Free Practice Questions for Time-and-work in Maths

Let the total work $60$ , (LCM of 12 and 20) , then

Efficiency of A $=\frac{60}{12} = 5$

Efficiency of B $=\frac{60}{20} = 3$

A and B worked for 6 days so they completed $6(5+3) = 48$

Remaining work $60-48=12$

Remaining work was completed by C in 12 days so,

Efficiency of C $=\frac{12}{12} = 1$

to complete $\frac{2}{3}$ of the total work , 

The time taken by A and C $=\frac{2}{3}\times\frac{60}{5+1} = \frac{2}{3}\times10 = 6\frac{2}{3}$ days

Let the total work be $60$ , (LCM of 12, 10 and 15)

So,

Efficiency of A $=\frac{60}{12} = 5$

Efficiency of B $=\frac{60}{10} = 6$

Efficiency of C $=\frac{60}{15} = 4$

They all worked together for 3 days so the work completed in 3 days $= 3(5+6+4) = 45$

Remaining work $=60-45 = 15$

Time taken by A to complete the remaining work $=\frac{15}{5} = 3$ Days

Let the wages of Men is $M$ and Women is $W$ , then

$5M+3W = \frac{8400}{4} = 2100$ Rs ..........(i)

& $9M+7W = \frac{28700}{7} = 4100$ Rs..............(ii)

Multiply equation (i) by 7 and equation (ii) by 3, so

$35M+21W = 14700$ ........(iii)

$27M+21W = 12300$ .......(iv)

Subtracting equation (iv) from equation (iii) - 

$8M = 2400$ Rs

$M = 300$ Rs 

Putting the value of $M$ in equation (i) 

$5\times300+3W = 2100$

$3W = 600$

$W=200$

So, Wages of $8M$ and $5W$ for 1 day $=8\times300+5\times200 = 2400+1000 = 3400$ Rs

Let $8M$ and $5W$ take $t$ days to earn $30600$ Rs , then

$3400\times t = 30600$

$t = 9$ days

 Time $\propto\frac{1}{\text{Efficiency}}$

Since the work of 8 hr is completed so,

Remaining work $= (24-8)\times 3 = 16\times 3 = 48$

Time taken to complete remaining work $=\frac{48}{8}= 6$ hours

Total time $= 8+6 = 14$ hours

 Time $\propto\frac{1}{\text{Efficiency}}$

Since the work of 7 hr is completed so,

Remaining work $= (28-7)\times 4 = 21\times 4 = 84$

Time taken to complete remaining work $=\frac{84}{7}= 12$ hours

Total time $= 12+7 = 19$ hours

X completes $\frac{1}{4} $ of the total work in 3 days.

So, to complete total work, the time taken by X $=3\times \frac{4}{1} = 12$ days

Y completes $\frac{1}{6} $ of the total work in 4 days.

So, to complete total work, the time taken by Y $=4\times \frac{6}{1} = 24$ days

So, Work completed by X and Y in 6 days $=(2+1)\times6 = 18$ units

Remaining work $=24-18 = 6$ units

6 units work was completed by Z in 8 days

So, Efficiency of Z $=\frac{6}{8} = \frac{3}{4}$

Efficiency of Y+Z $=1+\frac{3}{4} = \frac{7}{4}$

So, Time taken by them to complete total work $= \frac{24}{7/4} = \frac{96}{7} = 13\frac{5}{7}$ days

We know that - 

$\frac{M_1D_1H_1}{W_1} = \frac{M_2D_2H_2}{W_2}$

Where $M_1 = 12 , D_1 = 8,W_1 = 20 , D_2 = 6$ and $W_2 = 30$ 

$\frac{12\times8}{20} = \frac{M_2\times6}{30}$

$M_2 = \frac{96\times30}{20\times6}$

$M_2 = 24$

Total of 24 persons are required to complete the said work in said time.

Let the efficiency of Radha is 1, then efficiency of Kirti will be 2.

So, total work $=30\times1 = 30$

Then, time taken by both together to complete the work $=\frac{30}{1+2} = \frac{30}{3} = 10$ days

Let the efficiency of the Men is M and that of Women is W, then

$10\times15\times W = 15\times5\times M$

$\frac{M}{W} = \frac{2}{1}$

Total work $=10\times15\times1 = 150$

10 Women worked for 3 days so work completed by them in 3 days $=3\times10\times1 =30 $

Remaining work $=150-30 = 120$

The remaining work is to be completed by 5 Men and 5 Women 

So the time taken by  them $=\frac{120}{5\times1+5\times2}= \frac{120}{15} = 8$ days

Total time $=3+8 = 11$ days

Let the efficiency of Honey and Junnu is $\mathrm{H}$ and $\mathrm{J}$, respectively, then
$
\begin{aligned}
&20(H+J)=12 H+36 J \\
&8 H=16 J \\
&\frac{H}{J}=\frac{2}{1}
\end{aligned}
$
Total work $=20(2+1)=60$
Time taken by Junnu to complete total work $=\frac{60}{1}=60$ days

25 women can do a piece of work in 60 days

Let 15 more women join the work after x days

25 × 60 = 25 × x + 40 × (45 - x) 

1500 = 25x + 40 × 45 - 40x

15x = 1800 - 1500 

15x = 300 

x = 20 days

LCM of 10 and 15 = 30

Total Work done by (A+B) in 4 days  = (3+2)×4 = 20 units

Remaining Work =(30-20)=10 units 

The efficiency of C for completing the remaining work =10/12= (5/6) unit per day

Time is taken by C to complete 4/9 part of the original work =30×(4/9)×(6/5) =16 days

$\frac{1}{2}$ of total work is  =Time $\times$ number of people =$ 300 \times 60$

= 18000 

Total work is = 36000

Let number of remaining people is $=  x$

According to question 

Total work = $\frac{1}{2}$ of total work + work done in 120 days by 100 people + work done in 100 days by $x$ people  

$36000 = 18000 +  120 \times{100}  + 100\times{x}$    

$36000-30000 =100x$

$x= 60$

18×15×6=12× Number of working days  ×10

Number of working days =$\frac{18×15×6}{12×10}$

Number of working days =$\frac{27}{2}$

Number of working days =13$\frac{1}{2}$ days.

As per the question,
Since A and B worked for 5 days , So Work done by A and B together $=14 \times 5=70$
Remaining Work $=90-70=20$
Efficiencies of $\mathrm{A}$ and $\mathrm{C}=\frac{20}{3}$
Since efficiency of $\mathrm{A}=5$ then efficiency of $\mathrm{C}=\frac{20}{3}-5=\frac{5}{3}$
$\mathrm{C}$ alone can complete the original work $=\frac{90 \times 3}{5}$
$
=54 \text { days }
$

If $\mathrm{A}$ and $\mathrm{B}$ work for 6 days then work completed by them $=6 \times 8$ $=48$
Remaining work $=90-48=42$
Efficiency of $\mathrm{C}=\frac{42}{7}=6$
$\mathrm{A}, \mathrm{B}$ and $\mathrm{C}$ together can complete the same work $=\frac{90}{14} = 6 \frac{3}{7} \text { days }$

Hence, Share of $A=\frac{4}{9} \times 22500=$₹ $10000$

As the work done in both cases is equal, So 

$25 \times 36=12 \times D$
$D=75$ days

As per the question,
Since A and B worked for 6 days, So Work done by A and B together $=7 \times 6=42$
Remaining Work $=90-42=48$
Efficiencies of $\mathrm{A}$ and $\mathrm{C}=\frac{48}{3}=16$
Since efficiency of $\mathrm{A}=4$ then efficiency of $\mathrm{C}=12$
C alone can complete the original work $=\frac{90}{12}=7 \frac{1}{2} \text { days }$

Since Total work remains same.
ATQ,

$(\mathbf{x}+\mathbf{y}) \times 6=\left(\mathbf{3 x}+\frac{y}{4}\right) \times 4$
$\frac{x}{y}=\frac{5}{6}$,

Total work $=$ Efficiency $\times$ days
Total work done by $\mathrm{x}$ and $\mathrm{y}=11 \times 6 = 66$
$\mathrm{x}$ alone can complete the work $=\frac{66}{5}=13 \frac{1}{5}$ days