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Parin Sir > Quantitative Aptitude (Maths) > Time And Distance > Examples 11 to 15 of Time and Distance > Examples 11 to 15 of Time and Distance

Examples 11 to 15 of Time and Distance

August 9, 2020
Category: Examples 11 to 15 of Time and Distance

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(11) An motor cars moves along the four sides of a square at the speeds of 20, 40, 60 and 80 km/hr. Find the average speed of the motor car around the field.

Soln. Since each side of the square be x km is same, average speed = Harmonic

∴ Average Speed = H.M. = $\frac{n}{\frac{1}{a_1}+\frac{1}{a_2}+...+\frac{1}{a_n}}$

= $\frac{4}{\frac{1}{20}+\frac{1}{40}+\frac{1}{60}+\frac{1}{80}}$

∴ Average speed = 384 km/hr.

(12) Pratap goes to school at 10 kmph and reaches the school 4 minutes late. Next day, he covers this distance at 12 km per hour and reaches the school 6 minutes earlier than the scheduled time. What is the distance of his school from his house ?

Soln. Let the required distance be D km.

With the two speeds, the difference of time taken is 10 min.

Now, T₁ – T₂ = 10 min. = $\frac{10}{60}$ hrs.

∴ $\frac{D}{S_1} - \frac{D}{S_2} = \frac{10}{60}$

∴ $\frac{\mathrm{6D\ }-\mathrm{\ 5D}}{\mathrm{60}} = \frac{10}{60}$

∴ D = 10 km.

∴ Required distance = 10 km.

(13) A and B are two stations 620 km apart. A scooterist start from A at 8:00 a.m. and travels towards B at 60 kmph. Another scooterist starts from B at 10:00 a.m. and travels towards A at 40 kmph. At what time do they meet ?

Soln. Suppose both start at 10:00 a.m.,

then distance between them = 500 km.

( ∵ First covers 60 km. in 1 hr. ∴ 120 km. in 2 hrs. )

First scooterist covers 60 km. and another covers 40 km. towards each other in 1 hr.

∴ Both covers 100 km. in 1 hr.

∴ To cover 500 km. in 5 hr.

i.e. Both will meet at 10:00 a.m. + 5 hrs. = 3:00 p.m.

(14) A thief is spotted by a policeman from a distance of 500 metres. When the policeman starts the chase, the thief also starts running. If the speed of the thief be 8 m/sec and that of the policeman 10 m/sec, then when the policeman will catch the thief.

Soln. Relative speed of the policeman = ( 10 – 8 ) m/sec = 2 m/sec

∴ For gaining 2 meter, policeman has to run for 1 sec.

So, for gaining 500 meter, he has to run for $\frac{500\times 1}{2}$ = 250 sec.

∴ Policeman catch the thief in 250 sec. = 4 min. 10 sec.

☼ Formula ( Problems on Trains ) :

■ Relative Velocity :

↦ If two articles move in the same direction, their relative speed = difference of their speeds.

↦ If they move in opposite directions, their relative speed = sum of their speeds.

■ Let the length of the first train is x km and its speed is u kmph,

the length of the second train is y km and its speed v kmph,

the length of a stationary object is z km. then

↦ time taken in passing a signal post or pole or a standing man for the first train

= $\frac{x}{u}$ hour.

↦ time taken in passing a stationary object for the first train = $\frac{x+z}{u}$ hr.

time taken for two trains to cross each other, if they are moving in the same direction = $\frac{x+y}{u-v}$ hr.

↦ time taken for two trains to cross each other, if they are moving in the opposite direction = $\frac{x+y}{u+v}$ hr.

■ If two trains start at the same time from A and B towards each other and after crossing, they take a & b hours in reaching B and a respectively. Then,

A’s speed : B’s speed = $\sqrt b$ : $\sqrt a$