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Parin Sir > Quantitative Aptitude (Maths) > Fundamental (Numbers) > Time and Work Examples

Time and Work Examples

July 1, 2019
Category: Quantitative Aptitude (Maths) Fundamental (Numbers)

TIME & WORK

Examples :

(1) A can build a wall in 40 days, which B alone can build in 50 days. If they build it together and get a payment of Rs. 1125, what is B’s share ?

Soln.

Ratio of their Days = 40 : 50 = 4 : 5

$\therefore$ Ratio of their shares = Ratio of their work = 5 : 4.

$\therefore$ B’s share = Rs. $\left ( 1125\times \frac{4}{9} \right )$ = Rs. 500.

(2) A can finish a piece of work in 8 days while B can do it in 20 days. If both work at it together, what time will they take to do the work ?

Soln.

A’s 1 day’s work = $\frac{1}{8}$ ; B’s 1 day’s work = $\frac{1}{20}$ .

$\therefore$ ( A + B )’s 1 day’s work = $\frac{1}{8}$ + $\frac{1}{20}$ = $\frac{7}{40}$ .

$\therefore$ Time taken by both to finish the work = $\frac{40}{7}$ days = 5 $\frac{5}{7}$ days.

(3) Two pipes A and B can fill a tank in 12 hours and 16 hours respectively. Find the time taken to fill the tank when both the pipes are turned on simultaneously.

Soln.

A’s 1 hour’s work = $\frac{1}{12}$ & B’s 1 day’s work = $\frac{1}{16}$ .

$\therefore$ ( A + B )’s 1 day’s work = $\frac{1}{12}$ + $\frac{1}{16}$ = $\frac{7}{48}$ .

Thus, both the pipes together can fill the tank in $\frac{48}{7}$ = 6 $\frac{6}{7}$ hours.

(4) Pipe A can fill a tank in 10 hours while pipe B alone can fill it in 20 hours and pipe C can empty the full tank in 30 hours. If all the pipes are opened together, how much time will be needed to make the tank full ?

Soln.

Net part filled in 1 hour = $\left ( \frac{1}{10}+\frac{1}{20}-\frac{1}{30} \right )=\frac{7}{60}$ .

$\therefore$ The tank will be full in $\frac{60}{7}$ hrs = 8 $\frac{4}{7}$ hrs.

(5) A cistern has two taps which fill it in 10 minutes and 15 minutes respectively. There is also a waste pipe in the cistern. When all the three are opened, the empty cistern is full in 30 minutes. How long will the waste pipe take to empty the full cistern ?

Soln.

Let, the waste pipe empty the tank in x hrs., then

Net part filled in 1 hour = $\frac{1}{30}$ = $\left ( \frac{1}{10}+\frac{1}{15}-\frac{1}{x}\right )$

$\therefore \frac{1}{x} = \left ( \frac{1}{10}+\frac{1}{15}-\frac{1}{30}\right ) = \frac{4}{30}=\frac{2}{15}$

$\therefore x = \frac{15}{2}= 7 \frac{1}{2}$ hrs.

Thus, waste pipe empty the tank in 7.5 hrs.

(6) A is thrice as good a workman as B and is therefore able to finish a piece of work in 40 days less than B. Find the time in which they can do it, working together.

Soln.

Ratio of work done by A and B in same time = 3 : 1.

Ratio of times taken to finish the same work = 1 : 3.

A can finish the work in k days and while B can finish it in 3k days.

3 k – k = 40

$\therefore$ k = 20

A can finish the work in 20 days and while B can finish it in 60 days

$\therefore$ ( A + B )’s 1 day’s work = $\left ( \frac{1}{20}+\frac{1}{60} \right )$ = $\frac{4}{60}$ = $\frac{1}{15}$ .

Hence, both together can finish the work in $\frac{15}{1}$ = 15 days.

(7) A can do a piece of work in 40 days. B is 60 % more efficient than A. Find the number of days it takes B to do the same piece of work.

Soln.

The ratio of work of A and B = 100 : 160 = 5 : 8

The ratio for days for them = 8 : 5

For A, 8 k = 40

k = 5

Now, For B, 5 k = 5 ( 5 ) = 25 days.

(8) A & B can do a piece of work in 24 days; B & C in 30 days; C & A in 40 days. In how many days will they finish it together and separately ?

Soln.

(A + B)’s 1 day’s work = $\frac{1}{24}$ ; (B + C)’s = $\frac{1}{30}$ ; (C + A)’s work = $\frac{1}{40}$ .

$\therefore$ [ ( A + B ) + ( B + C ) + ( C + A ) ] ‘s 1 day’s work = $\frac{1}{24}$ + $\frac{1}{30}$ + $\frac{1}{40}$

$\therefore$ 2 ( A + B + C )’s 1 day’s work = $\frac{12}{120}$ = $\frac{1}{10}$ .

$\therefore$ ( A + B + C )’s 1 day’s work = $\frac{1}{20}$ .

So, they together can finish the work in 20 days.

A’s 1 day’s work

= [( A + B + C )’s 1 day’s work ] – [( B + C )’s work ]

= $\left ( \frac{1}{20}-\frac{1}{30} \right ) = \frac{1}{60}$

$\therefore$ A alone can finish the work in 60 days.

Similarly, B’s 1 day’s work = $\left ( \frac{1}{20}-\frac{1}{40} \right ) = \frac{1}{40}$ .

$\therefore$ B alone can finish the work in 40 days.

And, C’s 1 day’s work = $\left ( \frac{1}{20}-\frac{1}{24} \right ) = \frac{1}{120}$

$\therefore$ C alone can finish the work in 120 days.

(9) If two pipes function simultaneously, the reservoir will be filled in 18 hours. One pipe fills the reservoir 15 hours faster than the other. How many hours does it take the second pipe to fill the reservoir ?

Soln.

Let the reservoir be filled by first pipe in x hours.

Then, second pipe will fill it in ( x + 15 ) hours.

So, the second pipe will take ( 30 + 15 ) hrs i.e., 45 hrs. to fill the reservoir.

(10) A and B working separately can do a piece of work in 15 and 25 days respectively. If they work for a day alternately, (i) A beginning and (ii) B beginning, in how many days, the work will be completed ?

(11) A can do a piece of work in 10 days and B can finish it in 15 days. They work together for 2 days and then A goes away. In how many days will B finish the work ?

(12) Two pipes A and B can fill a cistern in 10 minutes and 15 minutes respectively. If both the pipes are opened together, then after how many minutes should A be closed so that the tank is full in 9 minutes ?

(13) Two pipes can fill a cistern in 21 hours and 24 hours respectively. The pipes are opened simultaneously and it is found that due to leakage in the bottom it took 48 minutes more to fill the cistern. When the cistern is full, in what time will the leak empty it ?

Soln.

Work done by the two pipes in 1 hour = $\left ( \frac{1}{21}+\frac{1}{24} \right )=\frac{15}{168}=\frac{5}{56}$