Free Practice Questions for Time-and-work in Maths

Let the total work be $30 x$ units and time taken by $Y$ be $y$ days.
Then, time taken by $X$ in completing the work $=(y-5)$ days
And efficiency of $X$ and $Y$ together $=30 x / 6$
$\Rightarrow$ Efficiency of $X+$ Efficiency of $Y=5 x$ units/day
$\Rightarrow \frac{30 x}{(y-5)}+\frac{30 x}{y}=5 x$
$\Rightarrow x\left(\frac{30 y+30 y-150}{y(y-5)}\right)=5 x$
$\Rightarrow 60 y-150=5 y^{2}-25 y$
$\Rightarrow 5 y^{2}-25 y-60 y+150=0$
$\Rightarrow y^{2}-5 y-12 y+30=0$
$\Rightarrow y^{2}-15 y-2 y+30=0$
$\Rightarrow y(y-15)-2(y-15)=0$
$\Rightarrow(y-15)(y-2)=0$
$\Rightarrow y=15$ or $y=2$
2 will make time of $X$ negative
Therefore, time taken by $\mathrm{Y}$ is 15 days.
Hence, option C is the correct answer. 

Let the total work $=$ LCM of 20,35 and $60=420$ units

Work done by $B$ in 8 days $=8 \times 12=96$ units
Work done by $C$ in 12 days $=12 \times 7=84$ units
Therefore, required time to complete the work $=\frac{420+96+84}{(21+12+7)}=\frac{600}{40}=15$ days.

A and $B$ can complete $36 \%$ of the work $=9$ days
A and $B$ can complete whole work in $=\frac{9}{36} \times 100=25$ days
B can complete whole work in $=30$ days
Now,
$
\frac{1}{A} =\frac{1}{(A+B)} -\frac{1}{B}
$
$
\Rightarrow \frac{1}{\mathrm{~A}}=\frac{1}{25}-\frac{1}{30}
$
$
\Rightarrow \frac{1}{\mathrm{~A}}=\frac{6-5}{150}
$
$
\Rightarrow \frac{1}{A}=\frac{1}{150}
$
$
\Rightarrow A=150 \text { days }
$
Therefore, A can complete the work in 150 days
Hence, option C is correct.

A can do $40 \%$ of a work in $=24$ days
A can complete $100 \%$ of work in $=\frac{24}{40} \times 100=60$ days
B can do $33 \frac{1}{3} \%$ of the work in $=15$ days
B can complete $100 \%$ of the work in $=\frac{15}{\left(\frac{100}{3}\right)} \times 100=45$ days
Let total work $=$ LCM of $(60,45)=180$ units
Work done by $\mathrm{A}$ in 1 day $=\frac{180}{60}=3$ units
Work done by $B$ in 1 day $=\frac{180}{45}=4$ units
Work done by $A$ and $B$ in 15 days $=15(3+4)=15 \times 7=105$ units
Remaining work $=180-105=75$ units
C alone completes the remaining work in 10 days,
Work done by C in 1 day $=\frac{75}{10}=7.5$ units
Therefore, required time to complete the $35 \%$ of the work by $A$ and $C=\frac{180 \times \frac{35}{100}}{(3+7.5)}$
$
\begin{aligned}
&=\frac{180 \times 35}{10.5 \times 100} \\
&=6 \text { days }
\end{aligned}
$
Hence, option 2 is correct.

Let the efficiency of $X$ be a units/ $h$
Efficiency of $Y=(a) \times \frac{(100+25)}{100}=\frac{125 a}{100}=\frac{5 a}{4}$
Efficiency of $Z=\left(\begin{array}{c}5 a \\ 4\end{array}\right) \times\left(\begin{array}{c}(100+28) \\ 100\end{array}=\left(\frac{5 a}{4}\right) \times \frac{128}{100}=\frac{8 a}{5}\right.$
Total work $=40 \times a=40 a$
Therefore, required time taken by $X, Y$ and $Z$ to complete the work $=\frac{40 a}{\left(a+\frac{5 a}{4}+\frac{8 a}{5}\right)}$
$
\begin{aligned}
&=\frac{40 a}{\left(\frac{20 a+25 a+32 a}{20}\right)} \\
&=\frac{40 a \times 20}{77 a} \\
&=\frac{800}{77} \text { days }
\end{aligned}
$
Hence, option A is correct.

Let the L.C.M of (16,24) be $48 $ units.

Remaining work (done by Z in $10 \frac{1}{2}$ days)=48 - 20 =28 units

Therefore, the efficiency of Z =$\frac{28}{21}\times2= \frac{8}{3} $ units/day

Combined efficiency of $Y$ and $Z=2 +8/3 =14/3 $ units/day

Time taken by $Y$ and Z to complete $\frac{7}{8}$ of the same work

$=(48 ) \times \frac{7}{8} \times \frac{3}{14 }=9 \text {days}$

Let the efficiency of a man be m units/day and that of a woman be w units/day.
As per question, total work in both cases
(5m + 2w) × 9 = (11m + 5w) × 4
⇒ 45m + 18w = 44m + 20w
⇒ 45m – 44m = 20w – 18w
⇒ m = 2w
Now, total work = (5m + 2w) × 9
= (5 × 2w + 2w) × 9
= (10w + 2w) × 9
= 12w × 9 = 108w
Number of women required to complete the work in 6 days
= 108w/6 = 18
Therefore, 18 women are required to complete the work in 6 days.

Let the number of persons required be x

We know that:

W = MDH

Now, according to the question,
$
\begin{aligned}
&\Rightarrow 20 \times 15 \times 8=(x) \times 4 \times 10 \\
&\Rightarrow x=\frac{20 \times 15 \times 8}{4 \times 10} \\
&\Rightarrow x=60
\end{aligned}
$
Therefore, the number of persons required is 60 .

Let the total work $=30$ units
One day work of $A=\frac{30}{10}=3$ units
One day work of $B=\frac{30}{15}=2$ units
One day work of $A, B$ and $C=\frac{30}{5}=6$ units
One day work of $C=6-3-2=1$ unit
Ratio of the efficiency of $A, B$ and $C=3: 2: 1$
Therefore, the amount which was paid to $B=\frac{42000}{(3+2+1)} \times 2=\frac{42000}{6} \times 2=14000$ Rs.
Hence, option A is correct.

The total number of workers whose daily wages are less than ₹550 = 20 + 25 + 30 = 75

The total number of workers whose daily wages are ₹550 and above = 25 + 40 + 60 = 125

Therefore, required percentage =  =  = 40%

Hence, option C is correct.

Let x persons were originally there

According to question,

Total work = Total work

 46 × (x) = (46 - 16) × (x + 8)

 46x = 30x + 240

 46x – 30x = 240

 16x = 240

x =

 x = 15

Therefore, 15 persons were originally there.

Let the total work = LCM of (15, 25) = 75 units

Work done by A and B in 5 days = 5 × (5 + 3) = 5 × 8 = 40 units

Remaining work = 75 – 40 = 35 units

Now, B was replaced by C and the remaining work was completed by A and C in 4 days

Work done by A in next 4 days = 4 × 5 = 20 units

Remaining work which was done by C in 4 days = 35 – 20 = 15 units

Work done by C in one day = units

Therefore, required time taken by C to complete the work alone == = 20 days

Let efficiency of one man $=\mathrm{M}$ unit/day
Efficiency of one woman = W unit/day
We know that
Total work = Efficiency $\times$ Time taken
15 men can do a job in 12 days and 18 women can do it in 15 days.
$
\Rightarrow 15 M \times 12=18 \mathrm{~W} \times 15
$
$
\Rightarrow 2 M=3 W
$
$
\Rightarrow M=\frac{3}{2} W
$
Let 5 men and 3 women can complete the job in $x$ days.
$
\begin{aligned}
&\Rightarrow(5 M+3 W) \times x=18 W \times 15 \\
&\Rightarrow\left[5\left(\frac{3}{2} W\right)+3 W\right] \times x=18 W \times 15 \\
&\Rightarrow\left[\frac{15}{2} W+3 W\right] \times x=18 W \times 15 \\
&\Rightarrow\left[\frac{21}{2} W\right] \times x=18 W \times 15 \\
&\Rightarrow x=\frac{540}{21}=\frac{180}{7}=25 \frac{5}{7} \text { days }
\end{aligned}
$
Hence, 5 men and 3 women can complete the job in $25 \frac{5}{7}$ days .

A can do one-fourth of a work in 5 days

Number of days taken by A to complete the work =

B can do two-fifth of the work in 10 days.

Number of days taken by B to complete the work =

Let total work = LCM (20,25) = 100 unit

Efficiency of A and B together = 5 + 4 = 9 unit/day

Number of days taken byA and B together to complete the whole work =

Let the time required to complete the same work = x

We know that:

W = MDH

According to question,

Work = work

⇒ 14 × 18 = 21 × (x)

⇒ x =

⇒ x = 12 days

Therefore, the time required to complete the same work = 12 days

Ratio of his wages = 8500 : 6050 = 170 : 121

The ratio of his actual days and working days = 170 : 121

Therefore, he was absent = 170 – 121 = 49 days

Work done by $A$ in 1 day $=\frac{1}{60}$
Work done by $A$ in 15 days $=\frac{1}{60} \times 15=\frac{1}{4}$
Remaining work $=1-\frac{1}{4}=\frac{3}{4}$
B can complete $\frac{3}{4}$ th of the work $=30$ days
B can complete whole work $=\frac{30}{3} \times 4=40$ days
Required, time taken by $A$ and $B$ to complete the whole work-
$
\Rightarrow \frac{1}{(A+B)}=\frac{1}{A}+\frac{1}{B}
$
$
\Rightarrow \frac{1}{(A+B)}=\frac{1}{60}+\frac{1}{40}
$
$
\Rightarrow \frac{1}{(A+B)}=\frac{2+3}{120}
$
$
\Rightarrow \frac{1}{(A+B)}=\frac{5}{120}
$
$
\Rightarrow(A+B)=24 \text { Days }
$
Therefore, $A$ and $B$ working together can finish the whole work in 24 days.

Let the total work $=$ LCM of $8,12, \frac{120}{8}=120$ units
Efficiency of $A=\frac{120}{12}=15$ units/hour
Efficiency of $B=\frac{120}{\left(\frac{40}{3}\right)}=10$ units $/$ hour
Efficiency of all three pipes $=\frac{120 \times 3}{40}=\frac{120}{16}=9$ units $/$ hour
Efficiency of outlet pipe $C=9-15-10=-16$ units/hour
Therefore, required time taken by $C$ to empty the full tank $=\frac{1}{2}=7 \div$ hours
Hence, option B is correct.

If 84 persons take 56 days to complete a certain work, then they will complete one-fourth of the
$
\text { task }=\frac{\text { 1 }}{\mathcal{4}}=14 \text { days }
$
Remaining work$=1-\frac{1}{4}=\frac{3}{4}$
If one-seventh of the workers left, then remaining worker$=84 \times\left(1-\frac{1}{7}\right)=84 \times \frac{6}{7}=72$
We know that:
W = MDH
According to question,,
$\Rightarrow \frac{\left(\frac{1}{4}\right)}{\left(\frac{3}{4}\right)}=\frac{84 \times 14}{72 \times(x)}$
$\Rightarrow \frac{1 \times 4}{4 \times 3}=\frac{84 \times 14}{72 \times(x)}$
$\Rightarrow x=49$ days
Therefore, required total time to complete the entire task = 49 + 14 = 63 days

Clearly, Ratio of efficency of $A, B$ and $C$ respectively

$ Total work =\text { LCM of } 8,10,12=120 $

$=\frac{120}{8}: \frac{120}{10}: \frac{120}{12}$

$=15: 12: 10 $

So, B's share $=\frac{5550}{(15a+12+10)} \times 12=150 \times$ 12 $= Rs.1800$