Pumping alone at their respective constant rates, one inlet pipe fills an empty tank to 1/2 of capacity in 3 hours and a second inlet pipe fills the same empty tank to 2/3 of capacity in 6 hours. How many hours will it take both pipes, pumping simultaneously at their respective constant rates, to fill the empty tank to capacity?
1 Explanation
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Brittany Hunter
(I teach 4th grade, so I tend to go the long way, so ignore if you want a shortcut)
We can use the formula provided in an earlier video. 1/s + 1/r = 1/t...Ignoring the letters and thinking about the numerator being the AMT OF WORK COMPLETED and the denominator being the TOTAL HOURS it took to complete, we can plug in and solve.
Pipe A: 1/2 capacity in 3 hrs
Pipe B: 2/3 capacity in 6 hours
?- How much time to fill the tank, so we are solving for TIME which is in the denominator.
1.) Change times to have the same denominator. 1/2= 3/6 and 2/3= 4/6
2.) Set up ratios 3/6 Divided by 3 Hrs PLUS 4/6 divided by 6 Hrs EQUALS 6/6 divided by X
3.) Give the Times the same denominator, so 3/6 divided by 3 becomes 6/6 divided by 6 Hours
3.) Cross Multiply and Solve: 6/6 + 4/6 =10/6 all over 6 hours= 6/6 divided by X
4.) 10/6 X = 6
5.) Multiply by the reciprocal to get rid of the fraction, so now you have x = 6/1 * 6/10
6.) You get x = 36/10
X = 3 6/10 and if you can convert the fraction into a decimal its 3.6 because the 6 is in the tenths place
1 Explanation