## Class 10 Mathematics Note

# Time and Work

**Chapter – 2**

**Time and work **

The **unitary method** is a technique in mathematics for solving a problem finding the value of single unit, and then finding the necessary value by multiplying the single unit value.

**Variation **

This shows the relation between the two quantities

**Types of variation**

**Direct variation:**

If the increase in one quantity results in increase in other quantities and decrease in one quantity results in decrease in other quantities, then the relation between the quantities is said to be in direct variation.

Egif the cost of 2 chair is Rs 400 then, the cost of 4 chair is Rs 800

**Indirect variation:**

If the increase in one quantity results in decreasein other quantities and decrease in one quantity results in increasein other quantities, then the relation between the quantities is said to be in indirect variation. Time and work

Eg

20 men can do a piece of the work in 40 days

1 man can do a piece of the work in 80 days

** Time and work**

Time and work vary directly or indirectly depending upon the condition.

**Some important formula **

a. A can do a piece of the work in x days

A can do $\frac{1}{{\rm{x}}}{\rm{\: }}$of the work in 1 days${\rm{\: }}$

b. No of days =$\frac{{{\rm{total\: work\: done}}}}{{{\rm{work\: done\: in\: }}1{\rm{\: days\: }}}}$

c. Time and work (direct variation)

More work →more time

Less work → less time

d. men andtime (indirect variation)

More men↑less time ↓

Lessmen ↓more time ↑

e. If A_{1} men can do W_{1 }work in D_{1}days and A_{2 }can do W_{2 }work in D_{2 }days

$\frac{{{{\rm{A}}_1}{{\rm{D}}_1}}}{{{{\rm{W}}_1}{\rm{\: }}}} = \frac{{{{\rm{A}}_2}{{\rm{D}}_2}}}{{{{\rm{W}}_2}{\rm{\: }}}}{\rm{\: }}$

In case of hours

$\frac{{{{\rm{A}}_1}{{\rm{D}}_{1{\rm{\: }}}}{{\rm{H}}_1}}}{{{{\rm{W}}_1}{\rm{\: }}}} = \frac{{{{\rm{A}}_2}{{\rm{D}}_2}{{\rm{H}}_2}}}{{{{\rm{W}}_2}{\rm{\: }}}}$

f. If A can do a work in ‘x’ days and B can do the same work in ‘y’ days, then the number of days required to complete the work if A and B work together is$\frac{{{\rm{xy}}}}{{{\rm{x\: }} + {\rm{\: y}}}}$

**Example 1**

A can do apiece of the work in 6 days and B can do same work in 12 days. A worked alone for 2 days and B for 3 days. In how many days will the remaining work be completed if they work together?

Solution

A can do 1 work in 8 days

A can do $\frac{1}{8}$ of the work in 1day

A can do $\frac{1}{8}{\rm{*}}2{\rm{\: }}$of the work in 2 day

= $\frac{1}{4}{\rm{\: }}$of the work in 2 days${\rm{\: }}$

B can do 1 work in 12 days

B can do $\frac{1}{{12}}$work in 1 days

B can do $\frac{1}{{12}}{\rm{*}}3$work in 3 days

=$\frac{1}{4}$work in 3 days

Remaining work = 1 -$\frac{1}{4}{\rm{\: }} - \frac{1}{4}$ =$\frac{1}{2}$

A + B can do$\frac{1}{8} + \frac{1}{{12}}$of the work in 1 day

A + B can do a piece $\frac{5}{{24}}$ of the work in 1 day

A + B can do 1 work in $\frac{{24}}{5}$ days

A + B can do$\frac{1}{2}$ work in $\frac{{24}}{5}{\rm{*}}\frac{1}{2}$ days = 12/5 days

**Examples 2**

A takes twice as much as B and thrice as much as C to finish a work; working together they can finish the work in 10 days. Find the time each will take to finish the work separately.

Soln

The working capacity of B = 2 times the working capacity of A

B = 2 A

The working capacity of C= 2 times the working capacity of A

C = 3 A

Now, A + B+ C can finish 1 work 10 days

A + 2A+ 3A can finish 1 work 10 days

6 A can finish 1 work in 10 days

A cando 1work in 60 days

B can do 1 work in $\frac{1}{2}{\rm{\: }}$time taken by A = $\frac{1}{2}$*60 = 30 days${\rm{\: }}$

C can do 1 work in$\frac{1}{3}{\rm{\: time\: taken\: by\: A\: }} = {\rm{\: }}\frac{1}{3}{\rm{*}}60{\rm{\: }} = {\rm{\: \: }}20{\rm{\: days}}$

**Examples 3**

A ,B and C can finish a piece of the work in 30, 40 , and 60 days respectively . After working together for 10 days, B leaves. Then A leaves 4 days after B left and C completes the remaining work . Find how many days C has worked together.

Solution

A can do 1 work in 30days

A can do $\frac{1}{{30}}$ work in 1 days

B can do 1 work in 40 days

B can do $\frac{1}{{40}}$work in 1 days

C can do 1 work in 60days

C can do $\frac{1}{{60}}$ work in 1 days

A+ B+C can do a ($\frac{1}{{30}} + \frac{1}{{40}}{\rm{\: }} + \frac{1}{{60}}$ ) of the work in 1 day

A+ B+C can do a $\frac{{18}}{{240}}$ of the work in 1 day

A+ B+C can do a $\frac{{18}}{{240}}{\rm{*}}10$ of the work in 10 day

=${\rm{\: }}\frac{3}{4}$ of the work in 10 day

Remaining work = 1 -${\rm{\: }}\frac{3}{4}$= $\frac{1}{4}$

Then B leaves. A+ C did remaining work together for 4 days

A+ C can do $\frac{1}{{30}} + \frac{1}{{60}}$ of the work for 1 day

A+ C can do $\frac{1}{{20}}$ of the work for 1 day

A+ C can do $\frac{1}{{20}}$ *4 of the work for 4 day

=${\rm{\: }}\frac{1}{5}$ of the work for 4 day

R .W = $\frac{1}{4} - \frac{1}{5}$ =$\frac{1}{{20}}$

C can do 1 work in 60days

C can do $\frac{1}{{20}}$ work in 60*$\frac{1}{{20}}$days = 3days

Total days C has worked = 10 +4+ 3 = 17 days