math
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stocktroll
- Posts: 1314
- Joined: Mon Mar 21, 2005 2:44 am
math
how do i solve this?:
it takes person A 5 hours to do task.
it takes person B 15 hours to do task.
how long will it take to do task together?
it takes person A 5 hours to do task.
it takes person B 15 hours to do task.
how long will it take to do task together?
10 hours..
Id assume anyways.. id "figure" its 5+15/2=10
Dunno tho, i suck at math
Id assume anyways.. id "figure" its 5+15/2=10
Dunno tho, i suck at math
[size=75][i]I once had a glass of milk.
It curdled, and then I couldn't drink it. So I mixed it with some water, and it was alright again.
I am now sick.
[/i][/size]
[img]http://img162.imageshack.us/img162/3631/171164665735hk8.png[/img]
It curdled, and then I couldn't drink it. So I mixed it with some water, and it was alright again.
I am now sick.
[/i][/size]
[img]http://img162.imageshack.us/img162/3631/171164665735hk8.png[/img]
Figure out the percentage of project they can do per hour individually, (1project/5hours and 1project/15hours) and add them until you reach 100%?
Hour-------1----------2-----------3--------------4
Person A--20%------40%------60%----------80%
Person B--6.5%-----13%------19.5%--------26%
Total-------26.5%---53%-------79.5%--------106%
1/15 is around 6.6666%
Project got done at around 3h45m?
I could be totally wrong though, I never do any problems like that.
edit: 54 wtf
Hour-------1----------2-----------3--------------4
Person A--20%------40%------60%----------80%
Person B--6.5%-----13%------19.5%--------26%
Total-------26.5%---53%-------79.5%--------106%
1/15 is around 6.6666%
Project got done at around 3h45m?
I could be totally wrong though, I never do any problems like that.
edit: 54 wtf
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stocktroll
- Posts: 1314
- Joined: Mon Mar 21, 2005 2:44 am
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[xeno]Julios
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well imagine it took A 5 hours to build 1 house and B 5 hours to build one house.
How long would it take them both to build one house?
With the addition of B, we have double the efficiency of A, so it'll take half the time (2.5 hours).
How can we arrive at this number mathematically?
let A's time = 5 hours = x
let B's time = 5 hours = y
let AB's time = z
how can we manipulate the number 5 to reach 2.5 (which we know intuitively is the correct answer)?
We divide by 2.
What can we do with x and y to make 2?
how about 5 divided by 5, added to 1
so x/y + 1 = 2.
why would this make intuitive sense though? On the surface it seems arbitrary.
well the combined efficiency would be the efficiency contributed by B, added to the A
so if be was 20% as efficient as A, we'd have 1 + 0.2
(this is sort of a botched explanation but work with me - i'm flying on the seat of my pants here!)
so if x = 5 and y = 15, we have z = 1+ 5/15 = 1.33333333
so i'm guessing the answer is 5 hours divided by 1 and a third which is 3.75
which is exactly what postal got
3 hours and three quarters
3:45
How long would it take them both to build one house?
With the addition of B, we have double the efficiency of A, so it'll take half the time (2.5 hours).
How can we arrive at this number mathematically?
let A's time = 5 hours = x
let B's time = 5 hours = y
let AB's time = z
how can we manipulate the number 5 to reach 2.5 (which we know intuitively is the correct answer)?
We divide by 2.
What can we do with x and y to make 2?
how about 5 divided by 5, added to 1
so x/y + 1 = 2.
why would this make intuitive sense though? On the surface it seems arbitrary.
well the combined efficiency would be the efficiency contributed by B, added to the A
so if be was 20% as efficient as A, we'd have 1 + 0.2
(this is sort of a botched explanation but work with me - i'm flying on the seat of my pants here!)
so if x = 5 and y = 15, we have z = 1+ 5/15 = 1.33333333
so i'm guessing the answer is 5 hours divided by 1 and a third which is 3.75
which is exactly what postal got
3 hours and three quarters
3:45
Last edited by [xeno]Julios on Thu Sep 08, 2005 5:42 am, edited 1 time in total.
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[xeno]Julios
- Posts: 6216
- Joined: Fri Dec 10, 1999 8:00 am
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stocktroll
- Posts: 1314
- Joined: Mon Mar 21, 2005 2:44 am
bah i just came up with that equation but thanks for the verification. But sadly thats just the easy part... now i have to explain it to a plant.mjrpes wrote:person a: 1x = 5 (takes 5 hours to get 1 task done)
person b: 1x = 15 (takes 15 hours to get 1 task done)
algebra:
x/5 = 1
x/15 = 1
together, to complete one task, the equation is
x/5 + x/15 = 1
3x/15 + x/15 = 1 (common denominator)
3x + x = 15 (multiply both side)
4x = 15
x = 15/4 = 3.75 hours
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[xeno]Julios
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- Joined: Fri Dec 10, 1999 8:00 am
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[xeno]Julios
- Posts: 6216
- Joined: Fri Dec 10, 1999 8:00 am
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[xeno]Julios
- Posts: 6216
- Joined: Fri Dec 10, 1999 8:00 am
Ok let's use a velocity metaphor, where distance is the number of houses built, time is the time it takes to build it, and velocity is the rate of house building.
(gotta go for a while - bbl)
will continue when i get back if nobody's picked up this thought
i think key is to quantify the efficiency of each using their "velocity"
houses per unit time
standard units - then easy to combine
(gotta go for a while - bbl)
will continue when i get back if nobody's picked up this thought
i think key is to quantify the efficiency of each using their "velocity"
houses per unit time
standard units - then easy to combine
Let's try starting with this:
(a) x/5 = y
(b) x/15 = y
Where each equation stands for the efficiency of each worker: x is a variable that stands for the amount of hours put in, and y stands for the number of tasks to complete. Let's plug in 1 for the number of tasks to complete:
(a) x/5 = 1
(b) x/15 = 1
Solving for x, we see it takes 5 hours for worker A to get one task done, and 15 hours for worker B to get a task done. Now let's go on to finding the equation that expresses their work efficiency to get one task done:
(c) x/5 + x/15 = 1
From here, use algebra to solve, and you get 3.75 hours in result.
(a) x/5 = y
(b) x/15 = y
Where each equation stands for the efficiency of each worker: x is a variable that stands for the amount of hours put in, and y stands for the number of tasks to complete. Let's plug in 1 for the number of tasks to complete:
(a) x/5 = 1
(b) x/15 = 1
Solving for x, we see it takes 5 hours for worker A to get one task done, and 15 hours for worker B to get a task done. Now let's go on to finding the equation that expresses their work efficiency to get one task done:
(c) x/5 + x/15 = 1
From here, use algebra to solve, and you get 3.75 hours in result.
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[xeno]Julios
- Posts: 6216
- Joined: Fri Dec 10, 1999 8:00 am
there's something fundamentally wrong with this formulation: there is an inconsistency in the equations you set out.mjrpes wrote:snip
a) x/5 = y
(b) x/15 = y
this is as logical as saying 1 = 2
If we use algebra to solve we get(c) x/5 + x/15 = 1
From here, use algebra to solve, and you get 3.75 hours in result.
2x + 20 = 1
2x = -19
x = -9.5
We need to define efficieny in quantitative terms:
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:
A's rate = 1 house divided by 5 hours = 1/5 houses per hour
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
This might be a little more clearer:
mjrpes wrote:Let's start with the following equations that express the overall work done per worker:
x * e = A
x * e = B
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. For worker A,
1 task = 5 hours
1/5 tasks per hour = e (for worker A)
For worker B,
1 task = 15 hours
1/15 tasks per hour = e (for worker B)
We can substitude the efficiency back into our original equations:
x * (1/5) = A
x * (1/15) = B
Simplified:
x/5 = A
x/15 = B
Now let's go on to finding the equation that expresses the amount of work done for a certain amount of tasks, if both work together:
A + B = t
Where t stands for the total number of tasks needed to complete. We know that t=1, so we now have,
A + B = 1
Now we can go and substitute in
(x/5) + (x/15) = 1
From here, use algebra to solve:
(3x/15) + (x/15) = 1
3x + x = 15
4x = 15
x = 3.75 hours
Last edited by mjrpes on Thu Sep 08, 2005 6:56 am, edited 1 time in total.