Question

A car covers a distance of 400 km at a certain speed. Had the speed been 12 km/h more, the time taken for the journey would have been 1 hour 40 minutes less. Find the original speed of the car.

No spams

Answer 1

Answer:

Answer is in picture

Explanation:

Hope its help ful

Answer 2

$\red{\bigstar}★ G I V E N$

A car covers a distance of 400 km at a certain speed. Had the speed been 12 kmph more, the time taken for the journey would have been 1 hour 40 minutes less.

$\orange{\bigstar}★ T OㅤF I N D$

Original speed of the car ?

$\green{\bigstar}★ S O L U T I O N$

Let original speed of car be v kmph

$\quad\footnotesize\bf{\dag}\:\underline{\frak{Using\:formula\:::}}$

$\qquad\odot\:{\underline{\boxed{\bf{\blue{Time = \dfrac{Distance}{Speed}}}}}}$

In first case ::

We have ::

Distance = 400 km

Speed = v kmph

$\quad\footnotesize\bf{\dag}\:\underline{\frak{Putting\:all\:values\:in\:formula\:::}}$

$\begin{gathered}\\ \dashrightarrow \: \sf Time = \dfrac{400}{v}\end{gathered}$

In second case ::

We have ::

Distance = 400 km

Speed = (v + 12) kmph

$\quad\footnotesize\bf{\dag}\:\underline{\frak{Putting\:all\:values\:in\:formula\:::}}$

$\begin{gathered}\\ \dashrightarrow \: \sf Time = \dfrac{400}{v + 12}\end{gathered}$

Now,

According to the Question, if speed of car has been 12 kmph more the time taken for the journey would have been 1 hour 40 minutes less.

★ Converting units of time taken ::

$\quad\footnotesize\bf{\dag}\:\underline{\frak{We\:know\:that\:::}}$

$\qquad\odot\:{\underline{\boxed{\bf{\blue{1\:hour = 60\:minutes}}}}}$

\:

Therefore,

$\begin{gathered}\\ \dashrightarrow \:\sf 1\:hour \: 40\:minutes = \big(60 + 40\big) minutes \end{gathered}$

$\begin{gathered}\\ \dashrightarrow \:\bf \pink{ 1\:hour \: 40\:minutes = 100\:minutes}\end{gathered}$

\:

★ According to the Question ::

$\begin{gathered}\\ \dashrightarrow \:\sf \dfrac{400}{v} - \dfrac{400}{v + 12} = \dfrac{100}{60} \end{gathered}$

$\begin{gathered}\\ \dashrightarrow \:\sf \dfrac{400}{v} - \dfrac{400}{v + 12} = {\cancel{\dfrac{100}{60}}} \end{gathered}$

\:

In right hand side, after cancelling 100 and 60, we get

$\begin{gathered}\\ \dashrightarrow \:\sf \dfrac{400}{v} - \dfrac{400}{v + 12} = \dfrac{5}{3} \end{gathered}$

$\begin{gathered}\\ \dashrightarrow \:\sf \dfrac{400\:\times\:\big(v + 12\big) - 400 \:\times\:v}{v\:\times\:\big(v + 12\big)} = \dfrac{5}{3} \end{gathered}$

$\begin{gathered}\\ \dashrightarrow \:\sf \dfrac{400v + 4800 - 400v}{v^2 + 12v} = \dfrac{5}{3} \end{gathered}$

$\begin{gathered}\\ \dashrightarrow \:\sf \dfrac{\cancel{400v} - \cancel{400v} + 4800}{v^2 + 12v} = \dfrac{5}{3} \end{gathered}$

$\begin{gathered}\\ \dashrightarrow \:\sf \dfrac{4800}{v^2 + 12v} = \dfrac{5}{3} \end{gathered}$

\:

By cross multiplication ::

$\begin{gathered}\\ \dashrightarrow \:\sf 4800\:\times\:3 = 5\:\times\:\big(v^2 + 12v\big) \end{gathered}$

$\begin{gathered}\\ \dashrightarrow \:\sf 14400 = 5v^2 + 60v \end{gathered}$

$\begin{gathered}\\ \dashrightarrow \:\sf 5v^2 + 60v - 14400 = 0\end{gathered}$

$\begin{gathered}\\ \dashrightarrow \:\sf 5\:\times\:\big(v^2 + 12v - 2880\big) = 0\end{gathered}$

$\begin{gathered}\\ \dashrightarrow \:\sf v^2 + 12v - 2880 = \dfrac{0}{5}\end{gathered}$

$\begin{gathered}\\ \dashrightarrow \:\sf v^2 + 12v - 2880 = 0\end{gathered}$

\:

Splitting middle term ::

$\begin{gathered}\\ \dashrightarrow \:\sf v^2 + \big(60v - 48v\big) - 2880 = 0\end{gathered}$

$\begin{gathered}\\ \dashrightarrow \:\sf v^2 + 60v - 48v - 2880 = 0\end{gathered}$

$\begin{gathered}\\ \dashrightarrow \:\sf v\big(v + 60\big) - 48\big(v + 60\big) = 0\end{gathered}$

$\begin{gathered}\\ \dashrightarrow \:\sf \big(v + 60\big)\: \big(v - 48\big) = 0\end{gathered}$

$\begin{gathered}\\ \dashrightarrow \:\sf v + 60 = 0\:\:or\:\:v - 48 = 0 \end{gathered}$

$\begin{gathered}\\ \dashrightarrow \:\bf \purple{v = -60\:\:or\:\:v = 48\:}\end{gathered}$

\:

Since, either speed of car is -60 kmph or 48 kmph.

We clearly know that speed can't be negative.

So, we will reject Speed (v) = -60 kmph!

\:

$\therefore\:{\underline{\sf{Hence,\:original\:speed\:of\:car\:is\:{\textit{\textbf{48\:kmph.}}}}}}$