Question -
Answer -
Solution:
Let the speed of therickshaw and the bus are x and y km/h, respectively.
Now, she has takentime to travel 2 km by rickshaw, t1 = (2/x) hr
Speed = distance/ time
she has taken time totravel remaining distance i.e., (14 – 2) = 12km
By bus t2 =(12/y) hr
By first condition,
t1 + t2 =½ = (2/x) + (12/y) … (i)
Now, she has takentime to travel 4 km by rickshaw, t3 = (4/x) hr
and she has taken timeto travel remaining distance i.e., (14 – 4) = 10km, by bus = t4 =(10/y) hr
By second condition,
t3 + t4 =½ + 9/60 = ½ + 3/20
(4/x) + (10/y) =(13/20) …(ii)
Let (1/x) = u and(1/y) = v
Then Equations. (i)and (ii) becomes
2u + 12v = ½ …(iii)
4u + 10v = 13/20…(iv)
[First,multiply Eq. (iii) by 2 and then subtract]
(4u + 24v) – (4u +10v) = 1–13/20
14v = 7/20
v = 1/40
Substituting the valueof v in Eq. (iii),
2u + 12(1/40) = ½
2u = 2/10
u = 1/10
x = 1/u = 10km/hr
y = 1/v = 40km/hr
Hence, the speed ofrickshaw = 10 km/h
And the speed of bus =40 km/h.