# faster and unconventional way of doing maths problem?

I just wanted to know is there any book or resource or perhaps online resource which could help me to do faster problem solving problems like below(so that i could do it mentally an perhaps a lot faster even if i write)

There are 12 pipes that are connected to a tank. Some of them are fill pipes and the others are drain pipes. Each of the fill pipes can fill the tank in 8 hours and each of the drain pipes can drain the tank completely in 6 hours. If all the fill pipes and drain pipes are kept open, an empty tank gets filled in 24 hours. How many of the 12 pipes are fill pipes?

Explanatory Answer
Let there be 'n' fill pipes attached to the tank.
Therefore, there will be 12 - n drain pipes attached to the tank
Each fill pipe fills the tank in 8 hours.
Therefore, each of the fill pipes will fill 1/8th of the tank in an hour.
Hence, n fill pipes will fill n/8 of the tank in an hour.
Each drain pipe will drain the tank in 6 hours.
Therefore, each of the drain pipes will drain 1/6th of the tank in an hour.
Hence, (12 - n) drain pipes will drain (12-n)/6 of the tank in an hour.

When all these 12 pipes are kept open,
it takes 24 hours for an empty tank to overflow. Therefore, in an hour
1/24th of the tank gets filled.

Hence, (n/8)-((12-n)/6) = 1/24

i.e.  or 7n - 48 = 1 => 7n = 49 or n = 7.


If A and B work together, they will complete a job in 7.5 days. However, if A works alone and completes half the job and then B takes over and completes the remaining half alone, they will be able to complete the job in 20 days. How long will B alone take to do the job if A is more efficient than B?

Explanatory Answer
Let 'a' be the number of days in which A can do the job alone. Therefore,
working alone, A will complete  of the job in a day.
Similarly, let 'b' be the number of days in which B can do the job alone.
Hence, B will complete  of the job in a day.

Working together, A and B will complete  of the job in a day.

The problem states that working together, A and B will complete
the job in 7.5 or 15/2 days. i.e they will complete 2/15th of the job in a day.
Therefore,  ...... (1)
From the question, we know that if A completes half the job
working alone and B takes over and completes the next half, they will take 20
days.

As A can complete the job working alone in 'a' days, he will
complete half the job, working alone, in  days.
Similarly, B will complete the remaining half of the job in  days.
Therefore,  => a + b = 40 or a = 40 - b ...... (2)

From (1) and (2) we get,  => 600 = 2b(40 - b)

=> 600 = 80b - 2b2
=> b2 - 40b + 300 = 0
=> (b - 30)(b - 10) = 0
=> b = 30 or b = 10.
If b = 30, then a = 40 - 30 = 10 or
If b = 10, then a = 40 - 10 = 30.
As A is more efficient than B, he will take lesser time to do the job alone.
Hence A will take only 10 days and B will take 30 days.

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It is true that formal, written manipulation of a problem is often not the fastest way to the solution. Moreover, no single method will be optimal for all problems.

For practical problems, taking an intuitive guess at the solution and then confirming it or working backwards is often much faster. Also, for constrained problems with not many possible solutions, attempting to confirm the potential solutions can be an effective technique.

For example, in your first problem there are only 13 possible answers since you are constrained to a positive integer number of pipes. You could observe that if there were 8 fill and 6 drain pipes then the tank would be in equilibrium. This sums to 14 pipes though, so you need to both 1) reduce by two pipes and 2) create an excess of 1/24. Removing one drain pipe creates an excess of 1/6 which is too much, so it is reasonable at that point to determine the excess from also removing one fill pipe: $\frac16 - \frac18 = \frac2{48} = \frac1{24}$ . Thus by finding a condition that was "close" to the solution it allowed the solution to be found by incremental refinement.

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nice observation @half-integer fan –  munish Dec 27 '12 at 15:42

Variants of the following method were formally taught from very early on in China, under a name that translates to something like "too much and not enough." It was also widely taught elsewhere.

Let us guess that $12$ fill pipes and therefore $0$ drain pipes will do the job. Check whether this works, we might get lucky. In $24$ hours, the $12$ fill pipes fill $36$ tanks, and there is no draining, so we get $36$ tanks. Oops, too much.

Let's trade in a fill pipe for a drain pipe. We lose $3$ tanks from the missing fill pipe, plus $4$ tanks from the new drain pipe, a total of $7$.

Any such trade leads to a loss of $7$ tanks. We need to lose $35$ tanks to get from $36$ to $1$. So we need to make $5$ trades.

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+++1wow a complete new way of looking at the same problem...nice –  munish Dec 28 '12 at 0:00
I applied the same approach to the other problem also,since A is more eficient so A would take less than 7.5 to complete half of the job. i assumed that A would take 5 days to complete half of the job and so 10 days to do full job. then b would take 20-5=15 days to complete half of the job and 30 days for full job. so they compelete the job together in (1/10 + 1/30) * 7.5 days which is equivalent to 1. –  munish Dec 28 '12 at 5:55
another way of solving the second problem is expessing 1/7.5=2/15=10+30/10.30 = 40/10.30 in the form x+y/xy i.e where xy can represent the time for doing the job twice(once by A and once by B) –  munish Dec 28 '12 at 12:10