math

[Canis]

I was guessing ~3.33 hours before scrolling down...damn I'm good.

[[xeno]Julios]

I think we both solved it from different angles. I don't know how you got x = -9.5 Jules

woops that was a major gaffe on my part.

reading your solution now.

[[xeno]Julios]

Let's start with the following equations that express the overall work done per worker:

x * e = A
x * e = B

unless A = B, you can't start off like this. x can't be two different values at once. You need two different variables x and y which represent the specific times for A and B respectively.

Your solution seems correct overall, but it's overly convoluted. Hard to follow the logic exactly because of the problem with the x's.

Where x is the number of hours worked, e is the efficiency of the worker, and A and B stand for the overall amount of work done over time (hours * efficiency = total work done)

We can define efficiency e as a coefficient that stands for how many 'tasks per hour' a worker can complete.

Better to just stick to simple consistent units. Instead of "tasks" just use houses. Instead of "work" just say houses.

Also, efficiency isn't a coefficient, it's a variable.

[mjrpes]

So let's clear up the problem with the x's. I don't want to switch to houses

STEP 1: The general equation

• Let's start with the general equation. We want to find the time needed for two people to complete one task. The equation can be worded as:
  
  "The total work output of both workers combined to complete one task"
  
  Which looks like:
  
  <Work Done by A> + <Work Done by B> = 1 Task
  
  We will define the work done as:
  
  <Work Done by Person A> = d1
  <Work Done by Person B> = d2
  
  So that:
  
  d1 + d2 = 1

STEP 2: Equation for work done per worker:

• Now we need to find equation for the work done portion of the general equation above. This can be written out as:
  
  time (hours) * efficiency (tasks/hour) = Work Done (tasks)
  
  Notice the units we are using:
  
  hours * tasks/hour = task
  
  This makes sense because the hours cancel out!!
  
  Let's set our variables. So, per worker, we will use variables
  
  <hours worked> = t1, t2
  <efficiency> = e1, e2
  
  Where Worker A uses t1, e1 and Worker B uses t2, e2
  
  Using these variables for each worker we get the equations:
  
  t1 * e1 = d1
  t2 * e2 = d2
  
  Substituting into the equation above, our new equation looks like
  
  (t1 * e1) + (t2 * e2) = 1

STEP 3: Get efficiency:

• We know that for worker A, it takes him 5 hours to do one task. And for worker B it takes 15 hours to do one task. So their rate is
  
  5 hours per task = 1/5 tasks per hour
  15 hours per task = 1/15 tasks per hour
  
  So we know that:
  
  e1 = 1/5
  e2 = 1/15
  
  Substituting this into our equation, we get
  
  t1 * (1/5) + t2 * (1/15) = 1

STEP 4: Deal with the amount of time worked for both

• We can't solve yet because we have two time variables, t1 and t2. But we know that t1 will be the same value as t2, because the problems explains that they will be working together from start to finish. The time has to be the same Thus,
  
  t1 = t2
  
  I will substitute in t1 for t2:
  
  t1 * (1/5) + t1 * (1/15) = 1

STEP 5: Solve for t1

• Using some algebra, we get:
  
  t1 = 3.75
  Tada!

I'm a bit worried that you said my explanation is convoluted. I hope this is better.

[phantasmagoria]

What is the task? It sounds like person B doesn't know what the fuck hes doing. person A would be the one doing all the work, while person B flops around like a mongaloid and screws shit up.

[Pext]

what a retarded question this is.

the question itself is not sufficient as a math question. what if none of the workers can give 100% if both are working together?


this is another example of the retarded idea to make maths more appealing to pupils by embedding it in 'real world' problems.

a shorter solution:

T=Ammount of 'Work' needed to finish the product.

W(P)='Work' per hour for person P
W(A)= T/5
W(B)= T/15
W(A)+W(B)=T(1/5+1/15)=T(4/15)

How long does it take both to produce an ammount of 'Work' equal to T?

T = x*(W(A)+W(B))= T*x*4/15
<=> 1 = x * 4/15
<=> 15/4 = x

[losCHUNK]

you cant add letters

x+x+b = 2 x's and a b innit

[Pext]

?

you don't add letters, you add numbers represented by letters.

[glossy]

okay i read the topic title as "meth" and was like ! but now i'm just like "nevermind "

[losCHUNK]

?

you don't add letters, you add numbers represented by letters.

well if the letters reprezentin a number then let the n00bi3 number stick up for himself, he should show himself from the get go if he thinks hes all that

also, i fucking suck with numbers, always have and always will

[plained]

i have no way of knowing but i guess good.

mmmm i so i say 3.65 hours

[[xeno]Julios]

I'm a bit worried that you said my explanation is convoluted. I hope this is better.

nice - seems consistent, but I still think my way is simpler!

Let X = A's house-building speed, or rate of housebuilding
let Y = B's rate of house building

We define rate of house building as number of houses built per hour

Houses built per house = # of houses divided by time taken

(just like velocity = distance divided by time)

So:

X = A's rate = 1 house divided by 5 hours = 1/5 houses per hour
Y = B's rate = 1 house divided by 15 hourse = 1/15 houses per hour

So combining A's rate and B's rate, we get 1/5 + 1/15 houses per hour:

1/5 + 1/15
= 3/15 + 1/15
= 4/15 houses per hour

Now if the combined rate is 4/15 houses per hour, then how long will it take to build 1 house, or 15/15 houses?

Just as time = distance divided by velocity, time = houses built divided by rate of house building


so time = unknown
houses built = 15/15
rate = 4/15

so time = 15/15 divided by 4/15

15/15 divided by 4/15
= 1 divided by 4/15
= 1 multiplied by 15/4

1/1 * 15/4 = 15/4

15/4 = 3 and three quarters = 3.75 hours

[Grandpa Stu]

there was no reason for this topic to continue after doomers post.

[mjrpes]

Speaking of math...

Proof that 1 = 0:

let x=1

x*x = x
(x*x) -1 = x-1
(x+1)(x-1) = x-1
x+1 = 1
x = 0
So 1 = 0.

[mjrpes]

Ignore the division by zero part

[[xeno]Julios]

did you short circuit the quadratic by doing that division, thus castrating off one of the two possible solutions (1 and 0)?

[mjrpes]

Something like that... I got it from another forum. Perhaps a mathematician can explain what happened more clearly.

[Nightshade]

did you short circuit the quadratic by doing that division, thus castrating off one of the two possible solutions (1 and 0)?

No, he simply factored (x^2-1) and then ignored that fact that unless you're in a modular number system, 1 cannot equal 0.

[menkent]

ffs, two pages for simple algebra. imagine if he'd asked a real math question.

[Pext]

Something like that... I got it from another forum. Perhaps a mathematician can explain what happened more clearly.

you divided something by 0...

[MKJ]

did you short circuit the quadratic by doing that division, thus castrating off one of the two possible solutions (1 and 0)?

No, he simply factored (x^2-1) and then ignored that fact that unless you're in a modular number system, 1 cannot equal 0.

its like blizzard's 0.999~ == 1 april's fools joke

[Pext]

did you short circuit the quadratic by doing that division, thus castrating off one of the two possible solutions (1 and 0)?

No, he simply factored (x^2-1) and then ignored that fact that unless you're in a modular number system, 1 cannot equal 0.

no... not really.

the 1=0 part only comes by dividing by x-1 = 1-1 = 0

[Pext]

did you short circuit the quadratic by doing that division, thus castrating off one of the two possible solutions (1 and 0)?

No, he simply factored (x^2-1) and then ignored that fact that unless you're in a modular number system, 1 cannot equal 0.

its like blizzard's 0.999~ == 1 april's fools joke

hell yes... ppl still think it's wrong.

[Nightshade]

did you short circuit the quadratic by doing that division, thus castrating off one of the two possible solutions (1 and 0)?

No, he simply factored (x^2-1) and then ignored that fact that unless you're in a modular number system, 1 cannot equal 0.

no... not really.

the 1=0 part only comes by dividing by x-1 = 1-1 = 0

If you're still dealing with the variable and not the number, what I said is correct. At that point it's still a valid algebraic operation.

[Underpants?]

What is the task? It sounds like person B doesn't know what the fuck hes doing. person A would be the one doing all the work, while person B flops around like a mongaloid and screws shit up.

assgasm0 speaks the truth this man is funny.