Free Practice Questions for Time-and-work in Maths

Let the efficiency of $Y=5$
Efficiency of $X=5 \times \frac{(100+60)}{100}=8$
Total work $=5 \times 80=400 $
Therefore, required time to complete $52 \%$ of the same work by $A$ and $B$ = $\frac{400 \times \frac{52}{100}}{8+5}=400 \times \frac{52}{100} \times \frac{1}{13}=16 \text { days }$
Hence, option A is correct.

Let the time required to complete the same work = x

We know that:

W = MDH

According to question,

Work = work

$\Rightarrow 14 \times 18=21 \times(x)$
$\Rightarrow x=\frac{14 \times 18}{21}$
$\Rightarrow x=12$ days

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

Hence, option D is correct.

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

Ratio of his actual days and working days = 170 : 121

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

Hence, option A is correct.

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.
Hence, option D is correct.

According to the question, 

Total work = 48

Now, 2(A + B + C) = 9

Efficiency of (A + B + C) = 4.5

So. they will take  = $\frac {48}{4.5}$

⇒ 10.67 days

∴ They will take 10.67 days to complete the work.

ATQ,
LCM (12, 30) = 60

(K+M) work together = 5 unit/day
Malti can do work = 2 unit/day
kriti alone complete the work = $\frac{60}{5-2}$
$=\frac{60}{3}=$ 20 day

ATQ,
2× 30 Men = 3 × 30 Women
$\frac{\text { Men }}{\text { Women }}$ = $\frac{3}{2}$
Total work = 2 × 3 × 30 = 180 unit
(6 Men + 1 Women) Complete the work
= $\frac{180}{6 \times 3+2 \times 1}$ = $\frac{180}{20}$ = 9 day

MATHS MIRROR APPROACH -

Let total work $=\operatorname{LCM}(80,96,32)=480$ units
Efficiency of $A=\frac{480}{80}=6$ unit / day
Efficiency of $B=\frac{480}{96}=5$ unit / day

Efficiency of $A+B+C=\frac{480}{32}=15$ unit / day

Efficiency of $C= 15-(6+5) = 4

Number of days taken by C to complete the whole work $=\frac{480}{4}=120$ days

BASIC APPROACH -

Number of days taken by A to complete the whole work $=80$ days
Number of days taken by B to complete the whole work = 96 days
Let total work $=\operatorname{LCM}(80,96)=480$ units
Efficiency of $A=\frac{480}{80}=6$ unit/day
Efficiency of $B=\frac{480}{96}=5$ unit / day
They completed the work in 32 days with the help of C.
Efficiency of $A, B$ and $C$ altogether $=\frac{480}{32}=15$ unit/day
Efficiency of $C=15-6-5=4$ unit/day
Number of days taken by C to complete the whole work $=\frac{480}{4}=120$ days

MATHS MIRROR APPROACH:

Time taken by $A = \frac{9}{5}$$\times $$($Time taken by $B)$

Distance =360 km (constant for both)

                $\mathrm{A}$        $\mathrm{B}$

$\mathrm{Time}$      $9 \ \ :  \ \ 5$

$\mathrm{Speed}$     $5 \ \ :  \ \ 9$     (Speed is inversely proportional to time when distance is constant)

When speed of train A doubles -

                $\mathrm{A}$        $\mathrm{B}$

$\mathrm{Speed}$     $10 \ \ :  \ \ 9$

$\mathrm{Time}$      $9 \ \ :  \ \ 10$ 

ATQ,  (10-9) unit = 36 min

 1 unit = 36 min

Time taken by train B to cover a distance of $360 \mathrm{~km} $$\Rightarrow 5$ units $= 180$ min $= 3$ hr

Required speed of B = $\frac {360}{3} = 120 \mathrm{~km} / \mathrm{hr}$

BASIC APPROACH:

Train A takes $1 \frac{4}{5}$ hr more than the time taken by train B to cover a distance of $360 \mathrm{~km}$.
Distance $=360 \mathrm{~km}$
Let speed of train $\mathrm{A}=\mathrm{A} \mathrm{km} / \mathrm{hr}$
Speed of train $\mathrm{B}=\mathrm{B} \mathrm{km} / \mathrm{hr}$
As we know,
$\text { Time }=\frac{\text { Distance }}{\text { Speed }}$
$\frac{360}{A}=\frac{360}{B}+\frac{9}{5}$........(1)
If the speed of train A doubles, it will take 36 min less than the time taken by train B to travel the same distance.
$\frac{360}{2 A}=\frac{360}{B}-\frac{36}{60}=\frac{360}{B}-\frac{3}{5}$........(2)
Subtract (2) from (1) -
$\frac{360}{2 A}=\frac{12}{5}$
=$A=180 \times \frac{5}{12}=75 \mathrm{~km} / \mathrm{hr}$
Put the value of $A$ in (1)
=$\frac{360}{75}=\frac{360}{B}+\frac{9}{5}$
=$\frac{360}{B}=\frac{24}{5}-\frac{9}{5}=\frac{15}{5}=3$
=$B=\frac{360}{3}=120 \mathrm{~km} / \mathrm{hr}$
Hence, Speed (in km/h) of train $B=120 \mathrm{~km} / \mathrm{hr}$

ATQ,
$M_{1} \times D_{1}=M_{2} \times D_{2}$
$\Rightarrow 200 \times 1024=256 \times M_{2}$
$\Rightarrow M_{2}=\frac{200 \times 1024}{256}$
$M_{2}$ = 800

ATQ,

$\therefore$ 6.30

Number of workers = 14

Number of days = 12

Total work = 14 × 12 = 168

Let the capacity of a worker = 1

So work done in 4 days = 14 × 4 × 1 = 56

Work remaining = 168 - 56 = 112

Number of people coming = 2

Total number of people = 14+2 = 16

Time taken to complete the remaining work = $\frac{112}{16}$ =7 Days

ATQ,
$\Rightarrow \mathrm{M}_{1} \times \mathrm{D}_{1}=\mathrm{M}_{2} \times \mathrm{D}_{2}$
$\Rightarrow 120 \times 6=x \times 4$
$\Rightarrow x=\frac{120 \times 6}{4}$
$x=180$

So, more men required =180-120= 60

ATQ,
$\Rightarrow \frac{\mathrm{M}_{1} \times \mathrm{D}_{1}}{\mathrm{~W}_{1}}=\frac{\mathrm{M}_{2} \times \mathrm{D}_{2}}{\mathrm{~W}_{2}}$
$\Rightarrow \frac{5 \times 24}{5}=\frac{4 \times \mathrm{D}_{2}}{3}$
$\mathrm{D}_{2}=18$ days

ATQ,

$(A+B)$ work together
$\Rightarrow \frac{24}{3+2} \mathrm{hrs}$
$\Rightarrow \frac{24}{5}$ hours $\Rightarrow \frac{24}{5}$

Let total work $=$ LCM of 4,6 and $12=12$ units
Efficiency of $A=\frac{12}{4}=3$ units
Efficiency of $B=\frac{12}{6}=2$ units
Efficiency of $C=\frac{12}{12}=1$ unit
Total time taken by A, B and $C=\frac{12}{3+2+1}=\frac{12}{6}=2$ days

$P$ can complete five-eighths of a work in 15 days.
Number of days taken by P to complete the whole work $=15 \times \frac{8}{5}=24$ days
Q can complete three-fourths of the same work in 30 days.
Number of days taken by $Q$ to complete the whole work $=30 \times \frac{4}{3}=40$ days

They worked together for 8 days and then P left.
Work completed by $\mathrm{P}$ and $\mathrm{Q}$ together in 8 days $=8 \times(5+3)=64$ units
Remaining work $=120$ units $-64$ units $=56$ units
Number of days taken by $Q$ to complete the remaining work =
$\frac{56}{3}=18 \frac{2}{3}$ days $=18$ days and 16 hours

Let total number of person $=x$
Now, according to question,
Total work $=$ total work
$
\Rightarrow(x) \times 54=(x-15) \times(54+18)
$
$\Rightarrow 54 x=(x-15) \times 72$
$
\Rightarrow 3 x=(x-15) \times 4
$
$
\Rightarrow 3 x=4 x-60
$
$
\Rightarrow x=60
$
Therefore, the number of person is 60.

Given, the efficiency of Ram is twice that of Rahim
Ram alone takes 6 days to complete a piece of work.
1-day work done by individual = Work/time
$\Rightarrow 1$-day work of Ram $=1 / 6$
The efficiency of Ram is twice that of Rahim
$\Rightarrow 1$ day work of Rahim $=1 / 2 \times(1$ day work of $A)=(1 / 2) \times(1 / 6)=1 / 12$
Time taken by Rahim $=12$ days