Tool By Zeal Point

Type 7: Late and Early

Q. If Reena walks at a speed of 8 km/h, she reaches a place 15 minutes late. If she walks at 12 km/h, she reaches the place 10 minutes early. Find the distance to the place.

Formula

If:

First speed = v_1,Second speed = v₂

Late time = L minutes, Early time = E minutes

Then,

$Distance =$ $v 1 \times v 2 \times (L+E) / 60(v 2 -v 1)$

Solution

Given:

v₁₌8 km/h, v₂=12 km/h

$L = 15 minutes, E = 10 minutes$

Therefore,

$Distance = 8\times12\times(15+10)/60(12-8)$

$= 96\times25/240$

$= 2400/240 =$ $10 km$

Ans: 10 km

Trick:
When one speed causes a delay and another causes an early arrival, use:

$D =$ v₁v₂(Late + Early) / 60(v₂ − v₁)

TYPE 7: Late & Early - Practice Set

Q1. A man travels at 6 km/h and reaches a place 20 minutes late. At 8 km/h, he reaches 10 minutes early. Find the distance.
(A) 4 km (B) 5 km (C) 6 km (D) 8 km

Q2. A student walks at 5 km/h and reaches school 12 minutes late. At 6 km/h, he reaches 8 minutes early. Find the distance.
(A) 2 km (B) 3 km (C) 4 km (D) 5 km

Q3. A person travels at 10 km/h and is 18 minutes late. At 15 km/h, he is 12 minutes early. Find the distance.
(A) 10 km (B) 12 km (C) 15 km (D) 18 km

Q4. If a person travels at 12 km/h, he reaches 15 minutes late. At 15 km/h, he reaches 9 minutes early. Find the distance.
(A) 18 km (B) 20 km (C) 24 km (D) 30 km

Q5. A boy travels at 4 km/h and reaches a destination 30 minutes late. If he travels at 5 km/h, he reaches 18 minutes early. Find the distance.
(A) 12 km (B) 14 km (C) 16 km (D) 18 km