# SSC Quantitative Aptitude Time And Work Work and Time

Work and Time

Category : SSC

Work and Time

Problems from this chapter are solved with the help of the concept used in elementary algebra. This concept is called unitary method.

Unitary method used for solving a class of problems in variation. It consists of altering one of the variable to a single unit i.e., 1 and then performing the operation necessary to alter it to the desired value.

BASIC CONCEPTS OF WORK AND TIME

(x) How to calculate the required number of persons to complete a particular work in a stipulated time period?

(y) How to calculate the required time to complete a particular work by certain number of persons?

To find (x) and (y) the 1st and foremost task is to keep following basic points in mind

(i) While solving problems, the work done is always supposed to be equal to 1.

(ii) If a person can do a piece of work in n days, then that person's 1 day's work$=\frac{1}{n}.$

(iii) If a person's 1 day's work$=\frac{1}{n},$then the person will complete the work in n days.

(iv) A person works equally everyday.

Some Useful Facts/Formulae

 If A can do a piece of work in a days and B can do the same work in b days, then both (A and B) of them working together will do the same work in $\left( \frac{ab}{a+b} \right)$days.

e.g., A can finish a piece of work by working alone in 6 days and B while working alone, can finish the same work in 12 days. If both of them work together, then in how many days, the work will be finished?

Sol. Here$a=6,$$b=12$

$\therefore$Required time to complete the work$=\frac{ab}{a+b}=\frac{6\times 12}{6+12}$

= 4 days

 If A, B and C, while working alone can complete a work in a, b and c days, respectively, then they will together complete the work in $\left( \frac{abc}{ab+bc+ca} \right)\text{days}$

e.g., If A, B and C while working alone can complete a work in 5, 6 and 4 days, respectively, then in how many days, the work will be finished ?

Sol. The required time$=\left( \frac{a\times b\times c}{ab\times bc\times ca} \right)\text{days}$

$=\left( \frac{5\times 6\times 4}{5\times 6\times 4\times 6\times 5} \right)\text{days}$



 If ${{m}_{1}}$ persons can do ${{w}_{1}}$work in ${{d}_{1}}$days working ${{t}_{1}}$   in a day and ${{m}_{2}}$persons can do ${{w}_{2}}$work in ${{d}_{2}}$days working ${{t}_{2}}$ h in a day, then we have a very basic and all in one relationship as$\left[ \frac{{{m}_{1}}{{d}_{1}}{{t}_{1}}}{{{w}_{1}}}=\frac{{{m}_{2}}{{d}_{2}}{{t}_{2}}}{{{w}_{2}}} \right]$

e.g. 16 men can do a piece of work in 10 days. How many men are needed to complete the work in 20 days?

Sol. Here ${{m}_{1}}=16=10,$${{w}_{2}}=1,$${{d}_{2}}=20,$${{m}_{2}}=?$

According to the formula,

$\frac{{{m}_{1}}{{d}_{1}}{{t}_{1}}}{{{w}_{1}}}=\frac{{{m}_{2}}{{d}_{2}}{{t}_{2}}}{{{w}_{2}}}$

$\frac{16\times 10\times T}{1}=\frac{{{m}_{2}}\times 20\times T}{1}$

${{m}_{2}}\times 20=16\times 10$

$\therefore$                  ${{m}_{2}}=\frac{16\times 10}{20}=8\,\,\text{men}$

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