In a 400 m race, A gives B a start of 5 seconds and beats him by 15 m. In another race of 400 m, A beats B by $$7\frac{1}{7}$$ seconds. Their respective speed are -
Correct Answer: 8 m/sec, 7 m/sec
Suppose A covers 400 m in t seconds
Then, B covers 385 m in (t + 5) seconds
$$\eqalign{ & \therefore {\text{ B covers 400 m in}} \cr & = \left\{ {\frac{{\left( {t + 5} \right)}}{{385}} \times 400} \right\}{\text{ sec}} \cr & = \frac{{80\left( {t + 5} \right)}}{{77}}{\text{ sec}} \cr & {\text{Also, B covers 400 m in}} \cr & = \left( {t + 7\frac{1}{7}} \right){\text{ sec}} \cr & = \frac{{\left( {7t + 50} \right)}}{7}{\text{ sec}} \cr & \therefore \frac{{80\left( {t + 5} \right)}}{{77}} = \frac{{7t + 50}}{7} \cr & \Rightarrow 80\left( {t + 5} \right) = 11\left( {7t + 50} \right) \cr & \Rightarrow \left( {80t - 77t} \right) = \left( {550 - 400} \right) \cr & \Rightarrow 3t = 150 \cr & \Rightarrow t = 50 \cr & \therefore {\text{ A's speed}} \cr & = \frac{{400}}{{50}}\,m/\sec \cr & = 8\,m/\sec \cr & \therefore {\text{B's speed}} \cr & = \frac{{385}}{{55}}\,m/\sec \cr & = 7\,m/\sec \cr} $$
Then, B covers 385 m in (t + 5) seconds
$$\eqalign{ & \therefore {\text{ B covers 400 m in}} \cr & = \left\{ {\frac{{\left( {t + 5} \right)}}{{385}} \times 400} \right\}{\text{ sec}} \cr & = \frac{{80\left( {t + 5} \right)}}{{77}}{\text{ sec}} \cr & {\text{Also, B covers 400 m in}} \cr & = \left( {t + 7\frac{1}{7}} \right){\text{ sec}} \cr & = \frac{{\left( {7t + 50} \right)}}{7}{\text{ sec}} \cr & \therefore \frac{{80\left( {t + 5} \right)}}{{77}} = \frac{{7t + 50}}{7} \cr & \Rightarrow 80\left( {t + 5} \right) = 11\left( {7t + 50} \right) \cr & \Rightarrow \left( {80t - 77t} \right) = \left( {550 - 400} \right) \cr & \Rightarrow 3t = 150 \cr & \Rightarrow t = 50 \cr & \therefore {\text{ A's speed}} \cr & = \frac{{400}}{{50}}\,m/\sec \cr & = 8\,m/\sec \cr & \therefore {\text{B's speed}} \cr & = \frac{{385}}{{55}}\,m/\sec \cr & = 7\,m/\sec \cr} $$