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**Example 1:**

A can do a piece of work in 5days, and B can do it in 6 days. How long will they take if both work together?

**Solution:**

A’s 1 day’s work =1/5th part of whole work and

B’s 1 day’s work =1/6th part of whole work

∴ (A+B)’s one day’s work = 1/5+1/6=11/30 th part of whole work.

So, both together will finish the work in 30/11 days = 28/11days.

**By Direct Formula**:

A+B can do the work on (5×6)/(5+6) days =30/11=2 8/11 days.

**Example 2:**

Two men, Vikas and Vishal, working separately can mow a field in 8 and 12 hours respectively. If they work in stretches of one hour alternately, Vikas beginning at 8 a.m., when will the mowing be finished?

**Solution:**

In the first hour, Vikas mows 1/8 of the field.

In the second hour, Vishal mows 1/12 of the field.

∴ In the first 2 hours, (1/8+1/12=5/24) of the field is mown.

∴ In 8 hours, 5/24×4 = 5/6 of the field is mown.

Now, (1-5/6)=1/6 of the field remains to be mown.

In the 9th hour, Vikas mows 1/8 of the field.

Remaining work = 1/6-1/8=1/24

∴ Vishal will finish the remaining work in (1/24+1/12) or 1/2 of an hour.

∴ The total time required (8 +91/2 = 171/2) or 5.30 pm.

**Example 3:**

A can do a piece of work in 36 days, B in 54 days and C in 72 days. All the three began the work together on the Dec. 15, 2014, but A left 8 days and B left 12 days before the completion of the work. If C took the reset for a week then in how many days, the work was finished from the day it started?

**Solution:**

Let the total time taken be x days.

According to the given condition

⇒(x-8)/36+(x-12)/54+x/72=1

⇒(6(x-8)+4(x-12)+3x)/216=1

⇒ (6x-48+4x-48+3x)/216 = 1 ⇒(13x-96)/216 = 1

⇒ 13x – 96= 216 ⇒ 13x = 216 +96 = 312 ⇒ x=312/13 = 24

Since, C takes the rest for a week, so the number of days in which the work was finished from one day it started = 31 i.e. on 14.01.2015.

**Example 4:**

A and B can do a certain piece of work in 8 days, B and C can do it in 12 days and C and A can do it in 24 days. How long would each take separately to do it?

**Solution:**

(A+B)’s one day’s work = 1/18,

(A+C)’s one day’s work = 1/24,(B+C)’s one day’s work = 1/12,

Now add up all three equations:

2 (A+B+C)’s one day’s work = 1/18+1/24+1/12=13/72

(A+B+C)’s one day’s work = 13/144

A’s one day’s work = (A+B+C)’s one day’s work -(B+C)’s one day’s work = 13/144-1/2=1/144

Since A completes of the work in 1 day, he will complete 1 work in 144/1 = 144 days

By similar logic we can find that B needs days and C will require 144/5 days.

- If A and B toether 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 XY/(Y-X) days

**Example 5:**

A and B together can do a piece of work in 6 days and A alone can do it in 9 days. In how many days can B alone do it?

**Solution:**

(A+B)’s 1 day’s work = 1/6 th part of the whole work.

A’s 1 day’s work = 1/9 th part of the whole work.

∴ B’s 1 day’s work = 1/6-1/9=(3-2)/18=1/18 th part of the whole work.

∴ B alone can do the work in 18th days.

By direct Formula:

B alone can do the whole work in(6×9)/(9-6)=54/3= 18 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

(Y(X+a))/(X+Y)If ‘A’ id ‘a’ times efficient than B and A can finish a work in X days, then working together, they can finish the work inaX/(a+1)days. - If A is ‘a’ times efficient than B and working together they finish a work in Z days then, time taken by A =(Z(a+1))/a 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 (√xy)days.

**WORK AND WAGES**

Wages are distributed in proportion to the work done in indirect proportion to the time taken by the individual

**Example 19:**

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. 1350. What is the share of B in that amount?

**Solution:**

A’s one day’s work = 1/6

B’s one day’s work = 1/8

C’s one day’s work = 1/12

A’s share : B’s share : C’s share

=1/6:1/8:1/12

Multiplying each ratio by the L.C.M. of their denominators, the ratio become 4: 3: 2

∴ B’s share = (1350×3)/9 = `450

**PIPE 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 (creak) to fill or empty a cistern respectively.

Inlet: A pipe connected with a tank (or a cistern or a reservoir is called in inlet, if it fills it.

**Outlet:**

- 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 = 1/x
- If a pipe can empty a tank in y hours, then the part filled in 1 full tank emptied in 1 hour =1/y.
- 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 = (1/x-1/y).

∴ Time taken to fill the tank, when both the pipes are opened =xy/(y-x). - If a pipe can fill a tank in x hours and another can fill the same tank in y hours, then time taken to fill the tank = xy/(y+x), 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 together, then net part filled in 1 hr = [1/x+1/y-1/z]

∴ Time taken to fill the tank= xyz/(yz+xz-xy)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 = xy/(y-x)hrs.
- 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 (X+Y+Z)/(Z-X)litres.
- 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

[PZ^2/(X^2+Y^2+Z^2 )] minutes. - 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 [nX/((n^2-1) )] minutes. A will fill the cistern in (X/(n-1)) minutes and B will taken to fill the cistern (nX/(n-1)) 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 [t=√xy]minutes.