Quantitative Aptitude - SPLessons

Time – Work Problems

Chapter 16

SPLessons 5 Steps, 3 Clicks
5 Steps - 3 Clicks

Time – Work Problems

shape Introduction

In this chapter, we will be studying about the concept of TIME AND WORK and some rules to solve the problems quickly.

Time: It is defined as a sequence of actions and events that take place in present, future, past and time is measured in seconds, minutes, hours, days, weeks, months and years.


Work: When a force acts upon an object to cause a displacement of the object, then it is said that a work has been done upon the object.


shape Methods

  • Broadly, WORK is the quantity of energy transferred from one system to another. Since we are focused on this topic, work is defined as the amount of job assigned or the amount of job actually completed.

  • Work is always considered as a whole.

  • There exists an analogy between the time-speed-distance problems and work.

    • Work = Distance

    • Rate at which work is done = speed

    • Number of days required to do the work = Time

  • Relationships:

    • Worker \(\propto\) Work ⇒ More workers can do more work and vice-versa.

    • Time \(\propto\) Work ⇒ In more time, more work can be done and vice-versa.

    • Worker \(\propto\) \(\frac{1}{Time}\) ⇒ More workers can do more work in less time and vice-versa.

  • Rules to Quickly Solve Problems Related To Time and Work:


    Rule 1: If number of days required to complete the work = n, then work that can be done per day = \(\frac{1}{n}\), and efficiency is = \(\frac{100}{n}\) %


    • e.g. If John can finish a work in 10 days, then he will finish \(\frac{1}{10}\) part of the work in 1 day.

    • e.g. If Mary can finishes \(\frac{1}{7}\) part of a work in 1 days, then she will finish the complete work in7 days.


    Rule 2: If \(M_{1}\) workers can finish \(W_{1}\) work in \(D_{1}\) number of days, and if \(M_{2}\) workers can finish \(W_{2}\) work in \(D_{2}\) number of days, then,


      \(\frac{M_{1}D_{1}}{W_{1}}\) = \(\frac{M_{2}D_{2}}{W_{2}}\)


    Example:
    If 3 workers are able to finish in \(\frac{1}{3}\) part of work in 6 days, then in how many days can 4 workers finish \(\frac{1}{2}\) part of the work?

    Solution:

      Here \(M_{1}\) = 3, \(W_{1}\) = \(\frac{1}{3}\), \(D_{1}\) = 6, and \(M_{2}\) = 4, \(w_{2}\) = \(\frac{1}{2}\). We have to find out \(D_{2}\). Using Rule 2, we have.

      \(\frac{M_{1}D_{1}}{W_{1}}\) = \(\frac{M_{2}D_{2}}{W_{2}}\) = \(\frac{3 \times 6}{\frac{1}{3}}\) = \(\frac{4 \times D_{2}}{\frac{1}{2}}\) ⇒ \(D_{2}\) = 6.75 Days.


    Rule 3: If \(M_{1}\) workers can finish \(W_{1}\) work in \(D_{1}\) number of days after working for \(T_{1}\) hours per day, and if \(M_{2}\) workers can finish \(W_{2}\) work in \(D_{2}\) number of days after working for \(T_{2}\) hours per day, then,

      \(\frac{M_{1}D_{1}T_{1}}{W_{1}}\) = \(\frac{M_{2}D_{2}T_{2}}{W_{2}}\)


    Example:
    If 4 men are able to finish in \(\frac{1}{5}\) part of work in 4 days after working for 3.5 hours per day, then in how much work can be finished by 2 workers after working 9 hours per day for 5 days?

    Solution:

      Here \(M_{1}\) = 4, \(W_{1}\) = \(\frac{1}{5}\), \(D_{1}\) = 4, \(T_{1}\) = 3.5, and \(M_{2}\) = 2, \(W_{2}\) = ?, \(T_{2}\) = 9, \(D_{2}\) = 5.

      We have to find out \(W_{2}\). UsinRule 3, we have

      \(\frac{M_{1}D_{1}T_{1}}{W_{1}}\) = \(\frac{M_{2}D_{2}T_{2}}{W_{2}}\) = \(\frac{4 \times 4 \times 3.5}{\frac{1}{5}}\) = \(\frac{4 \times 5 \times 9}{W_{2}}\) ⇒ \(W_{2}\) = 1.56 (or) 1 and \(\frac{5}{9}\) part more of work.


    Rule 4: If A completes a piece of work in ‘x’ days, and B completes the same work in ‘y’ days, then


      1. Work done by A in 1 day = \(\frac{1}{x}\), and work done by B in 1 day = \(\frac{1}{y}\)

      2. work done by A and B together in 1 day = \(\frac{1}{x}\) + \(\frac{1}{y}\)

      3. Total days taken to complete the work by A and B working together = \(\frac{xy}{x + y}\)


    Rule 5: If A alone can do a certain work in ‘x’ days and A and B together can do the same work in ‘y’ days, then B alone can do the same work in


      1. work done by A, B and C together in 1 day = \(\frac{1}{x}\) + \(\frac{1}{y}\) + \(\frac{1}{z}\)

      2. Total days taken to complete the work by A, B and C working together = \(\frac{xyz}{xy + yz + zx}\)


    Rule 6: If A alone can do a certain work in ‘x’ days and A and B together can do the same work in ‘y’ days, then B alone can do the same work in \(\frac{xy}{x – y}\)


    Rule 7: If A and B can do a work in ‘x’ days, B and C can do the same work in ‘y’ days, C and
    A can do the same work in ‘z’ days. Then total days taken, when A, B and C work together = \(\frac{2xyz}{xy + yz + zx}\)


    Rule 8:

    • Work of 1 day = \(\frac{Total \ work}{Total \ no.of \ working \ days}\)

    • Total work = (Work of 1 day) × (Total no. of working days)

    • Remaining work = 1 – (Work done)

    • Work done by A = (Work done in 1 day by A) × (Total no. of days worked by A)


    Rule 9: If A can finish \(\frac{m}{n}\) part of the work in D days, then total days taken to finish the work by A = \(\frac{n}{m} \times D\).


    Rule 10: If A can do a work in ‘x’ days and B can do the same work in ‘y’ days and when they
    started working together,


    • A left the work ‘m’ days before completion, then total days taken to complete the work is \(\frac{(x + m)y}{x + y}\)

    • B left the work ‘m’ days before completion, then total days taken to complete the work is \(\frac{(y + m)x}{x + y}\)


    Rule 11: A and B together can finish a certain work in ‘a’ days. They worked together for ‘b’ days and then B (or A) left the work. A (or B) finished the rest work in ‘d’ days, then total days taken by A (or B) alone to complete the work is \(\frac{ad}{a – b}\) (or \(\frac{bd}{a – b})\).


    Let’s see some examples for applying these rules:


    Example 1:
    A, B and C can do a work in 6, 8 and 12 days respectively. Doing that work together they get an
    amount of Rs. 1550. What is the share of B in that amount?

    Solution:

      A’s share : B’s share : C’s share

      = (B’s time × C’s time) : (A’s time × C’s time) : (A’s time × B’s time)

      = 96 : 72 : 48

      = 4 : 3 : 2

      ∴ B’s share = \(\frac{1550}{9} \times 3\)= Rs. 516.66.


    Example 2:
    Lita can complete a piece of work in 5 days, but with the help of her son she can do it in 3 days.
    Find the time taken by the son alone to complete the work.

    Solution:

      Use RULE 6! We have x = 5 and y = 3. Then time taken by the son alone to complete the work alone = \(\frac{xy}{x – y}\) = \(\frac{5 \times 3}{5 – 3}\) = 7.5days.


    Example 3:
    Two pipes can fill the cistern in 10 hrs and 12 hrs respectively, while the third pipe can empty it
    in 20 hrs. Simultaneously, if all the pipes are opened then the cistern will be filled in how many
    hours?

    Solution:

      Use RULE 5 (1)! We have x = 10, y = 12, and z = 20. Then time taken by the pipes
      when opened together = \(\frac{1}{\frac{1}{10} + \frac{1}{12} – \frac{1}{20}}\) = 7.5 hours.


    Example 4:
    If machine X can produce 1,000 bolts in 8 hours and machine Y can produce 1,000 bolts in 24 hours. In how many hours can machines X and Y, working together at these constant rates, produce 1,000 bolts?

    Solution:

      Total work of X and Y: \(\frac{1}{8}\) + \(\frac{1}{24}\) = \(\frac{1}{6}\), Therefore, Working together, machines X and Y can produce 1,000 bolts in 6 hours.

    shape Formulae

    1. If A can do a piece of work in n days, then A’s 1 day work = \(\frac{1}{n}\).

    2. If A’s 1 day work = \(\frac{1}{n}\), then A can finish the work in n days.

    3. If A is thrice as good a workman as B, then:

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

      Ratio of time taken by A and B to finish a work = 1 : 3


    4. Universal law:

      a) If \(M_{1}\) persons can do \(W_{1}\) work in \(D_{1}\) days and \(M_{2}\) persons can
      do \(W_{2}\) work in \(D_{2}\) days then, \(M_{1}W_{2}D_{1}\)= \(M_{2}W_{1}D_{2}\)

      b) If the persons work for \(T_{1}\) and \(T_{2}\) hours per day respectively
      then, \(M_{1}W_{2}D_{1}T_{1}\)= \(M_{2}W_{1}D_{2}T_{2}\)

      c)If the persons have efficiency of \(E_{1}\) and \(E_{2}\) respectively then,
      \(M_{1}W_{2}D_{1}T_{1}E_{1}\)= \(M_{2}W_{1}D_{2}T_{2}E_{2}\)


    5. If A can do a piece of work in \(D_{1}\) days and B can do the same work in \(D_{2}\) days then, A and B can do the same work together in \(\frac{D_1 * D_2}{D_1 + D_2}\) days.

    6. If A is as good a workman as B, then A will take half of the time taken by B to complete a piece of work.

    7. If A and B can do a piece of work together in X days, B and C can do it in Y days, and C and A can do it in Z days. Then the same work is done.

      By A alone in \(\frac{2XYZ}{XY + YZ – ZX}\)days

      By B alone in \(\frac{2XYZ}{YZ + ZX – XY}\)days

      By C alone in \(\frac{2XYZ}{ZX + XY – YZ}\)days

      By A, B, and C together in \(\frac{2XYZ}{XY + YZ + ZX}\)days


    8. If A can do a piece of work in \(D_1\) days, B can do in \(D_2\) days and C do in \(D_3\) days. Then together they can do the work in

      \(\frac{D_1 * D_2 * D_3} {D_1D_2 + D_2D_3 + D_3D_1}\)


    9. If the number of men are changed in the ratio m : n then the time taken to complete the work will change in the ratio n : m.

    shape Samples

    1. If A takes 6 hours to do a piece of work, B takes 12 hours to do the same work. How long should it take to both A and B, working together but independently, to do the same work?

    Solution:

      Given that,

      A can do the work in 6 hours ⇒ A’s 1 hour work = \(\frac{1}{6}\)

      B can do the work in 12 hours ⇒ B’s hour work = \(\frac{1}{12}\)

      Then, A’s an B’s work in 1 hour is

      (A + B)’s work = \(\frac{1}{6}\) + \(\frac{1}{12}\) = \(\frac{1}{2}\)

      Therefore, both can do the work in \(\frac{2}{1}\) days = 2 days


    2. P can do a piece of work in 8 days of 10 hours each and Q can do it in 6 days of 8 hours each. How long will they take to do it, working together 6\(\frac{2}{5}\) hours a day?

    Solution:

      P can do the work in 8 days of 10 hours ⇒ 8 x 10 ⇒ 80 hours

      Q can do the work in 6 days of 8 hours ⇒ 6 x 8 ⇒ 48 hours

      Now, P’s 1 hour work = \(\frac{1}{80}\)

      Q’s 1 hour work = \(\frac{1}{48}\)

      (P + Q)’s 1 hour work = \(\frac{1}{80}\) + \(\frac{1}{48}\) = \(\frac{3 + 5}{240}
      \) = \(\frac{8}{240}\) = \(\frac{1}{30}\)

      Both will finish the work in 30 hours

      Therefore, number of days of 6\(\frac{2}{5}\) hours each = 30 x \(\frac{32}{5}\) = 6 x 32

      = 192 days


    3. S and T can do a piece of work in 15 days, T and U can do it in 20 days, U and S can do it in 30 days. In how many days will S, T, and U finish it, working together and separately?

    Solution:

      Given

      S and T can do a work in 15 days = (S + T)’s work in 1 hour = \(\frac{1}{15}\)

      T and U can do a work in 20 days = (T + U)’s work in 1 hour = \(\frac{1}{20}\)

      U and S can do a work in 30 days = (U + S)’s work in 1 hour = \(\frac{1}{30}\)

      By adding them all,

      (S + T + T + U + U + S) = \(\frac{1}{15}\) + \(\frac{1}{20}\) + \(\frac{1}{30}\)

      ⇒ 2(S + T + U) = \(\frac{4 + 3 + 2}{60}\)

      ⇒ 2(S + T + U) = \(\frac{9}{60}\)

      ⇒ (S + T + U) = \(\frac{9}{60}\) x \(\frac{1}{2}\)

      ⇒ (S + T + U) = \(\frac{3}{20}\) x \(\frac{1}{2}\)

      ⇒ (S + T + U) = \(\frac{3}{40}\)

      Thus, S, T and U together can do the work in \(\frac{40}{3}\) days

      Now, S’s 1 day work = [((S + T + U)’s work in 1 hour] -[(T + U)’s work in 1 hour]

      ⇒ \(\frac{3}{40}\) – \(\frac{1}{20}\)

      ⇒ \(\frac{3 – 2}{40}\)

      ⇒ \(\frac{1}{40}\)

      Therefore, S alone can do the work in 40 days

      Similarly, T’s 1 day work = \(\frac{3}{40}\) – \(\frac{1}{30}\)

      ⇒ \(\frac{9 – 4}{120}\)

      ⇒ \(\frac{5}{120}\)

      ⇒ \(\frac{1}{24}\)

      Therefore, T alone can do the work in 24 days

      U’s 1 day work = \(\frac{3}{40}\) – \(\frac{1}{15}\)

      ⇒ \(\frac{9 – 8}{120}\)

      ⇒ \(\frac{1}{120}\)

      Therefore, U alone can do the work in 120 days


    4. 5 men can prepare 10 cycles in 6 days working 6 hours a day. Then in how many days can 12 men prepare 16 cycles working 8 hours a day?

    Solution:

      Let,

      Number of men \(M_1\) = 5

      Number of cycles \(W_1\) = 10

      Number of days \(D_1\) = 6

      Number of hours \(T_1\) = 6

      Number of men \(M_2\) = 12

      Number of cycles \(W_2\) = 16

      Number of hours \(T_2\) = 8

      Number of days \(D_2\) = ?

      Now, consider

      \(M_{1}W_{2}D_{1}T_{1}\)= \(M_{2}W_{1}D_{2}T_{2}\)

      substitute the given values, i.e.

      5 x 6 x 6 x 16 = \(D_{2}\) x 12 x 10 x 8

      ⇒ \(D_{2}\) x 12 x 10 x 8 = 2880

      ⇒ \(D_{2}\) = \(\frac{2880}{960}\)

      ⇒ \(D_{2}\) = 3 days

      Therefore, Required number of days = 3 days.


    5. If A can do a piece of work in 10 days and B can do it in 15 days then how long will they take if they work together?

    Solution:

      Given

      Number of work days of A = \(D_1\) = 10 days

      Number of work days of B = \(D_2\) = 15 days

      Now, consider

      \(\frac{D_1 * D_2}{D_1 + D_2}\)

      ⇒ \(\frac{10 * 15}{10 + 15}\)

      ⇒ \(\frac{150}{25}\)

      ⇒ 6

      Therefore, A and B can do the work together in 6 days.