Understanding the Speed and Time After Meeting Problem

This problem involves two individuals, A and B, starting simultaneously from two different points, X and Y, and moving towards each other. They meet at a point on the way, and we are given the time each takes to reach the other's starting point after they meet. We are also given a relationship between their speeds. Our goal is to find the unknown time taken by one person after meeting.

Analyzing the Given Information

• A starts from X and B starts from Y.
• They start at the same time and move towards each other.
• Speed of A (\(v_A\)) is 20% more than the speed of B (\(v_B\)).
  • This means \(v_A = v_B + 0.20 v_B = 1.2 v_B\).
  • The ratio of their speeds is \(\frac{v_A}{v_B} = 1.2 = \frac{12}{10} = \frac{6}{5}\).
• After meeting at a point (say, M), A takes \(t_A = 2\frac{1}{2}\) hours to reach Y.
• After meeting at M, B takes \(t_B = x\) hours to reach X.
• We need to find the value of x.

Applying the Relevant Formula for Time After Meeting

For problems where two objects start simultaneously from opposite ends and move towards each other, meeting at a point, there is a specific relationship between their speeds and the time they take to reach the destination *after* meeting. If \(v_A\) and \(v_B\) are their speeds, and \(t_A\) and \(t_B\) are the times they take to reach the opposite starting points after meeting, the relationship is:

\[ \frac{v_A}{v_B} = \sqrt{\frac{t_B}{t_A}} \]

This formula is derived from the fact that the distance covered before meeting is proportional to speed and time to meet, and the distance covered after meeting is also related to speed and time after meeting. Since they start at the same time and meet at the same time, the time taken to meet is the same for both. Let this time be \(t\). The distance covered by A before meeting is \(v_A \times t\), which is equal to the distance B covers after meeting (\(v_B \times t_B\)). Similarly, the distance covered by B before meeting is \(v_B \times t\), which is equal to the distance A covers after meeting (\(v_A \times t_A\)). Thus, we have:

• \(v_A \times t = v_B \times t_B\)
• \(v_B \times t = v_A \times t_A\)

Dividing the first equation by the second gives:

\[ \frac{v_A \times t}{v_B \times t} = \frac{v_B \times t_B}{v_A \times t_A} \] \[ \frac{v_A}{v_B} = \frac{v_B}{v_A} \times \frac{t_B}{t_A} \] \[ \left(\frac{v_A}{v_B}\right)^2 = \frac{t_B}{t_A} \] \[ \frac{v_A}{v_B} = \sqrt{\frac{t_B}{t_A}} \] This confirms the formula we will use.

Calculation Steps to Find x

We are given:

• \(\frac{v_A}{v_B} = \frac{6}{5}\)
• \(t_A = 2\frac{1}{2}\) hours \( = \frac{5}{2}\) hours
• \(t_B = x\) hours

Substitute these values into the formula \(\frac{v_A}{v_B} = \sqrt{\frac{t_B}{t_A}}\):

\[ \frac{6}{5} = \sqrt{\frac{x}{\frac{5}{2}}} \]

To solve for x, first square both sides of the equation:

\[ \left(\frac{6}{5}\right)^2 = \frac{x}{\frac{5}{2}} \] \[ \frac{36}{25} = \frac{x}{\frac{5}{2}} \]

Now, multiply both sides by \(\frac{5}{2}\) to isolate x:

\[ x = \frac{36}{25} \times \frac{5}{2} \]

Perform the multiplication. We can simplify before multiplying:

\[ x = \frac{36^{\cancel{18}}}{25^{\cancel{5}}} \times \frac{\cancel{5}^{1}}{\cancel{2}^{1}} \] \[ x = \frac{18 \times 1}{5 \times 1} \] \[ x = \frac{18}{5} \]

The value of x is \(\frac{18}{5}\) hours. We can convert this improper fraction to a mixed number:

\[ \frac{18}{5} = 18 \div 5 \]

18 divided by 5 is 3 with a remainder of 3. So, \(\frac{18}{5} = 3\frac{3}{5}\).

Thus, B takes \(3\frac{3}{5}\) hours to reach X after meeting A.

Final Answer

The value of x is \(3\frac{3}{5}\) hours.

Revision Table: Speed, Time, and Distance Concepts

Additional Information on Meeting Point Problems

Problems involving two bodies moving towards each other and meeting are common in quantitative aptitude. Understanding the concept of relative speed and the time taken to meet is crucial. If two bodies start at the same time from distance D apart and move towards each other with speeds \(v_A\) and \(v_B\), their relative speed is \(v_A + v_B\). The time taken to meet is \(T_{meet} = \frac{D}{v_A + v_B}\).

In this specific problem, the crucial piece of information is the time taken *after* meeting. The derived formula \(\frac{v_A}{v_B} = \sqrt{\frac{t_B}{t_A}}\) is a direct consequence of the fact that the distance covered by A before meeting is the same as the distance covered by B after meeting, and vice versa, coupled with the constant speed of each person.

• Distance XM (covered by A before meeting) = \(v_A \times T_{meet}\)
• Distance MY (covered by B before meeting) = \(v_B \times T_{meet}\)
• Distance MY (covered by A after meeting) = \(v_A \times t_A\)
• Distance XM (covered by B after meeting) = \(v_B \times t_B\)

From these, we get \(v_B \times T_{meet} = v_A \times t_A\) and \(v_A \times T_{meet} = v_B \times t_B\). Solving these simultaneous equations for \(T_{meet}\) and eliminating it leads back to the square root formula used.