There are no need to say that how important

**time and work arithmetic**math chapter and its shortcut formula in competitive exam. Earlier we have delivered you two tutorial based on time and works shortcut formula, aptitude question and tricky solution. Today, we also discuss the same topic but increasing a level. As you know that we should learn many shortcuts formula on every topic to crack our upcoming competitive examination. See the previous tutorial related on time and works, we have shown you how to use shortcut time and works formula when some questions are not matching precisely, we need to jump a level or require some extra calculation. However, today's tutorial is the 3^{rd}tutorial on the math chapter time and works. Here you will get some fantastic shortcut formula which is works like maths tricks, you could not imagine it until reading the full tutorial.**Formula condition:-**

When working 1

^{st}couple of worker__A and B__can do a piece of works in x time, 2^{nd}couple of worker__B and C__can do the same work in y time, and another couple of worker__C and A__can do the same work in z days. Then- Find how many time unit they need to complete the whole works if they work altogether. Only A can dothe work in ..?
- Only B can do the work in...?
- Only C can do the work in...?

Getting confused by looking the formula. Dont worry, we have alreay shown how to remember the shortcut formula. Again look at the formula image there are three pair of workers, and all pairs have a commn man from other two pairs. Also look that every numenetor is 2xyz and dumenetoer is diffrence while exucuting working time unit of A, B or C. In demenotor we take time value as common and it is either first pair or second pair or 3rd pair. But the common time value (x,y or z) is belongs from that pair where the worker is not belong. for example C can do ... in demenotor, we take common x. Because x belong from the the pair A and B whre C is not present. And then we put sum of other two pairs time unit and then finally we also substract its multiply value.

**Example question:-**

A and B can do a work in 20 days. Another two couple ‘B and C’ and ‘C and A’ can do the same work 20 and 40 days respectively. If the job works all together then how many days they will need to complete the whole works?

Is not it easy friends?

Look at another example:

A couple of worker A and B can do a work in 10 days. Another two couple of workers ‘B and C’ and ‘C and A’ can do the same works in 20 and 25 days respectively. If they have to complete the half of the work altogether then find the days they need to complete the work?

The solution with math tricks:

You have seen already how we have evaluated the correct answer this kind of math solution. Here only one thing have included that here we should find out the time to need to half of the work than the previous question. So we will find out the time which they need to complete the full work first then we will multiply by 1/2. They need time to complete the full work

2*10*20*25/ 10*20 + 20*25 + 25*10

=10000/ 200+ 500 + 250 = 10000/950 = Days. It's not the correct answer friends. Because they will only work ½ parts of the work. So they need ½ time for completing time.

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