QUESTION:- A train covered a certain distance at a uniform speed. If the train have been 10km/h faster, it would have taken 2hrs less than the scheduled time. And, if the train was slower by 10km/h ,it would have taken3 hrs more than the scheduled time. Find the distance covered by the train.

Distance = speed × time = xt

Distance = speed × time = xtCondition 1: When the train would have been 10 km/h faster, it would have taken 2 hours less than the scheduled time.

Distance = speed × time = xtCondition 1: When the train would have been 10 km/h faster, it would have taken 2 hours less than the scheduled time.(x + 10)(t - 2) = xt

Distance = speed × time = xtCondition 1: When the train would have been 10 km/h faster, it would have taken 2 hours less than the scheduled time.(x + 10)(t - 2) = xtxt - 2x + 10t - 20 = xt

Distance = speed × time = xtCondition 1: When the train would have been 10 km/h faster, it would have taken 2 hours less than the scheduled time.(x + 10)(t - 2) = xtxt - 2x + 10t - 20 = xt- 2x + 10t = 20 ....(1)

Distance = speed × time = xtCondition 1: When the train would have been 10 km/h faster, it would have taken 2 hours less than the scheduled time.(x + 10)(t - 2) = xtxt - 2x + 10t - 20 = xt- 2x + 10t = 20 ....(1)Condition 2: When the train would have been slower by 10 km/h, it would have taken 3 hours more than the scheduled time.

Distance = speed × time = xtCondition 1: When the train would have been 10 km/h faster, it would have taken 2 hours less than the scheduled time.(x + 10)(t - 2) = xtxt - 2x + 10t - 20 = xt- 2x + 10t = 20 ....(1)Condition 2: When the train would have been slower by 10 km/h, it would have taken 3 hours more than the scheduled time.(x - 10)(t + 3) = xt

Distance = speed × time = xtCondition 1: When the train would have been 10 km/h faster, it would have taken 2 hours less than the scheduled time.(x + 10)(t - 2) = xtxt - 2x + 10t - 20 = xt- 2x + 10t = 20 ....(1)Condition 2: When the train would have been slower by 10 km/h, it would have taken 3 hours more than the scheduled time.(x - 10)(t + 3) = xtxt + 3x - 10t - 30 = xt

Distance = speed × time = xtCondition 1: When the train would have been 10 km/h faster, it would have taken 2 hours less than the scheduled time.(x + 10)(t - 2) = xtxt - 2x + 10t - 20 = xt- 2x + 10t = 20 ....(1)Condition 2: When the train would have been slower by 10 km/h, it would have taken 3 hours more than the scheduled time.(x - 10)(t + 3) = xtxt + 3x - 10t - 30 = xt3x -10t = 30 ....(2)

Distance = speed × time = xtCondition 1: When the train would have been 10 km/h faster, it would have taken 2 hours less than the scheduled time.(x + 10)(t - 2) = xtxt - 2x + 10t - 20 = xt- 2x + 10t = 20 ....(1)Condition 2: When the train would have been slower by 10 km/h, it would have taken 3 hours more than the scheduled time.(x - 10)(t + 3) = xtxt + 3x - 10t - 30 = xt3x -10t = 30 ....(2)Adding equations (1) and (2), we obtain

Distance = speed × time = xtCondition 1: When the train would have been 10 km/h faster, it would have taken 2 hours less than the scheduled time.(x + 10)(t - 2) = xtxt - 2x + 10t - 20 = xt- 2x + 10t = 20 ....(1)Condition 2: When the train would have been slower by 10 km/h, it would have taken 3 hours more than the scheduled time.(x - 10)(t + 3) = xtxt + 3x - 10t - 30 = xt3x -10t = 30 ....(2)Adding equations (1) and (2), we obtain- 2x + 10t + 3x -10t = 20 + 30

Distance = speed × time = xtCondition 1: When the train would have been 10 km/h faster, it would have taken 2 hours less than the scheduled time.(x + 10)(t - 2) = xtxt - 2x + 10t - 20 = xt- 2x + 10t = 20 ....(1)Condition 2: When the train would have been slower by 10 km/h, it would have taken 3 hours more than the scheduled time.(x - 10)(t + 3) = xtxt + 3x - 10t - 30 = xt3x -10t = 30 ....(2)Adding equations (1) and (2), we obtain- 2x + 10t + 3x -10t = 20 + 30x = 50

x = 50 in equation (1), we obtain- 2 × 50 + 10t = 20

x = 50 in equation (1), we obtain- 2 × 50 + 10t = 20- 100 + 10t = 20

x = 50 in equation (1), we obtain- 2 × 50 + 10t = 20- 100 + 10t = 2010t = 120

x = 50 in equation (1), we obtain- 2 × 50 + 10t = 20- 100 + 10t = 2010t = 120t = 120/10

x = 50 in equation (1), we obtain- 2 × 50 + 10t = 20- 100 + 10t = 2010t = 120t = 120/10t = 12

x = 50 in equation (1), we obtain- 2 × 50 + 10t = 20- 100 + 10t = 2010t = 120t = 120/10t = 12Therefore, distance = xt = 50 × 12 = 600

x = 50 in equation (1), we obtain- 2 × 50 + 10t = 20- 100 + 10t = 2010t = 120t = 120/10t = 12Therefore, distance = xt = 50 × 12 = 600Hence, the distance covered by the train is 600 km.

yes   because as the functions are graphed they eventually will intersect unless otherwise

step-by-step explanation: