# Time and Distance Questions and Answers updated daily – Aptitude

Time and Distance Questions: Solved 288 Time and Distance Questions and answers section with explanation for various online exam preparation, various interviews, Aptitude Category online test. Category Questions section with detailed description, explanation will help you to master the topic.

## Time and Distance Questions

21. A pedestrian and a cyclist start simultaneously towards each other from Aurangabad and Paithan which are 40 km apart and meet 2 hours after the start. Then they resume their trips and the cyclist arrives at Aurangabad 7 hours 30 minutes earlier than the pedestrian arrives at Paithan. What of these could be the speed of the pedestrian?

SHOW ANSWER

Correct Ans:4 km/hr

Explanation:

Let the Speed of pedestrian = x km/h

Speed of cyclist = y km/h

Given, Distance between Aurangabad and Paithan = 40 km

Since they are moving in opposite direction, speed can be added.

=> Relative speed = Speed of pedestrian + Speed of cyclist = (x + y) km/h

Given, They meet 2 hours after the start.

W.K.T: Time * Relative Speed = Distance

---> 2 * (x + y) = 40

---> x + y = 20 -------> eqn (1)

According to second condition, i.e., cyclist arrives at Aurangabad 7 hours 30 minutes earlier than the pedestrian arrives at Paithan

---> Time taken by cyclist to complete his journey = Time taken by pedestrian to complete his journey - 7 hours 30 minutes

---> (40/y) = (40/x) - [7 + (30/60)] [Since, 30 minutes is converted into (30/60) hours]

---> (40/x) - (40/y) = [7 + (30/60)]

---> 40 [(1/x) - (1/y)] = 15/2

---> 8 [(y - x) / xy] = 3/2

---> 16 (y - x) = 3xy

From eqn (1), y = 20 - x, substitute this x value in the above equation, we get

---> 16 (20 - x - x) = 3x(20 - x)

---> 16 (20 - 2x) = 60x - 3x

---> 320 - 32x - 60x + 3x

---> 3x

By using the formula, x = {-b ± √[b

We can find the value of x,

Where, b = - 92

a = 3

c = 320

---> x = {92 ± √[(-92)

= {92 ± √[8464 - 3840]} / 6

= {92 ± √[4624]} / 6

= {92 ± 68} / 6

= {(92 + 68)/6}, {(92 - 68)/6}

= {160/6}, {24/6}

--->

Now, going through options, we get

--->

Speed of cyclist = y km/h

Given, Distance between Aurangabad and Paithan = 40 km

Since they are moving in opposite direction, speed can be added.

=> Relative speed = Speed of pedestrian + Speed of cyclist = (x + y) km/h

Given, They meet 2 hours after the start.

W.K.T: Time * Relative Speed = Distance

---> 2 * (x + y) = 40

---> x + y = 20 -------> eqn (1)

According to second condition, i.e., cyclist arrives at Aurangabad 7 hours 30 minutes earlier than the pedestrian arrives at Paithan

---> Time taken by cyclist to complete his journey = Time taken by pedestrian to complete his journey - 7 hours 30 minutes

---> (40/y) = (40/x) - [7 + (30/60)] [Since, 30 minutes is converted into (30/60) hours]

---> (40/x) - (40/y) = [7 + (30/60)]

---> 40 [(1/x) - (1/y)] = 15/2

---> 8 [(y - x) / xy] = 3/2

---> 16 (y - x) = 3xy

From eqn (1), y = 20 - x, substitute this x value in the above equation, we get

---> 16 (20 - x - x) = 3x(20 - x)

---> 16 (20 - 2x) = 60x - 3x

^{2}---> 320 - 32x - 60x + 3x

^{2}= 0---> 3x

^{2}- 92x + 320 = 0By using the formula, x = {-b ± √[b

^{2}- 4ac]} / 2aWe can find the value of x,

Where, b = - 92

a = 3

c = 320

---> x = {92 ± √[(-92)

^{2}- 4 * 3 * 320]} / 2 * 3= {92 ± √[8464 - 3840]} / 6

= {92 ± √[4624]} / 6

= {92 ± 68} / 6

= {(92 + 68)/6}, {(92 - 68)/6}

= {160/6}, {24/6}

--->

**x = 26.667, 4**Now, going through options, we get

**x = 4 km/h**--->

**Speed of pedestrian = 4 km/h**
Workspace

22. A steamer covers 80 km in 4 hrs in still water. If speed of boat is 50% more than that of stream then time required for steamer to cover this distance in upstream?

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Correct Ans:12 hrs

Explanation:

Speed of steamer in still water = 80/4

= 20 km/hr.

Speed of stream = 20*100 / 150

= 40/3 km/hr.

Speed of steamer in upstream = 20 - 40/3

= 20/3 km/hr.

Required time = (80*3)/20

= 12 hrs.

= 20 km/hr.

Speed of stream = 20*100 / 150

= 40/3 km/hr.

Speed of steamer in upstream = 20 - 40/3

= 20/3 km/hr.

Required time = (80*3)/20

= 12 hrs.

Workspace

23. A man makes four trips of equal distances. His speed on first trip was 60 km/hr and in each subsequent trip his speed was half of the previous trip.What is the average speed (in km/hr) of the man in these four trips?

SHOW ANSWER

Correct Ans:16

Explanation:

Let distance travelled in each trip = d km

Total distance travelled = 4d km

Speed in first trip = 60

Speed in second trip = 60/2 = 30

Speed in third trip = 15

Speed in fourth trip = 7.5

Time taken in each trip = distance/speed

= d/60 + d/30 + d/15 + d/7.5

= d/7.5 (1/8 + 1/4 + 1/2 + 1)

= d/7.5 * (1 + 2 + 4 + 8)/8

= d/7.5 * 15/8

= d/4

Average speed = total distance/total time

= 4d / (d/4) = (4d * 4)/d = 4*4 = 16

Total distance travelled = 4d km

Speed in first trip = 60

Speed in second trip = 60/2 = 30

Speed in third trip = 15

Speed in fourth trip = 7.5

Time taken in each trip = distance/speed

= d/60 + d/30 + d/15 + d/7.5

= d/7.5 (1/8 + 1/4 + 1/2 + 1)

= d/7.5 * (1 + 2 + 4 + 8)/8

= d/7.5 * 15/8

= d/4

Average speed = total distance/total time

= 4d / (d/4) = (4d * 4)/d = 4*4 = 16

Workspace

24. A boat goes 16 km upstream in 4 hours and travels the same distance downstream in 96 minutes. What is the speed of the boat in still water?

SHOW ANSWER

Correct Ans:7 km/hr

Explanation:

Let the speed of boat be x km/hr

Let the speed of stream be y km/hr

Time taken to go upstream = 16/(x-y) = 4

(x-y) = 4 km/hr

Time taken to go upstream = 16/(x+y) = 96/60

(x+y) = 10 km/hr

Solving these equations give,

x = 7 km/hr, y = 3 km/hr

Let the speed of stream be y km/hr

Time taken to go upstream = 16/(x-y) = 4

(x-y) = 4 km/hr

Time taken to go upstream = 16/(x+y) = 96/60

(x+y) = 10 km/hr

Solving these equations give,

x = 7 km/hr, y = 3 km/hr

Workspace

25. The wheel of a car made 200 rotations, how much distance did the car travel if the diameter of the wheel is 28 inches. (1 inch = 2.54cm)

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Correct Ans:447.04 m

Explanation:

Distance travelled in 1 rotation = 2Ï€r [circumference of the wheel]

Radius = 28/2 = 14 inches

Distance travelled in 200 rotations = 200 * 2Ï€r = (400*22*14)/7 inches

= 17600inches

Distance in cm = 17600*2.54 = 44704 cm

1 m = 100 cm

Distance in m = 447.04 m

Radius = 28/2 = 14 inches

Distance travelled in 200 rotations = 200 * 2Ï€r = (400*22*14)/7 inches

= 17600inches

Distance in cm = 17600*2.54 = 44704 cm

1 m = 100 cm

Distance in m = 447.04 m

Workspace

26. A car covers 1/5th the distance from A to B at the speed of 8 km/hr, 1/10 of the distance at 25 km per hour and the remaining at the speed of 20 km per hour. Find the average speed of the whole journey.

SHOW ANSWER

Correct Ans:15.625 km/hr

Explanation:

Time = Distance/Speed

Remaining distance = 1 - (1/5 + 1/10)

= 1 - 3/10 = 7/10

Total time = 1/(5*8) + 1/(10*25) + 7/(10*20)

= 1/40 + 1/250 + 7/200

= (25 + 4 + 35)/1000 = 64/1000

= 8/125 hrs

Average speed = Total distance/Time taken

= 125/8 = 15.625 km/hr

Remaining distance = 1 - (1/5 + 1/10)

= 1 - 3/10 = 7/10

Total time = 1/(5*8) + 1/(10*25) + 7/(10*20)

= 1/40 + 1/250 + 7/200

= (25 + 4 + 35)/1000 = 64/1000

= 8/125 hrs

Average speed = Total distance/Time taken

= 125/8 = 15.625 km/hr

Workspace

27. A car crosses a man walking at 6 km/hr. The man can see the things upto 450 m only in one direction due to fog. He sees the car which was going in the same direction for 4.5 minutes. What is the speed of the car?

SHOW ANSWER

Correct Ans:12 km/hr

Explanation:

Time = 4.5/60 = 45/600

Time = Distance / Speed

45/600 = (450/1000) / Speed

Speed = (45*600)/(100*45)

Speed = 6 km/hr

peed of car - speed of man = 6

x - 6 = 6

x = 12 km/hr

Time = Distance / Speed

45/600 = (450/1000) / Speed

Speed = (45*600)/(100*45)

Speed = 6 km/hr

peed of car - speed of man = 6

x - 6 = 6

x = 12 km/hr

Workspace

28. A motorboat went downstream for 36 km and immediately returned. It took the boat twice as long to make the return trip. If the speed of the river flow were twice as high, the trip downstream and back would take 648 minutes. Find the speed of the boat in still water and the speed of the river flow.

SHOW ANSWER

Correct Ans:12 km/hr, 4 km/hr

Explanation:

Given, Downstream Distance = 36 km

Let Speed of Boat = B km/hr

and Speed of River (stream) = R km/hr

W.K.T:

----> Downstream Speed = B + R km/hr

W.K.T:

----> Upstream Speed = B - R km/hr

Time taken for downstream = Distance / Downstream Speed = 36/(B + R)

Time taken for upstream = 36/(B - R)

Given, time taken for upstream is double the time taken for downstream

---> Time taken for upstream = 2 * Time taken for downstream

---> 36/(B - R) = 2 * [36/(B + R)]

---> 1/(B - R) = 2/(B + R)

---> B + R = 2(B - R)

---> B + R = 2B - 2R

--->

Given, if the speed of the River flow were twice as high,

--->

Then, Downstream Speed = B + 2R

----> Substituting B = 3R (from eqn (1)) in above expression

----> Downstream Speed = 3R + 2R = 5R km/hr

And Upstream speed = B - 2R = 3R - 2R = R km/hr

Given, if R = 2R, then, Downstream time + Upstream time = 648 minutes

---> (36/5R) + (36/R) = (648/60) hr

---> [36/R] [(1/5) + 1] = (648/60)

---> [36/R] [6/5] = (648/60)

---> 216/5R = 648/60

---> R = (216 * 60) / (5 * 648)

---> R = 12960 / 3240

--->

Hence,

From eqn (1),

Let Speed of Boat = B km/hr

and Speed of River (stream) = R km/hr

W.K.T:

**Speed of Downstream = speed of boat + Speed of stream**----> Downstream Speed = B + R km/hr

W.K.T:

**Speed of Upstream = speed of boat - Speed of stream**----> Upstream Speed = B - R km/hr

Time taken for downstream = Distance / Downstream Speed = 36/(B + R)

Time taken for upstream = 36/(B - R)

Given, time taken for upstream is double the time taken for downstream

---> Time taken for upstream = 2 * Time taken for downstream

---> 36/(B - R) = 2 * [36/(B + R)]

---> 1/(B - R) = 2/(B + R)

---> B + R = 2(B - R)

---> B + R = 2B - 2R

--->

**B = 3R**---- eqn (1)Given, if the speed of the River flow were twice as high,

--->

**R = 2R**Then, Downstream Speed = B + 2R

----> Substituting B = 3R (from eqn (1)) in above expression

----> Downstream Speed = 3R + 2R = 5R km/hr

And Upstream speed = B - 2R = 3R - 2R = R km/hr

Given, if R = 2R, then, Downstream time + Upstream time = 648 minutes

---> (36/5R) + (36/R) = (648/60) hr

---> [36/R] [(1/5) + 1] = (648/60)

---> [36/R] [6/5] = (648/60)

---> 216/5R = 648/60

---> R = (216 * 60) / (5 * 648)

---> R = 12960 / 3240

--->

**R = 4**Hence,

**Speed of River**= R =**4 km/hr**From eqn (1),

**Speed of Boat**= 3R = 3 * 4 =**12 km/hr**
Workspace

29. A plane left 30 min later than the scheduled time and in order to reach its destination 1500 km away in time, it has to increase its speed by 250 km/hr from its usual speed. Find its usual speed?

SHOW ANSWER

Correct Ans:750 km/hr

Explanation:

Let the usual speed of the plane be x km/hr.

New Speed of the plane = (x + 250) km/hr

Given, distance to reach the destination =1500 km

According to the question,

Scheduled (Usual) time - current time taken by the train to reach destination after 30 mins delay = 30 mins

----> (1500/x) - [1500/(x + 250)] = 30/60 [Since 30 mins is converted into hours]

---> {[1500(x + 250)] - 1500x} / x(x + 250) = 1/2

---> {[1500x + 375000] - 1500x} / x(x + 250) = 1/2

---> 375000 / x(x + 250) = 1/2

---> 375000 * 2 = x(x + 250)

---> 750000 = x

---> x

---> x

---> x(x + 1000) - 750(x + 1000) = 0

---> (x + 1000) (x - 750) = 0

---> x = -1000, 750

Hence,

New Speed of the plane = (x + 250) km/hr

Given, distance to reach the destination =1500 km

According to the question,

Scheduled (Usual) time - current time taken by the train to reach destination after 30 mins delay = 30 mins

----> (1500/x) - [1500/(x + 250)] = 30/60 [Since 30 mins is converted into hours]

---> {[1500(x + 250)] - 1500x} / x(x + 250) = 1/2

---> {[1500x + 375000] - 1500x} / x(x + 250) = 1/2

---> 375000 / x(x + 250) = 1/2

---> 375000 * 2 = x(x + 250)

---> 750000 = x

^{2}+ 250x---> x

^{2}+ 250x - 750000 = 0---> x

^{2}+ 1000x - 750x - 750000 = 0---> x(x + 1000) - 750(x + 1000) = 0

---> (x + 1000) (x - 750) = 0

---> x = -1000, 750

Hence,

**Usual speed of the plane = 750 km/hr**
Workspace

30. A train met with an accident 60 km away from Howrah station. It completed the remaining journey at 5/6 th of the previous speed and reached Cuttack station 1 hour 12 mins late. Had accident taken place 60 km further, it would have been only 1 hour late.

A) What is the normal speed of the train?

B) What is the distance between Howrah and Cuttack?

A) What is the normal speed of the train?

B) What is the distance between Howrah and Cuttack?

SHOW ANSWER

Correct Ans:60 km/hr, 420 km

Explanation:

Let

Then, Actual Time taken by train to reach destination = Distance/Speed = y/x hrs.

Distance covered by the train till accident point = 60 km

Till accident point, the train runs in normal speed ie., x km/hr

Time taken by the train till accident = 60/x hours

For the remaining journey, new speed of train = (5/6) * previous speed = (5x/6)

Remaining distance = y - 60 (ie., Total distance - distance covered till accident)

Then, Time taken = Remaining distance/new speed of train = (y - 60)/ (5x/6)

---> Time taken by the train for remaining journey = 6(y - 60)/5x

According to the question,

Time taken by the train till accident + Time taken by the train for remaining journey = Actual Time taken by train to reach destination + 1 hour 12 minutes

---> (60/x) + [6(y - 60)/5x] = (y/x) + (6/5) [--- Since, converted 1 hour 12 minutes into (6/5) hours]

Solving the above eqn, we get

---> (60/x) + [6(y - 60)/5x] = (y/x) + (6/5)

---> [300 + 6(y - 60)] / 5x = (5y + 6x) / 5x

---> [300 + 6y - 360] = (5y + 6x)

---> 6y - 60 = 5y + 6x

--->

Had accident taken place 60 km further.

Now, distance covered by the train till accident point = 60 + 60 = 120 km

And Speed of the train till accident happens = x km/hr

Then, Time taken till accident = (120 km/x km/hr) = (120/x) hr

For the remaining journey, new speed of train = (5/6) * previous speed = (5x/6)

Remaining distance = y - 120

Then, Time taken by the train for remaining journey = (y - 120)/(5x/6) = 6(y - 120)/5x

According to the question,

Time taken by the train till accident + Time taken by the train for remaining journey = Actual Time taken by train to reach destination + 1 hour

---> (120/x) + [6(y - 120)/5x] = (y/x) + 1

---> [120 * 5 + 6(y - 120)]/5x = (y + x)/x

---> [600 + 6y - 720]/5x = (y + x)/x

---> [6y - 120]/5 = (y + x)

---> 6y - 120 = 5y + 5x

--->

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

---> Eqn (1) - eqn (2), we get,

**normal speed**of the train be**x km/hr**and**distance**between Howrah and Cuttack be**y km**.Then, Actual Time taken by train to reach destination = Distance/Speed = y/x hrs.

**Case 1:**Distance covered by the train till accident point = 60 km

Till accident point, the train runs in normal speed ie., x km/hr

Time taken by the train till accident = 60/x hours

For the remaining journey, new speed of train = (5/6) * previous speed = (5x/6)

Remaining distance = y - 60 (ie., Total distance - distance covered till accident)

Then, Time taken = Remaining distance/new speed of train = (y - 60)/ (5x/6)

---> Time taken by the train for remaining journey = 6(y - 60)/5x

According to the question,

Time taken by the train till accident + Time taken by the train for remaining journey = Actual Time taken by train to reach destination + 1 hour 12 minutes

---> (60/x) + [6(y - 60)/5x] = (y/x) + (6/5) [--- Since, converted 1 hour 12 minutes into (6/5) hours]

Solving the above eqn, we get

---> (60/x) + [6(y - 60)/5x] = (y/x) + (6/5)

---> [300 + 6(y - 60)] / 5x = (5y + 6x) / 5x

---> [300 + 6y - 360] = (5y + 6x)

---> 6y - 60 = 5y + 6x

--->

**y - 6x = 60**------> eqn (1)**Case 2:**Had accident taken place 60 km further.

Now, distance covered by the train till accident point = 60 + 60 = 120 km

And Speed of the train till accident happens = x km/hr

Then, Time taken till accident = (120 km/x km/hr) = (120/x) hr

For the remaining journey, new speed of train = (5/6) * previous speed = (5x/6)

Remaining distance = y - 120

Then, Time taken by the train for remaining journey = (y - 120)/(5x/6) = 6(y - 120)/5x

According to the question,

Time taken by the train till accident + Time taken by the train for remaining journey = Actual Time taken by train to reach destination + 1 hour

---> (120/x) + [6(y - 120)/5x] = (y/x) + 1

---> [120 * 5 + 6(y - 120)]/5x = (y + x)/x

---> [600 + 6y - 720]/5x = (y + x)/x

---> [6y - 120]/5 = (y + x)

---> 6y - 120 = 5y + 5x

--->

**y - 5x = 120**------> eqn (2)On solving eqn (1) and eqn (2) we get,

---> Eqn (1) - eqn (2), we get,

**x = 60**km/hr (which is the normal speed of the train) and**y = 420**km (which is the distance between Howrah and Cuttack)
Workspace

31. Two buses start at same time from Chennai and Bangalore, which are 250 km apart. If the two buses travel towards each other, they meet after 1 hr and if they travel in same direction they meet after 5 hrs. What is the speed of the bus starts from Chennai if it is know that the one which started from Chennai has more speed than the other one?

SHOW ANSWER

Correct Ans:150 km/hr

Explanation:

Given: Distance - 250 km

WKT,

If the two buses travel towards each other, they meet after 1 hr ,

Speed of Chennai bus + Speed of Bangalore bus = 250/1 ...(i)

If they travel in same direction they meet after 5 hrs,

Speed of Chennai bus - Speed of Bangalore bus = 250/5

Speed of Chennai bus - Speed of Bangalore bus = 50 ...(ii)

By solving (i) and (ii),

Speed of Chennai bus = 150 km/hr.

WKT,

**Speed = Distance/Time**If the two buses travel towards each other, they meet after 1 hr ,

Speed of Chennai bus + Speed of Bangalore bus = 250/1 ...(i)

If they travel in same direction they meet after 5 hrs,

Speed of Chennai bus - Speed of Bangalore bus = 250/5

Speed of Chennai bus - Speed of Bangalore bus = 50 ...(ii)

By solving (i) and (ii),

Speed of Chennai bus = 150 km/hr.

Workspace

32. A boat can travel 352 km downstream and 112 km upstream in total 24 hours. If respective ratio of speed of boat in still water to speed of stream is 9: 2, then find total distance travelled by boat in 5 hours in downstream is what percent more than total distance travelled by boat in two hours in upstream?

SHOW ANSWER

Correct Ans:292(6/7)%

Explanation:

Given:

Ratio of speed of boat in still water to speed of stream = 9: 2

Let the speed of the boat in still water and speed of stream be 9X and 2X respectively.

WKT,

Given that a boat can travel 352 km downstream and 112 km upstream in total 24 hours.

[352/(9X + 2X)] + [112/(9X - 2X)] = 24

(352/11X) + (112/7X) = 24

1/X[32 + 16] = 24

48 = 24X

X = 2 km/hr

Speed of boat in still water = 9X = 9*2 = 18 km/hr

Speed of stream = 2X = 2*2 = 4 km/hr

WKT,

Total distance travelled by boat in 5 hours in downstream = (18 + 4) *5 = 110 km

Total distance travelled by boat in 2 hours in upstream = (18 - 4)*2 = 28 km

Required percentage = [(110 - 28)/28]*100 = 292(6/7)%.

Ratio of speed of boat in still water to speed of stream = 9: 2

Let the speed of the boat in still water and speed of stream be 9X and 2X respectively.

WKT,

**Time = Distance/Speed**Given that a boat can travel 352 km downstream and 112 km upstream in total 24 hours.

[352/(9X + 2X)] + [112/(9X - 2X)] = 24

(352/11X) + (112/7X) = 24

1/X[32 + 16] = 24

48 = 24X

X = 2 km/hr

Speed of boat in still water = 9X = 9*2 = 18 km/hr

Speed of stream = 2X = 2*2 = 4 km/hr

WKT,

**Distance = Speed x Time****Speed of downstream = (u + v)**

Speed of upstream = (u - v)Speed of upstream = (u - v)

Total distance travelled by boat in 5 hours in downstream = (18 + 4) *5 = 110 km

Total distance travelled by boat in 2 hours in upstream = (18 - 4)*2 = 28 km

Required percentage = [(110 - 28)/28]*100 = 292(6/7)%.

Workspace

33. A boat goes 28 km downstream and while returning covered only 75% of distance that covered in downstream. If boat takes 3 hr more to cover upstream than downstream then find the speed of boat in still water (km/hr) if speed of stream is (5/9) m/sec?

SHOW ANSWER

Correct Ans:5 km/hr

Explanation:

Given:

Speed of the stream = (5/9)m/sec

Converting m/sec to km/hr,

Speed of the stream = (5/9) x (18/5) = 2 km/hr

Let the speed of boat in still water be 'X'.

WKT,

As per the question,

[28 x (75/100)/(X - 2)] - [28/(X + 2)] = 3

[21/(X - 2)] - [28/(X + 2)] = 3

[21X + 42 - 28X + 56]/[X

-7X + 98 = 3X

3X

3X

3X(X - 5) + 22(X - 5) = 0

(3X + 22) (X - 5) = 0

X = -(22/3), 5

Therefore, speed of the boat in still water is 5 km/hr.

Speed of the stream = (5/9)m/sec

Converting m/sec to km/hr,

Speed of the stream = (5/9) x (18/5) = 2 km/hr

Let the speed of boat in still water be 'X'.

WKT,

**Time = Distance/Speed**As per the question,

**Distance/Speed in upstream - Distance/Speed in downstream = Difference in time**[28 x (75/100)/(X - 2)] - [28/(X + 2)] = 3

[21/(X - 2)] - [28/(X + 2)] = 3

[21X + 42 - 28X + 56]/[X

^{2}- 4] =3-7X + 98 = 3X

^{2}- 123X

^{2}+ 7X - 110 = 03X

^{2}- 15X + 22X - 110 = 03X(X - 5) + 22(X - 5) = 0

(3X + 22) (X - 5) = 0

X = -(22/3), 5

Therefore, speed of the boat in still water is 5 km/hr.

Workspace

34. A boat travels with the direction of the current from point A to B and then returns against the current and stop at point C. Distance between B and C is 75% of distance between A and B and boat takes 9 hours to cover total distance. If distance between A to B is 40 Km and Speed of current is 2 km/hr, then find speed of boat in still water ?

SHOW ANSWER

Correct Ans:8 km/hr

Explanation:

Let speed of boat in still water be x km/hr

Water Current speed = 2 km/hr

Distance between A and C = 40 km

Distance between B to C = 40 * 75/100 = 30 km

Speed of downstream = (x + 2) km/hr

Speed of upstream = (x - 2) km/hr

Time taken to cover total distance by boat = 9 hours

Time = Distance/Speed

9 = 40/(x + 2) + 30/(x - 2)

9 = [40(x - 2) + 30(x + 2)] / [(x + 2)(x - 2)]

9 = (40x - 80 + 30x + 60)/(x

9x

9x

9x

9x(x - 8) + 2(x - 8) = 0

x = 8, -2/9

So, the speed of boat in still water = 8 km/hr

Water Current speed = 2 km/hr

Distance between A and C = 40 km

Distance between B to C = 40 * 75/100 = 30 km

Speed of downstream = (x + 2) km/hr

Speed of upstream = (x - 2) km/hr

Time taken to cover total distance by boat = 9 hours

Time = Distance/Speed

9 = 40/(x + 2) + 30/(x - 2)

9 = [40(x - 2) + 30(x + 2)] / [(x + 2)(x - 2)]

9 = (40x - 80 + 30x + 60)/(x

^{2}- 4)9x

^{2}- 36 = 70x -209x

^{2}- 70x - 16 = 09x

^{2}- 72x + 2x - 16 = 09x(x - 8) + 2(x - 8) = 0

x = 8, -2/9

So, the speed of boat in still water = 8 km/hr

Workspace

35. Krishna covers a certain distance by train at 25 km/hr. and the equal distance on foot at 4 km/hr. If the time taken by him for the whole journey be 5 hrs and 48 minutes, how much total distance did he cover ?

SHOW ANSWER

Correct Ans:40 km

Explanation:

Total distance = x km

Distance by train = x/2 km

Distance by Feet = x/2 km

Time taken to cover x/2 by train = Distance/Speed = x/50 hours

Time taken to cover x/2 by foot = x/8 hours

x/50 + x/8 = 5(48/60)

(8x + 50x)/400 = 5(4/5)

58x/400 = 29/5

2x/80 = 1

x = 40 km

Distance by train = x/2 km

Distance by Feet = x/2 km

Time taken to cover x/2 by train = Distance/Speed = x/50 hours

Time taken to cover x/2 by foot = x/8 hours

x/50 + x/8 = 5(48/60)

(8x + 50x)/400 = 5(4/5)

58x/400 = 29/5

2x/80 = 1

x = 40 km

Workspace

36. Train A crosses a pole in 25 seconds and Train B crosses the same pole in 40 seconds. The length of Train A is half the length of Train B. What is the ratio of the speed of Train A to that of Train B?

SHOW ANSWER

Correct Ans:4 : 5

Explanation:

The length of Train A = (1/2)* The length of Train B

The length of Train A : The length of Train B = 1 : 2

Speed of Train A = Distance/Time = x/25

Speed of Train B = 2x/40 = x/20

Required ratio = (x/25) : (x/20) = 20 : 25 = 4 : 5

The length of Train A : The length of Train B = 1 : 2

Speed of Train A = Distance/Time = x/25

Speed of Train B = 2x/40 = x/20

Required ratio = (x/25) : (x/20) = 20 : 25 = 4 : 5

Workspace

37. Two trains pass each other on parallel lines. Each train is 100 metres long. When they are going in the same direction, the faster one takes 60 seconds to pass the other completely. If they are going in opposite directions they pass each other completely in 10 seconds. Find the speed of the slower train in km/hr.

SHOW ANSWER

Correct Ans:30 km/h

Explanation:

Let the speed of faster train = x and speed of slower train = y.

We know that if

Two trains are moving in the

Two trains are moving in the

Case 1: Two trains moving in same direction

Relative distance of two trains = Train 1 length + Train 2 length = 100 + 100 = 200 m

Time taken by faster train to pass slower train (t) = Relative distance/Relative speed

---> t = 200/(x - y)

Given t = 60 sec

---> 60 = 200/(x - y)

---> 60(x - y) = 200

---> 3(x - y) = 10

---> 3x - 3y = 10 ----------(1)

Case 2: Two trains moving in opposite direction

Relative distance = 200 m

Time taken by faster train to pass slower train (t) = 200/(x + y)

Given, t = 10 sec

---> 10 = 200/(x + y)

---> x + y = 20 -------------(2)

Now dividing eqn (1) by 3 and adding eqn (1) and (2), we get,

x - y = 10/3

x + y = 20

_________

2x = 70/3

x = 70/6

Substitute this x value in eqn (2), then y = 20 - (70/6)

---> y = 50/6

Hence,

Now converting m/s to km/hr

y = (50/6)*(18/5) (since 1 m/s = 18/5 km/h)

= 30 km/h.

Therefore the

We know that if

Two trains are moving in the

**same direction**at x m/s, y m/s then**relative speed = (x - y)**m/sTwo trains are moving in the

**opposite direction**at x m/s, y m/s then**relative speed = (x + y)**m/sCase 1: Two trains moving in same direction

Relative distance of two trains = Train 1 length + Train 2 length = 100 + 100 = 200 m

Time taken by faster train to pass slower train (t) = Relative distance/Relative speed

---> t = 200/(x - y)

Given t = 60 sec

---> 60 = 200/(x - y)

---> 60(x - y) = 200

---> 3(x - y) = 10

---> 3x - 3y = 10 ----------(1)

Case 2: Two trains moving in opposite direction

Relative distance = 200 m

Time taken by faster train to pass slower train (t) = 200/(x + y)

Given, t = 10 sec

---> 10 = 200/(x + y)

---> x + y = 20 -------------(2)

Now dividing eqn (1) by 3 and adding eqn (1) and (2), we get,

x - y = 10/3

x + y = 20

_________

2x = 70/3

x = 70/6

Substitute this x value in eqn (2), then y = 20 - (70/6)

---> y = 50/6

Hence,

**speed of slower train = y = 50/6 m/s**Now converting m/s to km/hr

y = (50/6)*(18/5) (since 1 m/s = 18/5 km/h)

= 30 km/h.

Therefore the

**speed of slower train = 30 km/h**
Workspace

38. The ratio of the speed of the boat in still water to the speed of stream is 17 : 6. A boat goes 15.5 km in 45 minute upstream, find the time taken by boat to cover the distance of 18.5 km downstream.

SHOW ANSWER

Correct Ans:25.8 minutes

Explanation:

Let the speed of the boat in still water = 17x,

speed of stream = 6x

Now, Speed upstream = 17x - 6x = 11x

W.K.T:- Speed = Distance/Time

So, Speed upstream = upstream Distance/upstream Time

---> 11x = (15.5/45)*60

(Here, time is converted from minutes to hour)

--->

Then, Speed of boat in still water = 17x = 17 * (1.87) = 31.79 km/h

Speed of stream = 6x = 6 * (1.87) = 11.22 km/h

So

---> Speed downstream = 31.79 + 11.22 = 43.01 km/h

= downstream distance / Speed downstream

= 18.5/43.01

= 0.43 hr

= (43/100)*60

=

speed of stream = 6x

__Formulae used:__**Speed upstream**= Speed of boat in still water - Speed of streamNow, Speed upstream = 17x - 6x = 11x

W.K.T:- Speed = Distance/Time

So, Speed upstream = upstream Distance/upstream Time

---> 11x = (15.5/45)*60

(Here, time is converted from minutes to hour)

--->

**x = 1.87**Then, Speed of boat in still water = 17x = 17 * (1.87) = 31.79 km/h

Speed of stream = 6x = 6 * (1.87) = 11.22 km/h

So

**Speed downstream**= Speed of boat in still water + Speed of stream.---> Speed downstream = 31.79 + 11.22 = 43.01 km/h

**Time taken by boat to cover downstream**= downstream distance / Speed downstream

= 18.5/43.01

= 0.43 hr

= (43/100)*60

=

**25.8 min**
Workspace

39. A steamer goes downstream from one port to another in 3 hours. It covers the same distance upstream in 4 hours. If the speed of stream is 5 km/h, the distance between the two ports is

SHOW ANSWER

Correct Ans:120 km

Explanation:

Time in upstream = 4 hours

Time in downstream = 3 hours

Speed of stream = 5 km/h

Let the distance from 1 port to another = d

Let the speed of boat = x km/h

We know that Speed (s) = Distance/Time taken.-----(1)

i) For downstream:

Speed downstream = x + 5 (from 2)

Speed downstream = Distance/Time in downstream (from 1)

x + 5 = d/3

d = 3(x + 5)

ii) For upstream:

Speed upstream = x - 5 (from 3)

Speed upstream = Distance/Time in upstream (from 1)

x - 5 = d/4

d = 4(x - 5)

Given it covers the same distance both in upstream and downstream ,so equating both the distance

3x + 15 = 4x - 20

4x - 3x = 20 + 15

x = 35

Therefore speed of boat = x = 35 km/hr.

Speed = Distance/Time taken

Either calculate with upstream or downstream equation we get distance,

Say calculating with upstream distance d = 4x - 20

= 4(35) - 20

= 120 km

Time in downstream = 3 hours

Speed of stream = 5 km/h

Let the distance from 1 port to another = d

Let the speed of boat = x km/h

We know that Speed (s) = Distance/Time taken.-----(1)

**Formulae used:**

Speed downstream = speed of boat in still water + speed of stream ---(2)

Speed upstream = speed of boat in still water - speed of stream---(3)

Speed downstream = speed of boat in still water + speed of stream ---(2)

Speed upstream = speed of boat in still water - speed of stream---(3)

i) For downstream:

Speed downstream = x + 5 (from 2)

Speed downstream = Distance/Time in downstream (from 1)

x + 5 = d/3

d = 3(x + 5)

**d = 3x + 15**ii) For upstream:

Speed upstream = x - 5 (from 3)

Speed upstream = Distance/Time in upstream (from 1)

x - 5 = d/4

d = 4(x - 5)

**d = 4x - 20**Given it covers the same distance both in upstream and downstream ,so equating both the distance

3x + 15 = 4x - 20

4x - 3x = 20 + 15

x = 35

Therefore speed of boat = x = 35 km/hr.

Speed = Distance/Time taken

Either calculate with upstream or downstream equation we get distance,

Say calculating with upstream distance d = 4x - 20

= 4(35) - 20

= 120 km

Workspace

40. A train leaves Chennai at 7.30 am and travels 40 km an hour, another train leaves Chennai at noon and travels 64 km an hour, when and where will the second train overtake the first?

SHOW ANSWER

Correct Ans:480 km, 2.30 pm

Explanation:

Given:

First train: Speed - 40 km/hr; 7.30 am

Second train : Speed - 64 km/hr; 12.00 pm

WKT,

Distance travelled by first train by 4(1/2) hours,

= 40 x 4(1/2)

= 40 x (9/2)

= 180 km

= 180/(64 - 40)

= 180/24

= 7.5 hour

So, time taken by second train to cross first train is 7.5 hour that is 7 hr 30 minutes.

Exact time when second train cross first train = 12.00 pm + 7.30 = 7.30 pm

Distance from chennai = 64 x 7.5 = 480 km

Therefore, second train meets the first train at 7.30 pm, 480 km.

First train: Speed - 40 km/hr; 7.30 am

Second train : Speed - 64 km/hr; 12.00 pm

WKT,

**Distance = Speed x Time**Distance travelled by first train by 4(1/2) hours,

= 40 x 4(1/2)

= 40 x (9/2)

= 180 km

**Time at which second train cross first train = Distance covered by first train/Difference of second train and first train**= 180/(64 - 40)

= 180/24

= 7.5 hour

So, time taken by second train to cross first train is 7.5 hour that is 7 hr 30 minutes.

Exact time when second train cross first train = 12.00 pm + 7.30 = 7.30 pm

Distance from chennai = 64 x 7.5 = 480 km

Therefore, second train meets the first train at 7.30 pm, 480 km.

Workspace

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