Question

A and B travel around in a circular path at uniform speeds in the opposite direction starting from the diagrammatically opposite point at the same time. They meet each other first after B has traveled 100 meters and meet again 60 meters before A complete one round. What is the circumference of the circular path?
A. $$240\text{m}$$
B. $$360\text{m}$$
C. $$480\text{m}$$
D. $$300\text{m}$$

Answer 1

Hint: Determine the distances travelled by A and B in both the cases. Take the ratio of these distances and equate them as their speed is uniform.

Complete step by step answer:Here, we have to calculate the circumference of the circular path travelled by A and B.

Let the half of the circumference of the circular path is $$x$$.

Initially, A and B start from the opposite points of any diameter of the circular track. They meet each other when B has travelled a distance of $$100\text{m}$$ on the half circumference.

Hence, the distance $$x_{B1}$$ travelled by B on the half circumference is $$100\text{m}$$ and distance travelled $$x_{A1}$$ by A is$$\left(x-100\right)\text{m}$$.
$$x_{B1}=100\text{m}$$
$$x_{A1}=\left(x-100\right)\text{m}$$

Take the ratio of $$x_{A1}$$ and $$x_{B1}$$.
$${\displaystyle \frac{x_{A1}}{x_{B1}}}={\displaystyle \frac{\left(x-100\right)\text{m}}{100\text{m}}}$$

A and B once again meet each when A is $$60\text{m}$$ before to complete one round.

Hence, the distance $$x_{A2}$$ travelled by A on the half circumference is $$\left(2x-60\right)\text{m}$$ and distance travelled $$x_{B2}$$ by B is$$\left(x+60\right)\text{m}$$.
$$x_{A2}=\left(2x-60\right)\text{m}$$
$$x_{B2}=\left(x+60\right)\text{m}$$

Take the ratio of $$x_{A2}$$ and $$x_{B2}$$.
$${\displaystyle \frac{x_{A2}}{x_{B2}}}={\displaystyle \frac{\left(2x-60\right)\text{m}}{\left(x+60\right)\text{m}}}$$

Since A and B travel with uniform speed on the circular track, the ratios $${\displaystyle \frac{x_{A1}}{x_{B1}}}$$ and $${\displaystyle \frac{x_{A2}}{x_{B2}}}$$ of the distances covered by A and B are equal.
$${\displaystyle \frac{x_{A1}}{x_{B1}}}={\displaystyle \frac{x_{A2}}{x_{B2}}}$$

Substitute $${\displaystyle \frac{\left(x-100\right)\text{m}}{100\text{m}}}$$ for $${\displaystyle \frac{x_{A1}}{x_{B1}}}$$ and $${\displaystyle \frac{\left(2x-60\right)\text{m}}{\left(x+60\right)\text{m}}}$$ for $${\displaystyle \frac{x_{A2}}{x_{B2}}}$$ in the above equation.
$${\displaystyle \frac{\left(x-100\right)\text{m}}{100\text{m}}}={\displaystyle \frac{\left(2x-60\right)\text{m}}{\left(x+60\right)\text{m}}}$$

Solve the above equation for $$x$$.
$$\left(x-100\right)\left(x+60\right)=\left(2x-60\right)100$$
$$\Rightarrow x^2+60x-100x-6000=200x-6000$$
$$\Rightarrow x\left(x-240\right)=0$$
$$\Rightarrow x=0$$ or $$x=240$$

Since, the value of the half circumference cannot be zero.
$$x=240\text{m}$$

Therefore, the half circumference of the circular path is $$240\text{m}$$.

Since the circumference is twice the half circumference, $$2x=2\left(240\text{m}\right)=480\text{m}$$.

Hence, the circumference of the circular path is $$480\text{m}$$.

Hence, the correct option is D.

Note:The value zero obtained when solved the quadratic equation should be neglected as the circumference cannot be zero for the present question.