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.

Answer 1

Answer:

Let the original speed of the car be x km/h

Case 1:-

Distance = 400 km

speed = x km/h

Time taken = 400/x hrs

Case 2:-

Distance = 400 km

speed = (x+12) kmph

Time taken = (400/x+12)

As per the question, the time taken is 1 hr 40 min lesser than the previous.

$\frac{480}{x} - \frac{5}{3} = \frac{400}{x + 12}$

$\frac{1200 - 5x}{3x} = \frac{400}{x + 12}$

$1200x- 5x {}^{2} + 14400 - 60x = 1200x$

$x {}^{2} + 12x - 2880 = 0$

$(x - 48)(x + 60)$ = 0

Discarding x = -60 as speed cannot be negative

Therefore speed of car is

48km/h

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Answer 2

$\red{\bigstar}$ G I V E N

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• 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.

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$\orange{\bigstar}$ T OㅤF I N D

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• Original speed of the car ?

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$\green{\bigstar}$ S O L U T I O N

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• Let original speed of car be v kmph

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

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

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• In first case ::

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We have ::

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• Distance = 400 km

• Speed = v kmph

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

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

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• In second case ::

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We have ::

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• Distance = 400 km

• Speed = (v + 12) kmph

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

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

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Now,

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• 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.

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★ Converting units of time taken ::

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

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

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Therefore,

$\\ \dashrightarrow \:\sf 1\:hour \: 40\:minutes = \big(60 + 40\big) minutes$

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

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★ According to the Question ::

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

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

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

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

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

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

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

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

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By cross multiplication ::

$\\ \dashrightarrow \:\sf 4800\:\times\:3 = 5\:\times\:\big(v^2 + 12v\big)$

$\\ \dashrightarrow \:\sf 14400 = 5v^2 + 60v$

$\\ \dashrightarrow \:\sf 5v^2 + 60v - 14400 = 0$

$\\ \dashrightarrow \:\sf 5\:\times\:\big(v^2 + 12v - 2880\big) = 0$

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

$\\ \dashrightarrow \:\sf v^2 + 12v - 2880 = 0$

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Splitting middle term ::

$\\ \dashrightarrow \:\sf v^2 + \big(60v - 48v\big) - 2880 = 0$

$\\ \dashrightarrow \:\sf v^2 + 60v - 48v - 2880 = 0$

$\\ \dashrightarrow \:\sf v\big(v + 60\big) - 48\big(v + 60\big) = 0$

$\\ \dashrightarrow \:\sf \big(v + 60\big)\: \big(v - 48\big) = 0$

$\\ \dashrightarrow \:\sf v + 60 = 0\:\:or\:\:v - 48 = 0$

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

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• 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!

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