/sci/ - Science & Math

[Anonymous No.3713869]

Jack and John have been employed to paint a room in the house. Jack estimates he could paint it alone in 3 hours, whilst John estimates it would take him 2 hours to do it alone. How long would it take if Jack and John painted it together?

How do I work this out?

[Anonymous No.3713876]

>mfw 90% of Muslims get this question wrong

[Anonymous No.3713872]

>mfw all Muslims worship a monolith

[Anonymous No.3713880]

muslims have taken episode three

[Anonymous No.3713894]

We don't know, because we're not sure if Jack and John's estimates are accurate.

[Anonymous No.3713905]

Do your own homework.

John starts painting from left to right.

Jack starts painting from right to left.

They will not meet in the middle.

Use the size of the room (20 ft across x 4 walls) and the rate at which they paint to draw a diagram and sketch out the problem.

[Anonymous No.3713906]

>>3713894
Assume they are.

[Anonymous No.3713912]

>>3713894
> We don't know
> Some psuedointellectual bullshit
> I can't answer this basic arithmetic question

[Anonymous No.3713916]

>>3713905
It needs to be solved using algebra with working shown. I'm not sure if I can assume the room has a set size because in the working it will be shown as a variable since it wasn't included in the question. Basically, I'm trying to do my own homework and I'm stumped and could use some help.

[Anonymous No.3713928]

Shit, to be honest, I forgot how to do these types of problems

[Anonymous No.3713930]

VJack = 1 room / 3 hours = 1/3 rooms per hour
VJohn = 1 room / 2 hours = 1/2 rooms per hour

VJack + VJohn = 1/3 + 1/2 = 5/6 rooms per hour

T(Jack + John) = 1 room / (5/6 rooms per hour)
= 6/5 hours.

[Anonymous No.3713943]

>>3713928
> implying you need to be taught a method for every distinct mathematical problem
> not just use common sense
> IQ 80 or lower

[Anonymous No.3713945]

>>3713930
Aha! So obvious now.

[Anonymous No.3713955]

>>3713912

I just enjoy pointing out technical mistakes in a question.

It's a bad habit, I know.

[Anonymous No.3713969]

>>3713930
i'm really dense.

I get everything up until
>T(Jack + John) = 1 room / (5/6 rooms per hour)
>= 6/5 hours.

Why did you invert the 5/6?

[Anonymous No.3713971]

>>3713943
>129 IQ according to online tests
>A in Calc 2
>94th percentile in SAT

[Anonymous No.3713973]

>>3713930
Ok thanks, it's the right answer.

/thread

[Anonymous No.3713985]

2+3 = 5
5/2 = 2.5 (average speed)
2.5/2 = 1.25 hours (amount of work is halved)

[Anonymous No.3713989]

>>3713985
Except that's wrong.

[Anonymous No.3713993]

Jack paints a room in 3 hours

John paints a room in 2 hours.

Two Jacks would take 1.5 hours to paint a room

Two Johns would take 1 hour to paint a room

Therefore, a Jack and a John would take 1.25 hours.

[Anonymous No.3714005]

(a/b) / (c/d) = (a/b) * (d/c)

(1/1) / (5/6) = (1/1) * (6/5)

[resident /sci/ economist !!0CqB7P/574e No.3714008]

I think these questions are really funny.

Because we dont actually know anything about the nature in which jack or john paint houses. Nor do we know how they will paint when together. But that doesnt stop people from coming to all kinds of conclusions.

[Anonymous No.3714012]

>>3713971
1. that's only just above average
2. and you still can't work out how to do extremely basic number problems using the excuse that you can't remember the trivially deducible method
3. yet you still seem to think you're intelligent

sure is dunning kruger in here

[Anonymous No.3714019]

>>3714012

>two standard deviations
>just above average

[Anonymous No.3714027]

>>3714008

The thing is there are no painters.

It's just an scenario to give us familiarity when performing math.

[Anonymous No.3714029]

>>3713869
When you do finally get around to doing your own homework OP, make sure you state your assumptions. As its stated the question is impossible to solve, you need to start with some (reasonable) assumptions.

[Anonymous No.3714030]

>>3714027
So math really IS useless!

[Anonymous No.3714038]

>>3713989

Why?

[Anonymous No.3714039]

>>3714019
> can't do baby maths
> 'you're a retard'

[Anonymous No.3714056]

Is the answer 6/5 or 1.25?

[Anonymous No.3714057]

>>3714019

Knowing how to 'do types of problems' is not problem solving. Your online IQ tests have lied to you if you honestly believe that you are of above average IQ and yet are unable to solve OP's trivial question. This is the problem with the education system around mathematics these days, you don't get to do any maths until you get to university, and it leaves stupid people believing they are smart because they could do a variety of 'types of problems'. There is no thinking involved in the use of formulas.

[Anonymous No.3714063]

>>3714027

The mathematically correct way to write this problem would be along the lines of:

There are two functions, Function A and Function B, which deal with Area over Time.

Function A (X) = 3 hours

Function B (X) = 2 hours.

Determine the time for Function AB (X)

[Anonymous No.3714067]

Two plumbers are fixing a drain. It takes Plumber A two hours to fix the drain. and a total of 3 hours to fix the drain with help from Plumber B. If they work together, taking two fifteen minute breaks, then why aren't you checking these doubles?

[Anonymous No.3714070]

>>3713872

They actually worship a meteorite kept inside there.

I'm not even joking.

[Anonymous No.3714072]

>>3714063
What are the actual steps involved?

I'm actually quite interested.

[Anonymous No.3714078]

>>3714056
i think it's 1.25, not 6/5

[Anonymous No.3714081]

>>3714063

How do you account for the painters interfering with each other, getting in each other's way, etc...?

[Anonymous No.3714085]

An engineer, a physicist, and a mathematician are shown a pasture
with a herd of sheep, and told to put them inside the smallest
possible amount of fence. The engineer is first. He herds the sheep
into a circle and then puts the fence around them, declaring, "A
circle will use the least fence for a given area, so this is the
best solution." The physicist is next. She creates a circular fence of
infinite radius around the sheep, and then draws the fence tight around
the herd, declaring, "This will give the smallest circular fence around
the herd." The mathematician is last. After giving the problem a little
thought, he puts a small fence around himself and then declares, "I
define myself to be on the outside!"

[Anonymous No.3714086]

>>3714063
confirmed for moron who thinks he's smart because he used the word 'function'

[Anonymous No.3714099]

>>3714085
1. not funny
2. doesn't even make sense

i don't think you got the joke in the first place and now you're giving us some fucked up version

[Anonymous No.3714103]

TWophyiscists are working down the street the first phyidicist says will u love me 4eva teh second says no and the first physicists is hit by and tachyon because they was sad and the second phycistis runs up 2 the others corpse and whipsters in there ear "wat i mean was, i will live u 5eva. tats more den 4eva"

repost this if it makes you sad everytime u read it

[Anonymous No.3714108]

>>3714078
Why?

[Anonymous No.3714114]

>>3714103

>premed students discuss physics 101

[Anonymous No.3714125]

>>3714103
>intellectualism on facebook.txt

[Anonymous No.3714144]

>>3714078
its 6/5 use distance (or area painted, measured in rooms in this case) = speed x time

you have the speed (jack does 1/3 of a room in an hour and john does 1/2 a room in an hour, adding them gives 5/6 rooms an hour) and you want the time so it becomes T = D/V which is 1/(5/6) which is 6/5

[Anonymous No.3714148]

>>3714144
nope i definitely got 1.25, i'm not sure where your mistake is

[Anonymous No.3714162]

>this thread.

[Anonymous No.3714166]

>>3714148
nowhere, as its correct

[Anonymous No.3714172]

>>3713869
>>3713869
Think about it like this...
Start by getting a feel for the problem;
When one guy is 1 hour in, he'll be halfway through, when the other guy is 1 hour in, he'll be 1/3 of the way through.

Now
Work done by the 2 hour guy will be given by
ct = w
c is constant, t is time, w is portion of work completed.
We need c so that when t is 2, w is 1
c.2 = 1
c = 1/2
So his work done will be given by w = t/2, test this to see if it 'seems' right. If he works for 1 hour we get w = 1/2, if it takes him two hours to do it all, 1 to do half seems about right, what if he's worked for half an hour? we get w = 1/4, quarter of the work in half an hour sounds right too, so this models our situation reasonably.

Doing the same for the other guy now
kt = w
need k so that when t is 3 w is 1
k.3 = 1
k = 1/3
Again check out to see if it seems right, from w=t/3 In one hour, one third of the work, in half an hour one sixth... Seems good.

More to come.

[Anonymous No.3714173]

>>3714172
So now we can combine our models to find the total work done
w = (work done by one man) + (work done by the other)
w = t/3 + t/2

Now we want to solve for the time taken to complete the work, so we set w = 1 and do a little algebra
1 = t/3 + t/2
Multiply both sides by 2x3 = 6 to make the calculations easier
6 = 2t + 3t
6 = 5t
t = 6/5

So in seventy two minutes (six fifths of an hour) the work should be complete, once again we should check if these seems right intuitively, never trust your models, at least not until you're much more competent (but then you'll be making much more complex models, so you still shouldn't trust them...)

We said earlier that in an hour one man would have done half, one would have done one third, which would be about 80% of the work, the extra 12 minutes is about a fifth of an hour, so about an extra fifth of 80% of the work, which is about 15% of the work, which makes up 95% of the work, since this was pretty rough 95% is close enough to 100% to give us some confidence that our model has worked.

[Anonymous No.3714193]

i got 1.2

I remember studying this for the SHSAT's and I never understood it. I just figured it out now and I'm p happy

[Anonymous No.3714204]

>>3714193

[Anonymous No.3714210]

>>3714172
>>3714173
Op here. This guy is right, stop arguing.