The Algebra Word Problem That Stumps Everyone – Can You Solve It?

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[gavindeane]

Dan can do two driveways in an hour and Jay can do three driveways in an hour.

So together they can do five in an hour. But they don't need to do five, they only need to do one. So that takes one fifth of an hour, which is 12 minutes.

[KenFullman]

Dan can do it twice in an hour, while while Jay can do it three times in an hour. So between them they can do the job 5 times in an hour. So since it only needs doing once, that'll be 12 minutes.

[tomtke]

You will find a wealth of problems when calculating series/parallel combinations of resistors. Current flow is similar to work rates.

When two resistors are in parallel their equivalent resistance is:

1/ [ 1/R.1 + 1/R.2 ]

this problem:
1/ [ 1/30min + 1/20min ]
1/ [ 20/600 + 30/600 ]
1/ [ 50/600 ]
600 / 50
60/5
12 min

[dondonny]

My approach was imagining Dan could shift 1 ton of snow in 30mins, ie 2 tons per hour, and Jay 3 tons per hour. That totals 5 tons per hour working together. 1 ton between them would therefore take 12 mins (60mins/5)

[Like_An_Eagle]

This type of problem ALWAYS gets me!!

My line of thought...
In 10 minutes one guy gets 1/2 the driveway done and the other gets 1/3 done. That's 5/6 (83.333%) of the driveway shoveled in 10 minutes.

Then I turn it into a proportion. If 10 min=83.333...% then "X" min=100%. (10/83.3=X/100)

You cross multiply and then divide to solve for "X", and you get X=12.

So, it takes 12 minutes for both guys to get 100% of the driveway shoveled.

This is probably the long way to solve the problem. But I'm pretty sure it's correct and it makes sense to me. 😁

That's the beauty in math. There are many roads to the same solution. 💯

[danielparsons]

I worked this out by surmising that Dan can do half of the driveway in 10minutes and in that same time Jay could shovel a third of the driveway. That means they can do 83.3333% together in 10 minutes. Divide it by 10 and you see they can do 8.333% of the driveway in a minute. Therefore 12*8.333 = 100% approximately.

[TheViennese]

That is what I got, doing it in my head. 12:30 half of 20 being 10, half of 30 being 15, so add 10 plus 15, and then divide by 2

[clmkc]

Find common denominator of 60, equals 2/60 + 3/60 = 5/60. Invert the fraction, 60/5 = 12 minutes.

[Stylux-zp]

Dan --> 30 min -> Dan's work rate : 30^-1
Jay --> 20 min -> Jay's work rate : 20^-1
working together :
D + J = [30^-1 + 20^-1]^-1
(1 ) / (1/30 + 1/20)
(1 ) / (2/60 + 3/60)
(1 ) / (5/60) = 1/ 1 · 60/5 = 12 min✅
Its like when calculating the total resistance (Rt) of parallel connected resistors

[Paughco]

Reminds me of the problem where you're flying an airplane that has a cruising airspeed of 100 mph. You are flying to a waypoint directly into a headwind of 25 mph. The waypoint is 50 miles away, then you turn around and fly back to your home field. How long will you be flying? The intuitive answer is wrong, because you'll spend way more time flying into the headwind with a 75 mph groundspeed than you'll spend on your return, which will be with a 125 mph groundspeed. This would take about 63.5 minutes. Sorry - I didn't do it with algebra. I used my AN5835-1 dead reckoning computer (precedes the E6-B).

[mikehopkins]

Remember this from 9th grade algebra class in 1972-73 school year. CR Herpitch was a great math teacher.

[eric]

It would have been far better if you had spent the time explaining how you arrived at the equation 1/t = 1/P1 + 1/P2,
which is what the question is really about - rather than spending time on simple algebra.
Just quoting a formula imo is not constructive.
How many students go on to just blindly applying formulas to solve things.. without ever understanding them ( good example is the quadratic formula)

[vincentrobinette]

I did it slightly different. Product/sum. 30x20=600
30+20=50
600/50=12.

[gottadomor]

Fun problem.
Every minute, 20 min guy does 1/20th of the job; 30 minute guy does 1/30th. Oui?
60 mins in an hour. converting: 3/60th & 2/60th, separately, together/combining, 5/60th of the job per minute. Oui?
Or 1/12th in one minute. Oui?
12/12ths in 12 minutes. Oui? Tres bien. ;-)

[doughoffman]

Dimensional analysis:
| 1dw 1dw |
| + | Xmin = 1 dw Solve for Xminutes, so divide both sides by the bracketed quantity on the left of Xmin
| 30min 20min |

1dw
Xmin = = 12 min
( 1/30 + 1/20 )

[davidwilliams]

Similar to the hole digging problem but I forgot how to solve, so thanks for this one.

[prestwig]

12 minutes 1/5 of an hour Dans rate is 2 shoveled payments per hour Jays rate is 3 shoveled payments per hour. working together the formula is (Jays rate x t)+ (Dans rate x t)= 1 shoveled payment. So 3t+2t=1 5t=1 t=1/5 hours=12 minutes

[stevendebettencourt]

There’s a simple way to solve problems like this, so let’s go!

Having read the problem, here are the important parts:

-1.) Dan can shovel a driveway in 30 minutes.
-2.) Jay can shovel the same driveway in 20 minutes.
-3.) Calculate how long to shovel if they work together.

Alright, if you have a list of t people who can do a task in x amount of time, the amount of time y it would take if they all work together is given by:

(1/x_1) + (1/x_2) + … + (1/x_t) = 1/y

It’s actually pretty easy to understand why this works. Let’s return to our example:

1/30 min + 1/20 min = 1/y min

If Dan takes 30 minutes to shovel, in one minute, he would complete 1/30th of the task. Jay does it in 20 minutes, so one minute gets him 1/20th of the task. Add these two together is what fraction of the task is done in one minute. Let’s get a common denominator on the left:

(1*2)/(30*2) + (1*3)/(20*3) = 1/y
(2/60 min) + (3/60 min) = 1/y min
(2+3)/60 min = 1/y min
5/60 min = 1/y min

You could simplify the left, but this will be done regardless after we cross multiply:

5y min = 1*60 min
Y min = 60 min/5
y min = 12 minutes

And 12 minutes makes sense as it is less than the 20 Jay takes alone and less than the 15 you’d need with two Dans, but more than the 10 it would take two Jays.

Any questions?

[RobertWhite]

I just took the LCM of 20 and 30 and supposed the driveway was 60 feet. This means Dan can do 2 feet per minute and Jay can do 3 feet per minute, both together doing 5 feet per minute at the same time, finishing the job in 12 minutes.

[yrb]

That presumes that the two workers continue to work at maximum pace until the entire job is done. However if the job is divided in equal halves this will not be the case and the total job would be completed in 15 minutes.