# 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

1. Train X starts at 7.00 a.m. from a certain station with A Km/h and train Y starts at 9.30 a.m. from the same station at B km/h. If B > A, then how many hours will train Y take to overtake train X?

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Correct Ans:5A/(2(B-A)) hrs

Explanation:

Explanation :

Let train y overtakes train x after t hours ATQ,

Now, Distance covaed by train = Distance covered by train Y

Let train y overtakes train x after t hours ATQ,

Now, Distance covaed by train = Distance covered by train Y

Workspace

2. Two places are 60 km apart. A and B start walking towards each other at the same time and meet each other after 6 hours. Had A travelled with 2/3rd of his speed and B travelled with double of his speed, they would have met after 5 hours. The speed of A is ?

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Correct Ans:6 km/hr

Explanation:

Workspace

3. A truck covers a distance of 376 km at a certain speed in 8 hours. How much time would a car take at an average speed which is 18 kmph more than that of the speed of the truck to cover a distance which is 14 km more than that travelled by the truck ?

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Correct Ans:6 hours

Explanation:

Explanation :

Speed of the truck = Distance/time

= 376/8 = 47 kmph

Now, speed of car = (speed of truck + 18) kmph

= (47 + 18) = 65 kmph

Distance travelled by car = 376 + 14 = 390 km

Time taken by car = Distance/Speed

= 390/65 =

Speed of the truck = Distance/time

= 376/8 = 47 kmph

Now, speed of car = (speed of truck + 18) kmph

= (47 + 18) = 65 kmph

Distance travelled by car = 376 + 14 = 390 km

Time taken by car = Distance/Speed

= 390/65 =

**6 hours.**
Workspace

4. A car traveled 80% of the distance from town A to B by traveling at T hours at an average speed of V km/h. The car travels at an average speed of S km/h for the remaining part of the trip. Which of the following expressions represents the average speed for the entire trip?

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Correct Ans:(5VS/(4S+V))

Explanation:

-----> 80% of the distance = VT km Total distance

-----> = (VT/80) * 100 = (5/4) VT km

-----> Remaining distance = (5/4) VT - VT = (1/4) VT km

-----> Remaining time = (VT/4 S) hour

-----> Total time = T + (VT/4 S)

-----> Average speed = (Total Distance/Total Time)

-----> = (5/4)VT/( T + (VT/4 S)) =

-----> = (VT/80) * 100 = (5/4) VT km

-----> Remaining distance = (5/4) VT - VT = (1/4) VT km

-----> Remaining time = (VT/4 S) hour

-----> Total time = T + (VT/4 S)

-----> Average speed = (Total Distance/Total Time)

-----> = (5/4)VT/( T + (VT/4 S)) =

**(5 VS/4S + V) (KM/H)**
Workspace

5. Two trains for Mumbai leave Delhi at 6 am and 6:45 am and travel at 100 kmph and 136 kmph respectively. Approx. how many km from Delhi will the two trains be together?

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Correct Ans:283 km

Explanation:

----> In 45 min,

----> 1st train will cover = 100*[45/60] = 75 km distance.

----> Relative speed of trains = 136-100 = 36 kmph.

----> 2nd train will catch up the 1st train in = (75/36) = (25/12) hours.

----> Distance from Delhi = [25/12]*136 = 283 km approx.

----> 1st train will cover = 100*[45/60] = 75 km distance.

----> Relative speed of trains = 136-100 = 36 kmph.

----> 2nd train will catch up the 1st train in = (75/36) = (25/12) hours.

----> Distance from Delhi = [25/12]*136 = 283 km approx.

Workspace

6. Two trains 200 m and 160 m long, run at the rate of 60 km/h and 100 km/h respectively on parallel rails. How long will it take a man sitting in the second train to pass the first train if they run in the opposite direction?

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Correct Ans:4.5 seconds

Explanation:

-----> Since both trains are moving in opposite direction so their speed must be added.

-----> As we know, speed = (distance/time) (where distance will be the train of the length)

-----> Time taken by man sitting in second train to pass the first train

-----> = (200/(60+100) * (5/18)) =

----->Note- To convert km/hours to m/sec we have to multiply it by 5/18

-----> As we know, speed = (distance/time) (where distance will be the train of the length)

-----> Time taken by man sitting in second train to pass the first train

-----> = (200/(60+100) * (5/18)) =

**4.5 sec**----->Note- To convert km/hours to m/sec we have to multiply it by 5/18

Workspace

7. The speed of a boat in still water is 24 km/hr and the speed of the stream is 8 km/hr. It takes a total 20 hours to row upstream from point X to Point Y and downstream from Point Y to Point Z. If the distance from X to Y is one third of the distance between Y and Z. What is the total distance travelled by the boat (both upstream and downstream)?

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Correct Ans:512 km

Explanation:

The speed of a boat in still water = 24 km/hr

The speed of the stream = 8 km/hr

Speed of upstream = 24 - 8 = 16 km/hr

Speed of downstream = 24 + 8 = 32 km/hr

The distance from X to Y is one third of the distance between Y and Z

Distance(X to Y) = 1/3 * Distance(Y to Z)

Distance (Y to Z) is,

x/(3*16) + x/32 = 20

x/48 + x/32 = 20

(2x + 3x)/96 = 20

5x/96 = 20

x = 4*96 = 384 km

Distance from X to Y = 384*(1/3) = 128 km

Total Distance = 384 + 128 = 512 km

The speed of the stream = 8 km/hr

Speed of upstream = 24 - 8 = 16 km/hr

Speed of downstream = 24 + 8 = 32 km/hr

The distance from X to Y is one third of the distance between Y and Z

Distance(X to Y) = 1/3 * Distance(Y to Z)

Distance (Y to Z) is,

x/(3*16) + x/32 = 20

x/48 + x/32 = 20

(2x + 3x)/96 = 20

5x/96 = 20

x = 4*96 = 384 km

Distance from X to Y = 384*(1/3) = 128 km

Total Distance = 384 + 128 = 512 km

Workspace

8. The ratio of time taken by a boat to cover (x + 4) km in upstream to that of time taken to cover (x + 20) km in still water is 4: 5. If the ratio of speed of boat in downstream to that of stream is 5: 1 and the sum of time taken by the boat to cover (x - 5) km in upstream to that in downstream is 8 hours then find the speed of boat in downstream?

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Correct Ans:5 km/hr

Explanation:

Let the downstream speed be 5p and speed of stream be p

Downstream = U + V; Upstream = U - V

Where, U = Boat and V = Stream

Then speed of boat in still water will be 5p - p = 4p

And speed of boat in upstream will be 4p - p = 3p

Time taken to cover (x + 4) km in upstream = (x + 4)/3p

And time taken to cover (x + 20) km in still water = (x + 20)/4p

[(x + 4)/3p] : [(x + 20)/4p] = 4 : 5

[(x + 4)*4] / [(x + 20)* 3] = 4/5

(x + 4) / (3x + 60) = 1/5

5x + 20 = 3x + 60

x = 20

Also,

The sum of time taken by the boat to cover (x - 5) km in upstream to that in downstream is 8 hours.

(x - 5) = 20 - 5 = 15 km

15/3p + 15/5p = 8

5/p + 3/p = 8

8/p = 8

p = 1 km/hr

The speed of boat in downstream = 5p = 5 * 1 = 5 km/hr

Downstream = U + V; Upstream = U - V

Where, U = Boat and V = Stream

Then speed of boat in still water will be 5p - p = 4p

And speed of boat in upstream will be 4p - p = 3p

Time taken to cover (x + 4) km in upstream = (x + 4)/3p

And time taken to cover (x + 20) km in still water = (x + 20)/4p

[(x + 4)/3p] : [(x + 20)/4p] = 4 : 5

[(x + 4)*4] / [(x + 20)* 3] = 4/5

(x + 4) / (3x + 60) = 1/5

5x + 20 = 3x + 60

x = 20

Also,

The sum of time taken by the boat to cover (x - 5) km in upstream to that in downstream is 8 hours.

(x - 5) = 20 - 5 = 15 km

15/3p + 15/5p = 8

5/p + 3/p = 8

8/p = 8

p = 1 km/hr

The speed of boat in downstream = 5p = 5 * 1 = 5 km/hr

Workspace

9. A man can row the boat x km downstream and the same distance upstream in 12 hours. Also he can row the boat can row 4x km downstream in 10 hours. In what time, he can row the x km distance in upstream?

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Correct Ans:9.5 hours

Explanation:

Let B be the speed of the boat in still water and W be the speed of the stream.

According to the question,

[x/(B+W)] + [x/(B-W)] = 12

4x/(B+W) = 10

x/(B+W) = 5/2

Then,

x/(B-W) = 12 - 5/2

x/(B-W) = 19/2 = 9.5 hours

According to the question,

[x/(B+W)] + [x/(B-W)] = 12

4x/(B+W) = 10

x/(B+W) = 5/2

Then,

x/(B-W) = 12 - 5/2

x/(B-W) = 19/2 = 9.5 hours

Workspace

10. Sum and difference of the speeds of two cars are 70km/hr & 30km/hr respectively. In a race, if the car with more speed finished 3 minutes earlier than the other. The racing distance was equal to?

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Correct Ans:5/3 km

Explanation:

Let speed of faster car â€˜Aâ€™ be x km/hr & slower car â€˜Bâ€™ be y km/hr

Then,

x + y = 70

x - y = 30

Adding them, 2x = 100, or, x = 50km/hr

And, y = 70 - 50 = 20, i.e., y = 20km/hr

Let racing Distance be D km,

Then,

time taken by A = D/50, and time taken by B = D/20,

(D/20) - (D/50) = 3/60 hr

(5D - 2D)/100 = 1/20 hr

3D/100 = 1/20

D = 100/(20*3) = 5/3 km

Then,

x + y = 70

x - y = 30

Adding them, 2x = 100, or, x = 50km/hr

And, y = 70 - 50 = 20, i.e., y = 20km/hr

Let racing Distance be D km,

Then,

time taken by A = D/50, and time taken by B = D/20,

(D/20) - (D/50) = 3/60 hr

(5D - 2D)/100 = 1/20 hr

3D/100 = 1/20

D = 100/(20*3) = 5/3 km

Workspace

11. Two trains each having a length of 200 meters moving in opposite direction crossed each other in 10 seconds. If first train crossed a 250 metre long platform in 30 seconds, then the ratio of their speeds is:

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Correct Ans:3 : 5

Explanation:

Let the speed of first train be ‘a’ m/sec and that of second train be ‘b’ m/sec.

Then, we get the below two equations,

10 = (200 + 200)/ (a + b)

a + b = 400/10

a + b = 40 ------- (i)

30 = (200 + 250)/a

a = 15 -------- (ii)

By substituting a = 15 in (i), we get b = 25

a = 15m/s, b = 25m/s

Required ratio = a : b = 15 : 25 = 3 : 5

Then, we get the below two equations,

10 = (200 + 200)/ (a + b)

a + b = 400/10

a + b = 40 ------- (i)

30 = (200 + 250)/a

a = 15 -------- (ii)

By substituting a = 15 in (i), we get b = 25

a = 15m/s, b = 25m/s

Required ratio = a : b = 15 : 25 = 3 : 5

Workspace

12. The distance between two schools of Madurai and Trichy is 150 km. A car starts from Madurai and moves towards Trichy at an average speed of 10 km/h. Another car starts from Trichy, 10 minutes earlier than the first car and moves towards Madurai at an average speed of 15 km/h. How far from Madurai and from Trichy will the two cars will meet respectively?

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Correct Ans:59km and 91km

Explanation:

Let t = time at which the two cars met each other.

If the car starting from Madurai have travelled for time â€œtâ€ hr, then the car starting from Trichy will have travelled for [t + (10/60)]hr.

Hence we get, (10 * t) + 15 * [t + (10/60)] = 150

(10 * t) + (15 * t) + (15 * 10/60) = 150

t = 59/10 hr.

Distance from Madurai when the two cars will met = 10 * 59/10 = 59km

Distance from Trichy when the two cars met = 150km - 59km = 91km

If the car starting from Madurai have travelled for time â€œtâ€ hr, then the car starting from Trichy will have travelled for [t + (10/60)]hr.

Hence we get, (10 * t) + 15 * [t + (10/60)] = 150

(10 * t) + (15 * t) + (15 * 10/60) = 150

t = 59/10 hr.

Distance from Madurai when the two cars will met = 10 * 59/10 = 59km

Distance from Trichy when the two cars met = 150km - 59km = 91km

Workspace

13. A bullet train D which is 323 metre long is moving with an average speed of 121 km/h crosses a bridge in 23 seconds. A farmer walking in the field crosses the same bridge in 4 minutes. Determine the speed of the farmer in m/sec?

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Correct Ans:1.87

Explanation:

---> Length of the train = 323 metre

---> Average speed of train = 121 km/h

---> Time taken to cross a platform = 23 seconds

---> Therefore, Speed = (Distance / Time)

---> = ((Length of train + Length of platform)/Time)

---> = ((323 + x)/23)

---> = (121 * (5/18))

---> x = ( 121 * (5/18) * 23 - 323)

---> A man crosses the same platform in 4 minutes

---> Therefore, Speed = (Distance / Time)

---> = (( 121 * (5/18) * 23 - 323)/4 * 60)

---> = ((121 * 5 * 23 - 323 * 18) / 4 * 60 * 18 )

---> = ( (13915 - 5814) / 4 * 60 * 18 )

---> = ( 8101/ 4320)

---> = 1.87

---> Thus, the speed of the man in (m/sec) = 1.87

---> Average speed of train = 121 km/h

---> Time taken to cross a platform = 23 seconds

---> Therefore, Speed = (Distance / Time)

---> = ((Length of train + Length of platform)/Time)

---> = ((323 + x)/23)

---> = (121 * (5/18))

---> x = ( 121 * (5/18) * 23 - 323)

---> A man crosses the same platform in 4 minutes

---> Therefore, Speed = (Distance / Time)

---> = (( 121 * (5/18) * 23 - 323)/4 * 60)

---> = ((121 * 5 * 23 - 323 * 18) / 4 * 60 * 18 )

---> = ( (13915 - 5814) / 4 * 60 * 18 )

---> = ( 8101/ 4320)

---> = 1.87

---> Thus, the speed of the man in (m/sec) = 1.87

**First option is the correct answer.**
Workspace

14. The average speed of a bus was slowed down by 15km/hr for bad weather in the journey. As a result, the bus reached the destination by 20 minutes late. If the total distance was 450 km, find the normal time required to reach the destination.

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Correct Ans:3 hours

Explanation:

---> Let the normal time required to reach the destination = x hours.

---> Total distance = 450 km

---> Now, according to the question, we can write,

---> (450/x) - (450/ x + (20/60)) = 15

---> ((450 ( 3x + 1) - 1350 x)/ x(3x+1) ) = 15

---> 450 + 1350x â€“ 1350x = 15 * x(3x + 1)

---> 450x = 45x

---> 3x

---> 3x

---> x(3x + 10) â€“ 3(3x + 10) = 0

---> (3x + 10)(x â€“ 3) = 0

---> Then, x = - (10/3) or x = + 3

---> We will ignore the negative value of x as the time cannot be negative.

--->

---> Total distance = 450 km

---> Now, according to the question, we can write,

---> (450/x) - (450/ x + (20/60)) = 15

---> ((450 ( 3x + 1) - 1350 x)/ x(3x+1) ) = 15

---> 450 + 1350x â€“ 1350x = 15 * x(3x + 1)

---> 450x = 45x

^{2}+ 15x---> 3x

^{2}+ x â€“ 30 = 0---> 3x

^{2}+ 10x â€“ 9x â€“ 30 = 0---> x(3x + 10) â€“ 3(3x + 10) = 0

---> (3x + 10)(x â€“ 3) = 0

---> Then, x = - (10/3) or x = + 3

---> We will ignore the negative value of x as the time cannot be negative.

--->

**The normal time required to reach the destination = 3 hours.**
Workspace

15. It takes eight hours for a 600 km journey if 120 km is done by train and the rest by car. It takes 20 minutes more if 200 km is done by train and the rest by car. The ratio of the speed of the train to that of the car is how much?

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Correct Ans:3 : 4

Explanation:

Let the speed of Train and Car be T and C respectively.

Time = Distance/Speed

120 km is done by train and the rest 480 by car

(120/T) + (480/C) = 8

(120C + 480T)/TC = 8

120C + 480T = 8TC

15C + 60T = TC --------- (i)

If 200 km is done by train and the rest by car

Again applying the same formula,

(200/T) + (400/C) = 8 + 20/60 = 25/3

(200C + 400T)/TC = 25/3

200C + 400T = 25TC/3

8C + 16T = TC/3

24C + 48T = TC ----------- (ii)

By solving (i) and (ii), we get

15C + 60T = 24C + 48T

12T = 9C

4T = 3C

T/C = 3/4 = 3 : 4

Time = Distance/Speed

120 km is done by train and the rest 480 by car

(120/T) + (480/C) = 8

(120C + 480T)/TC = 8

120C + 480T = 8TC

15C + 60T = TC --------- (i)

If 200 km is done by train and the rest by car

Again applying the same formula,

(200/T) + (400/C) = 8 + 20/60 = 25/3

(200C + 400T)/TC = 25/3

200C + 400T = 25TC/3

8C + 16T = TC/3

24C + 48T = TC ----------- (ii)

By solving (i) and (ii), we get

15C + 60T = 24C + 48T

12T = 9C

4T = 3C

T/C = 3/4 = 3 : 4

Workspace

16. Vishal travelled a certain distance in 2 hr and 15 min. He travelled this distance partly by car and partly by bus. The speed of the bus is 16 km/hr and the speed of car is 32 km/ hr. If the distance travelled by car is equal to distance travelled by bus, what is the total distance that he travelled?

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Correct Ans:48 km

Explanation:

Let, total distance travelled by vishal be 2x Km.

Distance travelled by car = x km.

Distance travelled by bus = x km.

Speed of car = 32 km /hr

Speed of bus = 16 km/hr

According to the question

x/32 + x/16 = 2 hr 15 min

(2x + 4x) / 64 = 2(15/60)

6x/64 = 2(1/4)

6x/64 = 9/4

x = 9/4 * 64 /6

x = 24

So, total distance travelled = 2x = 2*24 = 48

Distance travelled by car = x km.

Distance travelled by bus = x km.

Speed of car = 32 km /hr

Speed of bus = 16 km/hr

According to the question

x/32 + x/16 = 2 hr 15 min

(2x + 4x) / 64 = 2(15/60)

6x/64 = 2(1/4)

6x/64 = 9/4

x = 9/4 * 64 /6

x = 24

So, total distance travelled = 2x = 2*24 = 48

Workspace

17. A man takes 3 times as long to cover a distance against the stream as to cover the same distance downstream. What is the ratio of speed of stream to the speed of the man?

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Correct Ans:0.5

Explanation:

Let the speed of the stream be x km/h

Let the speed of the man be y km/h

Let the distance being covered be d km

Time taken to cover d km upstream = d/(y â€“ x)

Time taken to cover d km downstream = d/(y + x)

It is given that time taken to go upstream is three times the time taken to go downstream.

So, d/(y â€“ x) = 3d/(y + x)

y + x = 3y â€“ 3x

4x = 2y

x/y = 2/4 = 0.5

Let the speed of the man be y km/h

Let the distance being covered be d km

Time taken to cover d km upstream = d/(y â€“ x)

Time taken to cover d km downstream = d/(y + x)

It is given that time taken to go upstream is three times the time taken to go downstream.

So, d/(y â€“ x) = 3d/(y + x)

y + x = 3y â€“ 3x

4x = 2y

x/y = 2/4 = 0.5

Workspace

18. A train runs for 2 hrs at the speed of 40 km/h and then runs for 4(1/2) hrs at the speed of 60 km/h and then runs for 3(1/2) hrs, at the speed of 70 km per hour. Find the average speed of the train.

SHOW ANSWER

Correct Ans:59.5 km/h

Explanation:

Distance traveled by train in first 2 hrs = 40*2 =80 km.

Distance traveled by train in next 4(1/2) hrs.

= 60*4(1/2) = 60 * 9/2 = 270 km.

Distance traveled by train in last 3(1/2) hrs = 70*3(1/2)

= 70 * 7/2 = 245 km.

Total Distance traveled by train = 80 + 270 + 245 = 595 km.

Total time taken to travel = 2 + 4(1/2) + 3(1/2)

= 2 + 9/2 + 7/2 = 10 hrs.

Hence average speed = Distance/time =595/10 = 59.5 km/h.

Distance traveled by train in next 4(1/2) hrs.

= 60*4(1/2) = 60 * 9/2 = 270 km.

Distance traveled by train in last 3(1/2) hrs = 70*3(1/2)

= 70 * 7/2 = 245 km.

Total Distance traveled by train = 80 + 270 + 245 = 595 km.

Total time taken to travel = 2 + 4(1/2) + 3(1/2)

= 2 + 9/2 + 7/2 = 10 hrs.

Hence average speed = Distance/time =595/10 = 59.5 km/h.

Workspace

19. A motorcyclist covered (2/3)rd of a total journey at his usual speed. He covered the remaining distance at (3/4)th of his usual speed. As a result, he arrived 30 minutes later than the time he would have taken at usual speed. If the total journey was 180 km. What was his usual speed?

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Correct Ans:40 km/hr

Explanation:

Given, Total Distance = 180 km

(2/3)rd of a total journey = (2/3) * 180 = 120 km

Remaining Distance = 180 - 120 = 60 km

Let the usual speed be x km/hr.

Then, for the Remaining 60 km journey,

= 60/x

But, the motorcyclist covered the remaining distance (i.e., 60 km) at (3/4)th of x

--->

As a result, he arrived 30 minutes (i.e., 1/2 hour) later than the Usual time

--->

---> (60/x) + (1/2) = [60/(3x/4)]

---> [60/(3x/4)] - (60/x) = 1/2

---> [(60 * 4)/3x] - (60/x) = 1/2

---> (80/x) - (60/x) = 1/2

---> 20/x = 1/2

--->

Thus, the

(2/3)rd of a total journey = (2/3) * 180 = 120 km

Remaining Distance = 180 - 120 = 60 km

Let the usual speed be x km/hr.

Then, for the Remaining 60 km journey,

**Usual Time taken**by motorcyclist = Remaining Distance/ usual speed= 60/x

But, the motorcyclist covered the remaining distance (i.e., 60 km) at (3/4)th of x

--->

**Actual Time taken**by motorcyclist = 60/(3x/4)As a result, he arrived 30 minutes (i.e., 1/2 hour) later than the Usual time

--->

**Usual Time taken + (1/2) hour = Actual Time taken**---> (60/x) + (1/2) = [60/(3x/4)]

---> [60/(3x/4)] - (60/x) = 1/2

---> [(60 * 4)/3x] - (60/x) = 1/2

---> (80/x) - (60/x) = 1/2

---> 20/x = 1/2

--->

**x = 40**Thus, the

**usual speed of motorcyclist = 40 km/hr**
Workspace

20. Two racers start running towards each other, one from A to B and another from B to A. They cross each other after one hour and the first racer reaches B, 5/6 hour before the second racer reaches A. If the distance between A and B is 50 km. what is the speed of the slower racer?

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Correct Ans:20 km/hr

Explanation:

Let the second racer takes 'x' hr with speed S

and the first racer takes [x - (5/6)] hr with speed S

Given, Total distance = 50 km

Then, S

S

Total speed = S

= {50/[x - (5/6)]} + [50/x]

= 50 {1/[x - (5/6)] + [1/x]}

= 50 {x + [x - (5/6)] / [x - (5/6)]x}

= 50 {[2x - (5/6)] / [x

= 50 {[(12x - 5)/6] / [(6x

= 50 {[12x - 5] / [6x

As they cross each other in 1hr,

Now, Time = Distance/ Total Speed

---> 1 = 50 / (50 {[12x - 5] / [6x

---> 1 = 1 / {[12x - 5] / [6x

---> 1 = [6x

---> [12x - 5] = [6x

---> 6x

---> 6x

---> 6x

---> 3x(2x - 5) - 1(2x - 5) = 0

---> (2x - 5) (3x - 1) = 0

--->

[Neglecting x = 1/3]

Now, We have to find the speed of the slower racer ie., second racer.

Put x = 5/2 in S

= (50 * 2) / 5

=

Thus, the

_{2}and the first racer takes [x - (5/6)] hr with speed S

_{1}Given, Total distance = 50 km

Then, S

_{1}= 50/[x - (5/6)]S

_{2}= 50/xTotal speed = S

_{1}+ S_{2}= {50/[x - (5/6)]} + [50/x]

= 50 {1/[x - (5/6)] + [1/x]}

= 50 {x + [x - (5/6)] / [x - (5/6)]x}

= 50 {[2x - (5/6)] / [x

^{2}- (5x/6)]}= 50 {[(12x - 5)/6] / [(6x

^{2}- 5x)/6]}= 50 {[12x - 5] / [6x

^{2}- 5x]}As they cross each other in 1hr,

Now, Time = Distance/ Total Speed

---> 1 = 50 / (50 {[12x - 5] / [6x

^{2}- 5x]})---> 1 = 1 / {[12x - 5] / [6x

^{2}- 5x]}---> 1 = [6x

^{2}- 5x] / [12x - 5]---> [12x - 5] = [6x

^{2}- 5x]---> 6x

^{2}- 5x - 12x + 5 = 0---> 6x

^{2}- 17x + 5 = 0---> 6x

^{2}- 15x - 2x + 5 = 0---> 3x(2x - 5) - 1(2x - 5) = 0

---> (2x - 5) (3x - 1) = 0

--->

**x = 5/2, 1/3**[Neglecting x = 1/3]

Now, We have to find the speed of the slower racer ie., second racer.

Put x = 5/2 in S

_{2}, we get**S**= 50/(5/2)_{2}= (50 * 2) / 5

=

**20 km/hr**Thus, the

**speed of the slower racer = 20 km/hr**.
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