Question

a car travels a distance 840 kn at uniform speed if the speed of the car 10km/hr more it takes 2 hours less ti to cover the same distance.what was the original speed?​

Answer 1

$ANSWER$

Let the original speed of the car be x km/hr.

Time taken to move 840km

$\\ = \frac{distance}{speed} \\ \\ = \frac{840}{x}$

Also, it is given speed is going 10 km \hr more than the original speed i. e. x+10

Therefore,

The time taken in moving 840 km

=

$\\ \frac{840}{x + 10} hr \\ \\ \frac{840}{ x} - \frac{840}{ x + 10} = 2 \\ \\ 840( x + 10 ) - 840 \times x = 2x(x + 10)\\ \\ 840(x + 10) - 840x = 2x(x + 10) \\ \\ 2 {x}^{2} + 20x - 8400 = 0 \\ \\ {x}^{2} + 70x + \: -60x - 4200 = 0 \\ \\ (x + 70) (x - 60) = 0 \\ \\ x = 60 \: \: or \: x = - 70 \\ \\ Speed \: cannot \: be \: negative \\ \\ x = 60kmphr$

Therefore, original speed =x=60km/hr.

Answer 2

$\huge\underline{\bold{\red{Answer\::}}}$

$\underline{A \:car\: travels\: a\: distance \:840\: km \:at}$ $\underline{uniform \:speed.}$

$\textbf{Here distance is 840 km}$

$\underline{If \:the \:speed \:of\: the \:car \:10\: km/hr\: more\: it}$ $\underline{takes \:2 \:hours \:less\: to\: cover\: the\: same}$ $\underline{distance.}$

$\textbf{Let the speed of car be x km/hr}$ (original speed)

If the speed of the car 10 km/hr more then,

$\textbf{Speed is 10 + x km/hr}$

Also it takes 2 hours less.

$\textbf{time = 2 hrs}$

$\textbf{\red{A.T.Q.}}$

$Time$ = $\dfrac{Distance}{Speed}$

=> $\dfrac{840}{x}$ ___(1)

For case 2nd

$Time$ = $\dfrac{840}{x\:+\:10}$

(As the speed of the car increase 10 more than the original one)

Now

$\dfrac{840}{x}$ - $\dfrac{840}{x\:+\:10}$ = $2$

$(840)$ $\dfrac{1}{x}$ - $\dfrac{1}{x\:+\:10}$ = $2$

$(840)$ = $\dfrac{x\:+\:10\:-\:x}{ {x}^{2} \: + \: 10 }$ = $2$

$\dfrac{8400}{{x}^{2} \: + \: 10 }$ = $2$

$8400$ = (2) ${(x}^{2} \: + \: 10)$

$8400$ = ${2x}^{2} \: + \: 20x$

${2x}^{2} \: + \: 20x\:-\:8400\:=\:0$

${x}^{2} \: + \: 10x\:-\:4200\:=\:0$

(As 2 is common from them)

x² + 70x - 60x - 4200 = 0

x (x + 70) - 60 (x + 70) = 0

(x + 70) (x - 60) = 0

x + 70 = 0

x = -70 (Speed can never be in negative)

So, it is neglected.

Also

x - 60 = 0

x = +60

(We can't neglect it because it is in positive)

So,

$\textbf{The original speed of the car = x km/hr.}$

$\boxed{\boxed{x \:= \:60 \:km/hr}}$

(60 km/hr is the original speed of car)