Question

A man and a woman 63 miles apart from each other, start travelling toward each other at the same time, if the man covers 4 miles per hour and the woman 3 miles per hour. How far will the woman have travelled when they met?

Answer 1

Hint:
The term speed generally refers to the average speed of a journey and is defined as the ratio of distance covered to unit time.
Speed (S) = $\dfrac{{{\text{Distance (D)}}}}{{{\text{Time (T)}}}}$
Here, we will assume that after time ‘t’ they will meet. So, we can find out the distance travelled by both of them in time ‘t’ and the distance covered by both of them when they walk towards each other will be equal to total distance between them i.e. 63 miles

Complete step by step solution:
Let the time after they meet be t,
Speed of the man $ = 4\dfrac{{miles}}{{hour}}$
And speed of the woman $ = 3\dfrac{{miles}}{{hour}}$
∴In ‘t’ time distance travelled by man $ = 4 \times t = 4t{\text{ miles}}$
And in‘t’ time distance travelled by woman $ = 3 \times t = 3t{\text{ miles}}$
∴ Total travelled distance $ = 4t + 3t$
$ = 7t{\text{ miles}}$

But according to question
$\begin{gathered}
  7t = 63 \\
  \therefore t = \dfrac{{63}}{7} = 9hours \\
\end{gathered} $
Distance travelled by woman in 9 hours $ = 3 \times 9$
$ = 27miles$
Therefore, the woman had travelled 27 miles when they met.

Note:
 If two people travel from two points A and B towards each other, and they meet at point P. the total distance covered by them on the meeting will be AB. The time taken by both of them to meet will be the same. As the time is constant, distances AP and BP will be in the ratio of their speed.
If two people are walking towards each other from A and B, let’s say that the distance between A and B is d. Therefore, when they meet for the first time, they together cover a distance “d”. When they meet for the second time and third time, they together cover a distance “3d” and “5d” respectively.