Question

The winning relay team in a high school sports competition clocked $48$ minutes for a distance of $13.2$ km. Its runners A, B, C, and D maintained speeds of $15$ kmph, $16$ kmph, $17$ kmph, and $18$ kmph respectively. What is the ratio of the time taken by B to that taken by D?

• A. $5 : 16$
• B. $5 : 17$
• C. $9 : 8$
• D. $8 : 9$

Hint:

In relay problems, if not otherwise stated, assume equal distances for each runner. The ratio of times is inversely proportional to their speeds for equal distances.

Answer 1

The Correct Option is D

Step 1: Understanding the problem
The total time taken for the race is $48$ minutes, which is $\frac{48}{60} = 0.8$ hours.
The total distance covered is $13.2$ km, divided among the $4$ runners: A, B, C, and D. Each runs a certain segment at their given speed.
Let the distances run by A, B, C, and D be $d_A$, $d_B$, $d_C$, and $d_D$ km respectively.

Step 2: Time taken by each runner
The time taken by each runner is given by: \[ t_A = \frac{d_A}{15}, \quad t_B = \frac{d_B}{16}, \quad t_C = \frac{d_C}{17}, \quad t_D = \frac{d_D}{18} \] We know: \[ t_A + t_B + t_C + t_D = 0.8 \quad \text{hours} \] Also: \[ d_A + d_B + d_C + d_D = 13.2 \quad \text{km} \]
Step 3: Equal distance assumption in relay
In a standard relay, each runner runs the same distance. Therefore: \[ d_A = d_B = d_C = d_D = \frac{13.2}{4} = 3.3 \ \text{km} \]
Step 4: Calculate individual times
\[ t_B = \frac{3.3}{16} \ \text{hours}, \quad t_D = \frac{3.3}{18} \ \text{hours} \]
Step 5: Ratio of $t_B : t_D$
\[ t_B : t_D = \frac{3.3}{16} : \frac{3.3}{18} = \frac{1}{16} : \frac{1}{18} = 18 : 16 = 9 : 8 \]
Thus, the required ratio is $9 : 8$.