Shortcuts in Quantitative Aptitude for Competitive Exams – Time & Work
Shortcuts in Quantitative AptitudeReasoningEnglish
TIME AND WORK
In most of the problems on time and work, either of the following basic parameters are to be calculated:
- If A can do a piece of work in X days, then A’s one day’s work = part of whole work.
- If A’s one day’s work = part of whole work, then A can finish the work in X days.
- If A can do a piece of work in X days and B can do it in Y days then A and B working together will do the same work in days.
- If A, B and C can do a work in X, Y and Z days respectively then all of them working together can finish the work in days
- If (A + B) can do a piece of work in X days, (B + C) can do a place of work in Y days and
(C + A) can do a piece of work in Z days.
Then, (A + B + C) can do a piece of work in days
- If A and B together can do a piece of work in X days and A alone can do it in Y days, then B alone can do the work in days.
- If (A + B + C) can do a piece of work in X days and (B + C) can do a piece of work in Y days then
A can do a piece of work days.
- A and B can do a work in ‘X’ and ‘Y’ days respectively. They started the work together but A left ‘a’ days before completion of the work. Then, time taken to finish the work is
- lf ‘A’ is ‘a’ times efficient than B and A can finish a work in X days, then working together, they can finish the work in days.
- If A is ‘ a’ times efficient than B and working together they finish a work in Z days then, time taken by days, and time taken by B = Z(a +1) days.
- If A working alone takes ‘x’ days more than A and B together, and B working along takes ‘y’ days more than A and B together then the number of days taken by A and B working together is given by days.
- If a1 men and b1 boys can complete a work in x days, while a2 men and b2 boys can complete the same work in y days, then of 1 man
- If n men or m women can do a piece of work in X days, then N men and M women together can finish the work in days.
- A and B do a piece of work in a and b days, respectively. Both begin together but after some days, A leaves off and the remaining work is completed by B in x days. Then, the time after which A left is given by
- If ‘M1’ persons can do ‘W1’ works in ‘D1’ days and ‘M2’ persons can do ‘W2’ works in ‘D2’ days then M1D1W2 = M2D2W1
If T1 and T2 are the working hours for the two groups then M1D1W2T1 = M2D2W1T2
M1D1W2T1E1 = M2D2W1T2E2, where E1 and E2 are the efficiencies of the two groups.
- If the number of men to do a job is changed in the ratio a:b, then the time required to do the work will be in the ratio b: a, assuming the amount of work done by each of them in the given time is the same, or they are identical.
- A is K times as good a worker as B and takes X days less than B to finish the work. Then the amount of time required by A and B working together is days.
- If A is n times as efficient as B, i.e. A has n times as much capacity to do work as B, then A will take of the time taken by B to do the same amount of work.
WORK AND WAGES
Wages are distributed in proportion to the work done and in indirect proportion to the time taken by the individual.
PIPES AND CISTERNS
The same principle of Time and Work is employed to solve the problems on Pipes and Cisterns. The only difference is that in this case, the work done is in terms of filling or emptying a cistern (tank) and the time taken is the time taken by a pipe or a leak (crack) to fill or empty a cistern respectively.
A pipe connected with a tank (or a cistern or a reservoir) is called an inlet, if it fills it.
A pipe connected with a tank is called an outlet, if it empties it.
- If a pipe can fill a tank in x hours, then the part filled in 1 hour =
If a pipe can empty a tank in y hours, then the part of the full tank emptied in 1 hour =
- If a pipe can fill a tank in x hours and another pipe can empty the full tank in y hours, then the net part filled in 1 hour, when both the
pipes are opened =
.’. Time taken to fill the tank, when both the pipes are opened =
- If a pipe can fills or empties tank in x hours and another can fill or empties the same tank in y hours, then time taken to fill or empty the tank = when both the pipes are opened.
- If a pipe fills a tank in x hours and another fills the same tank is y hours, but a third one empties the full tank in z hours, and all of them are opened togrther,
then net part filled in 1hr =
.’. Time taken to fill the tank = hours
- A pipe can fill a tank in x hrs. Due to a leak in the bottom it is filled in y hrs. If the tank is full, the time taken by the leak to empty the tank
- A cistern has a leak which can empty it in X hours. A pipe which admits Y litres of water per hour into the cistern is turned on and now the cistern is emptied in Z hours. Then the capacity of the cistern is hours.
- A cistern is filled by three pipes whose diameters are X cm., Y cm. and Z cm. respectively (where X < Y < Z). Three pipes are running together. If the largest pipe alone will fill it in P minutes and the amount of water flowing in by each pipe is proportional to the square of its diameter, then the time in which the cistern will be filled by the three pipes is hours.
- If one filling pipe A is n times faster and takes X minutes less time than the other filling pipe B, then the time they will take to fill a cistern, if both the pipes are opened together, is minutes.A will fill the cistern in minutes and B will take to fill the cistern minutes.
Here, A is the faster filling pipe and B is the slower one.
- Two filling pipes A and B opened together can fill a cistern in t minutes. If the first filling pipe A alone takes X minutes more or less than t and the second fill pipe B along takes Y minutes more or less than t minutes, then t is given by minutes.
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