Free Practice Questions for Time-and-work in Maths

Let LCM of 60,30 and 20 be the total work.

Work done in 4 days by $A=4 \times 1=4$ unit
Work done in 5 th day $=1 \times(1+2+3)=6$ unit
Total work done in 5 days $=(4+6)$ unit $=10$ unit
Work done in $1 \mathrm{day}=\frac{10}{5}=2$ unit
Required days $=\frac{60}{2}=30$ days

Let the Lcm of 10,20 and 30 be the total work.

1st day $P$ and $Q$ will complete the work $=6+3=9$ unit
2nd day $P$ and $R$ will complete the work $=6+2=8$ unit

Total work done in two days $=9+8=17$ unit
So total work done in 6 days $=17 \times 3=51$
Remaining work $=60-51=9$ unit
Work done by $P$ and $Q$ in days $=\frac{9}{9}=1$ days
Total required days $=6+1=7$ days

Let efficiency of binni = 1 unit
So, anu's efficiency is $=4$ unit
Total work $=7 \times 5=35$ unit
Anu alone will complete the task $=\frac{35}{4}$ hours

A's one day's work $=\frac{1}{10}$
B's one day's work $=\frac{1}{15}$
Work done by $\mathrm{A}$ and $\mathrm{B}$ in one day $=\frac{1}{10}+\frac{1}{15}=\frac{3+2}{30}=\frac{1}{6}$
Total work done by $\mathrm{A}$ and $\mathrm{B}$ in 5 days $=5 \times \frac{1}{6}=\frac{5}{6}$
Remaining work $=1-\frac{5}{6}=\frac{1}{6}$

Let the efficiency of a boy is B and that of a girl is $\mathrm{G}$, then 

$15 \times B \times 10=12 \times G \times 15$
$
\frac{B}{G}=\frac{12}{10}=\frac{6}{5}
$
Total work $=15 \times 6 \times 10=900$
Total efficiency $=15 \times 6+12 \times 5=90+60=150$
Hence, time $=\frac{900}{150}=6$ days

$
20 \%=\frac{1}{5} \quad, 50 \%=\frac{1}{2}
$
As per the question,
$
\mathrm{A}=\frac{6}{5} \mathrm{~B} \quad, \mathrm{C}=\frac{1}{2} \mathrm{~B}
$
Efficiency ratio of $\mathrm{A}: \mathrm{B}: \mathrm{C}=12: 10: 5$
Total work $=27 \times 10=270$
B alone can complete the work $=\frac{270}{10}=27$ days

If A takes 15 days to complete $\frac{5}{7}$ of work .
So, A complete whole work $=\frac{15 \times 7}{5}=21$ days
Total work $=$ LCM of $(21$ and 12$)$
$
=84
$
Efficiency of $\mathrm{A}=\frac{\text { Total } \text { Work }}{\text { Total days }}=\frac{84}{21}=4$
Efficiency of $\mathrm{A}+\mathrm{B}=\frac{84}{12}=7$
So, Efficiency of $B=7-4=3$
B can complete the whole work $=\frac{84}{3}=28$ days

Time taken by A to complete total work $=\frac{4}{25 \%} \times 100 \%=16$ days
Time taken by B to complete total work $=\frac{12}{50 \%} \times 100 \%=24$ days
Time taken by $\mathrm{C}$ to complete total work $=32$ days
Total work $=96,(\mathrm{LCM}$ of 16,24 and 32$)$
Efficiency of $\mathrm{A}=\frac{96}{16}=6$
Efficiency of $\mathrm{B}=\frac{96}{24}=4$
Efficiency of $\mathrm{C}=\frac{96}{32}=3$
Let the total work was completed in $t$ days, then
$A(t-6)+B t+C \times 4=96$
$6(t-6)+4 t+3 \times 4=96$
$6 t-36+4 t=84$
$10 t=120$
$t=12$ days

Time taken by A to complete total work $=\frac{6}{40 \%} \times 100 \%=15$ days
Time taken by B to complete total work $=\frac{8}{33 \frac{1}{3} \%} \times 100 \%=24$ days
Total work $=120$ units (LCM of 15 and 24$)$
Efficiency of $\mathrm{A}=\frac{120}{15}=8$ units
Efficiency of $\mathrm{B}=\frac{120}{24}=5$ units
Work done by $\mathrm{A}$ and $\mathrm{B}$ in 8 days $=8(8+5)=104$ units
Remaining work $=120-104=16$ units
Remaining work was completed by $\mathrm{C}$ in 4 days, So
Efficiency of $\mathrm{C}=\frac{16}{4}=4$ units
Time taken by $\mathrm{A}$ and $\mathrm{C}$ to complete the total work $=\frac{120}{(8+4)}=10$ days

Total work $=\operatorname{LCM}$ of $(120,150$ and 200$)$$=600$ units
Efficiency of $\mathrm{A}+\mathrm{B}=\frac{600}{120}=5$ units $/$ day
Efficiency of $\mathrm{B}+\mathrm{C}=\frac{600}{150}=4$ units $/$ day
Efficiency of $\mathrm{A}+\mathrm{C}=\frac{600}{200}=3$ units/day
So efficiency of $\mathrm{A}+\mathrm{B}+\mathrm{C}=\frac{5+4+3}{2}=6$ units/day
Efficiency of $\mathrm{B}=$ Efficiency of $(\mathrm{A}+\mathrm{B}+\mathrm{C})$ $-$ Efficiency of $(\mathrm{A}+\mathrm{C})$
$=6-3=3$ units $/$ day
B can do $\frac{2}{5}$ th of the work $=\frac{2 \times 600}{5 \times 3}=80$ day

As per the question,
If $A$ can do $\frac{2}{5}$ th of work $=6$ days
Total work done by $A=6 \times \frac{5}{2}$
$=15$ days
B can do total work $=3 \times \frac{10}{3}=10$ days
Total work $=\mathrm{LCM}$ of $(15$ and 10$)$
$=30$ units
So efficiency of $\mathrm{A}=\frac{\text { Total Work }}{\text { Total Time }}=\frac{30}{15}=2$ units/day
Efficiency of $\mathrm{B}=\frac{\text { Total Work }}{\text { Total Time }}=\frac{30}{10}=3$ units $/$ day
$(\mathrm{A}+\mathrm{B})$ can complete the work in 4 days $=5 \times 4=20$
units Remaining Work $=30-20=10$ units
Efficiency of $\mathrm{C}=\frac{10}{6}=\frac{5}{3}$
$
\begin{aligned}
(\mathrm{B}+\mathrm{C}) \text { together can do } 70 \% \text { of work } &=\frac{30 \times \frac{70}{100}}{3+\frac{5}{3}} \\
&=\frac{21 \times 3}{14}=4 \frac{1}{2} \text { days }
\end{aligned}
$

Let the efficiency of Man and Woman is $M$ and W respectively.
According to question
$
5(4 M+4 W)=6(2 M+5 W)
$
$
\begin{aligned}
&(20-12) M=(30-20) W \\
&\frac{M}{W}=\frac{10}{8}=\frac{5}{4}
\end{aligned}
$
Total work $=20(5+4)=180$
Time taken by 1 woman $=\frac{180}{4}=45$
Time taken by 1 man $=\frac{180}{5}=36$
Hence, Required answer is $45$ and $36$.

Efficiency ratio of Machine to Man is $2: 1$, then
Total work $=60(10 \times 2+5 \times 1)=60 \times 25=1500$
Time taken by 10 men and 5 machines to complete the same work $=\frac{1500}{(10 \times 1+5 \times 2)}=\frac{1500}{20}=75$ days

As per the question,
$
\begin{aligned}
2 \mathrm{M} &=3 \mathrm{~W} \\
\frac{M}{W}=\frac{3}{2} \\
\Rightarrow \text { Total work } &=(10 \mathrm{M}+6 \mathrm{~W}) \times 5 \\
&=(10 \times 3+6 \times 2) \times 5 \\
&=210 \text { units }
\end{aligned}
$
$4 \mathrm{M}$ and $9 \mathrm{~W}$ can do the same the work $=\frac{210}{(4 \times 3+9 \times 2)}=\frac{210}{30}=7$ days

As per the question ,
Efficiency of $B=3 \times$ Efficiency of $A$
$
\frac{B}{A}=\frac{3}{1}
$
If $B$ complete the work in 28 days that means
Total work $=28 \times 3$
Work completed by both of them working together $=\frac{28 \times 3}{4}=21$ days

A takes time to complete work $=28$ hours
B takes time to complete work $=28 \times \frac{1}{4}=7$ hours
LCM of 28 and 7 be the total work $=28$ unit
So, efficiency of $A$ and $B$ is $1$ and $4$ respectively.
They will complete the work together $=\frac{28}{5}=5 \frac{3}{5}=5$ hour $36$ minute

Total efficiency of $A, B$ and $C=40$
B's 8 days work $=12 \times 8=96$
C's 12 days work $=7 \times 12=84$
Total work left $=96+84=180$
Total work $=420+180=600$
$\mathrm{A}, \mathrm{B}$ and $\mathrm{C}$ will together complete the work in days $=\frac{600}{40}=15$ days
All three work together $=15-12=3$ days

Total work is the L.C.M of 20,12 and 15 is 60

Total efficiency $=2($ Madhu $+$ shiney $+$ Rosy $)=3+5+4=12$
Madhu $+$ shiney $+$ Rosy $=\frac{12}{2}=6$
They will complete the work together $=\frac{60}{6}=10$ days

Efficiency of $A=6-3=3$
Efficiency of $\mathrm{C}=6-5=1$
Total efficiency $\mathrm{A}$ and $\mathrm{C}=3+1=4$
A and C complete the same work $=\frac{30}{4}=7 \frac{1}{2}$ days

According to the question
The ratio of efficiency of $\mathrm{A}: \mathrm{B}=7: 1$
The ratio of wages of $\mathrm{A}: \mathrm{B}=7: 1$
Daily wages of $\mathrm{A}$ and $\mathrm{B}$ together $=\frac{20800}{13}=1600$
Daily wages of $B=16000 \times \frac{1}{8}=200$