Wednesday, August 26, 2015

Time and work Tricks for Competitive Exams Part – 8th

In the previous post of Time and work we discussed about some questions of basic concepts of this chapter. Now we will see some of the some more various frequently asked important question of this chapter in competition like SSC, Bank (IBPS), and Railways. This chapter is one of the easiest chapters for students to solve with less calculations.
Type 3rd   :- Questions based One day work Concept with Efficiency:-
Q14. A is twice as good as workman B. Together they can do a work in 14 days. In how many days can it be done by each separately?

Sol. The work in this question is same means it is constant and always be considered as 1 (one) work only.
Worker                  A                              B                                              
Efficiency             2x                               x
In these questions the frequencies are interchanged to get the number of days of respective workers. Therefore the number of days in which workers “A” and “B” can do the same worker in following number of days.
No. of days           x                               2x
One day work    (1/x)                        (1/2x)
According to the question workers “A” and “B” can complete a work together in 14 days.
Therefore, the one day work done by “A” and “B” together = (1/14) work.
Now applying one day work concept:-
(1/x) + (1/2x) = (1/14)

Taking LCM and solving we get,
(3/2x) = 1/14.
x = 21
Hence the no. of days taken by worker “A” to complete the same work alone = x = 21 days.
And the no. of days taken by worker “B” to complete the same work alone = 2x = 2 × 21
                                                                                                                                     = 42days.
Hence “A” and “B” can do the same work in 21 days and 42 days respectively alone.
OR
Number of days taken by workers “A” and “B” together to complete the work = D = 14 days
As assumed “A” can do the same work in D1 = x days.
And “B” can do the same work in D2 = 2x days.
Now Using formula:-
D = (D1 × D2)/( D1 + D2) = (x × 2x)/ (x + 2x)
Putting the values,
14 = (2x2)/(3x)
14 = (2x/3).
x = (14 × 3)/2 = 21 days (No. of days taken by “A” alone.)
And no. of days taken by worker “B” to complete the same work alone = 2x = 2 × 21
                                                                                                                                     = 42days.
Hence “A” and “B” can do the same work in 21 days and 42 days respectively alone.

Q15. “A” is thrice as fast as “B” and together they can complete a work in 15 days. In how many days will B take to complete the same work alone?

Sol. The work in this question is same means it is constant and always be considered as 1 (one) work only.
Worker                  A                              B                                              
Efficiency             3x                               x
In these questions the frequencies are interchanged to get the number of days of respective workers. Therefore the number of days in which workers “A” and “B” can do the same worker in following number of days.
No. of days           x                               3x
One day work    (1/x)                        (1/3x)
According to the question workers “A” and “B” can complete a work together in 15 days.
Therefore, the one day work done by “A” and “B” together = (1/15) work.
Now applying one day work concept:-
(1/x) + (1/3x) = (1/15)
Taking LCM and solving we get,
(4/3x) = 1/15.
x = 20
Hence the no. of days taken by worker “B” to complete the same work alone = 3x = 3 × 20
                                                                                                                                     = 60days.
Hence “B” can do the same work in 60 days alone.
OR
Number of days taken by workers “A” and “B” together to complete the work = D = 15 days
As assumed “A” can do the same work in D1 = x days.
And “B” can do the same work in D2 = 3x days.
Now Using formula:-
D = (D1 × D2)/( D1 + D2) = (x × 3x)/ (x + 3x)
Putting the values,
15 = (3x2)/(4x)
15 = (3x/4).
x = (15 × 4)/3. = 20 days (No. of days taken by “A” alone.)
And no. of days taken by worker “B” to complete the same work alone = 3x = 3 × 20
                                                                                                                                     = 60 days.
Hence “B” can do the same work in 60 days alone in 60 days.

(Note you may say that this is so long but dear this is the only way to explain the whole concept of the question to you now its upto you that how fast and short handedly you can solve these type of questions)         
                                                              
The Various Questions related to this and other concept which are frequently asked in the examinations will be presented to your for self study on this blog soon, so keep visiting and sending your suggestions and requirements too to us through the forum.


More types of questions will be followed soon on this blog........................

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