SNAP Time Speed and Distance Questions PDF

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_ Time Speed and Distance Questions PDF

Time Speed and Distance Questions for SNAP

The Time Speed and Distance is an important topic in the Quant section of the SNAP Exam. You can also download this Free Time Speed and Distance Questions for SNAP PDF with detailed answers by Cracku. These questions will help you practice and solve the Time Speed and Distance questions in the SNAP exam. Utilize this PDF practice set, which is one of the best sources for practicing.

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

a) 27 km/hr

b) 49 km/hr

c) 72 km/hr

d) 70 km/hr

e) None of these

1) Answer (C)

Solution:

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

Question 2: A boat takes 2 hours to travel from point A to B in still water .To find out it’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) All are required

b) Even these we cannot found the answer

c) Only i,iii, and either ii or iv

d) Only i and iii

e) None of these

2) Answer (D)

Solution:

Time taken by boat to travel from point A to B in still water = 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)

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

a) 300

b) 200

c) 500

d) Cannot determined

e) None of these

3) Answer (C)

Solution:

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)

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

a) 72 km/hr

b) 60 km/hr

c) 68 km/hr

d) cannot determined

e) none of these

4) Answer (C)

Solution:

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)

Question 5: 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) 17 seconds

b) 3 seconds

c) 12 seconds

d) 8 seconds

e) None of these

5) Answer (A)

Solution:

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)

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Question 6: 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) 40

b) 8

c) 16

d) 24

e) 32

6) Answer (C)

Solution:

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 get :

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

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

Question 7: 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) 270

b) 225

c) 220

d) 480

e) 240

7) Answer (D)

Solution:

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

Question 8: 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) 12

b) 30

c) 18

d) 24

e) 36

8) Answer (C)

Solution:

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

Question 9: 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) 160

b) 260

c) 200

d) 310

e) 350

9) Answer (C)

Solution:

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

Question 10: 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) 32

b) 24

c) 16

d) 8

e) 38

10) Answer (C)

Solution:

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

Question 11: A completes ${5 \over 6}$ th of a given task in 10 days and is then replaced by B. The entire task is completed in 13 days. What is the respective ratio of the number of days in which A and B independently can complete the entire task?

a) 2 : 7

b) 3 : 8

c) 1 : 4

d) 2 : 3

e) 6 : 11

11) Answer (D)

Solution:

Let total work to be done = 6 units

=> A completes $\frac{5}{6} \times 6 = 5$ units in 10 days

=> A’s efficiency = $\frac{5}{10} = \frac{1}{2}$ units /day

Now, B finishes $\frac{1}{6}$th of the task in 3 days

=> B completes $\frac{1}{6} \times 6 = 1$ units in 3 days

=> B’s efficiency = $\frac{1}{3}$ units /day

Now, time taken by A alone to complete the entire task = $\frac{6}{\frac{1}{2}} = 12$ days

Time taken by B alone to complete the entire task = $\frac{6}{\frac{1}{3}} = 18$ days

$\therefore$ Required ratio = $\frac{12}{18} = 2 : 3$

Question 12: A boat takes six hours to travel a certain distance downstream and five hours to travel a certain distance upstream. The distance travelled upstream is half of the travelled downstream. If the speed of the current is 4 km/hr, what is the speed of the boat in still water? (in km/hr)

a) 16

b) 20

c) 24

d) 10

e) 18

12) Answer (A)

Solution:

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

Let distance travelled downstream = $2d$ km

=> Distance travelled upstream = $d$ km

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

=> $6 = \frac{2d}{x + 4}$ ———(i)

and $5 = \frac{d}{x – 4}$ ———(ii)

Dividing eqn(i) from (ii), we get :

=> $\frac{6}{5} = \frac{\frac{2d}{x + 4}}{\frac{d}{x – 4}}$

=> $\frac{6}{5} = \frac{2 (x – 4)}{x + 4}$

=> $6x + 24 = 10x – 40$

=> $10x – 6x = 24 + 40 = 64$

=> $x = \frac{64}{4} = 16$ km/hr

Question 13: At its usual speed, a 150 metre long train crosses a platform of length L metres in 24 seconds. AT 75% of its usual speed, the train crosses a vertical pole in 12 seconds. What is the value of L?

a) 250

b) 225

c) 240

d) 260

e) 280

13) Answer (A)

Solution:

Let speed of the train = $20x$ m/s

Now, 75% of the speed = $\frac{75}{100} \times 20x = 15x$ m/s

Length of train = 150 m

Time taken to cross the pole = 12 sec

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

=> $15x = \frac{150}{12}$

=> $x = \frac{10}{12} = \frac{5}{6}$

Length of platform = $l$ m

Acc. to ques, => $20x = \frac{150 + l}{24}$

=> $20 \times \frac{5}{6} = \frac{150 + l}{24}$

=> $\frac{50}{3} = \frac{150 + l}{24}$

=> $150 + l = 400$

=> $l = 400 – 150 = 250$ m

Question 14: A bus covers a distance of 2,924 kms. in 43 hours. What is the speed of the bus?

a) 72 kmph

b) 60 kmph

c) 68 kmph

d) Cannot be determined

e) None of these

14) Answer (C)

Solution:

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

= $\frac{2924}{43}$

= 68 kmph

Question 15: A train crosses a 300 metre long platform in 38 seconds while it crosses a signal pole in 18 seconds. What is the speed of the train in kmph ?

a) Cannot be determined

b) 72

c) 48

d) 54

e) None of these

15) Answer (D)

Solution:

Let the length of the train be $l$ metre and speed of the train be $x$ m/s

=> $\frac{l}{s} = 18$

=> $l = 18s$

Also, $\frac{l + 300}{s} = 38$

=> $18s + 300 = 38s$

=> $20s = 300$

=> $s = \frac{300}{20} = 15$ m/s

$\therefore$ Speed of train in kmph = $15 \times \frac{18}{5}$

= 54 kmph

Question 16: A car runs at the speed of 40 when not serviced and runs at 65 kmph when serviced. After servicing the car covers a certain distance in 5 hours. How much approximate time will the car take to cover the same distance when not serviced ?

a) 10

b) 7

c) 12

d) 8

e) 6

16) Answer (D)

Solution:

After servicing, the distance covered by car in 5 hours = 65 * 5 = 325 km

Without servicing, speed of car = 40 kmph

=> Required time = $\frac{Distance}{Speed}$

= $\frac{325}{40}$ = 8.125 hr

$\approx$ 8 hours

Question 17: A car runs at the speed of 50 kms per hour when not serviced and runs at 60 km./hr. when serviced. After servicing the car covers a certain distance in 6 hours. How much time will the car take to cover the same distance when not serviced?

a) 8.2 hours

b) 6.5 hours

c) 8 hours

d) 7.2 hours

e) None of these

17) Answer (D)

Solution:

After servicing, the distance covered by car in 6 hours = 60 * 6 = 360 km

Without servicing, speed of car = 50 kmph

=> Required time = $\frac{Distance}{Speed}$

= $\frac{360}{50}$ = 7.2 hr

Question 18: Yesterday Priti type an essay of 5000 words at the speed of 60 words per minute: Today she type the same essay faster and her speed was 15% more than yesterday. What is the approximate difference in the time she took to type yesterday and the time she took to type today?

a) 20 minutes

b) 30 minutes

c) 10 minutes

d) 40 minutes

e) 1 hour

18) Answer (C)

Solution:

Priti’s second day typing speed = $\frac{115}{100} \times 60$ = 69 wpm

Required difference in time

= $(\frac{5000}{60} – \frac{5000}{69})$

= $5000(\frac{69 – 60}{60 \times 69})$

$\approx$ 10 minutes

Question 19: The average speed of a train is 3 times the average speed of a car. The car covers a distance of 520 kms in 8 hours. How much distance will the train cover in 13 hours?

a) 2553 kms

b) 2585 kms

c) 2355 kms

d) 2535 kms

e) None of these

19) Answer (D)

Solution:

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

= $\frac{520}{8} = 65$ kmph

=> Speed of train = $65 \times 3 = 195$ kmph

$\therefore$ Distance covered by train in 13 hours

= $13 \times 195 = 2535$ km

Question 20: A boatman can row a boat downstream at 13 kmph and upstream at 9 kmph. What will be the speed of boat in still water ? (in kmph).

a) 12

b) 10.5

c) 11

d) 10

e) 11.5

20) Answer (C)

Solution:

Speed of boat in still water

= $\frac{1}{2}$ (downstream + upstream)

= $\frac{1}{2} (13 + 9) = \frac{22}{2}$

= $11$ kmph

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