PIPES & CISTERNS:

Inlet Pipe: A pipe which is used to fill the tank is known as Inlet Pipe.

Outlet Pipe: A pipe which can empty the tank is known as an Outlet Pipe.

• If a pipe can fill a tank in 'x' hours then the part filled per hour= 1/x
• If a pipe can empty a tank in 'y' hours, then the part emptied per hour= 1/y
• If pipe A can fill a tank 'x' hours and pipe can empty a tank in 'y' hours, if they are both active at the same time, then

                                The part filled per hour  =$$\frac{1}{x}-\frac{1}{y}$$(if y>x)

                                The part emptied per hour  =$$\frac{1}{y}-\frac{1}{x}$$(if x>y)

CLOCKS:

• In a well functioning clock, both hands meet after every 720 / 11 Mins.
• It is because the relative speed of the minute hand with respect to the hour hand = 11/2 degrees per minute.

BOATS & STREAMS

• If the speed of water is 'W' and speed of a boat in still water is 'B'

  1. Speed of the boat downstream is B+W
  2. Speed of the boat upstream is B-W

The direction along the stream is called downstream. And, the direction against the stream is called upstream.

• If the speed of the boat downstream is x km/hr and the speed of the boat upstream is y km/hr, then

                                                    Speed of boat in still water= $$\frac{x+y}{2}$$km/hr
                                                    Rate of stream= $$\frac{x-y}{2}$$km/hr

• While converting the speed in m/s to km/hr, multiply it by 3.6(18/5).

                                                    1m/s = 3.6 km/h

• While converting km/hr into m/sec, we multiply by 5/18

Distance = Speed$$\times$$Time

Speed = $$\frac{Distance}{Time}$$

Time = $$\frac{Distance}{Speed}$$

While covering the Speed in m/s to km/hr. multiply it by 3.6. It is because 1m/s = 3.6 km/hr

If the ratio of the speeds of A and B is a : b, then

• The ratio of the times taken to cover the same distance is 1/a : 1/b or b : a.
• The ratio of distance travelled in equal time intervals is a : b

                                       Average speed= $$\frac{Total Distance Travelled}{Total Time Taken}$$

If a part of a journey is travelled at speed $$S_{1}$$ km/hr in $$T_{1}$$ hours and the remaining part at speed $$S_{2}$$ km/hr in $$T_{2}$$ hours then.
                                       Total distance travelled= $$S_{1}T_{1}$$+$$S_{2}T_{2}$$ km
                                                      Average speed=$$\frac{S_{1}T_{1}+S_{2}T_{2}}{T_{1}+T_{2}}$$ km/hr

If $$D_{1}$$ km is travelled at speed of $$S_{1}$$ km/hr, and $$D_{2}$$ km is travelled at speed of $$S_{2}$$ km/hr then

                                                       Average Speed= $$\frac{D_{1}+D_{2}}{\frac{D_{1}}{S_{1}}+\frac{D_{2}}{S_{2}}}$$ km/hr

• In a journey travelled at different speeds, if the distance covered in each stage is constant, the average speed is the harmonic mean of the different speeds.
• Suppose a man covers a certain distance st x km/hr and an equal distance at y km/hr

Then the average speed during the whole journey is $$\frac{2xy}{x+y}$$ km/hr

• In a journey travelled with different speeds, if the time travelled in each stage is constant, the average speed is the harmonic mean of the different speeds.
• If a man travelled for a certain time at the speed of x km/hr and travelled for an equal amount of time at the speed of y km/hr then

Then the average speed during the whole journey is $$\frac{x+y}{2}$$ km/hr

Circular Tracks

If two people are running on a circular track with speeds in the ratio a:b where a and b are co-prime, then

• They will meet at a+b distinct points if they are running in the opposite directions.
• They will meet at |a-b| distinct points if they are running in the same direction.
  

If two people are running on a circular track having perimeter I, with speeds m and n,

• The time for their first meeting = $$\frac{I}{(m+n)}$$

(when they are running in opposite directions)

• The time for their first meeting = $$\frac{I}{(|m-n|)}$$

(when they are running in the same direction)

If a person P starts from A and heads towards B and another person Q starts from B and heads towards A and they meet after a time 't' then,

                                                        t = $$\sqrt{x*y}$$

where x = time taken (after the meeting) by P to reach B and y = time taken (after the meeting) by Q to reach A.

A and B started st a time towards each other. After crossing each other, they took $$T_{1}$$ hrs, $$T_{2}$$ hrs respectively to reach their destinations. If they travel at constant speeds $$S_{1}$$ and $$S_{2}$$ respectively all over the journey, Then
                                                     $$\frac{S_{1}}{S_{2}}$$=$$\sqrt{\frac{T_{2}}{T_{1}}}$$

TRAINS:

Two trains of length $$L_1$$ and $$L_2$$ travelling at speeds of $$S_1$$ and $$S_2$$ cross each other in

• $$\frac{L_1+L_2}{S_1+S_2}$$ if they are going in opposite directions

• $$\frac{L_1+L_2}{S_1-S_2}$$ if they are going in the same direction

• If X can do a work in 'n' days, the fraction of work X does in a day is $$\frac{1}{n}$$
• If X can do a work in 'x' days, and Y can do a work in 'y' days, the number of days taken by both of them together is $$\frac{x*y}{x+y}$$
• If $$A_1$$ men can do $$B_1$$ work in $$C_1$$ days and $$A_2$$ men can do $$B_2$$ work in $$C_2$$ days, then $$\frac{A_1 C_1}{B_1}$$ =$$\frac{A_2 C_2}{B_2}$$

• While converting the speed in m/s to km/hr, multiply it by 3.6. It is because 1 m/s = 3.6 km/hr

Work:

• If X can do a work in 'n' days, the fraction of work X does in a day us 1/n

• If X can do a work in 'x' days, and Y can do a work in 'Y' days,

                       The number of days taken by both of them together is $$\frac{x*y}{x+y}$$

• If $$M_{1}$$ men work for $$H_{1}$$ hours per day and worked for $$D_{1}$$ days and completed $$W_{1}$$ work, and if $$M_{2}$$ men work for $$H_{2}$$ hours per day and worked for $$D_{2}$$ days and completed $$W_{1}$$ work, then

                                                      $$\frac{M_{1}H_{1}D_{1}}{W_{1}}$$=$$\frac{M_{2}H_{2}D_{2}}{W_{2}}$$