Quantitative Aptitude

Time and Work

2,195 Questions
All Pairwise rates and staged schedules 440 Workers-days-hours chain rule 439 LCM units and efficiency ratios 409 Different worker categories 311 Combined rates and an unknown worker 257 Division of wages 219 Alternate work 110 Opposing rates 10

Time and Work Questions

Multiple choice business mathematics and statistics applications of calculus revenue functions from marginal revenue functions minimization of cost function and maximization of revenue function and profit function integral calculus – ii

A village road is to be constructed by a team of $250$ workers. After $12$ days it was found that only $2/7^{th}$ part of the work was complete. To complete the rest in another $25$ days, how many more workers should be employed?

  1. $53$
  2. $52$
  3. $55$
  4. $50$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Let the total unit of work be $7$ units.
According to question,
$\dfrac {M _{1}D _{1}}{W _{1}} = \dfrac {M _{2}D _{2}}{W _{2}} \Rightarrow \dfrac {250\times 12}{2} = \dfrac {M _{2}\times 25}{(7 - 2)} \Rightarrow M _{2} = 300$
Number of extra workers needed $= 300 - 250 = 50$.

Multiple choice maths direct proportion and inverse proportion inverse proportion rule of three types of proportions

Ten men, working for 6 days of 10 hours each, finish $ \dfrac {5}{21} $ of a piece of work. How many men working at the same rate and for the same number of hours each day, will be required to complete the remaining work in 8 days?

  1. 24

  2. 26

  3. 25

  4. 21
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Lets first find the number of working hours required to complete $\dfrac { 5 }{ 21 }$ of the work.
M = number of men
D = number of days
H = number of hours worked per day
N = work done
MDH = N
10$\times $6$\times $10 = $\dfrac { 5 }{ 21 }$
600 hours to complete $\dfrac { 5 }{ 21 }$ of work
Let the total number of hours required to complete the work be $x$ then
$x$ = $\dfrac { 5 }{ 21 }$ $\div $600
   = 600$\times $$\dfrac { 21 }{ 5
 }$
$x$   = 2520 hours
so the remaining work is $x$$-$  $\dfrac { 5 }{ 21 }$
               = $\dfrac {16 }{ 21 }$ $x$

               =$\dfrac {16 }{ 21 }$$\times $2520
               = 1920 hours required to complete remaining $\dfrac {16 }{ 21 }$ of the work
 now the equation we get is
MDH = N
MDH = 1920
M $\times $ 8$\times $10 = 1920
M =  $\dfrac { 1920 }{ 80 } $
M = 24
 Ans. 24 men

Multiple choice introduction to ratio and percentages comparing quantities maths

Sakshi can do a piece of work in $20$ days. Tanya is $25$% more efficient than Sakshi. The number of days taken by Tanya to do the same piece of work is:

  1. $15$
  2. $16$
  3. $18$
  4. $25$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Ratio of times taken by Sakshi and Tany $=125:100=5:4$.
Suppose Tanya takes $x$ days to do the work.
$5:4::20:x$ $\Rightarrow \left( \cfrac { 5\times 20 }{ 5 }  \right) $
$\Rightarrow$ $x=16$ days.
Hence, Tanya takes $16$ days to complete the work.

Multiple choice maths direct proportion and inverse proportion rule of three types of proportions direct proportion

$A$ can do a piece of work in $10$ days and $B$ in $15$ days. How long will they take together to finish it ? 

  1. $7$ days
  2. $3$ days
  3. $9$ days
  4. $6$ days
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Work done by A in 1 day $=\dfrac{1}{10}$


Work done by B in 1 day$=\dfrac{1}{15}$

Work done by A and B in 1 day$=\dfrac{1}{10}+\dfrac{1}{15}=\dfrac{25}{150}$

Working together they will complete the work in $\dfrac{150}{25}=6$ days

Multiple choice maths direct proportion and inverse proportion rule of three types of proportions direct proportion

$A, B$ and $C$ can finish a job working alone in $72, 24$ and $36$ days respectively. In how many days they can finish the job if they worked together?

  1. $12$
  2. $9$
  3. $15$
  4. $18$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Let the total work be $72$ units (LCM on $72, 24$ and $36$).


$A, B$ and $C's$ one day work is $1, 3$ and $2$ units respectively.

Required number of days $= \dfrac {72}{6} = 12$.


Alternate method
$(A+B+C)'s$ one day work =$\dfrac{1}{72}+\dfrac{1}{24}+\dfrac{1}{36}$

$=\dfrac{1+2+3}{72}=\dfrac{6}{72}$

Number of days required $= \dfrac {72}{6} = 12$ days to finish the work when 3 of them work together.

Multiple choice applications of quadratic equations solving (simple) problems word problems based on quadratic equations quadratic equation maths

 A can do a piece of work in $'x'$ days and $B$ can do the same work in $'x+16'$ days.If both working together can do it in $15$ days. Calculate $x$.

  1. $24$
  2. $25$
  3. $27$
  4. $None\ of\ the\ above$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given A do a piece of work in x days and B do work in x+16 days

Then A one day work = $\frac{1}{x}$ days
AND B one day work =$\frac{1}{x+16}$days
Then Both one day work =$\frac{1}{x}+\frac{1}{x+16}=\frac{x+16+x}{x(x+16)}=\frac{2x+16}{x^{2}+16x}$ 
So both do work in $\frac{x^{2}+16x}{2x+16} \ days$ 
But both do work in 15 days
$\therefore \frac{x^{2}+16x}{2x+16}=15$
$\Rightarrow x^{2}+16x=30x+240$
$\Rightarrow x^{2}-14x-240=0$
$\Rightarrow x^{2}-24x+10x-240=0$
$\Rightarrow x(x-24)+10(x-24)=0$
$\Rightarrow (x-24)(x+10)=0$
Then $x-24=0 , x=24$
And $x+10=0 , x=-10$
But work done is not negative
Then work done =$24$ days

Multiple choice exponent of a prime in n! factorial notation combinatorics and mathematical induction permutations and combinations maths

'$X$' completes a job in $2$ days and '$Y$' completes it in $3$ days and '$Z$' takes $4$ days to complete it. If they work together and get Rs. $3,900$ for the job, then how much amount does '$Y$' get?

  1. Rs. $1,800$
  2. Rs. $ 1,200$
  3. Rs. $ 900$
  4. Rs. $ 800$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$X$ do job in $2$ days, $Y$ completes it in $3$ days, $Z$ takes $4$ days.
If $X, Y, Z$ together can do in $1$ day, then 
$= \dfrac {1}{2} + \dfrac {1}{3} + \dfrac {1}{4} = \dfrac {13}{12}$ of work
Therefore, the whole work is done in $\dfrac {12}{13}$ of a day.
Daily wages of $Y = \dfrac {1}{3}\times $ Rs. $ 3,900 =$ Rs. $ 1,300$
$\therefore$ Amount of $Y = \dfrac {12}{13} \times$ Rs. $ 1,300 =$ Rs. $1,200$

Multiple choice maths parts and whole multiplication of a fraction multiplication of a fractions multiplication of fraction finding the whole when a fraction is given

Rehna works $2\cfrac { 1 }{ 2 } $ hours each day on her embroidery. She completes the work in $7$ days. How many hours did she take to complete her work?


Ans : $17\cfrac { 1 }{ 2 } $ hrs.

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Rehana works $2\dfrac{1}{2}$ hrs $= 150 $minutes


As she completed her work in $7$ days,

$150 \times 7 = 1050$ minutes = $\dfrac{1050}{60}$ hours

$=\dfrac{35}{2}$ hours $=17\dfrac{1}{2}$ hours


So, Rehana took $17\dfrac{1}{2}$ hours to complete her work.

Multiple choice maths parts and whole multiplication of a fraction multiplication of a fractions multiplication of fraction finding the whole when a fraction is given

Deepak can paint $\cfrac { 2 }{ 5 } $ of a house in one day. If he continuous working at this rate, how many days will he take to paint the whole house?

  1. $2\cfrac { 1 }{ 2 } $ days
  2. $1\cfrac { 1 }{ 2 } $ days
  3. $\cfrac { 1 }{ 2 } $ days
  4. $2\cfrac { 1 }{ 4 } $ days
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$\boxed{\text{Work(W)}\propto \text{Days(D)}\\dfrac{W _1}{D _1}=\dfrac{W _2}{D _2}}$

given,
$W _1=\dfrac25\quad D _1=1\,\text{day}\W _2=1\,\,\text{(painting the whole house)}$
to find, $D _2=?$

$\dfrac{2/5}{1}=\dfrac{1}{D _2}\Rightarrow D _2=\dfrac52$

$\therefore D _2=2\dfrac12 \,\text{days}$

Multiple choice maths calculations and mental strategies 1 equations from statements forming equations from statements writing mathematical statements

A can do a piece of work in $24$ days. If B is $60\%$ more efficient then the number of days required by B to do the twice as large as the earlier work is-

  1. $24$
  2. $36$
  3. $15$
  4. $30$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
According to the question-
A can do a piece of work in $24$ days.
$\therefore$ work done by A in $1$ day  $=\cfrac{1}{24}$
Given that, B is $60 \%$ more efficient, i.e., B can do $60 \%$ more work than work done by A in $1$ day-
Work done by B in $1$ day  $=\cfrac{1}{24} + \cfrac{60}{100} \times {1}{24} = \cfrac{1}{15}$
$\therefore$ number of days required by B to do the same work $= 15 $days
$\therefore$ number of days required by B to do the twice as large as the earlier work  $=2 \times 15 = 30$ days
Multiple choice maths calculations and mental strategies 1 equations from statements forming equations from statements writing mathematical statements

A and B can do a job in $12$ days, B and C can do it in $16$ days. After A has worked for $5$ days and B has worked for $7$ days, C can finish the rest in $13$ days. In how many days can C do the work alone?

  1. $16$ days
  2. $24$ days
  3. $36$ days
  4. $48$ days
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Solution:-

Let the amount of work done by $A, B,$ and $C$ per day be $x, y$ and $z$ respectively.
Case I:-

$A$ and $B$ can do the job in $12$ days.
$\therefore$ work done by $A$ and $B$ in one day  $=\cfrac{1}{12}$
$\Rightarrow \; x + y = \cfrac{1}{12}$
Case II:-
B and C can do the job in $16$ days.
$\therefore$ work done by B and C in one day  $=\cfrac{1}{16}$
$\Rightarrow \; y + z = \cfrac{1}{16}$
As given, A has worked for $5$ days and B has worked for $7$ days, C can finish the rest in $13$ days.
$5x + 7y + 13z = 1$
$\Rightarrow$ $5x + 5y + 2y + 2z + 11z = 1$
$\Rightarrow$ $5(x + y) + 2(y + z) + 11z = 1$
$\Rightarrow \; 5 \times \cfrac{1}{12} + 2 \times \cfrac{1}{16} + 11z = 1$
$\Rightarrow \; 11z = 1 - \cfrac{5}{12} - \cfrac{1}{8}$
$\Rightarrow \; 11z = \cfrac{22}{48}$
$\Rightarrow \; z = \cfrac{1}{24}$
Therefore, work done by $C$  $1$ day  $=\cfrac{1}{24}$
Hence, C alone can finish the work in $24$ days.

Multiple choice maths calculations and mental strategies 1 equations from statements forming equations from statements writing mathematical statements

If $12$ men complete a work in $20$ days. If only $8$ men are  employed, then the time required  to complete  the same work is

  1. $24$ days
  2. $25$ days
  3. $30$ days
  4. $35$ days
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

As $12$ men complete work in $20$ days, $1$ man will complete the same work in

$20\times12=240$ days.

Time required by $8$ men to complete the work=$\dfrac{240}{8}$=$30$ days..