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ALLU’s speed = 500/3% = 5/3 times Arjun’s.

Let the length of the RACE be x m. 

Let Arjun’s speed be 3s m/s and Allu’s be 5S m/s 

According to the QUESTION,

⇒ x/5 = (x - 200)/3 

⇒ x = 500 metres

Assume both trains meet after X hours after 7 am

DISTANCE covered by train starting from P in x hours = 20X KM

Distance covered by train starting from Q in (x - 1) hours = 25(x - 1) km (As it starts at 8 AM)

Total distance between P and Q = 110 km

⇒ 20x + 25(x - 1) = 110

⇒ 45x = 135

⇒ x = 3

∴ They meet after 3 hours after 7 am, i.e., they meet at 10 A.M.

LET the time taken by motorcyclist when travelling at 40 kmph be ‘t’ hours.

∴ Total distance travelled = (40 × t) km-----(1)

Now, after 30kms the motorcycle undergoes a problem and its speed reduces and it reaches the destination 20 mins late, that is, 1/3 hours late.

In this case, total time REQUIRED for journey = t + 1/3 hours

But time required to TRAVEL first 30kms = 30/40 = ¾ hours

∴ Time for which it travelled at 30 km/hr = t + 1/3 – ¾ = t – 5/12

∴ Total distance travelled = 30 + 30(t – 5/12) = 17.5 + 30t----(2)

From (1) and (2),

40t = 17.5 + 30t

∴ 10T = 17.5

t = 1.75 hours

QUANTITY A:

Total distance traveled = 200 × 5 + 100 × 10 + 50 × 2 + 25 × 5 km

⇒ 1000 + 1000 + 100 + 125 km = 2225 km

Total TIME = 5 + 10 + 2 + 5 = 22 hours

Required speed = 2225/22 = 101.14 kmph

Quantity B: 100

∴ Quantity A > Quantity B

Let the speed of the current be X and the man’s speed be y km/hr.

Distance = speed × time

16 = (x + y) × 4

⇒ x + y = 4 ----(1)

6 = (y - x) × 2

⇒ -x + y = 3 ----(2)

Adding (1) and (2)

⇒ 2Y = 7

⇒ y = 3.5

From (1)

Let the trains meet at a distance of ‘x’ km from B.

Distance of meeting POINT from A = 420 – x

To reach the meeting point, time taken by A = time taken by B

(Distance travelled by A/ speed of TRAIN from A) = (distance travelled by B/ speed of train from B)

⇒ (420 – x)/36 = x/24

⇒ 10080 – 24x = 36x

⇒ 24x + 36x = 10080

⇒ 60X = 10080

⇒ x = 10080/60 = 168 km.

Downstream SPEED = 45 + 9 = 54 km/hr

Time = 18 min = 18/60 hrs

Let the speed of current be x m/hr.

Speed of boat in still water = 60 m/hr

Speed of boat travelling upstream = (60 - x) m/hr

Time TAKEN to travel 10 m upstream = 10/(60 - x) hr

Speed of boat travelling downstream = (60 + x) m/hr

Time taken to travel 15 m downstream = 15/(60 + x) hr = 10/(60 - x) hr (GIVEN)

⇒ 900 - 15x = 600 + 10x

⇒ 25X = 300

⇒ x = 12 m/hr

∴ Speed of current is 12 m/hr

SPEED of train A, S_A = 36 km/h = 36 × (1000/3600) m/s = 10 m/s

Speed of train A, S_B = 54 km/h = 54 × (1000/3600) m/s = 15 m/s

Length of train A, L_a = 250 m

Time TAKEN for train B to overtake train A = 2 min = 120 s

Let the length of train B be L_b.

Time taken to overtake train A by train B = (L_a + L_b)/(S_B – S_A)

⇒ (250 + L_b)/(15 – 10) = 120

⇒ 250 + L_b = 600

⇒ L_b = 350 m

∴ length of train B is 350 m.

DOWNSTREAM speed = 10 + 4 = 14 KM/hr

UPSTREAM Speed = 10 - 4 = 6 km/hr 

Distance travelled Upstream = Speed of upstream × time = 6 × 10/60 = 1 km

∴ Distance travelled Upstream is 1 km

When 2 trains are traveling in opposite DIRECTION their relative speed is addition of their individual speeds.

So, relative speed of two trains = 60 + 48 = 108 kmph

$(= \frac{{108 \times 1000}}{{3600}})$

= 30 m/s

When trains cross each other they travel distance which is equivalent to sum of their lengths.

As per GIVEN data-

Trains cross each other in 30 SECONDS

& Length of 1 train is 400 meters.

Let the length of other train be x meters.

∴ Sum of lengths of trains = Relative distance traveled by trains in 30 seconds.

⇒ Sum of lengths of trains = Relative speed × time

⇒ 400 + x = 30 × 30

⇒ 400 + x = 900

⇒ x = 500 meters

⇒ LENGTH of BRIDGE = 1 km = 1000 m

⇒ Length of train = 1000/2 = 500 m

⇒ Total distance travelled = (1000 + 500) m = 1500 m

⇒ Time TAKEN = 2 min

Distance = Speed × Time ⇒ Time = Distance/Speed

If two trains are going toward each other their relative speed = Sum of their speed

Here, the train and the man is RUNNING towards each other. So, the relative speed will be

= (40 + 5) kmph = 45 kmph.

= 45 × (1000/3600) m/SEC

= (25/2) m/s

The length of the train = 120m

Quantity A:

TOTAL distance travelled by cycle in 4 hours is –

Distance COVERED by cycle in the first hour = 10 km

Distance travelled in second hour = 7 km

Distance travelled in third hour = 4 km

Distance travelled in forth hour = 1 km

Total distance covered by cycle in 4 hours =

= 10 + 7 + 4 + 1

= 22

Total distance travelled by cycle in 4 hours is 22 km.

Quantity B:

Difference between the time travelled by cycle = 12 min = 12/60h = 1/5h

Let the length of journey be a km.

Then,

(a/4) – (a/6) = 1/5

2a/24 = 1/5

a = 12/5 = 2.4 kms

Total journey is 2.4 kms

From above solution,

Let the distance travelled be x km

Let the speed of current be V km/hr

Let the speed of boat in still WATER be b km/hr

∴ Speed of boat upstream = (b – v) km/hr

⇒ Speed of boat downstream = (b + v) km/hr

∴ Time taken to travel x distance upstream = x/(b – v) hr

⇒ Time taken to travel x distance downstream = x/(b + v) hr

Now, $(\frac{{{\rm{x}}/\LEFT( {{\rm{b\;}}-{\rm{\;v}}} \right){\rm{\;}}}}{{{\rm{x}}/\left( {{\rm{b\;}} + {\rm{\;v}}} \right){\rm{\;}}}})$ = 3

⇒ b + v = 3(b – v)

⇒ b + v = 3B – 3v

⇒ 4v = 2b

⇒ b/v = 2 : 1

LET speed of car be x km/hr.

⇒ Speed of bus = (x - 25) km/hr

We KNOW that,

Time = Distance/Speed

According to question,

[500/(x - 25)] - (500/x) = 10

⇒ 500 × 25 = 10x(x - 25)

(x - 25)x = 50 × 25

⇒ x = 50 km/hr & x - 25 = 25 km/hr

∴ The DIFFERENCE between the speeds of the car and the bus car = 50 - 25 = 25

Quantity A:

LET the speed of stream be a km/hr.

Speed of boat in still water = 5 km/hr

Speed downstream = (a + 5)km/hr

Speed UPSTREAM = (5 – a)km/hr

(a + 5) × 2 = (5 – a) × 5/2

4a + 20 = 25 – 5a

9a = 5

a = 5/9

Speed of stream is 5/9 km/hr.

Quantity B:

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

Speed of stream = 6 km/hr

Time taken to row down and to row up are in the ratio of 3 : 1.

Then speeds are in the ratio of 1 : 3.

Then,

(x + 6)/3 = (x – 6)/1

x + 6 = 3X – 18

2x = 24

x = 12

Speed of boat in still water is 12 km/hr.

From above solution,

Relative SPEED between TWO trains = 72km/hr. or 20m/s

Let the length of train B be L m

Acc. to the question, $(\frac{{150 + \;L}}{{13}} = 20)$

Or, L = 110m

And, $(\frac{{190 + 110}}{{20}} = \;\frac{{300}}{{20}} = 15)$

GIVEN,

Total distance covered by car and motorcycle = 125 KM

Car's SPEED was 50% more than the motorcycle

Let the speed of the motorcycle be S₁ = x

Then, the speed of car S₂ = x + 50x/100 = 3x/2

Let the time TAKEN by both car and motorcycle be t hours.

We know that,

Distance = Speed × Time

⇒ S1 $× t + S2 $× t =125

⇒ xt + 3xt/2 = 125

⇒ xt = 50 km

∴ Distance covered by RAM by motorcycle = 50 km

Let speed of boat in still water is S_B, speed of stream is SC, and Speed of boat in downstream is S_D

Speed of boat in downstream, S_D = S_B + S_C = 8 + 2 = 10 km/hr

Average Speed of the boat $(= \frac{{40\; + \;36\; + \;45}}{{\frac{{40}}{{10}}\; + \;\frac{{36}}{{10\; + \;2}}\; + \;\frac{{45}}{{12\; + \;3}}}} = \frac{{121}}{{4\; + \;3\; + \;3}} = \frac{{121}}{{10}})$

∴ Average Speed of the boat = 12.1 km/hr

Total journey distance = 480 km

Ideal average speed = 80 kmph

Planned or time taken in ideal condition = 480/80 = 6 hour = 360 minutes

For actual time,

Distance in which average speed of 80 kmph is maintained,

Distance covered before break = 62.5% of 480 = (5/8) × 480 = 300 km

Time taken to COVER this distance = 300/80 = 3.75 hour = 225 minutes

After the break,

Distance need to be covered = 480 - 300 = 180 km

Time required to cover the distance = 180/60 = 3 hour = 180 minutes

Actual time of journey = 225 + 180 + 30 = 435 minutes

Let the first cyclist be A and second ONE be B

Distance covered by A and B TOGETHER in first hour = 8 + 4 = 12 km

Distance covered by A and B together in hour = 8 + 5 = 13 km and so on

Because B increases his speed by 1 km at every hour

So, in 9 hour they both will exactly cover 144 km

So A will meet B at half of the total distance = 1/2 × 144 = 72 km

Speed of 450 m long train = 117 kmph = 117 × (5/18) = 32.5 m/s

TOTAL distance covered by trains while crossing each other = Sum of their lengths = 650 m

Total TIME taken = 13 sec

RELATIVE speed = $(\frac{{Total\;distance\;covered}}{{Total\;time\;taken}})$

= 650/13 = 50 m/s

Distance travelled by boat in upstream = 32 km, time taken = 8 hr

SPEED of boat in upstream = 32/8 km/hr = 4 km/hr

And distance travelled by boat in DOWNSTREAM = 28 km, time taken = 4 hr

Speed of boat in downstream = 28/4 = 7 km/hr

Now, speed of the boat in still WATER

⇒ 1/2 × {speed of boat in upstream + speed of boat in downstream}

⇒ 1/2 × [4 + 7] = 1/2 × 11 = 5.5 km/hr

Speed of the current = 1/2 × {speed of boat in downstream - speed of boat in upstream}

⇒ 1/2 × [7 - 4] = 1.5 km/hr

Let the total distance traveled by Jay be D km.

As per given data-

Jay travels half journey by train at 40 kmph

∴ He travels D/2 km at 40 kmph SPEED.

∴ Time taken to travel D/2 km = $(\frac{{distance}}{{speed}})$

$({\rm{Time\;taken\;to\;travel\;}}\frac{{\rm{D}}}{2}{\rm{\;km}} = \frac{{\frac{D}{2}}}{{40}})$

⇒ Time taken to travel D/2 km = D/80 hr---- (1)

After travelling D/2 distance half of remaining distance he travels at 20 kmph.

∴ He travels D/4 distance at 20 kmph.

$(\therefore {\rm{Time\;taken\;to\;travel\;D}}/4{\rm{\;km}} = \frac{{distance}}{{speed}})$

Time taken to travel D/4 km $(= \frac{{\frac{D}{4}}}{{20}})$

Time taken to travel D/4 km = D/80 hr---- (2)

& Remaining distance he travels at 2.5 kmph.

∴ He travels D/4 distance at 2.5 kmph.

∴ Time taken to travel D/4 km = $({\rm{}}\frac{{distance}}{{speed}})$

Time taken to travel D/4 km = $(\frac{{\frac{D}{4}}}{{2.5}})$

Time taken to travel D/4 km = D/10 hr---- (3)

By ADDING results (1), (2) & (3) we get total time taken by Jay to travel distance D.

∴ Total time taken by RAJ to travel distance D = (D/80) + (D/80) + (D/10)

As we know,

$(Average\;speed = \frac{{Total\;distance}}{{Total\;time\;taken\;to\;travel}})$

⇒ Average speed = $(\frac{D}{{\frac{D}{{80}} + \frac{D}{{80}} + \frac{D}{{10}}}})$

⇒ Average speed = $(\frac{D}{{\frac{D}{{80}} + \frac{D}{{80}} + \frac{{8D}}{{80}}}})$(By keeping denominators equal)

⇒ Average speed = $(\frac{D}{{\frac{{10D}}{{80}}}})$

⇒ Average speed = 8 kmph.

Let, speed of STREAM be x km/h. 

∴ Speed of the ship along the stream = (75 + x) km/h 

∴ Speed of the ship against the stream = (75 – x) km/h 

According to question, 

⇒ 1.25 × (75 + x) = 125 

⇒ 93.75 + 1.25x = 125 

⇒ 1.25x = 32.75 

⇒ x = 25 

∴ Speed of stream = 25 km/h. 

∴ Speed of the ship against the stream = 75 – 25 = 50 km/h 

∴ REQUIRED time = 125/50 HOURS = 5/2 hours

In order to find the NUMBER of hours it will take for Akash and Baichung to meet together, LET us take the LCM of their SPEEDS 3 kmph and 4 kmph.

LCM (3, 4) = 12 KMS

∴ Akash will cover the distance of 12 kms in time = $(\frac{{12}}{3} = {\rm{\;}}4{\rm{\;hours}})$ 

∴ Baichung will cover the distance of 12 kms in time = $(\frac{{12}}{3} = {\rm{\;}}3{\rm{\;hours}})$

?Akash starts at 1 o’ clock and Baichung starts at 2 o’ clock, they will meet together at 5 o’ clock

Or we can say that Baichung catches Akash at 5 o’ clock

At 5 o’ clock, Charanjeet will have covered the distance of = 5kmph× 2 hrs = 10 kms

∴ Distance between Akash and Charanjeet is 2 kms now (12 kms - 10 kms), the time taken by Akash and Charanjeet to meet will be = $(\frac{2}{{\left( {5\; + \;3} \right)}}\;{\rm{hours\;}} = {\rm{\;}}\frac{1}{4}{\rm{\;hours\;}} = {\rm{\;}}15{\rm{\;mins}})$

Let the LENGTH of train be x m

⇒ Length of platform = 3x

⇒ Time TAKEN to CROSS the POLE = (x/25) sec

⇒ Time taken to cross the platform = (x + 3x)/25 = 4x/25

According to question,

⇒ (4x/25) – (x/25) = 25

⇒ 3x = 625

∴ x = 208.33 m ≈ 208 m$

Let the SPEED of TRAIN A be s m/s and that of train B be s - 20 m/s.

Distance COVERED by train A = distance covered by train B

∴ s × 45 = (s - 20) × 75

⇒ s = 50m/s

Distance travelled by train A = 50 × 45 = 2250 m

Let X be any number such that length of the trains is 2x and the length of tunnel is 3x.

Distance covered by train A = length of the train + length of the tunnel

∴ 2x + 3x = 2250

⇒ 5x = 2250

⇒ x = 450

∴ length of the tunnel, 3x = 1350 m

As per given data-

A man travels 30 mins at speed of 40 KMPH.

So, he travels 0.5 hr at speed of 40 kmph.

∴ Distance covered by man = speed of bike × time TRAVELED

⇒ Distance covered by man = 40 × 0.5

⇒ Distance covered by man = 20 km---- (1)

Next he travels 45 mins at 60 kmph

So, he travels 0.75 hr at speed of 60 kmph.

∴ Distance covered by man = speed of bike × time traveled

⇒ Distance covered by man = 60 × 0.75

⇒ Distance covered by man = 45 km---- (2)

& then he travels 45 mins at 100 kmph

So, he travels 0.75 hr at speed of 100 kmph.

∴ Distance covered by man = speed of bike × time traveled

⇒ Distance covered by man = 100 × 0.75

⇒ Distance covered by man = 75 km---- (3)

From above we can see total distance traveled by man is 20 + 45 + 75 = 140 kms.

& total time taken to travel = 0.5 + 0.75 + 0.75 = 2 hrs.

$(\THEREFORE Average\;speed\;of\;man = \frac{{Total\;distance\;traveles}}{{Time\;taken}})$

Average speed of man = 140/2

∴ Average speed of man = 70 kmph.

Distance in upstream = distance in DOWNSTREAM = 21 km

Journey time = 10 hour

Speed of boat = 5 kmph

Speed of upstream = (5 – x) kmph

Speed of downstream = (5 + x) kmph

Let assume, speed of STREAM as x kmph,

Journey time = Time in upstream + time in downstream

⇒ 10 = 21/(5 – x) + 21/(5 + x)

⇒ 10(25 – x²) = 21(5 + x) + 21(5 – x)

⇒ 250 – 10x² = 105 + 21x + 105 – 21x

⇒ 250 - 210 = 10x²

⇒ x² = 40/10

Let the SPEED of the 210 m LONG train = x km/h

As the trains are travelling in opposite directions,

Relative speed of the first train w.r.t the second train = Speed of train 1 + Speed of train 2

⇒ Relative speed = (70 + x)km/h = (70 + x) × 5/18 m/s---(1)

Thus,

Distance covered by train 1 = Length of Train 1 + Length of train 2

⇒ Distance covered by train 1 = 140 + 210 = 350 m

 Time TAKEN = 7 seconds

∴ Relative speed × (Time taken) = Total distance covered

⇒ (70 + x) × (5/18) × 7 = 350

⇒ (70 + x) × (5/18) = 50

⇒ (70 + x)/18 = 10

⇒ x = 110 km/h

Thus,

Let the length of PLATEFORM be P meters. So, length of train will be (800 – P) meters.

When train crosses platform, DISTANCE covered is sum of lengths of train and platform.

The train crosses the platform in 32 seconds.

We know, Time taken = Distance/speed

⇒ 32 = 800/speed of train

⇒ Speed of train = 25 m/sec

The train crosses a MAN STANDING on platform in 20 seconds.

When train crosses a man, distance covered is length of train.

⇒ 20 = (800 – P)/25

⇒ P = 800 – 500 = 300 meters

Speed of water in STILL water = ½ × (speed downstream + speed upstrem)

Speed upstream = 1/(15 / 60) = 4

Let distance covered in downstream be n km

Time REQUIRED = 1 + 45/60 = 7/4

Speed downstream = n/(7/4) = 4n/7

Speed of water in still water = 5

⇒ 5 = ½ × (4n/7 + 4)

⇒ 10 = (4n + 28)/7

⇒ 70 = 4n + 28

⇒ 4n = 42

⇒ n = 10.5km

In a 3000 m race AROUND a field having the circumference of 600 m, the top runner meets the last runner on the 5^th minute of the race and the top runner runs at twice the speed of the last runner. So, the relative DISTANCE between the top runner and last runner = 600 m

We know that,

Speed = Distance/TIME

And relative speed (Top runner’s speed - last runner’s speed) = (600/5) m/min = 120 m/min

If the top runner’s speed is twice the speed of the last runner, then Top runner’s speed = 2 × last runner’s speed.

So, (2 × last runner’s speed) - (last runner’s speed) = 120 m/min

⇒ Last runner’s speed = 120 m/min

The speed of top runner = 120 × 2 = 240 m/min

Time taken by top runner to FINISH the race = 3000/240 = 12.5 min

∴ Time taken by top runner to finish the race = 12.5 min

From the given data, we get

⇒ Speed of the train = 108 × 5/18 = 30 m/s

⇒ Length of the train = 30 × 10 = 300 m

Let us consider the length of the train be X metres

⇒ (x + 300)/18 = 30

⇒ x + 300 = 540

⇒ x = 540 - 300 = 240

∴ Length of the platform = 240 m

TOTAL DISTANCE = 45 + 45 = 90 miles

Total time = 1 + 2 = 3 hrs

Speed of each BOAT in still water = 8 km/hr

Suppose the speed of STREAM is v km/hr.

⇒ Speed of boat that goes upstream = (8 – v) km/hr

And, Speed of boat that goes downstream = (8 + v) km/hr

Suppose the boats meet after T hours.

We know, Distance = Speed × Time

⇒ Distance COVERED by boat moving upstream = (8 - v) km/hr × T hr = (8 - v)T km

⇒ Distance covered by boat moving downstream = (8 + v) km/hr × T hr = (8 + v)T km

Now, total distance covered by boats should be EQUAL to 20 km.

⇒ (8 – v)T + (8 + v)T = 20

⇒ 16T = 20

⇒ T = 1.25 hours = 1 hour + (0.25 × 60 MINUTES) = 1 hour and 15 minutes

∴ Time taken for the boats to meet = 1 hour and 15 minutes

Let the speed of the stream be V km/hr.

Upstream speed = (15 – V) km/hr

Downstream speed = (15 + V) km/hr

6 hours and 30 minutes is 6.5 hours.

To cover 78 upstream, 6.5 hours are required.

We know, Distance = Speed × Time

⇒ 78 = (15 – V) × 6.5

⇒ V = 15 – (78/6.5) = 15 – 12 = 3

⇒ Downstream speed = (15 + V) km/hr = 18 km/hr

∴ Time taken to cover 78 km downstream = 78/18 = 4.333 Time inhours = 4 hours + (1/3) hours = 4 hours and 20 minutes