[Part 4]Quantitative Aptitude Fast and Quick Shortcut Tricks in pdf Download | [TUTORIAL 4]

QuantitativeAptitude Shortcuts Tricks and Methods - Formulas and Handy Notes - Download

Quantitative-Aptitude-Fast-Quick-Shortcut-steps-Tricks-in PDF-Download | [TUTORIAL 4] 

 After Long search for tips and tricks you might be tired,Don't worry here is the solution for your questions.Please be passion on learning this tricks.Don't   Read   LEARN this post in hurry mode. 

  OK Lets Start our Tricks on Chapter TIME AND WORK  

 We can divided into two parts: 

1.First Set/First part of problem consists of below information

1.     Nos of men

2.     Nos of days

3.     Nos of hours

4.     Quantum of work

 2.Second set/Second parts may anything missing from above details and anyone would be common(refer problem no.1 ) which is actually the question to find out.

Formula for TIME AND WORK PROBLEMS

M1 x D1 x H1        M2 x D2 x H2
—————-     =     —————-
     W1                              W2

1.  36 men can do a piece of work in 25 days. In how many days can 30 men do it?

    Here , the work is same.. So,

                    

                    (M XD) / W = (m X d) / w 

        

                  = >       (M XD) / W = (m X d )/ w    [ Cancel W for both sides]

               Now, apply given values...

                   = >  36 X 25  = 30 X d 

                     = >  (36 X 25) / 30  = d  => d = 900/3  =  30

 2.If  6 men   dig 100 meters in 20 days working 8 hours  ,  how many men are required to dig 200 meters  in 10 days working 4 hours?

Sol:

6 x 20 x8          M2 x 10×4

————    =   —————— (solve this equation) Answer is 48 men

  100                    200

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Ok Lets work with some complex Problems:

3.A man engaged 10 laborers to make 320 toys in 5 days. After 3 days he found that only 120 toys were made. How many additional men should he engage to finish the work in time?

Problem Explanation 

10 laborers=1/5 days=>320 toys(engaged)     but   after 3days  --  eq1

10 laborers=1/3 days=>120 toys (only completed from engaged)--  eq2

M laborers =1/2 days=>200 toys (sub eq1- eq2) additional men 

Now put in our formula

till completed  = pending

        (10 X 3) / 120 = (10+M) x 2 / 200  

                                  = >   10+M = 25   =>   X = 15

4.A, B and C can do a job in 20 days, 30 days and 60 days respectively. If they work together, in how many days will the work be finished?

A work =1/20 days , B work =1/30 days        and C work = 1/60 days

A+B+C= (1/20)+(1/30)+(1/60) = (3+2+1)/60  =  6/60  = 1/10

               So, the number of days is = 10

Shortcut = product/sum i.e  Take L.C.M

   work=Days ---> L.C.M for A,B and C is 60

         A=20    ----> 3

         B=30    ----> 2

         C=60     ----> 1 

                                ---

                         (sum)6

                                ---

A+B+C------->6/60 = 10 answer

Note: above 6/60 is 6/L.C.M value(60)

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5.Two taps A and B can fill a tank in 10 hours and 15 hours respectively. a third tap C can empty the full tank in 12 hours. How many hours will be required if all of them are opened simultaneously to fill in an empty tank completely?

        Here, first two are Inlets and the last one is Outlet,

          So,      (1/10)+(1/15)-(1/12) =  (6+4-5)/60  =  5/60  = 1/12

               So, our answer is 12

6. A and B can do a job in 12 days. B and C in 15 days and C and A in 20 days. In how many days can they finish it if they work TOGETHER?

       A+B = 12

        B+C = 15

       C+A = 20

So here, A+B's One day's work = >   1/ (A+B) = 1/12

                B+C's one day's work  =>  1/ (B+C) =  1/15

                 C+A's one day's work = > 1/(C+A) = 1/20

              Just Add them  =>  1/( 2A+2B+2C)  =  12/60  =  1/5

                                  =>  1/2(A+B+C)  =  1/5

                                     =>1/(A+B+C)  =  1/10   [this is their one day's work TOGETHER]

                 

                      So, they can finish it in 10 days :)

7.  A and B can do a job in 12 days. B and C can do the same job in 15 days. C and A in 20 days. In how many days can A alone finish the whole task???

         A+B = 1/12

         B+C = 1/15

         C+A = 1/20

         Here we need A, so take a pair which is NOT HAVING A  and subtract it from the others, 

               so,  A+B-(B+C)+C+A  =  A+B-B-C+C+A =  2A   
         
But according to our Problem, 2A =  (1/12)-(1/15)+(1/20)
                                    =>A = 1/30 (this is A's one day work but we need A's total work)
                          
                                           => A = 30 Days

  8. A and B can do a piece of work in 20 days. A alone can do it in 30 days. In how many days can B alone do it?

    Per day work of A and B = 1/20

         Work done by A=  1/30

       So, B's one day work =  1/20  - 1/30  =  (3-2)/60  =  1/60  

                     =>  B's work is 60 Days

       Short Cut :   Product/diff  =  600/10  =  60  :)

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Aptitude Tricks

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