Question

A car made a run of $390km$ in $\text{'}x\text{'}$ hours. If the speed had been $4km/hr$ more, it would have taken $2hours$ less for the journey. Find $\text{'}x\text{'}$

Answer 1

Step 1: Given

Distance, $d=390km$

Speed, $$\begin{gathered} s_{1}km/hr \\ {s}_2=s_1+4km/hr \end{gathered}$$

Time, $$\begin{gathered} t_{1}hours \\ {t}_2=t_1-2hours \end{gathered}$$

Step 2: Determine the value of $s_1$

$$\begin{gathered} speed=\frac{dis\tan ce}{time} \\ {s}_1=\frac{390}{t_1}-\left(1\right) \end{gathered}$$

Step 3: Determine the value of $s_2$

$$\begin{gathered} speed=\frac{dis\tan ce}{time} \\ {s}_2=\frac{390}{t_2} \\ {s}_1+4=\frac{390}{t_1-2}-\left(2\right) \end{gathered}$$

Step 4: Solve the equations

Substitute (1) in (2),

$$\begin{gathered} \frac{390}{t_1}+4=\frac{390}{t_1-2} \\ \frac{390}{t_1}-\frac{390}{t_1-2}=-4 \\ {t_1}^2+4t_1-780=0 \\ {t}_1=-30,26 \end{gathered}$$

As time cannot be negative, $t_1=26hours$

Step 5: Determine the value of $s_1$

$$\begin{gathered} s_1=\frac{390}{26} \\ =15hours \end{gathered}$$

Therefore, the value of $\text{'}x\text{'}$ is $15hours$.