A train covered a certain distance at a uniform speed. if the train would have been 6 km/h faster, it would have taken 4 hours lessthan the scheduled time. and, if the train were slower by 6 km/hr ; it would have taken 6 hours more than the scheduled time. find the length of the journey

Let the original speed be x kmph and let the time taken to complete the journey be y hours. ∴ Length of the whole journey = (xy) km Case I: When the speed is (x + 5) kmph and the time taken is (y – 3) hrs: Total journey = (x + 5) (y – 3) km ⇒ (x + 5) (y – 3) = xy ⇒ xy + 5y – 3x – 15 = xy ⇒ 5y – 3x = 15 ………(i) Case II: When the speed is (x – 4) kmph and the time taken is (y + 3) hrs: Total journey = (x – 4) (y + 3) km ⇒ (x – 4) (y + 3) = xy ⇒ xy – 4y + 3x – 12 = xy ⇒ 3x – 4y = 12 ………(ii) On adding (i) and (ii), we get: y = 27 On substituting y = 27 in (i), we get: 5 × 27 – 3x = 15 ⇒135 – 3x = 15 ⇒3x = 120 ⇒x = 40 ∴ Length of the journey = (xy) km = (40 × 27) km = 1080 km Categories • • (31.9k) • (8.8k) • (764k) • (248k) • (10.0k) • (5.6k) • (36.4k) • (7.5k) • (10.7k) • (11.8k) • (11.2k) • (6.8k) • (4.9k) • (5.3k) • (2.8k) • (19.9k) • (959) • (2.9k) • (5.2k) • (664) • (121k) • (72.1k) • (3.8k) • (19.6k) • (1.4k) • (14.2k) • (12.5k) • (9.3k) • (7.7k) • (3.9k) • (6.7k) • (63.8k) • (26.6k) • (23.7k) • (14.6k) • (25.7k) • (530) • (84) • (766) • (49.1k) • (63.8k) • (1.8k) • (59.3k) • (24.5k)

Important Important Important Important Important Important Important Important Important Important Important Important Important Important Important Important Important Important You are here Important Important Important Important Important Important Important Important Important Important Important A train covered a certain distance at a uniform speed. If the train would have been 6 km/h faster, it would have taken 4 hours lessthan the scheduled time. And, if the train were slower by 6 km/hr ; it would have taken 6 hours more than the scheduled time. Find the length of the journey. This question is Similar to Transcript Question 28 (Choice 1) A train covered a certain distance at a uniform speed. If the train would have been 6 km/h faster, it would have taken 4 hours less than the scheduled time. And, if the train were slower by 6 km/hr ; it would have taken 6 hours more than the scheduled time. Find the length of the journey. Let Speed of train = x km/h & Time taken = y hours. We know that, Speed = 𝐷𝑖𝑠𝑡𝑎𝑛𝑐𝑒/𝑇𝑖𝑚𝑒 Distance = Speed × Time Distance = xy If the train would have been 6 km/h faster I.e. Speed = x + 6 It would have taken 4 hours less i.e. Time = y − 4 Now, Distance = Speed × time Distance = (x + 6) (y − 4) Putting Distance = xy from Equation (1) xy = (x + 6) (y − 4) xy = x (y − 4) + 6 (y − 4) xy = xy − 4x + 6y − 24 4x − 6y + 24 = xy − xy 4x − 6y + 24 = 0 2(2x − 3y + 12) = 0 2x − 3y + 12 = 0 Also, If the train were slower by 6km/h Speed = x − 6, it would have t...

A train covered a certain distance at a uniform speed . If the train would have been 10 km / h faster , it would have taken 2 hours less than the scheduled time . And , if the train were slower by 10 km / h : it would taken 3 hours more than the scheduled time .Find the distance covered by the train .

Let the speed of the train be x km/hr. Let the time taken to travel certain distance be y hrs. We know that, speed × time = distance ∴ Distance = xy km According to the first condition, if the train would have been 6 km/h faster, it would have taken 4 hours less than the scheduled time. ∴ (x + 6)(y – 4) = xy ∴ xy – 4x + 6y – 24 = xy ∴– 4x + 6y – 24 = 0 ∴ 2x – 3y = –12......(i) According to the second condition, if the train was slower by 6 km/hr, it would have taken 6 hours more than the scheduled time. ∴ (x – 6)(y + 6) = xy ∴ xy + 6x – 6y – 36 = xy ∴ 6x – 6y – 36 = 0 ∴ x – y = 6......(ii) Multiplying both sides by 2, we get 2x – 2y = 12......(iii) Subtracting equation (iii) from (i), we get 2x – 3y = –12 2x – 2y = 12 – + – – y = – 24 ∴ y = 24 Substituting y = 24 in equation (ii), we get x – 24 = 6 ∴ x = 30 ∴ Length of the journey = xy = 30 × 24 = 720 km

A train covered a certain distance at a uniform speed. If the train would have been 10 km/h faster, it would have taken 2 hours less than the scheduled time. And, if the train were slower by 10 km/h,it would have taken 3 hours more than the scheduled time. Find the distance covered by the train. Speed is equal to the distance traveled divided by time. Answer: The distance covered by the train will be 600 km. Let's find the distance covered by the train. Explanation: Let's consider the time taken to be 't' Let the speed be 'x' and distance be 'd' Distance = Speed× Time By substituting in the above equation we get, d = xt ----------------- (1) Case 1: When speed is increased by 10 km/h and time is decreased by 2 hours. New speed = x + 10, New time = t - 2 Substituting the values we get, d = (x + 10)(t – 2) d = xt – 2x + 10t - 20 d = d – 2x + 10t - 20 [since, d = xt from equation (1)] 10t – 2x = 20 ---------------------- (2) Case 2: When speed is reduced by 10 km/h, time increases by 3 hours. New speed = x - 10, New time = t + 3 Substituting the values we get, d = (x – 10)(t + 3) d = xt + 3x -10t – 30 d = d + 3x -10t – 30 [since, d = xt from equation (1)] 3x -10t = 30 ------------------------- (3) Adding equation (2) and (3) 10t -2x + 3x -10t= 20 + 30 x = 50 To calculate the value of time, substitute the value of x in equation (2) 10t – (2 × 50) = 20 10t = 20 + 100 10t = 120 t = 12 hours Now, to calculate the distance, substitute the value of t and x in equation (1) d = xt d ...

| a train covered A certain distance at a uniform speed if the train would have been 10 km per hour faster it would have taken 2 hours less than the scheduled time and the train was lower by 10 km per hour it would have taken 3 hours more than the scheduled time find the distance covered by the train SOLUTION Let the speed of the train by x km/hr And the time taken by the train be ‘y’ hrs Distance traveled by the train is x.y = xy Case I If the train would have been 10 km/hr faster, it would have taken 2 hours less than the scheduled time. ⇒ (x + 10) (y - 2) = xy ⇒ -2x + 10y = 20 ⇒ x – 5y = -10 ---------------- (1) Case II If the train were slower by 10 km/hr, it would have taken 3 hours more than the scheduled time ⇒ (x - 10) (y + 3) = xy ⇒ 3x – 10y - 30 = 0 ⇒ 3x – 10y = 30 ----------------(2) Solving the equations (1) and (2), we get y = 12, x = 50 Therefore, the distance of the journey = 50 × 12 = 600 km.

A train, after travelling 70 km from a station A towards a station B , develops a fault in the engine at C and covers the remaining journey to B at 3/4 of its earlier speed and arrives at B 1 hour and 20 minutes late. If the fault had developed 35 km further on at D , it would have arrived 20 minutes sooner. Find the speed (in km / hr) of the train. A train, after travelling 7 0 km from a station A towards a station B, develops a fault in the engine at C and covers the remaining journey to B at 4 3 ​ of its earlier speed and arrives at B 1 hour and 2 0 minutes late. If the fault had developed 3 5 km further on at D, it would have arrived 2 0 minutes sooner. Find the speed (in km / hr) of the train. ∴ 3 y 2 1 0 + 4 x ​ = y 7 0 + x ​ + 1 + 3 1 ​ ⇒ 3 y 2 1 0 + 4 x ​ = 3 y 2 1 0 + 3 x + 3 y + y ​ ⇒ 2 1 0 + 4 x = 2 1 0 + 3 x + 4 y ⇒ x − 4 y = 0 ......... (i) Case II : Total time = y 7 0 + 3 5 ​ + 4 3 ​ y x − 3 5 ​ = y 7 0 + 3 5 ​ + 3 y 4 x − 1 4 0 ​ = 3 y 1 0 5 × 3 + 4 x − 1 4 0 ​ = 3 y 3 1 5 − 1 4 0 + 4 x ​ = 3 y 1 7 5 + 4 x ​ ∴ 3 y 1 7 5 + 4 x ​ = y 7 0 + x ​ + 1 ⇒ 3 y 1 7 5 + 4 x ​ = y 7 0 + x + y ​ ⇒ 1 7 5 + 4 x = 2 1 0 + 3 x + 3 y ⇒ 4 x − 3 x = 2 1 0 − 1 7 5 ⇒ x − 3 y = 4 5 ............. (ii) We have, (i) ⇒ x − 4 y = 0 (i) ⇒ x − 3 y = 4 5 x − 4 y = 0 x − 3 y = 0 ____________ − y = − 4 5 ⇒ y = 4 5 km/h ∴ Speed of train = 45 km/h After covering a distance of 30 km with a uniform speed there is some defect in a train engine and therefore, its speed is reduced to 4/5 of its or...