Question

Starting from his house one day, a student walks with a speed of $2\tfrac{1}{2}$ km/h and reaches school 6 minutes late. Next day he increases his speed by 1 km/h and reaches 6 minutes early. How far is the school from the house?

• A. $1$ km
• B. $1\tfrac{1}{2}$ km
• C. $1\tfrac{3}{4}$ km
• D. $2$ km

Hint:

In “late/early” problems, set up two time equations and subtract to eliminate the schedule time. The difference equals the sum of early and late minutes (in hours).

Answer 1

The Correct Option is C

Step 1: Set variables and convert minutes to hours.
Let distance be $d$ km and on-time duration be $T$ hours.
$6$ minutes $= \dfrac{6}{60}=0.1$ hours.
Step 2: Write the two time equations.
Day 1 speed $v_1=2.5$ km/h: $t_1=\dfrac{d}{2.5}=T+0.1$.
Day 2 speed $v_2=2.5+1=3.5$ km/h: $t_2=\dfrac{d}{3.5}=T-0.1$.
Step 3: Eliminate $T$ using the difference.
$t_1-t_2=0.2 ⇒ \dfrac{d}{2.5}-\dfrac{d}{3.5}=0.2$.
\[ d\!\left(\frac{1}{2.5}-\frac{1}{3.5}\right) = d\!\left(\frac{3.5-2.5}{2.5 3.5}\right) = d\!\left(\frac{1}{8.75}\right)=0.2. \] Step 4: Solve for $d$.
\[ \frac{d}{8.75}=0.2 \ ⇒\ d=8.75\times 0.2=1.75\ \text{km} = 1\tfrac{3}{4}\ \text{km}. \] \[ \boxed{1\tfrac{3}{4}\ \text{km}} \]