# Time Speed Distance Questions for XAT 2022 – Download [PDF]

Download Time Speed Distance Questions for XAT PDF â€“ XAT Time Speed Distance questions pdf by Cracku. Top 10 very Important Time Speed Distance Questions for XAT based on asked questions in previous exam papers.

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**Question 1:Â **A 260 meter long train crosses a 120 meter long wall in 19 seconds .What is the speed of the train?

a)Â 27 km/hr

b)Â 49 km/hr

c)Â 72 km/hr

d)Â 70 km/hr

e)Â None of these

**Question 2:Â **A boat takes 2 hours to travel from point A to B in still water .To find out itâ€™s speed up-stream ,which of the following information is needed.

i. Distance between point A and B

Ii.Time taken to travel down stream from B to A

iii. Speed of the stream of the water

iv. Effective speed of Boat while traveling Downstream from B to A

a)Â All are required

b)Â Even these we cannot found the answer

c)Â Only i,iii, and either ii or iv

d)Â Only i and iii

e)Â None of these

**Question 3:Â **A train running at speed of 120 kmph crosses a signal in 15 seconds .What is the length of the train in meters?

a)Â 300

b)Â 200

c)Â 500

d)Â Cannot determined

e)Â None of these

**Question 4:Â **A bus covers a distance of 2,924 km,in 43 hours .what is the bus speed?

a)Â 72 km/hr

b)Â 60 km/hr

c)Â 68 km/hr

d)Â cannot determined

e)Â none of these

**Question 5:Â **A 240-metre long train running at the speed of 60 kmph will take how much time to cross another 270-metre long train running in opposite direction at the speed of 48 kmph?

a)Â 17 seconds

b)Â 3 seconds

c)Â 12 seconds

d)Â 8 seconds

e)Â None of these

**Question 6:Â **A man takes 2.2 times as long to row a distance upstream as to row the same distance downstream. If he can row 55 km downstream in 2 hours 30 minutes, what is the speed of the boat in still water? (in km/h)

a)Â 40

b)Â 8

c)Â 16

d)Â 24

e)Â 32

**Question 7:Â **At 60% of its usual speed, a train of length L metres crosses a platform 240 metre long in 15 seconds. At its usual speed, the train crosses a pole in 6 seconds. What is the value of L (in metre)?

a)Â 270

b)Â 225

c)Â 220

d)Â 480

e)Â 240

**Question 8:Â **A boat takes a total time of twelve hours to travel 105 kms upstream and the same distance downstream. The speed of the boat in still water is six times of the speed of the current. What is the speed of the boat in still water? (in km/hr)

a)Â 12

b)Â 30

c)Â 18

d)Â 24

e)Â 36

**Question 9:Â **At its usual speed, a train of length L metres crosses platform 300 metre long in 25 seconds. At 50% of its usual speed, the train crosses a vertical pole in 20 seconds. What is the value of L?

a)Â 160

b)Â 260

c)Â 200

d)Â 310

e)Â 350

**Question 10:Â **A boat takes a total time of eight hours to travel 63 kms upstream and the same distance downstream. The speed of the current is ${1 \over 8}$th of the speed of the boat in still water. What is the speed of the boat in still water? (in km/hr)

a)Â 32

b)Â 24

c)Â 16

d)Â 8

e)Â 38

**Answers & Solutions:**

**1)Â AnswerÂ (C)**

Length of the train is 260 metres

Length of the wall is 120 metres

total is 260+120 = 380 metres

Time taken is 19 seconds.

Hence, the speed is 380/19 = 20 m/s = 72 Km/hr

Answer is option C

**2)Â AnswerÂ (D)**

Time taken by boatÂ to travel from point A to B in still waterÂ = 2 hours

To find the upstream speed, we definitely need the speed of stream, thus statement (iii) is mandatory.

Also, the distance between points A and B or the speed of boat in still water is needed.

Thus, statements (i) and (iii) are required to find the upstream speed of the boat.

=> Ans – (D)

**3)Â AnswerÂ (C)**

Speed of train = 120 kmph

= $(120 \times \frac{5}{18})$ m/s = $\frac{100}{3}$ m/s

Let length of the train = $l$ meters

Using, speed = distance/time

=> $\frac{100}{3} = \frac{l}{15}$

=> $l=\frac{100}{3} \times 15$

=> $l=100 \times 5=500$ meters

=> Ans – (C)

**4)Â AnswerÂ (C)**

Let speed of bus = $s$ km/hr

Distance covered = 2924 km

Time taken = 43 hours

Using speed = distance/time

=> $s=\frac{2924}{43}=68$ km/hr

=> Ans – (C)

**5)Â AnswerÂ (A)**

Length of first train = 240 m and second train = 270 m

Total length of the two trains = 240 + 270 = 510 m

Speed of first train = 60 kmph and second train = 48 kmph

Since, the trains are moving in opposite direction, thus relative speed = 60 + 48 = 108 kmph

= $(108 \times \frac{5}{18})$ m/s = $30$ m/s

Let time taken = $t$ seconds

Using, time = distance/speed

=> $t=\frac{510}{30}=17$ seconds

=> Ans – (A)

**6)Â AnswerÂ (C)**

Let speed of boat in still water = $x$ km/hr

=> Speed of current = $y$ km/hr

Let distance travelled = $d$ km

Acc. to ques, => $2.2 (\frac{d}{x + y}) = \frac{d}{x – y}$

=> $2.2x – 2.2y = x + y$

=> $2.2x – x = y + 2.2y$

=> $3x = 8y$ ————-(i)

Also, the man takes 2 hrs 30 mins in travelling 55 km downstream.

=> $\frac{55}{x + y} = 2 + \frac{1}{2}$

=> $\frac{55}{x + y} = \frac{5}{2}$

=> $x + y = 22$

Multiplying both sides by 8, and using eqn(i), we getÂ :

=> $8x + 3x = 22 \times 8$

=> $x = \frac{22 \times 8}{11} = 16$ km/hr

**7)Â AnswerÂ (D)**

Let speed of the train = $10x$ m/s

Length of train = $l$ m

Time taken to cross the pole = 6 sec

Using, $speed = \frac{distance}{time}$

=> $10x = \frac{l}{6}$

=> $x = \frac{l}{60}$

Now, 60% of the speed = $\frac{60}{100} \times 10x = 6x$ m/s

Length of platform = 240 m

Acc. to ques, => $6x = \frac{240 + l}{15}$

=> $6 \times \frac{l}{60} = \frac{240 + l}{15}$

=> $\frac{l}{10} = \frac{240 + l}{15}$

=> $15l = 2400 + 10l$

=> $15l – 10l = 5l = 2400$

=> $l = \frac{2400}{5} = 480$ m

**8)Â AnswerÂ (C)**

Let speed of current = $x$ km/hr

=> Speed of boat in still water = $6x$ km/hr

Acc. to ques, => $\frac{105}{7x} + \frac{105}{5x} = 12$

=> $\frac{15}{x} + \frac{21}{x} = 12$

=> $\frac{36}{x} = 12$

=> $x = \frac{36}{12} = 3$

$\therefore$ Speed of boat in still water = $6 \times 3 = 18$ km/hr

**9)Â AnswerÂ (C)**

Let usual speed of the train = $10x$ m/s

Now, 50% of the speed = $\frac{50}{100} \times 10x = 5x$ m/s

Length of train = $l$ m

Time taken to cross the pole = 20 sec

Using, $speed = \frac{distance}{time}$

=> $5x = \frac{l}{20}$

=> $x = \frac{l}{100}$

Length of platform = 300 m

Acc. to ques, => $10x = \frac{300 + l}{25}$

=> $10 \times \frac{l}{100} = \frac{300 + l}{25}$

=> $\frac{l}{10} = \frac{300 + l}{25}$

=> $25l = 3000 + 10l$

=> $25l – 10l = 15l = 3000$

=> $l = \frac{3000}{15} = 200$ m

**10)Â AnswerÂ (C)**

Let speed of current = $x$ km/hr

=> Speed of boat in still water = $8x$ km/hr

Acc. to ques, => $\frac{63}{9x} + \frac{63}{7x} = 8$

=> $\frac{7}{x} + \frac{9}{x} = 8$

=> $\frac{16}{x} = 8$

=> $x = \frac{16}{8} = 2$

$\therefore$ Speed of boat in still water = $8 \times 2 = 16$ km/hr

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