Question

Four persons walk from Point A to Point D following different routes. The one following ABCD takes 70 minutes. Another person takes 45 minutes following ABD. The third person takes 30 minutes following ACD. The last person takes 65 minutes following ACBD. If all walk at the same speed, how long will it take to go from point B to point C?

• A. 10 min.
• B. 20 min.
• C. 30 min.
• D. 40 min.
• E. Cannot be answered as the angles are unknown.

Hint:

When speeds are equal, treat unknown segment {times} as variables. Write equations for each path’s total time and use simple additions/subtractions to eliminate segments.

Answer 1

The Correct Option is C

Let the common speed be $v$ (same for all). Then the time on any segment is proportional to its length, so we can work purely with times. Let \[ a=\text{time}(AB),\quad b=\text{time}(BC),\quad c=\text{time}(CD),\quad d=\text{time}(BD),\quad e=\text{time}(AC). \] From the given routes: \[ \begin{aligned} \text{ABD: }& a+d=45 \quad &(1)
\text{ACD: }& e+c=30 \quad &(2)
\text{ABCD: }& a+b+c=70 \quad &(3)
\text{ACBD: }& e+b+d=65 \quad &(4) \end{aligned} \] Step 1: Eliminate $a,d$ using (1) and (3).
Subtract (1) from (3): \[ (a+b+c)-(a+d)=b+c-d=70-45=25 \quad \Rightarrow \quad b+c-d=25. \tag{5} \] Step 2: Eliminate $e,c$ using (2) and (4).
Subtract (2) from (4): \[ (e+b+d)-(e+c)=b+d-c=65-30=35 \quad \Rightarrow \quad b+d-c=35. \tag{6} \] Step 3: Solve for $b$.
Add (5) and (6): \[ (b+c-d)+(b+d-c)=2b=25+35=60 \;\Rightarrow\; b=30. \] Hence, the time from $B$ to $C$ is \[ \boxed{30\ \text{minutes}}. \] \fbox{\parbox{0.97\linewidth}{ \centering Time to go from $B$ to $C$ $= \boxed{30\ \text{min}}$. }}