math q

[dentalplan]

Joe can paint a fence in 12 hours. Mack can paint it in 6 hours, how long will it take them to do it together?

i know the answer is 4, because someone told me to do 12 x 6, then divide by (12 + 6)

72/18 =4

Streetwolf, or anyone else, please explain the logic here. Why are you multiplying 12 and 6, then dividing by their sum?

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[jay47]

Joe can paint a fence in 12 hours. Mack can paint it in 6 hours, how long will it take them to do it together?

i know the answer is 4, because someone told me to do 12 x 6, then divide by (12 + 6)

72/18 =4

Streetwolf, or anyone else, please explain the logic here. Why are you multiplying 12 and 6, then dividing by their sum?

This is a typical question that you have to use logic on to figure out. I had to think about it for a while though myself. I find it best to apply arbitrary numbers to the situation.

Okay, let's say the fence is 100 m long (but you can use any length) If it takes Joe 12 hours to do, that means 100/12= 8.33 m/hour painting rate. Mack, 100/6=16.66 m/hour. Together, they could paint at 25 m/hour. To find the time it takes to paint 100 m, (100m) / (25 m/hour)= 4 Hours.

If you get some of these problems with no numbers, just try and apply some numbers to the situation and it usually helps.

The reason your method works is b/c 12x6 is essentially equal to the total rate of painting. 12+6 = the total time it takes to paint. To get the number of hours it takes to paint the whole thing:

(total rate of painting)/(total distance painted)= # hours

Hope this helps.

 

[shankin]

Joe can paint a fence in 12 hours. Mack can paint it in 6 hours, how long will it take them to do it together?

i know the answer is 4, because someone told me to do 12 x 6, then divide by (12 + 6)

72/18 =4

Streetwolf, or anyone else, please explain the logic here. Why are you multiplying 12 and 6, then dividing by their sum?

the easiest way to do this in my opinion is to think about it like this

joe paints at a speed of 1/12 fence per hour
mack paints at a speed of 1/6 fence per hour

therefore

1/12 + 1/6 = fence per hour painted together

so 1/12 + 2/12 = 3/12 or 1/4 or (25%) fence per hour together

25%*4= 100% of the fence painted

since your multiple is 4 to get 100% obviously it takes 4 hours to paint the whole fence

the logic behind the other thingy is a modified formula of speed=distance/time
speed is how fast they paint the fool thing together
distance= how long does guy 1 take to complete the fence distance* how long does guy 2 take to complete the fence distance
and time is time 1 + time 2

or speed= 12*6/12+6 which is 72/18 or naturally 4

 

[doc3232]

the easiest way to do this in my opinion is to think about it like this

joe paints at a speed of 1/12 fence per hour
mack paints at a speed of 1/6 fence per hour

therefore

1/12 + 1/6 = fence per hour painted together

so 1/12 + 2/12 = 3/12 or 1/4 or (25%) fence per hour together

25%*4= 100% of the fence painted

since your multiple is 4 to get 100% obviously it takes 4 hours to paint the whole fence

the logic behind the other thingy is a modified formula of speed=distance/time
speed is how fast they paint the fool thing together
distance= how long does guy 1 take to complete the fence distance* how long does guy 2 take to complete the fence distance
and time is time 1 + time 2

or speed= 12*6/12+6 which is 72/18 or naturally 4

Never thought of it like this.

 

[Hugh Mannity]

I do it like shankin. I remember this type of logic from resistances of capacitors from physics II (1/R = 1/R1 + 1/R2). If one thing can do a job at this rate and a second can do it at another rate, the combined rate must be less than the lesser of the 2 individual ones, and they are added inversely. (# of jobs) / (1st rate or time) + (# of jobs) / (2nd rate or time) = (# of jobs) / (combined rate or time). the easiest is when the $ of jobs is 1 but variations can exist to make it more complex and frustrating. Solving then obviously uses common denominators and cross multiplication.

 

[Streetwolf]

You get ab/a+b because that's how it works when you consider how fast they each go. As far as I can tell, it's just a short-cut and does not have anything to do with the distance/rate formula.

You have two rates: 12 hours per fence (1 fence / 12 hours or 1/12 fence/hour) and 6 hours per fence (1 fence / 6 hours or 1/6 fence/hour). You want to combine those into one rate. So you add them. If one person does 1/12 fence each hour and the other does 1/6 fence each hour, they both do 1/4 fence per hour combined. They need to complete 1 fence. So 1 fence / (1/4 fence per hour) = 4 hours.

More generally if person A needs 'a' hours to paint a fence (1/a fence per hour) and person B needs 'b' hours to paint a fence (1/b fence per hour), then combined they paint (1/a) + (1/b) = (a+b)/ab fence/hour. Since there's just 1 fence, they need 1 fence / (a+b)/ab (fence/hour) = ab/(a+b) hours to paint the fence together.

So (a+b)/ab is the rate and ab/(a+b) is the time.

 

[Hugh Mannity]

You get ab/a+b because that's how it works when you consider how fast they each go. As far as I can tell, it's just a short-cut and does not have anything to do with the distance/rate formula.

You have two rates: 12 hours per fence (1 fence / 12 hours or 1/12 fence/hour) and 6 hours per fence (1 fence / 6 hours or 1/6 fence/hour). You want to combine those into one rate. So you add them. If one person does 1/12 fence each hour and the other does 1/6 fence each hour, they both do 1/4 fence per hour combined. They need to complete 1 fence. So 1 fence / (1/4 fence per hour) = 4 hours.

More generally if person A needs 'a' hours to paint a fence (1/a fence per hour) and person B needs 'b' hours to paint a fence (1/b fence per hour), then combined they paint (1/a) + (1/b) = (a+b)/ab fence/hour. Since there's just 1 fence, they need 1 fence / (a+b)/ab (fence/hour) = ab/(a+b) hours to paint the fence together.

So (a+b)/ab is the rate and ab/(a+b) is the time.

Although this works for this question, I would be skeptical to just to use that formula to plug and chug. Variations will exist if it gives you different times as well as different number of tasks accomplished. Like with that equation you assume its only 1 job in a or b amount of time, but it becomes different when one is 2 jobs and the other is 3 jobs for example. I'm just getting at that you should try to critically think about the problem and them solve and never to just blindly use an equation unless it applies to all situations.

 

[dentalstudent00]

simple question. J=12 M=6; J+M = X (the time it takes for them to do it together); (1/12) + (1/6) = (1/x); find the lowest common denominator which is 12x; multiply across to get (x+2x) = 12; solve for x; x=4