Question

A man travels a total distance of 30 km. 2/3^rd of his total journey, he travelled by foot and remaining by bicycle. He travelled 165 min more time on foot as compared to bicycle. Had he travelled same equal distance on foot as well as on bicycle, average speed of his whole journey would have been \(6\frac{2}{{13}}\) km/h. What was the ratio of his speed?
1. 8 ∶ 5
2. 5 ∶ 7
3. 5 ∶ 8
4. 3 ∶ 2

Answer 1

Correct Answer - Option 3 : 5 ∶ 8

Given:

Total distance = 30 km

Total distance travelled on foot = 2/3^rd of 30 = 20 km,

Distance travelled by Bicycle = 10 km

He travelled 165 min more time on foot as compared to bicycle

Average speed, when he travelled equal distance in foot as well as by Bicycle = 80/13 km/h

Formula Used:

1.) Time = Distance/Speed

2.) When a person travels equal distance with speed V₁ and V₂, respectively. Average speed of whole journey becomes 2V₁V₂/ (V₁ + V₂)

Calculations:

Let speed of man on foot and by bicycle be V₁ and V₂ respectively

Time difference = 165 min = 165/60 h

20/V₁ – 10/V₂ = 165/60

(4/V₁ – 2/V₂) = 33/60 (divide by 5 on both sides)

V₁V₂ = (20/11) × (4V₂ – 2V₁)     ----(1)

Now, Average speed = 2V₁V₂/ (V₁ + V₂) = 80/13

⇒ 2V₁V₂/ (V₁ + V₂) = 80/13      ----(2)

By putting the value of V₁V₂ from equation (1) in (2)

⇒ 40(4V₂ – 2V₁)/11(V₁ + V₂) = 80/13

⇒ 13(4V₂ – 2V₁) = 22(V₁ + V₂)

⇒ 52V₂ – 26V₁ = 22V₁ + 22V₂

⇒ 30V₂ = 48V₁

⇒ V₁ ∶ V₂ = 5 ∶ 8

∴ Ratio of speed on foot and by bicycle is 5 ∶ 8