Practice questions here, for every subject and every exam. Unlimited questions for unlimited attempts, given with answers and explanations.
3 boys and 3 girls can do a piece of work in $\frac{6}{5}$ hour. How many hours will 1 boy and 1 girl need to complete the work?
ATQ,
$(3 \mathrm{~B}+3 \mathrm{G}) \times \frac{6}{5}=1$ unit work
$(B+G) \times \frac{18}{5}=1$ unit work
$(\mathrm{B}+\mathrm{G})$ complete in work $=\frac{18}{5} \mathrm{hrs}$
Alternate Method,
Let, the efficiency of a boy $=x$ hours
and the efficiency of a girl $=y$ hours
$(3 x+3 y) \times \frac{6}{5}=1$
$3(x+y)=\frac{5}{6}$
$x+y=\frac{5}{18}$
In 1 hour one boy and one girl can do complete the work $=\frac{5}{18}$ hours
$\therefore 1$ boy and 1 girl can complete the work in $\frac{18}{5}$ hours
ATQ,
$
\text { L.C.M }=(3.7)=21
$
$(A+B)$ work together 2 days
$
=(7+3) \times 2=20 \text { unit }
$
Remaining work $=(21-20)=1$ unit
Part of the work will be left
$\Rightarrow \frac{1}{21}$
X can finish a job in 131 days. He worked for 47 days alone and the remaining work was completed by $Y$ in 84 days. How many days would both te
Time taken by $X$ to complete the work$ \rightarrow 131$ days
Let Total work $=131$ unit
Efficiency of $X=1$ unit/day
Work = Efficiency $\times$ Time
$T_{1} E_{1}+T_{2} E_{2}=T_{3} E_{3}$
$47 x+84 y=131 x$
$84 x=84 y$
$x=y$
So, Efficiency of $X=$ Efficiency of $Y=1$
Required Time by A & B to complete the work $=\frac{131}{1+1}$
$ \begin{aligned} &=\frac{131}{2} \\ &=65.5 \end{aligned} $
Additional Information:
Direction: Study the graph and answer the question given below.
The given graph represents the number of workers getting daily wages (in₹).
Total number of workers = 200
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.
Hence, option C is correct.
A and B can do a work in 15 days and 25 days, respectively. They worked together for 5 days, after which B was replaced by C and the remaining work was completed by A and C in the next 4 days. In how many days will C alone complete the same work?
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
Hence, option B is correct.
$\mathrm{P}$ and $\mathrm{Q}$ together can finish a work in 6 days. $Q$ alone can do the same work in 10 days. In how many days can P alone do the same work?
ATQ
$\operatorname{L.C.M}$ of $(6,10)$
P can do done work finish
$\Rightarrow \frac{30}{5-3}$
$\Rightarrow \frac{30}{2}=15$
4 trained laborers take 8 days to build a pond whereas 8 untrained laborers take 12 days to build it. In how many days will it take 3 trained and 15 untrained laborers to build the pond?
ATQ
$
\begin{aligned}
&M_{1} \times D_{1}=M_{2} \times D_{2} \\
&T \times D_{1}=U \times D_{2} \\
&4 T \times 8=8 U \times 12 \\
&\frac{T}{U}=\frac{3}{1}
\end{aligned}
$
Total work $=4 \times 3 \times 8=96$ units
$(3 \mathrm{~T}+15 \mathrm{U})$ can build the pond
$\Rightarrow \frac{96}{(3 \times 3+15 \times 1)}=\frac{96}{24}=4$ days
If 16 people can build a wall in 15 days, how many people would be needed to build it in 8 days?
Given,
$ \mathrm{M} 1=16 \text { and } \mathrm{D} 1=15 \text { days } $
$ \mathrm{M} 2=? \text { and } \mathrm{D} 2=8 \text { days } $
We know,
$ \mathrm{M} 1 \times \mathrm{D} 1=\mathrm{M} 2 \times \mathrm{D} 2 $
$ \Rightarrow 16 \times 15=\mathrm{M} 2 \times 8 $
$ \Rightarrow \mathrm{M} 2=30 $
Hence 30 people needed to build it in 8 days.
7 experts and 5 trainees can complete a job in 9 days while 4 experts and 15 trainees can complete it in 12 days. How many days will 5 experts and 6 trainees need to complete the job?
Let efficiency of expert be E and efficiency of trainees be T
⇒ (7E + 5T) × 9 = (4E + 15T) × 12
⇒ 63E + 45T = 48E + 180T
⇒ 15E = 135T
⇒ E/T = 9/1
So, efficiency of experts is 9 and trainees is 1.
Then, total work
⇒ (7E + 5T) × 9
⇒ (7 × 9 + 5) × 9 = 612
Then, 5 Expert and 6 trainees need to complete 612 unit work is
⇒ (5E + 6T) × Days = 612
⇒ (5 × 9 + 6) × Days = 612
⇒ 51 × Days = 612
⇒ Days = 612/51 = 12
∴ 5 experts and 6 trainees need to complete the job in 12 days
Brij alone can paint a wall in $7.2$ days while Madhu takes $10.8$ days to do the same work. How many days will it take to paint $\frac{5}{6}$ of the wall working together?
ATQ,
Work done by Brij $=3$ unit / day
Work done by Madhu $=2$ unit/day
Work done by them in one day together =$\frac{21.6}{5}$
Total work to be done $=\frac{5}{6}$
Time taken $=\frac{216}{50} \times \frac{5}{6}$$=3.6$ days
ATQ,
$\mathrm{H}_{1} \times \mathrm{D}_{1}=\mathrm{M}_{2} \times \mathrm{D}_{2}$
$\Rightarrow 8 \times 18=12 \times \mathrm{D}_{2}$
$\mathrm{H}_{2}=\frac{8 \times 18}{12}$
$\mathrm{H}_{2}=12$ hours
Working alone, Rakesh and Ranjit can complete a piece of work in 4 and 6 days respectively. Ranjit started the work alone and Rakesh joined after 2 days. In how many days will they need to complete the remaining work?
ATQ,
Work completed by ranjit in 2 days $=2 \times 2=$ 4
Work left $=12-4=8$ unit
Time taken by ranjit and rakesh to complete the remaining work $=\frac{8}{2+3}=\frac{8}{5}=1.6$ days.
Rashid can complete a job by himself in 15 days while Kausik can do the same work alone in 21 days. If they work together how many days will it take them to complete the work?
Given:
R can do work in = 15 days
K can do work in = 21 days
Total work = LCM of 15 and 21 = 105
Efficiency of R = $\frac {105}{15}$ = 7
Efficiency of K = $\frac {105}{21}$ = 5
The efficiency of R ∶ K = 7 ∶ 5
Then efficiency of R+K = 7 + 5 = 12
Total time taken by R and K = $\frac {105}{12}$ = 8$\frac {3}{4}$
∴ Working together they completed the work in 8$\frac {3}{4}$ days.
Zubin and Pravin can individually complete a job in $36 \mathrm{~min}$ and $60 \mathrm{~min}$ respectively. Starting with Zubin, they work alternately for a minute each till the work is completed, only the one working at the end allowed to work for less than a minute in his final turn. How long will it take to complete the job?
According to the question,
Time taken by Zubin to complete a job $=36 \mathrm{~min}$
Time taken by Parvin to complete a job $=60 \mathrm{~min}$
Ratio of the time $=36: 60=3: 5$
Ratio of the efficiency $=5: 3$
Then total work $=36 \times 5=180$
Work done by Zubin and Parvin in the first two min. $=5+3=8$
8 units work $=2 \mathrm{~min}$
176 units work $=2 \times 22=44 \mathrm{~min}$
Time taken in next 4 units work by Zubin $=\frac{4}{5} \mathrm{~min}=48$ seconds
Total time taken to complete the job $=44 \mathrm{~min} 48$ seconds
Govinda usually leaves his home every day at 8 am and travels at a certain speed to reach office on time. One day, he travelled at $\frac{4}{5}$ of the usual speed and hence arrived $22.5$ min late. How much time does Govinda usually take to reach his office?
Given:
Let usual Speed be 5 unit
⇒ New Speed = ${5}\times$ $\frac {4}{5}$ = 4 unit
Ratio of Usual and New speeds = 5 ∶ 4
⇒ Ratio of Time taken = 4 ∶ 5
(∵ Time taken is inversely proportional to Speed)
Extra time = 1 unit = 22.5 mins
∴ Usual time = 4 units = 90 mins = 1 hr 30 min
ATQ,
$3 \operatorname{man} \times 15=5$ Boy $\times 15$
6 man $+3$ man $=5$ Boy $+7$ Boy
$\frac{\operatorname{Man}}{\mathrm{Boy}}=\frac{9}{12}=\frac{3}{4}$
$3 \operatorname{Man}=4$ Boy
The work of 4 boys is equal to the work of 3 man
Twenty persons take 15 days to complete a certain work, working 8 hours a day. To complete the same work in 4 days, working 10 hours a day, the number of persons required is:
Let the number of persons required be $x$
We know that:$\mathrm{W}=\mathrm{MDH}$
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.
'A' alone can do a piece of work in 10 days and 'B' alone in 15 days. 'A' and 'B' undertook to do it for Rs. 42,000. With the help of 'C', they completed the work in 5 days. How much (in Rs.) is to be paid to 'B'?
Let the total work $=30$ units
One day work of $A, B$ and $C=\frac{30}{5}=6$ units
One day work of $C=6-3-2=1$ unit
The 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=\mathrm{Rs}14000 $.