Speed, Time and Distance
- A man decides to travel 80 km in 8 h party by foot and party on a bicycle. If his speed on foot is 8 km/h and on bicycle 16 km/h, what distance would he travel on foot ?
-
View Hint View Answer Discuss in Forum
Here, S1 = 8 km/h
and S2 = 16 km/h
∴ d1 =s1 x t1 = 8t1 ...(i)
and d2 = s2 x t2 = 16t2 ...(ii)
We know that,
t1 + t2 = 8 ...(iii)
and
d1 + d2 = 80 (given) ...(iv)Correct Option: C
Here, S1 = 8 km/h
and S2 = 16 km/h
∴ d1 =s1 x t1 = 8t1 ...(i)
and d2 = s2 x t2 = 16t2 ...(ii)
We know that,
t1 + t2 = 8 ...(iii)
and
d1 + d2 = 80 (given) ...(iv)
From Eqs.(i) and (ii) put the value of d1 and d2 in Eq. (iv), we get
d1 + d2 = 80
8t1 + 16t2 = 80
⇒ 8t1 + 8t2 + 8t2 = 80
8(t1 + t2) + 8t2 = 80
From Eq (iii),
8 x 8 + 8t2 = 80
8t2 = 80 - 64 = 16
⇒ t2 = 16/8 = 2h
∴ t1 = 8 - 2 = 6 h
∴ Distance travelled by foot = d1 = 8 x 6 = 48 km.
- A car driver covers a distance between two cities at a speed of 60 km/h and on the return his speed is 40 km/h. He goes again from the 1st to the 2nd city at twice the original speed and returns at half the original return speed. Find his average speed for entire journey ?
-
View Hint View Answer Discuss in Forum
Required average speed = 4/(1/60 + 1/40 + 1/120 +1/20)
Correct Option: D
Required average speed = 4/(1/60 + 1/40 + 1/120 +1/20)
= (4 x 120)/12 = 40 km/h
