Solution

Let \( x \) be the speed of the car to reach office in 75 minutes. \[ \begin{array}{|c|c|} \hline \text{Speed (km/hr)} & 50 & x \\ \hline \text{Time (minutes)} & 90 & 75 \\ \hline \end{array} \]

As the time is reduced, the speed of the car will increase to reach the office. It is a case of inverse variation

\[ \begin{align*} \implies a \times b &= k \text{ (constant)} \\ \\ 50 \times 90 &= x \times 75 \\ \\ x &= \frac{\cancel{50}^2 \times \cancel{90}^{30}}{\cancel{75}_{\cancel3_1}} \\ \\ x &= 2 \times 30 \\ \color{green} x &= \color{green} 60 \end{align*} \]

To reach the office on time, the speed of the car should be \( \color{red} 60 \ km/hr \)

\[ \begin{align*} \color{green} Distance & = \color{green} Speed \times Time \\ & = 50 \times 1.5 \\ Distance & = 75 \ km \\ \\ \text{Total Distance} & = 75 \ km + 75 \ km \\ & = 150 \ km \end{align*} \]

Answer Speed \( = \color{red} 60 \ km/hr \) , Distance \( = \color{red} 150 \ km \)