Question

A car travel from p to q with constant speed. If its speed were increased by $10$ km/hr, it would have taken $1$ hr lesser to cover the distance. It would have taken further $45$ min lesser if the speed was further increased by $10$ km/hr. Find distance?

Answer 1

Let the Distance Between $p$ and $q$ be $D$

Using Given Information we can Formulate $2$ Equations

$\Rightarrow \frac{D}{V}-\frac{D}{V+10}=1...................................\left(1\right)$

$\Rightarrow \frac{D}{V}-\frac{D}{V+20}=1+\frac{3}{4}$

$\Rightarrow \frac{D}{V}-\frac{D}{V+20}=\frac{7}{4}..................................\left(2\right)$

Subtracting $Eq\left(2\right)-Eq\left(1\right)$

$\Rightarrow \frac{D}{V+10}-\frac{D}{V+20}=\frac{3}{4}..................................\left(3\right)$

Simplifying $Eq\left(3\right)$ we get

$\Rightarrow 3(V+20)(V+10)=40D................................\left(4\right)$

We can Find value of $D$ in terms of $V$ by Simpifying $Eq\left(1\right)$

$\Rightarrow D=\frac{V(V+10)}{10}.....................................\left(5\right)$

Putting Value of $D$ in $Eq\left(4\right)$ we get

$\Rightarrow 3(V+20)(V+10)=\frac{40(V+10)V}{10}$

$\Rightarrow 3V+60=4V$

$\Rightarrow V=60$

Putting value of V in $Eq\left(5\right)$ we get

$\Rightarrow D=420$

Therefore Answer is $420$