a person walks at 4km/hr for a particular duration t1 and 3km/hr for another duration t2 and covers a total distance of 36km. if he walks at 4km/hr for the duration t2 and at 3km/hr for the duration t1 then he covers only 34km. what willl be the time taken by him to cover the one of the legs?
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Case 1 :
d₁ = distance traveled at 4 km/hr in time t₁
distance traveled at 4 km/hr is given as
d₁ = 4 t₁
d₂ = distance traveled at 3 km/hr in time t₂
distance traveled at 3 km/hr is given as
d₂ = 3 t₂
total distance is given as
d₁ + d₂ = 36
4 t₁ + 3 t₂ = 36
t₁ = (36 - 3 t₂)/4 eq-1
Case 2 :
d₁ = distance traveled at 3 km/hr in time t₁
distance traveled at 3 km/hr is given as
d₁ = 3 t₁
d₂ = distance traveled at 4 km/hr in time t₂
distance traveled at 4 km/hr is given as
d₂ = 4 t₂
total distance is given as
d₁ + d₂ = 36
3 t₁ + 4 t₂ = 34
using eq-1
3( (36 - 3 t₂)/4) + 4 t₂ = 34
t₂ = 4 hr
using eq-1
t₁ = (36 - 3 (4))/4
t₁ = 6 hr
time taken to cover one leg is
t = t₁ + t₂ = 6 + 4 = 10 hr
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