A has covered A has covered 9 20 7 10 19 Sure! Let's use another approach: - A covered \(\frac{1}{3}\) of the total distance on his scooter and \(\frac{2}{3}\) of the distance on foot. Let the total distance be \(D\), scooter speed be \(S_r\), and walking speed be \(S_w\). If we denote the time taken to ride the scooter as \(t\), the distance covered by scooter is: \[ S_r \times t = \frac{D}{3} \] The time taken to walk is 22 times the riding time: \[ t_{\text{walk}} = 22t \] And the distance covered by walking is: \[ S_w \times 22t = \frac{2D}{3} \] By setting up the ratio of distances covered: \[ \frac{S_r \times t}{S_w \times 22t} = \frac{\frac{D}{3}}{\frac{2D}{3}} \] \[ \frac{S_r}{S_w \times 22} = \frac{1}{2} \] Solving for \( \frac{S_r}{S_w} \): \[ S_r = 11S_w \] So, the riding speed is 11 times the walking speed. Therefore, the correct answer is **4) 10**.Answers (2)
of total distance when his scooter failed. he parked it and cover the remaining distance by foot walking 22 times as much time as riding. how many times his riding speed more then his walking speed?
- Walking took 22 times as long as riding.