# Speed and Distance Advance Level

The problem in the Speed and distance chapter sometimes becomes very complex and needs to be solved with the help of equations. If you understand the depth of this chapter, then it is easy for you to solve the questions. This chapter is closely related to physics and its concepts are also used to solve problems like motion in a straight line, circular motion, relative motion etc. In SSC, 2 to 4 questions are asked regularly in prelims exams.

**Important concepts:**

**1. Motion:**

The rate at which distance is covered during the motion is called speed. Sometimes in questions, the word Speed is given as rate/velocity and it is measured in distance per unit of time.

The unit of speed are- metre/sec, km/hr, km/min, metre/min, etc.

The basic formula to calculate Distance= Speed × Time

**2. Conversion of Unit:**

1 kmph or 1km/hr

= 100 m/60×60sec

= 5m/18sec= 5/18 m/s

So, to convert km/hr into m/s, we multiply by 5/18 and similarly to convert m/s into km/hr we divide by 5/18.

When one of the terms is constant in the distance, speed or time, then the proportionality is:

S×T=D

When the speed of an object is constant, then the time is directly proportional to distance. So the proportionality follow is-

t1/t2=d1/d2

Same when time is constant, speed is directly proportional to distance.

s1/s2=d1/d2

But in case, when distance is constant, speed is inversely proportional to time.

s∝ 1/t

s1t1=s2t2

**3. Average speed:**

Average speed formula= Total covered distance/ Total time taken

It is the ratio of the total covered distance to the total time taken.

If a person travels somewhere p1, p2, p3,p4…..pn distance in time q1, q2, q3, q4….qn with different speeds r1, r2, r3, ….rn. then its average speed is-

Average speed=(p1+p2+p3+p4+⋯.pn)/(p1/r1+p2/r2+p3/r3+p4/r4+⋯..pn/rn)

**Note:**

We know according to average that when two different speeds (r1 and r2) covered the same distance.

then average speed= [2(r1+r2)]/(r1×r2)

If we want to calculate the average speed with three different speeds that covered the same distance, then the formula of average speed=

(3r1.r2.r3)/(r1.r2+r2.r3+r3.r1)

**Questions related to the article:**

**Q1: Govind travels from Allahabad to Bareilly at a speed of 15 km/hr, from Bareilly to Chandausi at 30 km/hr, and from Chandausi to Dehradun at 40 km/hr. If the distance travelled is the same in all cases then find the average speed of Govind?**

A. 18 km/hr

B. 24 km/hr

C. 16 km/hr

D. 10 km/hr

**Answer: Option B****Explanation:**

Suppose the distance travelled is 120 km.

let distance be x km Avg speed

= x+x+x/(x/15+x/30+x/40) km/hr

= 3x/(15x/120)

= (3*120)/15

= 24km/hr

**Q2: Ravindra wants to go to Delhi so, he divided the distance of his journey into 3 parts 20 km, 25km, 45 km. He travelled these distances at different speeds of 5 km/h, 10 km/h, 15 km/h. Find its average speed of total journey?**

A. 900/95 km/hr

B. 78 km/hr

C. 465/17 km/hr

D. 123 km/hr

**Answer: Option A****Explanation:**

Average speed= Total Distance/ total time

= (20+25+45)/(20/5+25/10+45/15)

=90/(4+5/2+3)

=90/9.5 = 900/95 km/hr

**Q3: In a competition, Rahul is running at a speed of 50 km/hr. if he runs at a speed of 40 km/hr then, find the percentage decrease in its speed?**

A. 20%

B. 50%

C. 78%

D. 10%

**Answer: Option A****Explanation:**

The required increase in the percentage

Percentage decrease= Reduced time/original time ×100

=(50-40)/50×100

=10/50×100= 20%

**Q4: The distance from Prayagraj to Mughal Sarai Junction is 1176 km. This distance is covered by a superfast train in 5 hours less than an express train while the average speed of the express train is 70 km/hr less than that of the superfast train. Find the time taken by express train?**

A. 46 km/hr

B. 35 km/hr

C. 98 km/hr

D. 17 km/hr

**Answer: Option C****Explanation:**

Assume, the speed of express train= x km/hr

Speed of superfast train= (x+70) km/hr

We know that, the formula to calculate distance is

Distance= speed×time

Distance= x(x+70)/70×5

1176= x(x+70)/70×5

x(x+70)=1176×14

x^2+70x-16464=0

x^2+168x-98x-16464=0

on solving x=98 km/hr

**Q5: Ravi in still water can row a boat at 8 kmph. He took thrice as much time to row up than as to row down the same distance in the river. Find the speed of the current?**

A. 3km/hr

B. 8 km/hr

C. 4 km/hr

D. 10 km/hr

**Answer: Option C****Explanation:**

Speed of boat = 8 kmph and x = speed of current

Speed in downstream(S1) = (x +8)

Speed in upstream (S2)= (8 – x )

Here, Distance is constant.

So, S1 × T2 = S2 × T2

=> (x +8) × 1 = (8-x) × 3

=> x + 8 = 24 – 3x

=> 4x = 16

:. x = 4

This , speed of current = 4 km/hr

**Q6: Duronto express with a speed of 105 km/hr starts from Prayagraj at 7:00 AM and reached Orissa. There it stopped for 45 minutes. While returning from Orissa Duronto speed was reduced by 20% and it reaches Prayagraj at 5:12 PM on the same day. Find the distance travelled between Prayagraj to Orissa?**

A. 300 km

B. 843 km

C. 441 km

D. 230 km

**Answer: Option C****Explanation:**

Let, D be the distance between Prayagraj to Orissa

Total time of journey including stoppage time at Prayagraj= 10 hours 12 minutes

Total time of journey excluding stoppage time at Prayagraj = 9 hours 27 minutes

9 hrs 27 minutes= 9.45 hours

Reduced speed = 80/100 of 105 = 84 km/hr

From the question:

(D/105) + (D/84) = 9.45

Taking LCM of 105, 84 we get,

(4D + 5D)/420 = 9.45

9D = 9.45 × 420

D = 9.45 × 420/9

D = 441 km

**Q7: If an E-rickshaw increases its speed by 20 kmph, then, 3 hours less time will be taken by him to cover the distance between P and Q and if the E-ricksaw reduce the speed by 20 kmph, then 4.5 hours more will be taken to cover the same distance. Calculate the distance between P and Q?**

A. 1800 km

B. 1313 km

C. 1900 km

D. 2300 km

**Answer: Option A ****Explanation:**

Let, the distance between P and Q = x

Initial Speed of E-ricksaw = y

When he takes 3 hr less time,

((x/y)-x/(y+20)) = 3

On taking LCM from denominator, we get

20x/(y^2+ 20y) = 3——-1

When he takes 4.5 hrs more,

x/((y-20) ) – x/y = 4.5

On taking LCM of denominator, we get

20x/((y^2-20y) ) = 4.5——–2

From eq.1 and 2 we get,

3y^2-60y = 4.5y^2+ 90

1.5y = 150

y = 100 km/hr

Now, put y=100 in Eq. 1 we get,

20/(100 × 100 + 20 × 100) = 3

x = 1800 km

**5) the First cab with a speed of 84 km/h leaves from Delhi and after 8 hours, the second cab leaves and they both meet after 6 hours. What is the speed of the 2nd cab?**

A. 12 kmph

B. 44 kmph

C. 36 kmph

D. 21 kmph

**Answer: Option C****Explanation:**

Let the speed of the 2nd cab be x km/h.

And where they meet the distance will be the same,

According to question,

(8 + 6)x= 84 × 6

=> x= 36

Hence, speed of 2nd car= 36 km/h.

**Q9: A special train departs from Mysore travelling at a speed of 25 kmph at 9 am and another mail train departs at 2 pm by travelling at a speed of 35 kmph departing in the same direction. They together travel how much distance?**

A. 21 3/7 km

B. 843/17 km

C. 437 1/2 km

D. 230 km

**Answer: Option C****Explanation:**

To travel together, the special train has started early and travel for 5 hrs with a speed of 25kmph= 25×5=125km

Now, both the train is at the same position and the mail train will get a gain of 10kmph= (35 – 25)=10 km per hour.

The second train gains the time= D/S = 125/10 or 12 ½ hrs.

The required distance they together travel= 12 ½ ×35= 437 ½ km.

**Q10: Avanish travelled on a scooter from Tamilnadu to Coimbatore at the speed of (x + 10) kmph and returned from Coimbatore to Tamilnadu at the speed of (x – 10) kmph. The distance between these two cities is 600 km and total time taken is 25 hrs for whole journey. If x be the usual speed, then find the value of x.**

A. 50

B. 84

C. 44

D. 23

**Answer: Option A****Explanation:**

Let Usual speed = x

600/((x+10) )+600/((x-10) )=25

Taking 25 common from LHS,

25[24/((x+10) )+24/((x-10) )]= 25

Taking LCM of denominator, we get;

24(x – 10) + 24(x + 10) = (x^2 – 100)

24x – 240 + 24x + 240 = x^2 – 100

x^2 – 48x – 100 = 0

on factorising we get,

x^2 – 50x + 2x – 100 = 0

x(x – 50) + 2(x – 50) = 0

(x + 2)(x – 50) = 0

x = -2, 50(negative value neglected)

**Q11: Tanya come from the school at 5 pm travelling at the speed of 75% of her usual speed and after 40 minutes she stop at a confectionery shop and bought some chocolates, then travel towards home. If the time taken by Tanya to reach the school is 1 hour at the usual speed, then find the distance travelled by Tanya between the confectionery shop and home is approximately what percent of the distance between school and home?**

A. 30%

B. 82%

C. 50%

D. 23%

**Answer: Option C****Explanation:**

Suppose, the speed be x km/hr

Complete distance between school and Home = x × 1 hr = x km

75% of the original speed = x × 75/100 = 3x/4

With new speed the distance covered in 40 minutes = 3x/4 × 40/60

= 3x/4 × 2/3 = x/2

Now, the distance between confectionery shop and home = x – x/2 = x/2

Required percentage = ((x/2))/x × 100

= 1/2 × 100 = 50%

**Q12: The sum of the length of the Rajdhani and Shatabdi trains is 700. If Rajdhani cross Shatabdi in opposite and same direction in 10 sec and 22.5 sec respectively, then find the Speed of Rajdhani train.**

A. 70 kmph

B. 202 kmph

C. 115 kmph

D. 182 kmph

**Answer: Option D****Explanation:**

We know the formula of Distance = time ×speed

Let the speed of Rajdhani and Shatabdi be R and S, then

700 = (R + S) ×5/18× 10

R + S = 252 —– (1)

700 = (R – S) ×5/18× 22.5 (R cross S so R has higher speed)

R – S = 112 —– (2)

On, solving (1) and (2) eq, we get

R = 182 kmph

so, the speed of the Rajdhani train is 182 kmph

**Q13: A fruit seller sells fruits by travelling by cycle. He travels the first 1/4 distance at 10km/hr,the second 1/4 distance at 20km/hr, the third 1/4 distance at 30km/hr and the last 1/4 distance at 40km /hr. Find the average velocity of the cycle?**

A. 19

B. 43

C. 41

D. 30

**Answer: Option A****Explanation:**

Let total distance= 80 km

He travelled ¼ km each, it means he travelled for 20 km each part.

Avg velocity= (Total distance)/(total time taken)

8/[(20/10)+(20/20)+(20/40)]

=80/((2+1+2/3+1/2) )

=(80/[(12+6+4+3)] )/6

=80×6/25=19 kmph

**Q14: A book is typed by Mr Modi in which he takes 5 hours to type 5 pages while Amit shah also type the same book in which he takes 4 hours to type 80 pages. find the time taken by both of them working together on another computer to type a consignment of 150 pages.**

A. 3 hrs

B. 5 hrs

C. 4 hrs

D. 6 hrs

**Answer: Option B ****Explanation:**

In 1 hr Mr Modi type no. of pages= 50/5=10

In 1 hr Amit Shah type no. of pages= 80/4=20

Here distance is 150 and the speed are 10 and 20, so we use the distance formula here,

For 150 pages they both take,

(10+20)×T=150

T= 5 hrs

**Q15: A tourist bus was taken from UP to Haridwar. The family goes from UP to Haridwar in 12 hours, while returning from Haridwar the tourist bus driver increases the speed of the bus by 20%. If the total distance between UP and Haridwar is 300 km, then find Average speed to travel from UP to Haridwar?**

A. 35.7 km/hr

B. 27.27 km/hr

C. 43.5 km/hr

D. 10.2 km/hr

**Answer: Option B ****Explanation:**

The speed of tourist bus, from UP to Haridwar = 300/12 = 25 km/h

The speed of tourist bus, from Haridwar to UP = 25 × (120/100) = 30 km/h

Average speed of the total journey = (Total distance)/(Total time taken)

Average speed = ((2×300))/( [(300/25) + (300/30)] )

Average speed =(2×300)/(12 + 10)

=(2×300)/22

=300/11

= 27.27 km/h

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