Free Practice Questions for Time-and-work in Maths

Let efficiency of manisha be $=1$ unit
Madhur is thrice as good as Manisha
So, Efficiency of Madhur $=3$ unit
Total work $=4 \times 15=60$
Madhur alone will complete the task in days $=\frac{60}{3}=20$ days

Let total work be the LCM of 10,15 and $20=60$

Efficiency of $2(\mathrm{~A}+\mathrm{B}+\mathrm{C})=6+4+3=13$
$
(\mathrm{A}+\mathrm{B}+\mathrm{C})=\frac{13}{2}
$
Efficiency of $\mathrm{C}=(\mathrm{A}+\mathrm{B}+\mathrm{C})-(\mathrm{A}+\mathrm{B})$
$
=\frac{13}{2}-6=\frac{1}{2}
$
$\mathrm{C}$ alone finish the work in days $=\frac{60}{\frac{1}{2}}=120$ days

  Let the total work be the LCM of 16,18 and 24 = 144.

Efficiency of $\mathrm{A}=(\mathrm{A}+\mathrm{B}+\mathrm{C})-(\mathrm{B}+\mathrm{C})=9-6=3$
Efficiency of $\mathrm{C}=(\mathrm{A}+\mathrm{B}+\mathrm{C})-(\mathrm{A}+\mathrm{B})=9-8=1$
Total efficiency of $\mathrm{A}+\mathrm{C}=3+1=4$
A and C complete the same work $=\frac{144}{4}=36$ days

8 taps fill the tank together in 27 minutes.
8 taps fill the tank together in 3 minutes $=\frac{27}{3}=\frac{1}{9}$
Remaining fills in the tank $=1-\frac{1}{9}=\frac{8}{9}$
We know, $\frac{M_1 \times D_1 \times H_1}{W_1}=\frac{M_2 \times D_2 \times H_2}{W_2}$
$
\begin{aligned}
&\Rightarrow \frac{8 \times 27}{1}=\frac{6 \times H}{\frac{8}{9}} \\
&\Rightarrow \mathrm{H}=\frac{8 \times 27 \times 8}{6 \times 9}=32 \text { minutes }
\end{aligned}
$

A's one day's work $=\frac{1}{15}$
B's one day's work $=\frac{1}{20}$
One day's work of both together $=\frac{1}{15}+\frac{1}{20}=\frac{4+3}{60}=\frac{7}{60}$

Let the efficiencies of Kelvin and Kim are $x$ and $y$ respectively,
According to question
$25(x+y)=15(x+y)+12 x$
$10(x+y)=12 x$
$10 y=2 x$
$\frac{x}{y}=\frac{10}{2}=\frac{5}{1}$
Total work $=25(5+1)=150$
Time taken by Kelvin to complete the total work, working alone $=\frac{150}{5}=30$ days

Let ${x}$ be the required number of days to complete the work.
According to question
$6$ Men $\times 9$ days $=3$ Men $\times x$ days
${x}=18$ days

Work done by $A$ in 4 days $=15 \times 4=60$
Remaining work $=240-60=180$
Total efficiency of $\mathrm{A}+\mathrm{B}=15+16=31$
Required days to finished the work by $\mathrm{A}$ and $\mathrm{B}=\frac{180}{31}=5 \frac{25}{31}$ days

Let the total work be $6, ($LCM of $3$ and $6)$
Efficiency of $\mathrm{A}$ and $\mathrm{B}=\frac{6}{3}=2$
Efficiency of $\mathrm{C}$ and $\mathrm{D}=\frac{6}{6}=1$
Time taken by $\mathrm{A}, \mathrm{B}, \mathrm{C}$ and $\mathrm{D}$ working together $=\frac{6}{(1+2)}=\frac{6}{3}=2$ days

Let the efficiencies of Men and Women are $M$ and W respectively, then
$
\begin{aligned}
&4(5 W+2 M)=3(6 W+3 M) \\
&20 W+8 M=18 W+9 M \\
&2 W=1 M \\
&\frac{M}{W}=\frac{2}{1}
\end{aligned}
$
Total work $=4(5 W+2 M)=4(5+4)=36$
Time taken by one woman and one man working together to complete the total work $=\frac{36}{(1+2)}=\frac{36}{3}=12$ days

Total work = LCM of $(8 , 12$ and $4 ) = 24$
Efficiency of $A=\frac{24}{8}=3$
Efficiency of $\mathrm{B}=\frac{24}{12}=2$
Efficiency of $\mathrm{A}+\mathrm{B}+\mathrm{C}=\frac{24}{4}=6$
So, Efficiency of $\mathrm{C}=($ Efficiency of $\mathrm{A}+\mathrm{B}+\mathrm{C})-$ (Efficiency of $(\mathrm{A}+\mathrm{B})=6-5=1$
$C$ complete the work $=\frac{24}{1}=24$ days

Let $x$ men decided to do the work.
By MDH formula ,
$
\begin{aligned}
&\frac{M_1 \times D_1 \times H_1}{W_1}=\frac{M_2 \times D_2 \times H_2}{W_2} \\
&x \times 20=(x-5) \times 24 \\
&4 x=120 \\
&x=30
\end{aligned}
$
So originally there were 30 men.

A + B can do a work = 36 days
B+ C can do the same work = 40 days
A+B+C can do it = 30 days
LCM of (36, 40 , 30) = 360
So, Total work = 360
Efficiency of $\mathrm{A}+\mathrm{B}=\frac{360}{36}=10$
Efficiency of $\mathrm{B}+\mathrm{C}=\frac{360}{40}=9$
Efficiency of $\mathrm{A}+\mathrm{B}+\mathrm{C}=\frac{360}{30}=12$
So, efficiency of $\mathrm{C}=($ Efficiency of $\mathrm{A}+\mathrm{B}+\mathrm{C}) - $(Efficiency of $\mathrm{A}+\mathrm{B})$
$
=12-10=2
$
Efficiency of $\mathrm{B}=9-2=7$
B alone can do $\frac{7}{8}$ part of the original work $=\frac{360 \times \frac{7}{8}}{7}$
$=45$ days

By MDH formula,
$
\frac{M_1 \times D_1 \times H_1}{W_1}=\frac{M_2 \times D_2 \times H_2}{W_2}
$
Let the total work be 3 units.
$\frac{75 \times 20}{1}=\frac{30\times[M_2]}{2}$
$
M_2=100
$
So more workers required $=100-75=25$.

Let the total work be 40 unit , then

Efficiency of X $= \frac{40}{40} = 1$

And Efficiency of X+Y $= \frac{40}{8} = 5$

So Efficiency of Y $=5-1=4$

Time taken by Y to complete total work $=\frac{40}{4}=10$ Days

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

Efficiency of $X, Y$ and $Z$ $=\frac{5 +4 +3}{2}=\frac{12}{2} = 6$
Therefore, required time taken by $X, Y$ and $Z=\frac{12}{6}=2$ days.

Work done by $A$ and $B$ in 3 days $=3 \times(12+7)=57$ units
Remaining work $=84-57=27$ units
Efficiency of $C=\frac{27}{9}=3$ units $/$ day
Required time taken by C to complete the whole work $=\frac{84}{3}=28$ days

ATQ,
$\begin{array}{cccc} & \mathrm{A} & & \mathrm{B} \\ \text { Days } & 3 & : & 5 \\ \text { Efficency } & 5 & : & 3\end{array}$
$(A+B)$ together complete work $=5 \frac{5}{8}=\frac{45}{8}$ $\times 8$
$
=45 \text { days }
$
A alone do twice the work $=\frac{45}{5} \times 2=18$ days

Given:
$A$ and $B$ can complete $45 \%$ of a work in $18$ days.
$A$ and $B$ can complete the whole work in $=\frac{18}{45} \times 100=40$ days
A alone can complete the whole work $=60$ days
Let total work $=\mathrm{LCM}$ of $40,60=120$ units
Work done by $A$ in a day $=\frac{120}{60}=2$ units
Work done by $A$ and $B$ in 1 day $=\frac{120}{40}=3$ units
Work done by B in 1 day $=3-2=1$ units
Work done by $A$ and $B$ in 16 days $=16 \times 3=48$ units
Remaining work $=120-48=72$ units
Therefore, required time taken by B to complete remaining work $=\frac{72}{1}=72$ days
Hence, option C is correct.