A person goes to a market between 4 p.m. and 5 p.m. When he comes back he finds that the hour hand and the minute hand of the clock have interchanged their positions. For how much time (approximately) was he out of his house?

Question

Prepp Team · Accepted Answer

Solving the Clock Hands Interchanged Problem

This is a classic clock problem where the hour hand and minute hand swap their positions between a departure time and a return time. The person leaves between 4 p.m. and 5 p.m. and returns to find the hands interchanged. For the hands to swap positions like this, if the departure is between 4 and 5, the return must be between 5 and 6.

Understanding Clock Hand Angles

To solve this, we need to calculate the angles of the hour and minute hands from the 12 o'clock position (measured clockwise). The speeds of the hands are constant:

The minute hand moves 360 degrees in 60 minutes, which is \(360/60 = 6\) degrees per minute.
The hour hand moves 360 degrees in 12 hours (720 minutes), which is \(360/720 = 0.5\) degrees per minute.

The angle of the minute hand \(M\) minutes past the hour is \(6M\) degrees.

The angle of the hour hand \(M\) minutes past hour \(H\) is \(30H + 0.5M\) degrees.

Setting Up Times and Angles

Let the departure time be 4 hours and \(M_1\) minutes past 4 p.m. (i.e., 4:\(M_1\)).

Hour hand angle at departure: \(A_{H,dep} = 30(4) + 0.5M_1 = 120 + 0.5M_1\) degrees.
Minute hand angle at departure: \(A_{M,dep} = 6M_1\) degrees.

Let the return time be 5 hours and \(M_2\) minutes past 5 p.m. (i.e., 5:\(M_2\)).

Hour hand angle at return: \(A_{H,ret} = 30(5) + 0.5M_2 = 150 + 0.5M_2\) degrees.
Minute hand angle at return: \(A_{M,ret} = 6M_2\) degrees.

Formulating Equations from Swapped Positions

The condition that the hour and minute hands have interchanged their positions means that the hour hand's position at departure is the same as the minute hand's position at return, and the minute hand's position at departure is the same as the hour hand's position at return.

From the first condition (\(A_{H,dep} = A_{M,ret}\)):

\[120 + 0.5M_1 = 6M_2\] Multiply by 2 to eliminate the decimal: \[240 + M_1 = 12M_2 \quad (Eq. 1)\]

From the second condition (\(A_{M,dep} = A_{H,ret}\)):

\[6M_1 = 150 + 0.5M_2\] Multiply by 2 to eliminate the decimal: \[12M_1 = 300 + M_2 \quad (Eq. 2)\]

Solving for M₁ and M₂

We have a system of two linear equations:

\[ M_1 - 12M_2 = -240 \] \[ 12M_1 - M_2 = 300 \]

From Equation 1, we can express \(M_1\) as \(M_1 = 12M_2 - 240\). Substitute this into Equation 2:

\[12(12M_2 - 240) - M_2 = 300\] \[144M_2 - 2880 - M_2 = 300\] \[143M_2 = 3180\] \[M_2 = \frac{3180}{143}\]

Now substitute the value of \(M_2\) back into the expression for \(M_1\):

\[M_1 = 12\left(\frac{3180}{143}\right) - 240\] \[M_1 = \frac{38160}{143} - \frac{240 \times 143}{143}\] \[M_1 = \frac{38160 - 34320}{143}\] \[M_1 = \frac{3840}{143}\]

So the departure time is 4 hours and \(3840/143\) minutes past 4 p.m., and the return time is 5 hours and \(3180/143\) minutes past 5 p.m.

Calculating the Time Duration

The time the person was out of the house is the difference between the return time and the departure time.

Departure time is at \(4 + \frac{M_1}{60}\) hours past 12 p.m.

Return time is at \(5 + \frac{M_2}{60}\) hours past 12 p.m.

Time out = (Return time) - (Departure time)

Time out = \(\left(5 + \frac{M_2}{60}\right) - \left(4 + \frac{M_1}{60}\right)\) hours

Time out = \(1 + \frac{M_2 - M_1}{60}\) hours

Convert to minutes:

Time out = \(60 \times \left(1 + \frac{M_2 - M_1}{60}\right)\) minutes

Time out = \(60 + (M_2 - M_1)\) minutes

Substitute the calculated values of \(M_1\) and \(M_2\):

\[\text{Time Out} = 60 + \left(\frac{3180}{143} - \frac{3840}{143}\right) \text{ minutes}\] \[\text{Time Out} = 60 + \left(\frac{3180 - 3840}{143}\right) \text{ minutes}\] \[\text{Time Out} = 60 + \left(\frac{-660}{143}\right) \text{ minutes}\] \[\text{Time Out} = \frac{60 \times 143 - 660}{143} \text{ minutes}\] \[\text{Time Out} = \frac{8580 - 660}{143} \text{ minutes}\] \[\text{Time Out} = \frac{7920}{143} \text{ minutes}\]

Now, we approximate the numerical value:

\[\frac{7920}{143} \approx 55.384615... \text{ minutes}\]

Approximating to two decimal places, the time out is 55.38 minutes.

Conclusion

The person was out of his house for approximately 55.38 minutes.

Revision Table: Clock Problem Key Points

Concept	Detail	Value
Minute Hand Speed	Degrees per minute	6°/min
Hour Hand Speed	Degrees per minute	0.5°/min
Departure Time	Minutes past 4 p.m.	\(M_1 = \frac{3840}{143}\)
Return Time	Minutes past 5 p.m.	\(M_2 = \frac{3180}{143}\)
Time Out Duration	Calculated difference	\(\frac{7920}{143}\) min \(\approx\) 55.38 min

Additional Information: Solving Swapped Hand Problems

Problems where clock hands swap positions require careful application of the formulas for hand angles at different times. The core idea is that the angular position of the hour hand at the first time equals the angular position of the minute hand at the second time, and vice versa. This typically leads to a system of two linear equations. Solving these equations gives the exact minute values for the departure and return times. The duration is then simply the difference between these two times. These problems are common in quantitative aptitude tests and require a good understanding of the relative speeds and positions of clock hands.

When the departure is between H and H+1, and the return is between H+1 and H+2, there is always one such time pair where hands swap positions. The logic derived here applies generally to such scenarios.