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 1st
:- Fundamental formulae based Questions:-
In these
type of questions you just have to read the questions carefully to analyse the
given data in the question and then calculate the required answer with the help
of formulae given in the last post of this blog.
Q1. 38 men,
working 6 hours a day can do a piece of work in 12 days. Find the number of
days in which 57 men working 8 hours a day can do twice the work. Assume that 2
men of the first group do as much work in 1 hour as 3men of the second group do
in 1
hours.
Solution:-
First of all you should read this question carefully and analyse it carefully
in such manner:-
The work in
this question is not same means its not constant and number of men, days,
number of articles (work) and number of hours per day are given for the first
group to complete the work along with the number of men, number of articles
(work) and number of hours per days for the double amount of work and number of
days is required for the second group of men, So the formula for this type of
question is :-
M1 ×
D1 × H1 × E1 × W2 = M2 × D2
× H2 × E2 × W1.
Let Efficiency of men of first group
= E1 while Efficiency of men of second group = E2
Data given :- M1 =
38, D1 =12, H1 =
6, W1 = 1, M2 = 57, D2. = ?, H2 = 8
& W2 = 2
And work done by 2 men of first
group in 1 hour = work done by 3 men of second group in 1.5 hours (→ 2×1 men of
first group = 3×1.5 men of second group)
In these type of questions number
men and their efficiency are inversely proportional to each other
Therefore E1 : E2
= 3×1.5 : 2×1 = 4.5 : 2 (This
implies that E1 = 4.5 and E2
= 2)
So as per our formula 38 × 12 × 6 × 4.5 ×2 = 57 × D2
× 8 × 2 × 1
On calculating we get 3D2 = 81
D2 = 81/3
D2 = 27 days.
Hence the
required answer of the question is 27 days.
Notes:- No
need to write unnecessary steps, just avoid them as per your requirement. Just
substitute the values in the formula direct and do the required calculations.
Type 2nd :- One day work Concept: -
The “one –
day work concept” is one of the most important concept to be understood in this
chapter. Here I am going to explain you this concept. Hope you get it right:-
1. If worker “A” can do a piece of work
in “x” days then the one day work of “A” = 1/x
2. If one day work of worker “A” is “1/x”
then he finish the work in “x” days.
3. If worker “A” can do a piece of work
in “x” days and worker “B” can do the same work in “y” days then
One
day work of worker “A” = 1/x
One
day work of worker “B” = 1/y
One
day work of workers “A” and “B” together:-
(A + B) = [(1/x) + (1/y)] = (x+y)/(x × y)
While
Worker “A” and worker “B” can finish the same work, working together in :-
[(x
× y)/(x + y)] days.
4. If worker “A” can do a piece of work
in “x” days, worker “B” can do the same work in “y” days and worker “C” can do
the same work in “z” days then
One
day work of worker “A” = 1/x
One
day work of worker “B” = 1/y
One
day work of worker “C” = 1/z
One
day work of workers “A” , “B” and “C” together:-
(A + B + C) = [(1/x) + (1/y) + (1/z)] = (xy + yz
+ zx)/(x × y × z)
While
Worker “A”, worker “B” and worker “C” can finish the same work, working
together in :-
[(x
× y × z)/(xy + yz + zx)] days.
The Various
Questions related to this 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.
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