Question

A student goes to school from his residence at a speed of \(2\frac{1}{2}\) km/h and reaches school 6 minutes late. If he travels at a speed of 3 km/h, he reaches 10 minutes before time. What is the distance of his school from his residence ?

• A. 4 km
• B. \(3\frac{1}{4}\) km
• C. 1 km
• D. \(3\frac{1}{2}\) km

Answer 1

The Correct Option is A

To determine the distance from the student's residence to the school, let us denote the distance as \(d\) km and the scheduled travel time as \(t\) hours. We are given two scenarios:

1. Traveling at \(2\frac{1}{2}\) km/h results in a 6-minute delay.

2. Traveling at 3 km/h leads to arriving 10 minutes early.

Convert time discrepancies into hours: 6 minutes = \(\frac{1}{10}\) hour and 10 minutes = \(\frac{1}{6}\) hour.

Formulate two equations from these conditions:

1. \( t + \frac{1}{10} = \frac{d}{2.5} \)

2. \( t - \frac{1}{6} = \frac{d}{3} \)

Re-write the equations:

\( \frac{d}{2.5} - \frac{1}{10} = t \)

\( \frac{d}{3} + \frac{1}{6} = t \)

Equate the two expressions for \(t\):

\( \frac{d}{2.5} - \frac{1}{10} = \frac{d}{3} + \frac{1}{6} \)

Clear fractions by multiplying through by 30:

\( 12d - 3 = 10d + 5 \)

Rearranging terms results in \( 12d - 10d = 5 + 3 \)

\( 2d = 8 \)

So \( d = 4 \) km.

Therefore, the distance of the school from the student's residence is 4 km.