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?

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.