A cyclist travels from town A to town B, a distance of 182 km. For the first 84 km, he rides at a certain speed. For the remaining distance (98 km), he increases his speed by 7 km/h and therefore reaches 42 minutes earlier than the time he would have taken if he had travelled the entire distance at the original speed. Find the original speed.
28 km/hr
Let the original speed be v km/h. Time at original speed for the full 182 km minus time at split speeds equals 42 minutes \((=0.7\text{ hr})\).
This simplifies to: \(98\left(\dfrac1v-\dfrac1{v+7}\right) = 0.7 \Rightarrow \dfrac{98\times7}{v(v+7)} = 0.7 \Rightarrow v(v+7) = 980\).
\(v^2+7v-980=0\), solving gives \(v = 28\).
Hence, the original speed of the cyclist is 28 km/hr.
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