all the times in minutes (or fractional hours)
4 hrs 20 min = 260 min = 13/3 hrs
5 hrs 40 min = 340 min = 17/3 hrs
6 hrs = 360 min

Then

in 1 min Jurgen loads 1/260 part of luggage

in 1 min Manfred loads 1/340 part of luggage

in 1 min Phillip loads 1/360 part of luggage

Now what fraction of luggage will be loaded - when they work together = 1/260 + 1/340 + 1/360

This calculation is messy - specially without calculator. But can be done...

#### Denis

##### Senior Member
Subhotosh Khan said:
in 1 min Jurgen loads 1/260 part of luggage

** in 1 min Manfred loads 1/340 part of luggage

** in 1 min Phillip loads 1/360 part of luggage
** oh oh !

To student: are you sure you posted correctly?
Answer will be 104.5466491458.... minutes: close but not quite 1 3/4 hr.

#### soroban

##### Elite Member
Hello, ss_0001!

The airport workers Jurgen, Manfred and Phillip are assigned to load luggage onto the aircraft.
If Jurgen does the job alone he would need 4 hrs 20 mins. Manfred would need 5 hrs 40 mins to do it on his own, and Phillip 6 hrs.
How long does it take if they all load the aircraft together?

The answer shown was 1 hr, 45 mins. . Approximately!

$$\displaystyle \text{Jurgens can do the job in }4\tfrac{1}{3} \,=\,\tfrac{13}{3}\text{ hours.}$$
$$\displaystyle \text{In one hour, he can do: }\:\frac{1}{\frac{13}{3}} \,=\,\tfrac{3}{13}\text{ of the job.}$$
$$\displaystyle \text{In }x\text{ hours, he can do: }\:\tfrac{3x}{13}\text{ of the job.}$$

$$\displaystyle \text{Manfred can do the job in }5\tfrac{2}{3} \,=\,\tfrac{17}{3}\text{ hours.}$$
$$\displaystyle \text{In one hour, he can do: }\:\frac{1}{\frac{17}{3}} \,=\,\tfrac{3}{17}\text{ of the job.}$$
$$\displaystyle \text{In }x\text{ hours, he can do: }\:\tfrac{3x}{17}\text{ of the job.}$$

$$\displaystyle \text{Philip can do the job in }6\text{ hours.}$$
$$\displaystyle \text{In one hour, he can do: }\:\tfrac{1}{6}\text{ of the job.}$$
$$\displaystyle \text{In }x\text{ hours, he can do: }\:\tfrac{x}{6}\text{ of the job.}$$

$$\displaystyle \text{Working together for }x\text{ hours, they can do: }\:\tfrac{3x}{13} + \tfrac{3x}{17} + \tfrac{x}{6}\text{ of the job.}$$

$$\displaystyle \text{But in }x\text{ hours, we expect them to compete the job (one job).}$$

$$\displaystyle \text{There is our equation: }\;\;\frac{3x}{13} + \frac{3x}{17} + \frac{x}{6} \;=\;1$$

$$\displaystyle \text{Multiply by the LCM,}\,1326\!:\;306x + 234x + 221x \:=\:1326 \quad\Rightarrow\quad 761x \:=\:1326$$

. . $$\displaystyle x \;=\;\frac{1326}{761} \;=\;1.742444152\text{ hours} \;=\;1\text{ hr, }44.54664915\text{ min} \;\approx\;1\text{ hr, }44\text{ min, }32.8\text{ sec}$$