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 \[{x_{B1}}\].
\[\dfrac{{{x_{A1}}}}{{{x_{B1}}}} = \dfrac{{\left( {x - 100} \right)\,{\text{m}}}}{{100\,{\text{m}}}}\]

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 \[{x_{B2}}\].
\[\dfrac{{{x_{A2}}}}{{{x_{B2}}}} = \dfrac{{\left( {2x - 60} \right)\,{\text{m}}}}{{\left( {x + 60} \right)\,{\text{m}}}}\]

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

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

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.