A train covered a certain distance at an uniform speed. If the train would have been 10 Km/h faster it would have taken 2 hours less than the scheduled time but if the train were 10 Km/h slower then it would have taken 3 hours more than the scheduled time. Find the distance covered by the Train.
#Just_For_Chill
Answers
Answer:Let the speed of the train be xkm/h and the time taken by train to travel the given distance be t hours and the distance to travel be dkm. We know that,
⇒Speed=
Time
Distance
⇒x=
t
d
∴d=xt.......(i)
Case 1
⇒(x+10)×(t−2)=d
⇒xt+10t−2x−20=d
⇒d+10t−2x−20=d
⇒$$− 2x + 10t = 20$$...... (ii)
Case 2
⇒(x−10)×(t+3)=d
⇒xt−10t+3x−30=d
⇒d−10t+3x−30=d
⇒$$3x − 10t = 30$$......... (iii)
Adding equations (ii) and (iii), we gets
⇒x=50
Substitute the value of x in (ii) we gets
$$\Rightarrow (−2) \times (50) + 10t = 20$$
$$\Rightarrow −100 + 10t = 20$$
⇒10t=120
⇒t=12 hours
Substitue the value of t and x in equation (i), we gets
Distance to travel =d=xt
⇒d=12×50=600Km
Hence, the distance covered by the train is 600km.
Let the time taken be t,
and speed be x.
i.e., distance = xt
In case 1
time = t1
time = xt / x + 10 (time = distance / speed)
xt / x + 10 = t - 2
xt = (t - 2 ) (x + 10)
xt = xt + 10t - 2x - 20
xt and xt get cancelled
10t - 2x - 20 = 0 ... equation 1
In case 2
t2 = xt / x- 10
xt / x - 10 = t + 3
xt = (t + 3) (x - 10)
xt = xt - 10t + 3x - 30
xt and xt get cancelled
-10t + 3x - 30 = 0 .... equation 2
now u can solve this by using any 1 method i.e, substitution, elimination or cross-multiplication.