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Relative speed of the policeman with respect to that of thief = (9 – 8) km/hr = \(1\times\frac{5}{18}\) m/sec

⇒ Time taken by the policeman to catch the thief = \(\frac{200}{\frac{5}{18}}\) sec = 720 seconds 

∴ The distance run by the thief = \(\bigg(8\times\frac{5}{18}\times720\bigg)\) metres = 1600 metres

(a) 30 km/hr

Let the speed of the faster train be x km/hr and that of the slower train be y km/hr. 

Relative speed when both move in same direction = (x – y) km/hr 

Relative speed when both move in opposite directions = (x + y) km/hr 

Total distance travelled = Sum of lengths of both the trains = 200 m

Given, \(\frac{200}{(x-y)\times\frac{5}{18}}=60\) and \(\frac{200}{(x+y)\times\frac{5}{18}}=10\)

⇒ \(\frac{3600}{5(x-y)}=60\) and \(\frac{3600}{5(x+y)}=10\)

⇒ x - y = \(\frac{3600}{300}=12\)                 ....(i)

and x + y = \(\frac{3600}{50}=72\)              ....(ii)

Adding eqn (i) and eqn (ii), we get 

2\(x\) = 84 ⇒ \(x\) = 42 km/hr 

∴ From (i), y = 30 km/hr.

(b) 42 m/s, 38 m/s 

Let the speeds of the faster train be x m/s and that of the slower train be y m/s. Then, When the two trains are moving in the same direction, Relative speed = (x – y) m/s

∴ \(\frac{130\,m+110\,m}{x-y}=60\)

⇒ \(\frac{240}{x-y}=60\)     ⇒ x – y = 4               …(i) 

When the two trains are moving in opposite direction, 

Relative speed = (x + y) m/s

∴ \(\frac{130\,m+110\,m}{x+y}=3\)           ⇒ x – y = 80              …(ii) 

Adding eqn (i) and (ii), we get 2x = 84 

⇒ x = 42 m/s 

∴ Putting in (i), we get y = 38 m/s.

(d) 200 m

Relative speed = (96 – 60) km/hr = 36 km/hr 

=\(\bigg(36\times\frac{5}{18}\bigg)\) m/s = 10 m/sec

As we know D = s × t 

∴ Length of faster train = 10 × 20 = 200 m.

Since the trains travel in the same direction, relative speed of the trains = (50 – 32) km/hr = 18 km/hr.

= \(\bigg(18\times\frac{5}{18}\bigg)\) m/x = 5 m/s

∴ Length of the faster train = Speed × Time 

= (5 × 15) m = 75 m.

Ratio of the downstream speed to upstream speed = 4 : 3 

∴ Ratio of times taken to cover a certain distance downstream and upstream = 3 : 4 

Given 3\(x\) + 4\(x\) = 14 hrs ⇒ \(x\) = 2

∴ The man took 6 hours to row 48 km downstream and 8 hours to row 48 km upstream.

∴Speed downstream = \(\frac{48}6\) km/hr = 8 km/hr 

Speed upstream = \(\frac{48}8\) km/hr = 6 km/hr 

∴ Rate of stream = \(\frac12\)(8 - 6) km/hr = 1 km/hr.

Let the speed of the boat in still water be \(x\) km/hr and speed of the stream = y km/hr 

Then, speed of boat downstream = (\(x\) + y) km/hr 

Speed of boat upstream = (\(x\) – y) km/hr

Given (\(x\) + y) × 1 = 20 and (\(x\) – y) × 2 = 20 

⇒ \(x\) + y = 20            …(i)           and \(x\) – y = 10        …(ii) 

Adding eqn (i) and (ii), we get 

2\(x\) = 30  ⇒ \(x\) = 15 km/hr.

(c) 80 km

Speed of boat in still water be u km/hr. Then, 

Distance covered downstream = (u + 1) × 8 = 8u + 8 

Distance covered upstream = (u – 1) × 10 = 10u – 10 

Since the distance both ways is the same,

8u + 8 = 10u – 10 ⇒ 2u = 18 ⇒ u = 9 km/hr

∴ Distance covered one way = (9 + 1) × 8 km = 80 km.

(c)  \(7\frac12\) min,  3km

Distance covered by hare in 2 min

\(=\frac{24}{60}\times\)1000 x 2 = 800 m

Now to overtake the hare, the dog has to cover a distance of (800 + 200) = 1000 m with a relative speed = (32 – 24) km/hr = 8 km/hr

∴ Time taken = \(\frac18\) hrs = \(\bigg(\frac18\times60\bigg)\) min = \(7\frac12\) min

 ∴ Distance travelled by hare in \(\frac18\) hrs

= \(\bigg(\frac18\times24\bigg)\) km = 3km.

(c) 12 km/hr

Let the distance travelled be x km. Then

\(\frac{x}{10}-\frac{x}{15}\) = 2 ⇒ 3x – 2x = 60 ⇒ x = 60 km 

Time taken to travel 60 km at 10 km/hr = \(\big(\frac{60}{10}\big)\) hrs = 6 hrs.

So Robert started 6 hours before 2 P.M., i.e., at 8 P.M.

∴ Speed required to reach A at 1 P.M. = \(\big(\frac{60}{5}\big)\) km/hr = 12 km/hr.

(d) 30 km

Let the distance between P and Q be x km. 

Then, Total time taken by Ram in the round trip – Total time taken by Shyam in the round trip = 12 min

⇒ \(\bigg(\frac{x}{10}+\frac{x}{15}\bigg)-\)\(\bigg(\frac{x}{12.5}+\frac{x}{12.5}\bigg)\) = \(\frac{12}{60}\)

⇒ \(\bigg(\frac{3x+2x}{30}\bigg)\) - \(\frac{2x}{12.5} = \)\(\frac{12}{60}\)

⇒ \(\frac{x}{6}-\frac{2x}{12.5}= \) \(\frac{12}{60}\) ⇒ \(\frac{12.5x-12x}{75}=\) \(\frac{12}{60}\)

⇒ 0.5x = 75 x \(\frac{12}{60}\) ⇒ \(x\) = \(\frac{75\times12}{0.5\times60}=\) 30 km.

(d) 8 km/hr

Let the speed of the car be x km/hr. Then, Relative speed of the pedestrian with respect to the car = (x – 2) km/hr (∴ Both move in the same direction)

∴ \(\frac{0.6}{(x-2)}=\frac{6}{20}\) ⇒ 6 = x – 2 ⇒ x = 8 km/hr.

(c) 5 : 1

Let the speed downstream be u km/hr and speed upstream be v km/hr. 

Given, 2u = 3v ⇒ u = \(\frac{3v}{2}\)

Required ratio = \(\frac12(u+v):\frac12(u-v)\)

= \(\bigg(\frac{3v}{2}+v\bigg):\bigg(\frac{3v}{2}-v\bigg)\)

= \(\frac{5v}{2}:\frac{v}2=5:1.\)

(c) 8 : 3

Let the distance covered in one direction be x km. 

Then, speed upstream = \(\frac{x}{\frac{44}{5}}\) km/hr

(8 hrs 48 min = \(8\frac{48}{60}=8\frac45\) hrs) = \(\frac{5x}{44}\) km/hr

Speed downstream =\(\frac{x}4\) km/hr

∴ Speed of boat in still water = \(\frac12\bigg(\frac{5x}{44}+\frac{x}{4}\bigg)\)

= \(\frac{8x}{44} = \frac{2x}{11}\) km/hr

Speed of current = \(\frac12\bigg(\frac{x}{4}-\frac{5x}{44}\bigg)=\frac{3x}{44}\) km/hr

∴ Required ratio = \(\frac{2x}{11}:\frac{3x}{44}=8:3.\)

Total time taken from A to D = \(\frac{12}{x}+x+\frac{12}{2x}+2x+\frac{12}{4x}\)

Given, \(\frac{21}{x}+3x=16\)

⇒ 21 + 3x² – 16x = 0 ⇒ 3x² – 16x + 21 = 0 ⇒ 3x² – 9x – 7x + 21 = 0 

⇒3x(x – 3) – 7(x – 3) = 0 ⇒ (x – 3) (3x – 7) = 0 

⇒ x = 3 or x = \(\frac73\) 

∴ x = 3 hrs or \(\frac73\) hrs.

(c) 990 km

Let the length of the journey be x km. Then,

\(\frac{x}{45}-\frac{x}{60}=\frac{11}{2}\) ⇒ \(\frac{4x-3x}{180}=\frac{11}{2}\) ⇒ \(\frac{x}{180}= \frac{11}{2}\)

⇒ x = \(\frac{11\times180}{2}\) = 990 km.

Time taken to cover AC = \(\frac{3x}{5\times3a}=\frac{x}{5a}hr\)

Time taken to cover CB = \(\frac{2x}{5\times2b}=\frac{x}{5b}hr\)

Time taken to cover BA and back AB = \(\frac{2x}{5c}\)

Given, \(\frac{x}{5a}+\frac{x}{5b}=\frac{2x}{5c}\) ⇒ \(\frac1a+\frac1b=\frac2c\)

(a) 25.2 km/hr

Let the length of the train be x metres. Then, Speed of the train in passing through the platform = \(\frac{x+84}{21}\) m/sec and speed of the train in passing the man = \(\frac{x}{9}\) m/sec

Since both the speeds are the same,

\(\frac{x+84}{21}\) = \(\frac{x}{9}\) ⇒ 9\(x\) + 756 = 21\(x\)

⇒ 12\(x\) = 756 

⇒ \(x\) = \(\frac{756}{12}\) = 63 m

∴ Speed of the train = \(\frac{(63+84)}{21}\) m/sec = 7 m/sec

= \(7\times\frac{18}{5}\) km/hr = 25.2 km/hr.

(c)  6 hrs 50 min

Let the speed of the motorboat be 36x km/hr and that of the current of water be 5x km/hr. 

Speed downstream = (36x + 5x) km/hr = 41x km/hr 

Speed upstream = (36x – 5x) km/hr = 31x km/hr 

Time taken to travel downstream = 5 hrs 10 min

= \(5\frac{10}{60}\) hrs = \(\frac{31}{6}\) hrs

∴ Distance covered downstream = \(\big(41x\times\frac{31}6\big)\) km

⇒ Distance covered upstream =   \(\big(41x\times\frac{31}6\big)\) km

∴ Time taken to travel upstream = \(\big(41x\times\frac{31}6\times\frac1{31x}\big)\) hrs

= \(\frac{41}{6}\) hrs = \(6\frac56\) hrs = 6 hrs 50 min.

(c) 200 metres

Let the length of the train be x metres.

Then, \(\frac{800+x}{100} = \frac{400+x}{60}\)

⇒ 2400 + 3\(x\) = 2000 + 5\(x\)

⇒ \(x\) = 200 metres.

The direction against the stream is called upstream.

Let the speed of the second train be x km/hr. Then, 

Relative speed of trains = (\(x\) + 50) km/hr

= (\(x\) + 50) x \(\frac{5}{18}\) m/s

Total time taken = \(\frac{\text{Sum of lengths of both the trains}}{\text{Relative speed}}\)

= \(\frac{108\,m\,+112\,m}{(x\,+50)\times\frac{5}{18}}\)

⇒ \(\frac{220\times18}{(x\,+50)\times5}\) = 6 

⇒ 44 × 3 = x + 50 ⇒ 132 = x + 50 ⇒ x = 132 – 50 = 82 km/hr.

(c) 5 hrs

Time taken to travel 50 km = \(\frac{50}{12.5} hrs =4\,hrs.\)

Number of stoppages = \(\frac{50}{12.5}-1=3\)

∴ Total duration of stoppages = (3 × 20) min = 60 min = 1 hrs 

∴ Total time to complete the whole distance = 5 hrs.

(d) 25 seconds

Since the trains are moving in the same direction the relative speed of the faster train w.r.t. the slower train = Difference of their speeds

= (68 – 50) km/hr = 18 km/hr = \(\bigg(18\times\frac{5}{18}\bigg)\) m/sec

= 5 m/sec

Distance travelled = Sum of the lengths of the two trains = (50 + 75) m = 125 m

∴ Required time = \(\frac{125}{5}\) seconds = 25 seconds.

(a) 85 km/hr

Let the speed of the first train be x km/hr. Then, 

Relative speed of first train w.r.t. second train

= (\(x\) - 30) km/hr =   (\(x\) - 30) x \(\frac{5}{18}\) m/sec

Total distance travelled = (66 + 88) m = 154 m

Time taken = 0.168 min = (0.168 × 60) sec = 10.08 sec

∴  (\(x\) - 30) x \(\frac{5}{18}\) x 10.08 = 154

⇒ \(x\) - 30 = \(\frac{154\times18}{5\times10.08}\) ⇒ 55 

⇒ \(x\) = 85 km/hr.

(c) 120 km/hr

Let the speed of the car be x km/hr 

Then, speed of the train = \(\frac{150x}{100}\) km/hr = \(\frac{3x}{2}\) km/hr

Time taken by car to reach point B = \(\frac{75}{x}\) hrs

Time taken by train to reach point B = \(\frac{75}{\frac{3}{2}x}\) hrs

Given, \(\frac{75}{x}\) - \(\frac{75}{\frac{3}{2}x}\) = \(\frac{12.5}{60}\)

⇒ \(\frac{75}{x}\) - \(\frac{50}{x}\) = \(\frac{125}{10\times60}\) = \(\frac{5}{24}\) ⇒ \(\frac{25}{x}\) = \(\frac{5}{24}\)

⇒ x = \(\frac{25\times24}{5}\) = 120 km/hr.

(c) 2 km/hr

Speed downstream = \(\big(\frac16\times60\big)\) = 10 km/hr 

Speed upstream = \(\big(\frac1{10}\times60\big)\) = 6 km/hr 

∴ Speed of the current = \(\frac12\) (10 – 6) km/hr = 2 km/hr.

In water, the direction along the stream is called downstream but the direction against the stream is called upstream.

(b) 700 m

Let the length of the first train be x metres. 

Then, length of the second train = \(\frac{x}{2}\) metres 

Relative speed = (36 + 54) km/hr = 90 km/h

= \(\big(90\times\frac5{18}\big)\) m/s = 25 m/s

∴ \(\frac{x+\frac{x}2}{25}=12\)  ⇒ \(\frac{3x}{2} = 300 \)  ⇒ \(x\) = 200.

∴ Length of the first train = 200 m. 

Let the length of the platform be y metres. 

Speed of the first train = \(\big(36\times\frac5{18}\big)\) m/s = 10 m/s

∴ (200 + y) x \(\frac1{10}\) = 90

⇒ 200 + y = 900 ⇒ y = 700 m.

(b) 60 km/hr

Let the speed of the second train = x km/hr 

Since both the trains are running in opposite directions. Their relative speed = (30 + \(x\)) km/hr = (30 + \(x\)) x \(\frac{5}{18}\) m/sec

∴ Time taken to pass another train = \(\frac{\text{Sum of the lengths of both the trains in metres}}{\text{Relative speed in m/sec}}\)

⇒ 10 = \(\frac{(150+100)}{(30+x)\times\frac{5}{18}}\)

⇒ (30 + \(x\)) = \(\frac{250\times18}{10\times5}\) = 90

⇒ \(x\) = (90 – 30) km/hr = 60 km/hr.

(d) 22.92 m (approx) abd 22.5 km/hr

Let the length of the train be l m and its speed be x km/hr. 

Then, relative speed of train w.r.t.1st man 

= (x – 6) km/hr = (x - 6) x \(\frac{5}{18}\) m/s

Relative speed of train w.r.t. 2nd man 

= (x – 7.5) km/hr = (x - 7.5) x \(\frac{5}{18}\) m/s

Length of the train = Distance travelled in both the cases

⇒  (x - 6) x \(\frac{5}{18}\) x 5 = (x - 7.5) x \(\frac{5}{18}\) x 5.5

⇒  (x - 6) x 5 = (x - 7.5) x 5.5

⇒ 5x – 30 = 5.5x – 41.25 ⇒ 0.5x = 11.25

⇒ \(x\) = \(\frac{11.25}{0.5}\) = 22.5 km/hr

∴ Length of the train = \(\bigg((22.5-6)\times\frac{5}{18}\times5\bigg)\) m

= 22.92 m (approx)

Let the distance covered on the bicycle was x km. Then,

\(\frac{x}{10}+\frac{35}{60}+x=6\,hrs\) ⇒ \(\frac{6x+35+60x}{60}\) = 6 ⇒ 66x + 35 = 360

⇒ 66x = 325 ⇒ x = \(\frac{325}{66}\) km = \(4\frac{61}{66}\) km.

(a) 11.9 m

Ratio of distances covered by A and B = 800 : 760 = 20000 : 190000 

Ratio of distances covered by B and C = 500 : 495 = 190000 : 18810 

∴ Ratio of distances covered by A and C = 20000 : 18810 = 200 : 188.1 

Hence A will beat C in a 200 m race by 200 m – 188.1 m = 11.9 m.

Speed upstream = \(\frac{13}{5}\) km/hr = 2.6 km/hr 

Speed downstream = \(\frac{28}{5}\) km/hr = 5.6 km/hr 

∴ Velocity of current = \(\frac12(5.6-2.6)\) km/hr 

= \(\big(\frac12\times3\big)\) km/hr = 1.5 km/hr.

(c) 7.2 sec

Speed of the train = 50 km/hr = \(50\times\frac{5}{18}\) m/sec

= \(\frac{125}{9}\) m/sec

Distance travelled = Length of train = 100 m

∴ Time taken to cross the pillar = \(\frac{100}{\frac{125}{9}}\) seconds

= \(\frac{36}{5}\) seconds = 7.2 sec.

(b) 105 min

Speed of the aeroplane in still air = 320 km/hr Speed of wind = 40 km/hr 

∴ Aeroplane will travel with the wind at (320 + 40) = 360 km/hr and Aeroplane will travel against the wind at (320 – 40) = 280 km/hr 

Suppose the distance to be travelled = x km Then,

\(\frac{x}{280}=\frac{135}{60}\) hrs = \(\frac94\) hrs

⇒ x = \(\frac{280\times9}{4}\) = 630 km

∴ Time taken to cover a distance of 630 km at 360 km/hr is = \(\bigg(\frac{630}{360}\times60\bigg)\) min= 105 min.

(d) 2 km/hr

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

Speed of boat in still water = 10 km/hr 

∴ Speed of boat downstream = (10 + x) km/hr 

Speed of boat upstream = (10 – x) km/hr

Given, \(\frac{36}{(10-x)}-\frac{36}{(10+x)}=\frac{90}{60}\)

⇒ \(\frac{36(10+x)-36(10-x)}{(10-x)(10+x)}=\frac32\)

⇒ \(\frac{72x}{100-x^2}=\frac32\) ⇒ 144x = 300 – 3x²

⇒ 3x² + 144x – 300 = 0 

⇒ 3x² + 150x – 6x – 300 = 0 

⇒ 3x(x + 50) –6(x + 50) = 0 

⇒ (3x – 6) (x + 50) = 0 

⇒ x = 2 or –50. 

Since speed is not negative, x = 2 km/hr.

(c) 1 hr. 15 min

Speed of the boatman upstream = \(\frac21\) km/hr = 2 km/hr 

Speed of the boatman downstream = \(\frac{1}{\frac{10}{60}}\) km/hr = 6 km/hr

∴ Speed of boatman in stationary water = \(\frac12(6+2)\) km/hr = 4 km/hr

∴ Time taken to cover 5 km in stationary water = \(\frac54\) hrs = \(1\frac14\) hrs = 1 hr. 15 min.

(c) 2.4 km

Let the distance of the destination be x km. 

Speed of boat downstream = (5 + 1) km/hr = 6 km/hr 

Speed of boat upstream = (5 – 1) km/hr = 4 km/hr

Then, \(\frac{x}6+\frac{x}4=1\) ⇒ \(\frac{2x+3x}{12}=1\)

⇒ x = \(\frac{12}5\) = 2.4 km.

Speed downstream = 11 km/hr 

Speed upstream = 5 km/hr 

∴ Speed of boat in still water = \(\frac12\) (11 + 5) km/hr = 8 km/hr.

Ratio of the distances covered by A and B = 190 : 200 = 1710 : 1800 

Ratio of the distances covered by B and C = 180 : 200 = 1800 : 2000

⇒ Ratio of the distances covered by A and C = 1710 : 2000 = 171 : 200 

Hence C will give a start of (200 m – 171 m) = 29 m to A in the same race.

(d) 105 metres

Let the length of train be l metres and its speed be x m/s.

Then, \(\frac{l}{x} = 1.5 ⇒l=1.5x\)              .....(i)

and \(\frac{l}{(x-10)}=1.75 ⇒ l = 1.75(x-10)\)

⇒ l = 1.75x – 17.5                ...(ii)

From (i) and (ii), 

1.5x = 1.75x – 17.5 ⇒ 0.25x = 17.5

⇒ x = \(\frac{17.5}{0.25}\) = 70 m/s

∴ Length of the train = (1.5 × 70) metres = 105 metres.

(d)   \(1\frac12\) km/hr

Speed of the flow of water = 2 km/hr 

Speed of the boat downstream = \(3\frac12\) km/hr 

Let the speed of the boat in still water = \(x\) km/hr 

Then, \(x\) + 2 = \(3\frac12\) ⇒ \(x\) = \(1\frac12\) km/hr.

(c) 2hrs

Let the distance to be covered be x km at y km/hr.

∴ \(\frac{x}{y}=\frac{x}{\frac23y}-1\) ⇒ \(\frac{x}{y}-\frac{3x}{2y}=-1\) ⇒ \(\frac{2x-3y}{2y}=-1\)

⇒ –x = – 2y ⇒ x = 2y. 

∴ With his usual speed the time taken to cover a distance x km. = \(\frac{x}{y}=\frac{2y}{y}\) hrs = 2hrs.

(d) 250 m

Since the man is walking in the opposite direction of the moving train, 

Relative speed of the train = (45 + 5) km/hr

= \(\bigg(50\times\frac{5}{18}\bigg)\) m/sec = \(\frac{250}{18}\) m/sec

Length of the train = Relative speed x Time taken in crossing the man

= \(\bigg(\frac{250}{18}\times18\bigg)\) m = 250 m.

(a) 3 : 2

Let the speed of A is x km/hr and speed of B is y km/hr. 

Let them meet after t hours. Then, 

xT + yT = Total distance 

After the meeting, distance left for A = yT and the distance left for B = xT 

Now, yT = 4x and xT = 9y

∴ \(\frac{yT}{xT} = \frac{4x}{9y}⇒\frac{x^2}{y^2}=\frac94⇒\frac{x}{y}=\frac32\) ⇒ x : y = 3 : 2

 (c) 25 km/hr

Distance covered by the first person in 10 seconds = \(\bigg[\bigg(3\times\frac{5}{18}\bigg)\times10\bigg]\) m = \(\frac{25}{3}\) m 

Distance covered by the second person in 11 seconds =  \(\bigg[\bigg(5\times\frac{5}{18}\bigg)\times11\bigg]\) m = \(\frac{275}{18}\) m 

∴ The train travels a distance = \(\frac{275}{18} - \frac{25}{3}= \frac{128}{8}\) m in (11 – 10) = 1 second.

∴ Speed of the train = \(\bigg(\frac{125}{18}\times\frac{18}{5}\bigg)\) km/hr = 25 km/hr.

(c) 240 km

Distance covered by goods train before express train leaves = 40 × 2 = 80 km 

Relative speed = (60 – 40) km/hr = 20 km/hr 

∴ Time taken to cross each other = \(\frac{80}{20}\) hrs = 4 hrs 

∴ Distance from Delhi where the express train is expected to overtake the goods train = (4 × 60) km = 240 km.

Let the original average speed of the cyclist be x km/hr.

Then, \(\frac{14}{(x-1)}-\frac{14}{x}=\frac13\) ⇒ \(\frac{14x-14(x-1)}{x(x-1)}\) = \(\frac13\)

⇒ \(\frac{14}{x^2-x}=\frac13\) ⇒ x² – x – 42 = 0 ⇒ x² – 7x + 6x – 42 = 0 

⇒ x(x – 7) + 6(x – 7) = 0 ⇒ (x – 7) (x + 6) = 0 ⇒ x = 7 or –6 

Since speed cannot be negative, x = 7 km/hr

(a) 5 am on the next day

Distance covered by train A before train B leaves 

Ludhiana = 60 × 3 = 180 km 

Relative speed = (72 – 60) km/hr = 12 km/hr 

∴ Time taken to cross each other = \(\frac{180}{12}\) hrs = 15 hrs

∴ Required time = 2 pm + 15 hrs = 5 am on the next day.