Q46. Ram travels a certain distance at 3 km/hr and reaches 15 mins late. If he travelled at 4 km/hr, he reaches 15 mins earlier. Find his total travel distance.
Show Answer
$\Rightarrow$ Let total distance be = $x$ kms
$\Rightarrow$ Time taken to $x$ kms at 3 km/hr = $\dfrac{x}{3}$
$\Rightarrow$ Time taken to $x$ kms at 4 km/hr = $\dfrac{x}{4}$
$\because$ Travel time difference = $15 + 15 = 30$
$\Rightarrow$ Converting min. into hrs. = $\dfrac{30}{60} = \dfrac{1}{2}$
As such, we have the equation:
$\Rightarrow \dfrac{x}{3} - \dfrac{x}{4} = \dfrac{1}{2}$
$\Rightarrow 4x - 3x = 6 \text{(L.C.M of 3, 4 and 2 = 12)}$
$\therefore x = 6$
Hence, required total distance = $6$ kms Ans
$\Rightarrow$ Time taken to $x$ kms at 3 km/hr = $\dfrac{x}{3}$
$\Rightarrow$ Time taken to $x$ kms at 4 km/hr = $\dfrac{x}{4}$
$\because$ Travel time difference = $15 + 15 = 30$
$\Rightarrow$ Converting min. into hrs. = $\dfrac{30}{60} = \dfrac{1}{2}$
As such, we have the equation:
$\Rightarrow \dfrac{x}{3} - \dfrac{x}{4} = \dfrac{1}{2}$
$\Rightarrow 4x - 3x = 6 \text{(L.C.M of 3, 4 and 2 = 12)}$
$\therefore x = 6$
Hence, required total distance = $6$ kms Ans
Q47. A man cycles from A to B, a distance of 21 kms, in 1 hour 40 minutes. He covers first 13 kms at the speed of 15 km per hour and the of the distance at the speed of $x$ km per hour. Find $x$.
Show Answer
$\Rightarrow$ Given, distance = 21 kms
$\Rightarrow$ Given, time = 1 hr. 40 mins.
$\Rightarrow$ Converting mins into hrs:
$\Rightarrow 1 + \dfrac{40}{60} = \dfrac{60 + 40}{60} = \dfrac{100}{60} = \dfrac{10}{6} = \dfrac{5}{3}$
$\Rightarrow$ Given, distance covered at 15 km/hr = 13 kms
$\Rightarrow$ Time taken to cover 13 kms = $\dfrac{13}{15}$
$\Rightarrow$ Remaining distance = $21 - 13 = 8$ kms
$\Rightarrow$ and, remaining time = $\dfrac{5}{3} - \dfrac{13}{15}$
$\Rightarrow \dfrac{25 - 13}{15} = \dfrac{12}{15} = \dfrac{4}{5}$
$\therefore$ Speed for covering the remaining distance = $\dfrac{8}{\dfrac{4}{5}}$
$\Rightarrow \dfrac{8 \times 5}{4} = \dfrac{40}{4} = 10$ km/hr Ans
$\Rightarrow$ Given, time = 1 hr. 40 mins.
$\Rightarrow$ Converting mins into hrs:
$\Rightarrow 1 + \dfrac{40}{60} = \dfrac{60 + 40}{60} = \dfrac{100}{60} = \dfrac{10}{6} = \dfrac{5}{3}$
$\Rightarrow$ Given, distance covered at 15 km/hr = 13 kms
$\Rightarrow$ Time taken to cover 13 kms = $\dfrac{13}{15}$
$\Rightarrow$ Remaining distance = $21 - 13 = 8$ kms
$\Rightarrow$ and, remaining time = $\dfrac{5}{3} - \dfrac{13}{15}$
$\Rightarrow \dfrac{25 - 13}{15} = \dfrac{12}{15} = \dfrac{4}{5}$
$\therefore$ Speed for covering the remaining distance = $\dfrac{8}{\dfrac{4}{5}}$
$\Rightarrow \dfrac{8 \times 5}{4} = \dfrac{40}{4} = 10$ km/hr Ans
Q48. A distance of 550 km is covered in 10 hours, partly by car and partly by train. If the speed of the car is 50 km per hour and speed of the train is 60 km per hour, find the distance covered by car and train.
Show Answer
$\Rightarrow$ Given, distance = 550 kms
$\Rightarrow$ Given, time = 10 hrs
$\therefore$ Overall speed = $\dfrac{550}{10} = 55$ km/hr
$\Rightarrow$ Given, speed of car = 50 km/hr
$\Rightarrow$ Difference between overall speed and speed of car = $55 - 50 = 5$ km/hr
$\Rightarrow$ Distance covered by car = $50 \times 5 = 250$ kms Ans
$\Rightarrow$ Difference between overall speed and speed of train = $60 - 55 = 5$ km/hr
$\Rightarrow$ Distance covered by train = $60 \times 5 = 300$ kms Ans
$\Rightarrow$ Given, time = 10 hrs
$\therefore$ Overall speed = $\dfrac{550}{10} = 55$ km/hr
$\Rightarrow$ Given, speed of car = 50 km/hr
$\Rightarrow$ Difference between overall speed and speed of car = $55 - 50 = 5$ km/hr
$\Rightarrow$ Distance covered by car = $50 \times 5 = 250$ kms Ans
$\Rightarrow$ Difference between overall speed and speed of train = $60 - 55 = 5$ km/hr
$\Rightarrow$ Distance covered by train = $60 \times 5 = 300$ kms Ans
Q49. A man covers a distance of 15 km in 3 hours, partly by walking and partly by running. If he walks at 3km/hr and runs at 9 km/hr, find the distance he covers by running.
Show Answer
$\Rightarrow$ Given, distance = 15 kms
$\Rightarrow$ Given, time = 3 hrs
$\therefore$ Overall speed = $\dfrac{15}{3} = 5$ km/hr
$\Rightarrow$ Difference between overall speed and walking = $5 - 3 = 2$ km/hr
$\Rightarrow$ Distance covered by walk = $3 \times 2 = 6$ kms
$\therefore$ Distance covered by running = $15 - 6 = 9$ kms Ans
$\Rightarrow$ Given, time = 3 hrs
$\therefore$ Overall speed = $\dfrac{15}{3} = 5$ km/hr
$\Rightarrow$ Difference between overall speed and walking = $5 - 3 = 2$ km/hr
$\Rightarrow$ Distance covered by walk = $3 \times 2 = 6$ kms
$\therefore$ Distance covered by running = $15 - 6 = 9$ kms Ans
Q50. Two boys A and B are cycling at the rate of 10 km per hour and 6 km per hour respectively. Find the distance between them after 3 hours 30 minutes; if:
(i) They are cycling in the same direction.
(ii) They are cycling in the opposite directions.
Show Answer
$\Rightarrow$ Given, time = 3 hrs. 30 mins.
$\Rightarrow$ Converting time into minutes = $3 + \dfrac{30}{60}$
$\Rightarrow \dfrac{180 + 30}{60} = \dfrac{210}{60} = \dfrac{21}{6} = \dfrac{7}{2}$
(i) Relative speed = $10 - 6 = 4$ km/hr
$\therefore$ Distance between 2 cyclist = $\dfrac{7}{2} \times 4 = 7 \times 2 = 14$ kms Ans
(ii) Relative speed = $10 + 6 = 16$ km/hr
$\therefore$ Distance between 2 cyclist = $\dfrac{7}{2} \times 16 = 7 \times 8 = 56$ kms Ans
$\Rightarrow$ Converting time into minutes = $3 + \dfrac{30}{60}$
$\Rightarrow \dfrac{180 + 30}{60} = \dfrac{210}{60} = \dfrac{21}{6} = \dfrac{7}{2}$
(i) Relative speed = $10 - 6 = 4$ km/hr
$\therefore$ Distance between 2 cyclist = $\dfrac{7}{2} \times 4 = 7 \times 2 = 14$ kms Ans
(ii) Relative speed = $10 + 6 = 16$ km/hr
$\therefore$ Distance between 2 cyclist = $\dfrac{7}{2} \times 16 = 7 \times 8 = 56$ kms Ans
Q51. A train 120 metres long, travelling at 90 km/hr overtakes another train travelling in the same direction at 72 km/hr and passes it completely in 50 seconds. Find:
(i) The lenght of the second train.
(ii) The time they would have taken to pass one another, if they had been travelling at these speeds in opposite directions.
Show Answer
(i) Let lenght of second train be = $x$ meter
$\Rightarrow$ Given, lenght of first train be = 120 meter
$\Rightarrow$ Time taken to pass one another = 50 seconds
$\Rightarrow$ Relative speed = $90 - 72 = 18$ km/hr
$\Rightarrow$ Converting speed to m/sec = $18 \times \dfrac{5}{18} = 5$ m/sec
So, we have the equation:
$\Rightarrow \dfrac{120 + x}{5} = 50$
$\Rightarrow 120 + x = 50 \times 5$
$\Rightarrow 120 + x = 250$
$\Rightarrow x = 250 - 120 = 130$
$\therefore$ lenght of second train = 130 meters Ans
(ii) Relative speed = $90 + 72 = 162$
$\Rightarrow$ Converting speed to m/sec = $162 \times \dfrac{5}{18} = 9 \times 5 = 45$ m/sec
$\Rightarrow$ Distance = $130 + 120 = 250$ meters
$\therefore$ Time taken to pass one another = $\dfrac{250}{45} = \dfrac{50}{9} = 5\dfrac{5}{9}$ seconds Ans
$\Rightarrow$ Given, lenght of first train be = 120 meter
$\Rightarrow$ Time taken to pass one another = 50 seconds
$\Rightarrow$ Relative speed = $90 - 72 = 18$ km/hr
$\Rightarrow$ Converting speed to m/sec = $18 \times \dfrac{5}{18} = 5$ m/sec
So, we have the equation:
$\Rightarrow \dfrac{120 + x}{5} = 50$
$\Rightarrow 120 + x = 50 \times 5$
$\Rightarrow 120 + x = 250$
$\Rightarrow x = 250 - 120 = 130$
$\therefore$ lenght of second train = 130 meters Ans
(ii) Relative speed = $90 + 72 = 162$
$\Rightarrow$ Converting speed to m/sec = $162 \times \dfrac{5}{18} = 9 \times 5 = 45$ m/sec
$\Rightarrow$ Distance = $130 + 120 = 250$ meters
$\therefore$ Time taken to pass one another = $\dfrac{250}{45} = \dfrac{50}{9} = 5\dfrac{5}{9}$ seconds Ans
Q52. The driver of a car, driving at the speed of 60 km/hr, locates a truck 60 metres ahead him. After 30 seconds the truck is 90 metres behind. Find the speed of the truck, if both are travelling in the same direction?
Show Answer
$\Rightarrow$ Given speed of car = 60 km/hr
$\Rightarrow$ Converting speed to m/sec
$\Rightarrow 60 \times \dfrac{5}{18} = 30 \times \dfrac{5}{9} = \dfrac{10 \times 5}{3} = \dfrac{50}{3}$ m/sec
$\Rightarrow$ Distance covered by in 30 seconds = $30 \times \dfrac{50}{3} = 10 \times 50 = 500$ meters
$\Rightarrow$ Let initial position of car be = A
$\Rightarrow$ Let final position of car be = D
$\Rightarrow$ Let initial position of truck be = B
$\Rightarrow$ Let final position of truck be = C
As such, we have the following distances:
$\Rightarrow$ AD = 500 meters
$\Rightarrow$ BC = ? meters
$\Rightarrow$ AB = 60 meters
$\Rightarrow$ CD = 90 meters
So, we have the equation:
$\Rightarrow AD = AB + BC + CD$
$\therefore BC = AD - AB - CD$
$\Rightarrow BC = 500 - 60 - 90 = 350$ meters
$\Rightarrow$ Now, speed of truck = $\dfrac{350}{30} = \dfrac{35}{3}$
$\therefore$ Converting speed to km/hr:
$\Rightarrow \dfrac{35}{3} \times \dfrac{18}{5} = 7 \times 6 = 42$ km/hr Ans
$\Rightarrow$ Converting speed to m/sec
$\Rightarrow 60 \times \dfrac{5}{18} = 30 \times \dfrac{5}{9} = \dfrac{10 \times 5}{3} = \dfrac{50}{3}$ m/sec
$\Rightarrow$ Distance covered by in 30 seconds = $30 \times \dfrac{50}{3} = 10 \times 50 = 500$ meters
$\Rightarrow$ Let initial position of car be = A
$\Rightarrow$ Let final position of car be = D
$\Rightarrow$ Let initial position of truck be = B
$\Rightarrow$ Let final position of truck be = C
As such, we have the following distances:
$\Rightarrow$ AD = 500 meters
$\Rightarrow$ BC = ? meters
$\Rightarrow$ AB = 60 meters
$\Rightarrow$ CD = 90 meters
So, we have the equation:
$\Rightarrow AD = AB + BC + CD$
$\therefore BC = AD - AB - CD$
$\Rightarrow BC = 500 - 60 - 90 = 350$ meters
$\Rightarrow$ Now, speed of truck = $\dfrac{350}{30} = \dfrac{35}{3}$
$\therefore$ Converting speed to km/hr:
$\Rightarrow \dfrac{35}{3} \times \dfrac{18}{5} = 7 \times 6 = 42$ km/hr Ans
Q53. A train 100 metres long meets a man going in opposite direction at the rate of 5 km/hr and passes him in 7.2 seconds. At what rate is the train going?
Show Answer
$\Rightarrow$ Given lenght of train = 100 metres
$\Rightarrow$ Time taken by train to pass the man = 7.2 seconds
$\therefore$ Speed of train relative to man:
$\Rightarrow \dfrac{100}{7.2} = \dfrac{100 \times 10}{72} = \dfrac{100 \times 5}{36} = \dfrac{50 \times 5}{18} = \dfrac{25 \times 5}{9} = \dfrac{125}{9}$
$\Rightarrow$ Converting speed to km/hr:
$\Rightarrow \dfrac{125}{9} \times \dfrac{18}{5} = 25 \times 2 = 50$ km/hr
As such, we have the equation:
$\Rightarrow x + 5 = 50$
$\Rightarrow x = 50 - 5 = 45$ km/hr Ans
$\Rightarrow$ Time taken by train to pass the man = 7.2 seconds
$\therefore$ Speed of train relative to man:
$\Rightarrow \dfrac{100}{7.2} = \dfrac{100 \times 10}{72} = \dfrac{100 \times 5}{36} = \dfrac{50 \times 5}{18} = \dfrac{25 \times 5}{9} = \dfrac{125}{9}$
$\Rightarrow$ Converting speed to km/hr:
$\Rightarrow \dfrac{125}{9} \times \dfrac{18}{5} = 25 \times 2 = 50$ km/hr
As such, we have the equation:
$\Rightarrow x + 5 = 50$
$\Rightarrow x = 50 - 5 = 45$ km/hr Ans
Q54. A train 130 meters in lenght and travelling at 45 km/hr crosses a bridge in 30 seconds. Find lenght of the bridge?
Show Answer
$\Rightarrow$ Given, lenght of train = 130 meters
$\Rightarrow$ Speed of train = 45 km/hr
$\Rightarrow$ Converting speed to m/sec = $45 \times \dfrac{5}{18} = \dfrac{25}{2}$
$\Rightarrow$ Time taken by train to pass the bridge = 30 seconds
$\Rightarrow$ Let lenght of bridge be = $x$ kms
So, we have the equation:
$\Rightarrow \dfrac{130 + x}{30} = \dfrac{25}{2}$
$\Rightarrow 2(130 + x) = 25 \times 30$
$\Rightarrow 260 + 2x = 750$
$\Rightarrow 2x = 750 - 260 = 490$
$\Rightarrow x = \dfrac{490}{2} = 245$
Hence, required lenght of bridge = 245 meters Ans
$\Rightarrow$ Speed of train = 45 km/hr
$\Rightarrow$ Converting speed to m/sec = $45 \times \dfrac{5}{18} = \dfrac{25}{2}$
$\Rightarrow$ Time taken by train to pass the bridge = 30 seconds
$\Rightarrow$ Let lenght of bridge be = $x$ kms
So, we have the equation:
$\Rightarrow \dfrac{130 + x}{30} = \dfrac{25}{2}$
$\Rightarrow 2(130 + x) = 25 \times 30$
$\Rightarrow 260 + 2x = 750$
$\Rightarrow 2x = 750 - 260 = 490$
$\Rightarrow x = \dfrac{490}{2} = 245$
Hence, required lenght of bridge = 245 meters Ans
Q55. Two trains running in opposite direction crosses a man standing on the platform in 27 seconds and 17 seconds respectively. They cross each other in 23 seconds. Find ratio of their speed.
Show Answer
$\Rightarrow$ Let speed of two trains be = $x$ and $y$
$\therefore$ Lenght of 1st train = $27x$
$\Rightarrow$ Lenght of 2nd train = $17y$
$\Rightarrow$ Relative speed = $x + y$
$\Rightarrow$ Given, time taken to cross each other = 23 seconds
So, we have the equation:
$\Rightarrow \dfrac{27x + 17y}{x + y} = 23$
$\Rightarrow 27x + 17y = 23(x + y)$
$\Rightarrow 27x + 17y = 23x + 23y$
$\Rightarrow 27x - 23x = 23y - 17y$
$\Rightarrow 4x = 6y$
$\Rightarrow \dfrac{y}{x} = \dfrac{6}{4} = \dfrac{3}{2}$
Hence, the required ratio = $3 : 2$ Ans
$\therefore$ Lenght of 1st train = $27x$
$\Rightarrow$ Lenght of 2nd train = $17y$
$\Rightarrow$ Relative speed = $x + y$
$\Rightarrow$ Given, time taken to cross each other = 23 seconds
So, we have the equation:
$\Rightarrow \dfrac{27x + 17y}{x + y} = 23$
$\Rightarrow 27x + 17y = 23(x + y)$
$\Rightarrow 27x + 17y = 23x + 23y$
$\Rightarrow 27x - 23x = 23y - 17y$
$\Rightarrow 4x = 6y$
$\Rightarrow \dfrac{y}{x} = \dfrac{6}{4} = \dfrac{3}{2}$
Hence, the required ratio = $3 : 2$ Ans
Q56. Two places P and Q are 324 kms apart. A man leaves P for Q and at the same time another man leaves Q for P. The two men meet at the end of 12 hrs. If the man from P to Q travels 8 km/hr faster than the other; find the speeds of the two men.
Show Answer
$\Rightarrow$ Given, total distance = 324 kms
$\Rightarrow$ Time travelled by both men = 12 hrs.
$\Rightarrow$ Let speed of both men be = $x$ km/hr
$\Rightarrow$ Given, difference in speed of both men = 8 km/hr
$\because$ Distance travelled by 1st man = $12(x + 8) = 12x + 96$
$\because$ And, distance travelled by 2nd man = $12x$
We have the following equation:
$\Rightarrow 12x + 96 + 12x = 324$
$\Rightarrow 24x + 96 = 324$
$\Rightarrow 24x = 324 - 96 = 228$
$\Rightarrow x = \dfrac{228}{24}$
$\Rightarrow \dfrac{228}{24} = \dfrac{114}{12} = \dfrac{57}{6} = \dfrac{19}{2} = 9.5$ km/hr
$\therefore$ Speed of 2nd man = $9.5$ km/hr Ans
$\Rightarrow$ And, speed of 1st man = $9.5 + 8 = 17.5$ km/hr Ans
$\Rightarrow$ Time travelled by both men = 12 hrs.
$\Rightarrow$ Let speed of both men be = $x$ km/hr
$\Rightarrow$ Given, difference in speed of both men = 8 km/hr
$\because$ Distance travelled by 1st man = $12(x + 8) = 12x + 96$
$\because$ And, distance travelled by 2nd man = $12x$
We have the following equation:
$\Rightarrow 12x + 96 + 12x = 324$
$\Rightarrow 24x + 96 = 324$
$\Rightarrow 24x = 324 - 96 = 228$
$\Rightarrow x = \dfrac{228}{24}$
$\Rightarrow \dfrac{228}{24} = \dfrac{114}{12} = \dfrac{57}{6} = \dfrac{19}{2} = 9.5$ km/hr
$\therefore$ Speed of 2nd man = $9.5$ km/hr Ans
$\Rightarrow$ And, speed of 1st man = $9.5 + 8 = 17.5$ km/hr Ans
Q57. Two cyclist start at the same time from opposite ends of a course that is 45 miles long. One cyclist is riding at 14 miles/hr and the other is riding at 16 miles/hr. How after they begin will they meet?
Show Answer
$\Rightarrow$ Given, total time = 45 miles/hr
$\Rightarrow$ Speed of 1st cyclist = 14 miles/hr
$\Rightarrow$ Speed of 2nd cyclist = 16 miles/hr
$\Rightarrow$ Let time taken by both cyclist be = $x$ hrs
$\because$ Distance covered by 1st cyclist = $14x$
$\Rightarrow$ And, distance covered by 2nd cyclist = $16x$
So, we have the following equation:
$\Rightarrow 14x + 16x = 45$
$\Rightarrow 30x = 45$
$\Rightarrow x = \dfrac{45}{30} = \dfrac{9}{6} = \dfrac{3}{2} = 1.5$
Hence, the two cyclist will meet after 1.5 hrs. Ans
$\Rightarrow$ Speed of 1st cyclist = 14 miles/hr
$\Rightarrow$ Speed of 2nd cyclist = 16 miles/hr
$\Rightarrow$ Let time taken by both cyclist be = $x$ hrs
$\because$ Distance covered by 1st cyclist = $14x$
$\Rightarrow$ And, distance covered by 2nd cyclist = $16x$
So, we have the following equation:
$\Rightarrow 14x + 16x = 45$
$\Rightarrow 30x = 45$
$\Rightarrow x = \dfrac{45}{30} = \dfrac{9}{6} = \dfrac{3}{2} = 1.5$
Hence, the two cyclist will meet after 1.5 hrs. Ans
Q58. A boat travels for 3 hrs. with a current speed of 3 miles/hr and then returns the same distance against the current in 4 hrs. What is the boats' speed in calm water? Find, how far did the boat travel one way?
Show Answer
$\Rightarrow$ Given, speed of current = 3 miles/hr
$\Rightarrow$ Let speed of boat be = $x$ kms
$\Rightarrow$ Time taken to travel downstream = 3 hrs.
$\therefore$ Speed of boat downstream = $3(x + 3) = 3x + 9$
$\Rightarrow$ Time taken to travel upstream = 4 hrs.
$\therefore$ Speed of boat upstream = $4(x - 3) = 4x - 12$
The distance being same, we have the following equation:
$\Rightarrow 3x + 9 = 4x - 12$
$\Rightarrow 3x - 4x = (-12) - 9$
$\Rightarrow - x = - 21$
$\Rightarrow x = 21$
$\Rightarrow$ Given, Speed of boat in calm water = 21 miles/hr
$\therefore$ Distance travelled by boat one way = $3(21 + 3)$
$\Rightarrow 3(21 + 3) = 63 + 9 = 72$ miles/hr Ans
$\Rightarrow$ Let speed of boat be = $x$ kms
$\Rightarrow$ Time taken to travel downstream = 3 hrs.
$\therefore$ Speed of boat downstream = $3(x + 3) = 3x + 9$
$\Rightarrow$ Time taken to travel upstream = 4 hrs.
$\therefore$ Speed of boat upstream = $4(x - 3) = 4x - 12$
The distance being same, we have the following equation:
$\Rightarrow 3x + 9 = 4x - 12$
$\Rightarrow 3x - 4x = (-12) - 9$
$\Rightarrow - x = - 21$
$\Rightarrow x = 21$
$\Rightarrow$ Given, Speed of boat in calm water = 21 miles/hr
$\therefore$ Distance travelled by boat one way = $3(21 + 3)$
$\Rightarrow 3(21 + 3) = 63 + 9 = 72$ miles/hr Ans
Q59. With the wind, an airplane travels 1120 miles in 7 hrs. Against the wind, it takes 8 hrs. Find the rate of speed of airplane in still air and the velocity of the wind?
Show Answer
$\Rightarrow$ Given, distance travelled by airplane = 1120 meters
$\Rightarrow$ Let speed of airplane be = $x$ km/hr
$\Rightarrow$ Let speed of wind be = $y$ km/hr
$\Rightarrow$ Given, time taken by airplane to travel with the wind = 7 hrs.
$\therefore$ Speed of airplane with the wind = $7(x + y)$
$\Rightarrow$ Given, time taken by airplane to travel against the wind = 8 hrs.
$\therefore$ Speed of airplane with the wind = $8(x - y)$
So, we have the equation:
$7(x + y) = 1120$......(i)
$8(x - y) = 1120$......(ii)
From (i) we have:
$\Rightarrow 7(x + y) = 1120$
$\Rightarrow (x + y) = \dfrac{1120}{7} = 160$
$\therefore$ Subtracting $y$ from either side we get:
$\Rightarrow x = 160 - y$
Upon substituting $x$ in equation (ii), we have:
$\Rightarrow 8[(160 - y) - y] = 1120$
$\Rightarrow 8[(160 - 2y] = 1120$
$\Rightarrow 1280 - 16y = 1120$
$\Rightarrow - 16y = 1120 - 1280$
$\Rightarrow - 16y = - 160$
$\Rightarrow y = \dfrac{160}{16} = 10$
$\therefore$ Hence, wind velocity = 10 miles/hr Ans
$\Rightarrow$ Now, speed of airplane in still air:
$\Rightarrow$ Speed of airplane with the wind $-$ Wind speed
$\Rightarrow 160 - 10 = 150$ miles/hr Ans
$\Rightarrow$ Let speed of airplane be = $x$ km/hr
$\Rightarrow$ Let speed of wind be = $y$ km/hr
$\Rightarrow$ Given, time taken by airplane to travel with the wind = 7 hrs.
$\therefore$ Speed of airplane with the wind = $7(x + y)$
$\Rightarrow$ Given, time taken by airplane to travel against the wind = 8 hrs.
$\therefore$ Speed of airplane with the wind = $8(x - y)$
So, we have the equation:
$7(x + y) = 1120$......(i)
$8(x - y) = 1120$......(ii)
From (i) we have:
$\Rightarrow 7(x + y) = 1120$
$\Rightarrow (x + y) = \dfrac{1120}{7} = 160$
$\therefore$ Subtracting $y$ from either side we get:
$\Rightarrow x = 160 - y$
Upon substituting $x$ in equation (ii), we have:
$\Rightarrow 8[(160 - y) - y] = 1120$
$\Rightarrow 8[(160 - 2y] = 1120$
$\Rightarrow 1280 - 16y = 1120$
$\Rightarrow - 16y = 1120 - 1280$
$\Rightarrow - 16y = - 160$
$\Rightarrow y = \dfrac{160}{16} = 10$
$\therefore$ Hence, wind velocity = 10 miles/hr Ans
$\Rightarrow$ Now, speed of airplane in still air:
$\Rightarrow$ Speed of airplane with the wind $-$ Wind speed
$\Rightarrow 160 - 10 = 150$ miles/hr Ans
Q60. A spike is hammered into a train rail. You are standing at the other end of the rail and you hear the sound of the hammer strike both through air and through rail itself. These sounds arrive at your point 6 seconds apart. You know that sound travels through air at 1100 feet/second and through steel at 16,500 feet/second. Calculate, how far is that spike (Round to one decimal place)?
Show Answer
$\because$ However long the sound takes to travel through air, it would take 6 seconds less to propogate through steel, since the speed of sound through steel is faster than air.
$\Rightarrow$ Let time taken by sound to travel through steel \& air be = $x$ hrs
$\therefore$ Distance travelled by sound through air = $1100 \times x = 1100x$ hrs
$\Rightarrow$ And, distance travelled by sound through steel = $16500(x - 6)$
$\because$ Distances are same, we have the equation:
$\Rightarrow 1100x = 16500(x - 6)$
$\Rightarrow 1100x = 16500x - 99000$
$\Rightarrow 1100x = 16500x - 99000$
$\Rightarrow 1100x - 16500x = - 99000$
$\Rightarrow - 15400x = - 99000$
$\Rightarrow x = \dfrac{99000}{15400}$
$\Rightarrow x = \dfrac{990}{154} = \dfrac{45}{7}$
$\Rightarrow$ Now distance = $1100 \times \dfrac{45}{7} = \dfrac{49500}{7} = 7071.42$
Hence the spike is at a distance of = 7071.42 feet Ans
$\Rightarrow$ Let time taken by sound to travel through steel \& air be = $x$ hrs
$\therefore$ Distance travelled by sound through air = $1100 \times x = 1100x$ hrs
$\Rightarrow$ And, distance travelled by sound through steel = $16500(x - 6)$
$\because$ Distances are same, we have the equation:
$\Rightarrow 1100x = 16500(x - 6)$
$\Rightarrow 1100x = 16500x - 99000$
$\Rightarrow 1100x = 16500x - 99000$
$\Rightarrow 1100x - 16500x = - 99000$
$\Rightarrow - 15400x = - 99000$
$\Rightarrow x = \dfrac{99000}{15400}$
$\Rightarrow x = \dfrac{990}{154} = \dfrac{45}{7}$
$\Rightarrow$ Now distance = $1100 \times \dfrac{45}{7} = \dfrac{49500}{7} = 7071.42$
Hence the spike is at a distance of = 7071.42 feet Ans
Q61. Walking at $\dfrac{3}{4}$ of his usual speed, a man is late is by $2\dfrac{1}{2}$ hrs. Find his usual time.
Show Answer
$\because$ The man walks $\dfrac{3}{4}$ of his usual speed, time taken is $\dfrac{4}{3}$ of his usual time. As such, we have the following equation:
$\Rightarrow (\dfrac{4}{3})$ of usual time = usual time $+ 2\dfrac{1}{2}$ hrs
$\Rightarrow (\dfrac{4}{3} - 1)$ of usual time = $\dfrac{5}{2}$ hrs
$\Rightarrow (\dfrac{1}{3})$ of usual time = $\dfrac{5}{2}$ hrs
$\therefore$ Usual time = $\dfrac{3}{1} \times \dfrac{5}{2} = \dfrac{15}{2} = 7.5$ hrs Ans
$\Rightarrow (\dfrac{4}{3})$ of usual time = usual time $+ 2\dfrac{1}{2}$ hrs
$\Rightarrow (\dfrac{4}{3} - 1)$ of usual time = $\dfrac{5}{2}$ hrs
$\Rightarrow (\dfrac{1}{3})$ of usual time = $\dfrac{5}{2}$ hrs
$\therefore$ Usual time = $\dfrac{3}{1} \times \dfrac{5}{2} = \dfrac{15}{2} = 7.5$ hrs Ans
Q62. Walking at $\dfrac{3}{5}$ of his usual speed, a man is late is by 40 minutes. Find his usual time.
Show Answer
$\because$ The man walks $\dfrac{3}{5}$ of his usual speed, time taken is $\dfrac{5}{3}$ of his usual time. As such, we have the following equation:
$\Rightarrow$ Converting minutes into hours = $\dfrac{40}{60} = \dfrac{2}{3}$
$\Rightarrow (\dfrac{5}{3})$ of usual time = usual time $+ \dfrac{2}{3}$ hrs
$\Rightarrow (\dfrac{5}{3} - 1)$ of usual time = $\dfrac{2}{3}$ hrs
$\Rightarrow (\dfrac{2}{3})$ of usual time = $\dfrac{2}{3}$ hrs
$\therefore$ Usual time = $\dfrac{3}{2} \times \dfrac{2}{3} = 1$ hour Ans
$\Rightarrow$ Converting minutes into hours = $\dfrac{40}{60} = \dfrac{2}{3}$
$\Rightarrow (\dfrac{5}{3})$ of usual time = usual time $+ \dfrac{2}{3}$ hrs
$\Rightarrow (\dfrac{5}{3} - 1)$ of usual time = $\dfrac{2}{3}$ hrs
$\Rightarrow (\dfrac{2}{3})$ of usual time = $\dfrac{2}{3}$ hrs
$\therefore$ Usual time = $\dfrac{3}{2} \times \dfrac{2}{3} = 1$ hour Ans
Q63. John has to reach a place 400 km away. he covers 160 km by bus at 40 km/hr, then travels by a fast train for the rest of the journey. However, if he had travelled by a fast train for the first 160 km and the remaining distance by bus; he would have taken an hour longer to complete the whole journey. Find the speed of fast train.
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$\Rightarrow$ Given, total distance = 400 kms
$\Rightarrow$ Time taken to complete first 160 kms by bus = $\dfrac{160}{40} = 4$ hrs
$\Rightarrow$ Time taken to complete first 160 kms by train = $4 + 1 = 5$ hrs
$\therefore$ Speed of train = $\dfrac{400}{5} = 80$ km/hr Ans
$\Rightarrow$ Time taken to complete first 160 kms by bus = $\dfrac{160}{40} = 4$ hrs
$\Rightarrow$ Time taken to complete first 160 kms by train = $4 + 1 = 5$ hrs
$\therefore$ Speed of train = $\dfrac{400}{5} = 80$ km/hr Ans