**Introduction**

**Important Formulae:**

- Speed=Distance/Time
- Time=Distance/speed
- Distance = speed*time
- 1km/hr=5/18m/s
- 1m/s=18/5Km/hr
- If the ratio of the speed of A and B is a:b,then the ratio of the time taken by them to cover the same distance is 1/a:1/b or b:a
- Suppose a man covers a distance at x kmph and an equal distance at y kmph, then the AVERAGE SPEED during the whole journey is (2xy/x+y) kmph

Out of time, speed and distance we can compute any one of the quantities when we happen to know the other two. For example, suppose we drive for 2 hours at 30 miles per hour, for a total of 60 miles.

If we know the time and the speed, we can find the distance:

2 hour * 30 miles/hour=60 miles

If we know the time and the distance, we can find the speed:

60 miles/2 hours=30miles/hour

**For a non-uniform motion**, Average speed= Total distance travelled/Total time taken

When the body travels at ‘u’ m/s for t1 seconds and ‘v’ m/s for t2 seconds, then**Average speed**= (ut1+vt2)/(t1+t2)

When the body travels l distance at ‘u’ m/s and ‘m’ distance at ‘v’ m/s;

**Average speed** = (mu+lv) / (l+m)**Relative Speed**: Speed of a moving body w.r.t. another moving body is called relative speed.

**Relative Speed:**

Case1: Two bodies are moving in opposite directions at speed V1 & V2 respectively. The relative speed is defined as Vr=V1+V2

Case2: Two bodies are moving in same directions at speed V1 & V2 respectively. The relative speed is defined as Vr=|V1–V2|

**Train Problems:**

The basic equation in train problem is the same Speed=Distance/Time

The following things need to be kept in mind while solving the train related problems.

1-When the train is crossing a moving object, the speed has to be taken as the relative speed of the train with respect to the object.

2-The distance to be covered when crossing an object, whenever trains crosses an object will be equal to: Length of the train + Length of the object

**BOATS AND STREAM**

Let U= Velocity of the boat in still water

V=Velocity of the stream.

While moving in upstream, distance covered, S=(U−V)T

In case of downstream, distance covered ,S=(U+V)T

**Key points on Trains:**

When a train is crossing a pole distance travelled by the train= length of train

When a train of length l is crossing a bridge of length b; the distance travelled by train=l+b

When a train of length l is crossing a platform of length p; then distance travelled by train=l+p

When a train of length l1 is crossing/ overtaking another train l2; then distance travelled = l1+l2

**Linear races:**

Imagine, X and Y are two contestants in a race:

- Before the start of the race, if X is at the starting point and Y is ahead of X by 10 meters, then X is said to give Y a start of 10 meters. 10 meter here can be called as start distance or distance at start
- In a 100m race, If it is written “X can give Y 20 m start” or “X beats Y by 20 m”, it means that in the time X runs 100 m, Y runs 80 m. 20 meter here can be called as beat distance called as beat distance.
- Similarly, If it is written “X can give Y 20 second start” or “X beats Y by 20 seconds”, it means if the given distance is covered by X in a seconds, then Y will take (a – 20)
- Winner’s distance – (start distance + beat distance) = loser’s distance
- Winner’s time + (start time + beat time) = loser’s time. (Remember here that the winner’s time is less than loser’s time, so something has to be added to equate)
- A dead heat means the contestants reached the end point at same time

**Circular races**

**Circular Races :** Circular races are on circular tracks where one can meet other person more than once.

When two persons A & B starts from same point at same time on a circular track then we can find

I. after how much time they meet for first time :- they meet for first time when one covers one more lap than other person. Relative distance would be length of track & using relative speed, time taken can be found.

II. After how much time they will meet for first time at starting point : this can be find out by taking LCM of time taken by individual to cover one lap.

**CLOCKS**

**Minute Spaces:**

The face or dial of watch is a circle whose circumference is divided into 60 equal parts, called minute spaces.

**Hour Hand and Minute Hand:**

A clock has two hands, the smaller one is called the hour hand or short hand while the larger one is called minute hand or long hand.

In 60 minutes, the minute hand gains 55 minutes on the hour on the hour hand.

In every hour, both the hands coincide once.

The hands are in the same straight line when they are coincident or opposite to each other.

When the two hands are at right angles, they are 15 minute spaces apart.

When the hands are in opposite directions, they are 30 minute spaces apart.

Angle traced by hour hand in 12 hrs = 360°

Angle traced by minute hand in 60 min. = 360°.

If a watch or a clock indicates 8.15, when the correct time is 8, it is said to be 15 minutes too fast.

On the other hand, if it indicates 7.45, when the correct time is 8, it is said to be 15 minutes too slow.

Problems in angles

**Method :1**

Before we actually start solving problems on angles, we need to know couple of basic facts clear:

Speed of the hour hand = 0.5 degrees per minute (dpm)

Speed of the minute hand = 6 dpm

At ‘n’ o’ clock, the angle of the hour hand from the vertical is 30n

The questions based upon these could be of the following types

**Example : 1 **

What is the angle between the hands of the clock at 7:20

At 7 o’ clock, the hour hand is at 210 degrees from the vertical.

In 20 minutes,

Hour hand = 210 + 20*(0.5) = 210 + 10 = 220 {The hour hand moves at 0.5 dpm}

Minute hand = 20*(6) = 120 {The minute hand moves at 6 dpm}

Difference or angle between the hands = 220 – 120 = 100 degrees

**Method : 2**

**Example :2 **

Find the reflex angle between the hands of a clock at 05.30?

The above problem are solved by the bellow formula

Angle between X and Y =|(X*30)-((Y*11)/2)|

Angle between hands at 5:30

Step 1: X=5 , Y=30

Step 2: 5*30=150

Step 3: (30*11)/2 = 165

Step 4: 165-150=15

Thus, angle between hands at 5:30 is 15 degrees.

**Method : 3**

**Example : 3 **

At what time 3&4’o clock in the hands of clock together.

Approximately we know at 03:15 hands of the clock together

So 15*60/55=16.36 min

**TIME, DISTANCE AND SPEED**

**Distance is directly proportional to Velocity when time is constant:**

**Example**: A car travels at 30km/hr for the first 2 hrs & then 40km/hr for the next 2hrs. Find the ratio of distance travelled

S1/S2=V1/V2=3/4

**Example**: Two cars leave simultaneously from points A & B (100km apart) & they meet at a point 40 km from A. What is

Va/Vb?

Time is constant so

V1/V2=S1/S2=40/60=4/6

**Example**: A train meets with an accident and moves at (3/4)th its original speed. Due to this, it is 20 min late. Find the original time for the journey beyond the point of accident?

Method1: Think about 2 diff. situations, 1st with accident and another w/o accident. As distance in both the cases is constant

So

V1/V2=T2/T1

=>V1/[3/4*V1]=T1+20/T1

=> 4/3=T1+20/T1

=>T1=60

**Method 2**: Velocity decreases by 25% (3/4 of original speed => decrement by 1/4) so time will increase by 33.3% (4/3 of original time => increment by 1/3)

now, 33.3%=20 min =>100%=60 min

**SOLVED EXAMPLES**

**1. How many minutes does Aditya take to cover a distance of 400 m, if he runs at a speed of 20 km/hr**

A-115min B-215min C-315min D-415min

Explanation:

We know that,

Time=Distance/Speed

Speed=20 km/hr=20∗518m/sec=509m/sec

Time =(400∗9/50)=72sec=115min

Answer: Option A

**2. A car moves at 80 km/hr. What is the speed of the car in meters per second ?**

A-2029msec B-2229msec C-2429msec D-2629msec

Explanation:

Speed=(80∗5/18)m/sec=2009m/sec=2229msec

Answer: Option B

**3. An athlete runs 200 meters in 24 seconds. His speed is ?**

A-10 km/hr B-17 k/hr C-27 km/hr D-30 km/hr

Explanation:

Speed=Distance/Time=200/24m/sec=25/3m/sec

253∗18/5km/hr=30km/hr

Answer: Option D

**4. A person crosses a 600 meter long street in 5 minutes. What is the speed in Km/hr**

A-6.2 km/hr B-7.2 km/hr C-8.2 km/hr D-9.2 km/hr

Explanation:

Two things to give attention on this question.

First time is in minutes, we need to change it to seconds to get speed in m/sec, then we need to get the final answer in km/hr.

So lets solve this.

Speed=Distance/Time

Distance=650meter Time=5 minutes=300sec

Speed=600/300=2m/sec=>2∗18/5km/hr=7.2km/hr

Answer: Option B

**5. A man is walking at the rate of 5 km/hr crosses a bridge in 15 minutes. The length of the bridge is**

A-1000 meters B-1050 meters C-1200 meters D-1250 meters

Explanation:

We need to get the answer in meters. So we will first of change distance from km/hour to meter/sec by multiplying it with 5/18 and also change 15 minutes to seconds by multiplying it with 60.

Speed=5∗5/18=25/18m/sec

Time=15∗60seconds=900seconds

Distance=Time∗Speed

Distance=2518∗900=1250meter

Answer: Option D

**6. A train covers a distance in 50 minutes, if it runs at a speed of 48kmph on an average. Find the speed at which the train must run to reduce the time of journey to 40 minutes.**

A-50 km/hr B-60 km/hr C-65 km/hr D-70 km/hr

Explanation:

We are having time and speed given, so first we will calculate the distance. Then we can get new speed for given time and distance.

Lets solve it.

Time = 50/60 hr = 5/6 hr

Speed = 48 mph

Distance = S*T = 48 * 5/6 = 40 km

New time will be 40 minutes so,

Time = 40/60 hr = 2/3 hr

Now we know,

Speed = Distance/Time

New speed = 40*3/2 kmph = 60kmph

Answer: Option B

**7. Two boys starting from the same place walk at a rate of 5kmph and 5.5kmph respectively. What time will they take to be 8.5km apart, if they walk in the same direction?**

A-15 hours B-16 hours C-17 hours D-18 hours

Explanation:

In this type of questions we need to get the relative speed between them,

The relative speed of the boys = 5.5kmph � 5kmph

= 0.5 kmph

Distance between them is 8.5 km

Time = Distance/Speed

Time= 8.5km / 0.5 kmph = 17 hrs

Answer: Option C

**8. Excluding stoppages, the speed of a bus is 54 kmph and including stoppages, it is 45 kmph. For how many minutes does the bus stop per hour ?**

A-8 minutes B-10 mintues C-12 minutes D-14 minutes

Explanation:

Due to stoppages, it covers 9 km less.

Time taken to cover 9 km = (9/54) hour

= (1/6)*60 minutes

= 10 minutes

Answer: Option B

**9. 2 trains starting at the same time from 2 stations 200 km apart and going in opposite direction cross each other at a distance of 110 km from one of the stations. What is the ratio of their speeds ?**

A-11:9 B-13:9 C-17:9 D-21:9

Explanation:

We know total distance is 200 Km

If both trains crossed each other at a distance of 110 km then one train covered 110 km and other 90 km [110+90=200km]

So ratio of their speed = 110:90 = 11:9

Answer: Option A

**10. A man complete a journey in 10 hours. He travels first half of the journey at the rate of 21 km/hr and second half at the rate of 24 km/hr. Find the total journey in km.**

A-200 Km B-222 Km C-224 Km D-248 Km

Explanation:

Let time taken to travel the first half = x hr

Then time taken to travel the second half = (10 – x) hr

Distance covered in the the first half = 21x [because, distance = time*speed]

Distance covered in the the second half = 24(10 – x)

Distance covered in the the first half = Distance covered in the the second half

So,

21x = 24(10 – x)

=> 45x = 240

=> x = 16/3

Total Distance = 2*21(16/3) = 224 Km [multiplied by 2 as 21x was distance of half way]

Answer: Option C

**11. A Man travelled a distance of 61 km in 9 hours. He travelled partly on foot at 4 km/hr and partly on bicycle at 9 km/hr. What is the distance travelled on foot?**

A-16 km B-14 km C-12 km D-10 km

Explanation:

Let the time in which he travelled on foot = x hour

Time for travelling on bicycle = (9 – x) hr

Distance = Speed * Time, and Total distance = 61 km

So,

4x + 9(9-x) = 61

=> 5x = 20

=> x = 4

So distance traveled on foot = 4(4) = 16 km

Answer: Option A

**12. In a flight of 600 km, an aircraft was slowed down due to bad weather. Its average speed for the trip was reduced by 200 km/hr and the time of flight increased by 30 minutes. What is the duration of the flight ?**

A-3 hours B-2.4 hours C-1.4 hours D-1 hour

Explanation:

Let the duration of the flight be x hours.

Then

600x−600x+12=200600x−12002x+1=200x(2x+1)=32×2+x−3=0=>(2x+3)(x−1)=0

Neglecting the negative value for x we get x = 1

Answer: Option D

**13. A man on tour travels first 160 km at 64 km/hr and the next 160 km at 80 km/hr. Find the average speed for first 320 km of tour.**

A-70.11 km/hr B-71.11 km/hr C-72.11 km/hr D-73.11 km/hr

Explanation:

We know Time = Distance/speed

So total time taken =

(16064+16080)=92hours Time taken for 320 Km = 320∗29=71.11km/hr

Answer: Option B

**14. Robert is travelling on his cycle and has calculated to reach point A at 2 P.M. if he travels at 10 kmph, he will reach there at 12 noon if he travels at 15 kmph. At what speed must he travel to reach A at 1 P.M.?**

A-9 km/hour B-10 km/hour C-11 km/hour D-12 km/hour

Explanation:

We need to calculate the distance, then we can calculate the time and finally our answer.

Lets solve this,

Let the distance travelled by x km

Time = Distance/Speed

x10−x15=2[because, 2 pm – 12 noon = 2 hours]3x−2x=60x=60.Time=DistanceSpeedTime@10km/hr=6010=6hours

So 2 P.M. – 6 = 8 A.M

Robert starts at 8 A.M.

He have to reach at 1 P.M. i.e, in 5 hours

So, Speed = 60/5 = 12 km/hr

Answer: Option D

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