Free Practice Questions for Time-and-work in Maths

They work alternatively

The wage will be distributed on the basis of part of work done by them.

P will work for 7 days

Work done by $\mathrm{P}=3 \times 7=21$

$Q$ will work for 7 days

work done by $Q=2 \times 7=14$

$\mathrm{R}$ will work for $6 \frac{1}{9}$ days

Work done by $R=9 \times \frac{55}{9}=55$

Ratio of work done

$\mathrm{P} \quad \mathrm{Q} \quad \mathrm{R}$

$21: 14: 55$

90 unit $= Rs 5400$ 

$\mathrm{R}=55\times60$ unit $=$ Rs $.3300$

Let the time taken by $A$ and $B$ to finish the work when working together be ' $x$ ' days

According to the question,

$ \frac{1}{x+8}+\frac{1}{x+18}=\frac{1}{x} $

$\Rightarrow 2 \mathrm{x}^2+26 \mathrm{x}=\mathrm{x}^2+26 \mathrm{x}+144 $

$ \Rightarrow \mathrm{x}^2=144$

$\Rightarrow \mathrm{x}=12$ days

time is taken for both to work together

 =12 days

Let, total work = 50 unit
A do $4 / 5$ th work $=4 / 5 \times 50=40$ unit
Efficiency of $A=\frac{40}{20}=2$
So,
Work left $=50-40=10$ unit
A's 3 days work $=3 \times 2=6$ unit
Left work $=10-6=4$ unit
Efficiency of $B=\frac{4}{3}$
' $\mathrm{B}$ ' alone take to do two times the original work =
$
\frac{100}{\frac{4}{3}}=\frac{300}{4}=75 \text { days }
$

Clearly, in 7 days working together A$\&$B will do $\frac{7}{16}$ part of the work
$ \frac{9}{16}$ work is done by A alone in 21 days
Total work is done by A alone in $21 \times \frac{16}{9}$ $=\frac{112}{3}$ days
Now,

B's 1 day work $=7-3=4$ unit
Required time $=\frac{112}{4}=28$ days

According to question

$42 \times 63=x \times 54$

$x=\frac{42 \times 63}{54}=49$

Hence, $4 x=4 \times 49=196$

A's 1 days' work $=\frac{1}{45}$, B's 1 days' work $=\frac{1}{40}$

A left the work after $x$ days of the start of the work.

Then, Work done by A in $\mathrm{x}$ days $=\frac{\mathrm{x}}{45}$

Work done by B in $(x+23)$ days $=\frac{(x+23)}{40}$

Given, 

$\frac{\mathrm{x}}{45}+\frac{(\mathrm{x}+23)}{40}=1$

$\Rightarrow \frac{(8 \mathrm{x}+9 \mathrm{x}+207)}{360}=1$

$\Rightarrow 17 \mathrm{x}=360-207=153 $

$\Rightarrow \mathrm{x}=\frac{153}{17}=9$ days

$\therefore$ A left the work after 9 days.

$Amlesh$ completes $25 \%$ of the total work in 15 days, so

Time taken to complete whole work $=\frac{15}{25} \times 100=60$ days

$\mathrm{Bablu}$ completes $75 \%$ of the total work in 30 days, So

Time taken to complete whole work $=\frac{30}{75} \times 100=40$ days

Let the total work be $120,(\mathrm{LCM}$ of 60 and 40$)$

Efficiency of $\mathrm{Amlesh}=\frac{120}{60}=2$

Efficiency of $\mathrm{Bablu}=\frac{120}{40}=3$

So, time taken to complete $30 \%$ of total work $=\frac{30}{100} \times \frac{120}{(2+3)}=\frac{36}{5}=7\frac{1}{5}$ days

Let the total work be 180 . (LCM of 12,15 and 18 )
Efficiency of $\mathrm{A}=\frac{180}{12}=15$
Efficiency of $\mathrm{B}=\frac{180}{15}=12$
Efficiency of $\mathrm{C}=\frac{180}{18}=10$
They worked together, So
Ratio of efficiencies $=$ Ratio of wages
Total wages $=15+12+10=37$ unit
Given that
$37 \rightarrow $₹ 6660
$1 \rightarrow $₹ 180
So, Share ofC
$10 \rightarrow $₹ 1800

Let the total work be 144 , (LCM of 48 and 72), So
Efficiency of $\mathrm{A}=\frac{144}{48}=3$
Efficiency of $\mathrm{B}=\frac{144}{72}=2$
They worked for 12 days
So, Work completed in 12 days $=12 \times(3+2)=60$
Remaining work $=144-60=84$
So, $25 \%$ of 84 work is to be completed by B
Time $=\frac{25}{100} \times \frac{84}{2}=\frac{21}{2}=10 \frac{1}{2}$ days
Or 10 days $\frac{1}{2} \times 24 h r=10$ days 12hours

Let the total work be $76,(\mathrm{LCM}$ of 38,76 and 19 )

So, Efficiency of $\mathrm{X}=\frac{76}{38}=2$

Efficiency of $\mathrm{Y}=\frac{76}{76}=1$

And Efficiency of $\mathrm{Z}=\frac{76}{19}=4$

Let the total time taken by $\mathrm{X}$ to complete the whole work be t, then

$2 \times t+1 \times(t-2)+4 \times(t-8)=76$

$2 t+t-2+4 t-32=76$

$7 t=110$

$t=\frac{110}{7} \mathrm{hr}$

Ratio of efficiency of $\mathrm{A}$ and $\mathrm{B}=50 \%: 100 \%=1: 2$
Total work $=16 \times(1+2)=16 \times 3=48$
So, Time taken by A to complete total work $=\frac{48}{1}=48$ days

According to question
Given, $6 \mathrm{~A}=8 \mathrm{C}$
$
\frac{A}{C}=\frac{4}{3}
$
given, $6 \mathrm{~B}=4 \mathrm{C}$
$
\frac{B}{C}=\frac{2}{3}
$
We get, $\mathrm{A}: \mathrm{B}: \mathrm{C}=4: 2: 3$
Let total work $=54 \times 4=216$ unit
B require time to finish work $=\frac{216}{2}=108$ days

Lcm of 10 and 15 be the total work.

Total efficiency of 1 st and 2 nd $\mathrm{AC}=3+2=5$ unit
Required time to cool the hall together $=\frac{30}{5}=6 \mathrm{~min}$

One day work for $\mathrm{A}$ and $\mathrm{B}=3+2$
$=5$
A and work for 4 days $=5 \times 4$
$=20$

$\begin{aligned} \text { Remainng work } &=36-20 \\ &=16 \end{aligned}$
$A$ and $C$ completed remaining work in 4 days

One day work for $A$ and $C=\frac{16}{4}$

$=4$
Then efficiency of $\mathrm{C}=4$ - efficiency of $\mathrm{A}$

$=4-3$

$=1$

one-third of the work $\begin{aligned} &=\frac{1}{3} \times 36 \\ &=12 \end{aligned}$
$\begin{aligned} \text { C can complete one-third of the same work } &=\frac{12}{1} \text { days } \\ &=12 \text { days } \end{aligned}$

$A+B=30$ days
$A=50$ days
$L C M$ of 50 and 30 will be 150
Then efficiency of $A+B=5$ unit/days
Efficiency of $A=3$ unit/days
Efficiency of $B= 5-3 = 2$ unit/days
B can do alone  complete the work $=\frac{\text { total work }}{\text { efficiency of B }}$
$=\frac{150}{2}$
$=75$ days

Let $x$ be the required machine.
According to question
56 Machine $\times 144$ Days $=x \times 96$ Days
$
\begin{aligned}
&\mathrm{x}=\frac{96 \text { Days }}{56 \text { Machine } \times 144 \text { Days }} \\
&\mathrm{x}=84 \text { Machine }
\end{aligned}
$

Let efficiency of $A=10$ unit/day
So, efficiency of B $=9$ unit/day
efficiency of $\mathrm{C}=12 \mathrm{unit} /$ day
Total efficiency of $\mathrm{A}+\mathrm{B}+\mathrm{C}=10+9+12=31$ unit
Total work $=10 \times 15=150$ unit
Work done by $\mathrm{A}+\mathrm{B}+\mathrm{C}$ in 3 days $=31 \times 3=93$ unit
Remaining work 150-93= 57 unit
Work done by D in days $=150 \times \frac{19}{2} \times \frac{1}{57}=25$ days

According to question
$
\begin{aligned}
&42 \times 63=x \times 54 \\
&x=\frac{42 \times 63}{54}=49
\end{aligned}
$
Hence, $3 x=3 \times 49=147$

B is $25 \%$ more efficient than A
So, $\frac{A}{B}=\frac{4}{5}$
$\mathrm{C}$ is $20 \%$ more efficient than $\mathrm{B}$
So, $\frac{B}{C}=\frac{5}{6}$
Comparing $\mathrm{A}, \mathrm{B}$ and $\mathrm{C}$ we get
$\mathrm{A}: \mathrm{B}: \mathrm{C}=4: 5: 6$
Total work $=20 \times 4=80$ unit
$\mathrm{A}$ and $\mathrm{C}$ work together done the work $=(4+6) \times 5=50$
Remaining work $=80-50=30$ unit
B alone will complete the remaining work in days $=\frac{30}{5}=6$ days

Let the Lcm of 8 and 12 be the total work $=24$ unit

Efficiency of Geeta $=3-2=1$
Gita alone will complete the job $=\frac{24}{1} = 24$ hours