The distance from Rhetwicks house to the laundromat is exactly 5 mile

Question

The distance from Rhetwick’s house to the laundromat is exactly 5 miles, and he has only two options for transit to and from the facility, either to walk the distance at 2 miles per hour or drive at 20 miles per hour, but should he choose to drive, he must pay $5 per half hour of parking.

Provided that Rhetwick sets his opportunity cost for doing laundry at the $15 per hour he could have earned performing online exam proctoring, that he allocates a maximum of 12 hours for completing his laundry, and that each load of laundry takes an hour to complete, select the maximum number of laundry loads that would make Walking more cost effective for Rhetwick and the maximum number of laundry loads that would make Driving more cost effective for him. 
 Make only two selections, one in each column.

Deepakjhamb · Accepted Answer

lets make the equations here:

lets take the case of walking now speed = 2miles per hour , total d = 5*2 = 10 miles so time for commuting = d/s = 10/2 = 5 hours

now maximum hours that can be allocated for completing laundry = 12 hours , out of which 5 is gone in walking commute alone so we have hours left to be devoted for laundry as the most = 12-5 = 7

as 1 load takes 1 hour we can have atmost 7 loads of laundry from walking case analysis

now lets take drive analysis , d= 10 , s= 20 m/h so time = 10/20 = .5 hrs

now as each load takes 1 hour to finish , in 12-.5 = 11.5 hrs we can have at most 11 loads done

but as we need to compare both driving and walking we have loads <=7 from walking and <=11 from driving

so definitely taking common area loads < = 7 overall

now lets equate costs

walking cost = oppt cost of laundry * time spent commuting when walking

driving cost = oppt cost of laundry * time spent commuting when driving

let x be the maximum laundry loads while walking which is cost effective than driving then we have

15 *5 + x*15 < .5*15 + x*15 + 5*2 *x ( each half an hour 5 $ charge , so while he does laundry in driving case he needs to shellout 2* 5* x )

so we get 75 + 15x < 7.5+15x+10x

or we get 25x-10x > 75-7.5

so 10 x > 67.5

or x> 6.75

now x has a maximum value = 7 , so for walking case its cost effective if we have 7 loads

now for driving case being cost effective

eq will be 75 + 15x > 7.5+15x+10x

we get 25x- 15x < 75 -7.5

10x < 67.5

x< 6.75

so x max = 6

so for driving cost effective we can have maximum of 6 laundry loads

we can check also , if we have 6 laundry loads then cost for walking laundry is 75 +15*6 = 165

in case we of driving for 6 laundry loads we have cost = .5*15+15x+10x = 7.5+ 25x = 7.5 + 150= 157.5

so 157.5 < 165

now lets take x = 7 , then we have cost while walking = 75 +15*7 = 180

cost while driving = 25x+7.5 = 25*7+7.5 = 182.5

so cost for driving > cost for walking for 7 loads

but we cannot have greater than 7 loads of laundry when comparing walking and driving as maximum time for laundry would be exceeded

so answer is

maximum loads for laundry for which cost while walking is more cost effective = 7 laundry loads

maximum loads for laundry for which cost while driving is more cost effective = 6 laundry loads

Sajjad1994 · Answer

This is a hard question, OA is: Walking:7 Driving: 6, good work  Deepakjhamb

Good Luck guys for today’s questions, I will post the points table in an hour or two.

Thank you

TarunKumar1234 · Answer

Given, The distance from Rhetwick’s house to the laundromat is exactly 5 miles, and he has only two options for transit to and from the facility, either to walk the distance at 2 miles per hour or drive at 20 miles per hour, but should he choose to drive,  he must pay $5 per half hour of parking.

Provided that Rhetwick sets his  opportunity cost for doing laundry at the $15 per hour  he could have earned performing online exam proctoring, that he allocates a  maximum of 12 hours for completing his laundry , and that each load of laundry takes an hour to complete, select the maximum number of laundry loads that would make Walking more cost effective for Rhetwick and the maximum number of laundry loads that would make Driving more cost effective for him. Make only two selections, one in each column.

Let's solve it, Total distance for up and down = 5+5= 10 miles
Time consumed in walking = Time = Distance/Speed = 10/2 = 5 hrs.
Time consumed in driving = Time = Distance/Speed = 10/20 = 0.5 hrs.

Total time available for Laundry = 12 hrs. and there is an opportunity cost = $15 per hr.
So, he should save maximum time from Laundry time, balancing others cost, if any (parking cost for driving)

Time saved from Laundry time, if he walks = 12-5= 7 hrs and Time saved from Laundry time, if he drives = 12-0.5= 11.5 hrs

If Laundry load = 3; Laundry time = 3 hrs. 
If he walks, time remained for other purpose = 7-3 = 4 hrs.; Money earned = 4*15 = $60
If he drives, time remained for other purpose = 11.5-3 = 8.5 hrs; Money earned = 8.5*15 -3*2*5 =$97.5

If Laundry load = 4; Laundry time = 4 hrs. 
If he walks, time remained for other purpose = 7-4 = 3 hrs.; Money earned = 3*15 = $45
If he drives, time remained for other purpose = 11.5-4 = 7.5 hrs; Money earned = 7.5*15 -4*2*5 =$72.5

If Laundry load = 5; Laundry time = 5 hrs. 
If he walks, time remained for other purpose = 7-5 = 2 hrs.; Money earned = 2*15 = $30
If he drives, time remained for other purpose = 11.5-5 = 6.5 hrs; Money earned = 6.5*15 -5*2*5 =$47.5

If Laundry load = 6; Laundry time = 6 hrs. 
If he walks, time remained for other purpose = 7-6 = 1 hrs.; Money earned = 1*15 = $15
If he drives, time remained for other purpose = 11.5-6 = 5.5 hrs; Money earned = 5.5*15 -6*2*5 =$22.5

If Laundry load = 7; Laundry time = 7 hrs. 
If he walks, time remained for other purpose = 7-7 = 0 hrs.; Money earned = 0*15 = $0
If he drives, time remained for other purpose = 11.5-7 = 4.5 hrs; Money earned = 4.5*15 -7*2*5 = -$2.5 (negative)

If Laundry load = 8; Laundry time = 8 hrs. 
If he walks, time remained for other purpose = 7-8 = Not possible
If he drives, time remained for other purpose = 11.5-8 = 3.5 hrs; Money earned = 3.5*15 -8*2*5 = -$27.5 (negative)

So,  Best case for driving will be "3" loads of Laundry . (as maximum earnings)
And,  Best case for walking will be "7" loads of Laundry . (as loss from driving)

rocky620 · Answer

Maximum Opportunity Cost = Maximum hour completing his laundry*Amount required for each load.

Maximum hour completing his laundry = 12
Amount Required for each load is fixed = 15

Should he chose to Walk:

Distance = 5+5 (to and fro) = 10
Speed = 2 mph
Transit Time =  \frac{Distance}{Speed}  =  \frac{10}{2}  = 5 hrs

So the time remaining with him for doing laundry = 12-5 = 7 hours.
So he can do maximum of 7 loads should he choose to walk.

Should he chose to drive:

Distance = 5+5 (to and fro) = 10
Speed = 20 mph
Transit Time =  \frac{Distance}{Speed}  =  \frac{10}{20}  =  \frac{1}{2}  hrs

So the time remaining with him for doing laundry = 12- \frac{1}{2}  = 11.5 hours. But Each load takes an hour to complete so he will have only 11 hours with him.

But he will have to pay 10$ per hour for parking, in addition to the 15$ cost for per load of laundry.

Let the maximum hours he can spent doing laundry be x hours (x is an integer)

Parking Cost = 10x
Laundry Cost = 15x
Total Cost = 10x+15x = 25x

So, 25x = 180 (maximum amount that can be spent) = 7.25

But since x has to be an integer so it will 7.

So he can do maximum of 7 loads should he choose to drive.

IMO 7 loads for both

Bambi2021 · Answer

This is a good question. And quite funny as well. Imagine this guy walking 2,5 hrs with laundry enough to have the washing machines working for 12 hrs. I would say walking would never be more cost effective as he will be so tired from carrying everything that he wont have energy left for anything else than sleeping, if he finally makes it home.

Joking aside, I reacted to the wording in the prompt. "The maximum loads of laundry that will make walking/driving more cost effective". More cost effective than what? The maximum number of loads that will make walking  more  cost effective than driving is 12 hrs. I think the wording should be something like "the number of loads that is most cost effective for either of the transit alternatives".

hrithiksindwani · Answer

Feel free to correct me but this is where IMO some questions fail to draw the boundaries of how much logic should one apply.

The Question talks about Opportunity cost i.e. Cost of the next best opportunity or money not earned by not choosing to be an exam practitioner in the laundry hours.

Now, as per the answer we got, on 7 Loads Driving becomes expensive the total cost is 182.5. But why did we all choose to not account for the money he makes in the time saved i.e. about 4.5 hours. His opportunity cost for the 4.5 hours is more than $60. On the other hand, he spends the entire 12 hours in the Laundry (travel + the deed).

Any rational and logical man thinking about Opportunity cost will not walk if the concept of opportunity cost is in his mind. This is where I think are we all finding logic or adjusting the answers to the given solution.

Yes, I know the question does not state he will use the remaining hours to make money but speaks loudly about Opportunity cost without even using the concept which made my brain overthink the question more than I should have.

pudu · Answer

how it is possible to get 0.5 hr for driving? it is 30 minutes. so available time is 11 hr and 30 minutes...not 11 hr and 50 minutes

pudu · Answer

for walking you need 5 hours for round trip. So now you have 12-5=7 hrs with you.

15x<=7*15
x=7

For driving you need 30 minutes. Now you have 11 hours and 30 minutes with you. You need 1 hr for one lot laundry work. So your maximum cost is 15*11=165.

15x+10x(1 hour parking cost)<= 165.
x<=165/25=33/5= 6 hr 36 minutes.

so 6 hour

praveengmat4 · Answer

Don't you think this is too hard to solve it in <2.5 minutes

haianh · Answer

Q's is tough as details are easy to miss.
 
 Distance: 5 miles 
 Walking speed: 2m/h or 5 hours round trip. 
 Driving speed: 20m/h or 0.5 hour round trip.  
 Cost per hour: standard $15/hour + $10/hour if parking.

Total cost:  
 
 Walking: 5*15 + 15*loads = 75 + 15L 
 Driving: 0.5*15 + 15*loads + 10*loads = 7.5 + 25L

Compare cost efficiency:  
 
 Walking better: 75 + 15L < 7.5 + 25L  \Leftrightarrow  6.75 < L. Total time is capped at 12 so for walking, max loads can only be 7. 
 Driving better: opposite to walking, meaning L < 6.75, thus max = 6.

The distance from Rhetwicks house to the laundromat is exactly 5 mile

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The distance from Rhetwicks house to the laundromat is exactly 5 mile

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