Question

Please help me with the complete solution of this sum. Please don't spam and give wrong answers.

NOTE: this sum is based on SSC Board portion. (Chapter 2. Quadratic Equations)

Correct answers will be marked as brainliest for sure.

I'll report and ban the one who spams and gives wrong answers unnecessarily.​

Answer 1

Given:-

distance=240km

let the speed be x

then,

$time1 = \frac{240}{x} - - - (1)$

and,

when speed is incresed

then,

speed=x+20

distance=240

$time2 = \frac{240}{x + 20} - - - (1)$

ACTQ,

and by using equ. 1 and 2,

$time1 - time2 = 2$

$\frac{240}{x} - \frac{ 240 }{x + 20} = 2$

$\frac{240(x + 20) - 240x}{(x)(x + 20)} = 2$

$\frac{240x + 240 \times 20 - 240x}{(x)(x + 20)} = 2$

$240 \times 20 = 2 \times (x)(x + 20)$

$240 \times 10 = {x}^{2} + 20x$

${x}^{2} + 20x - 2400 = 0$

splitting the middle term,

${x}^{2} + 60x - 40x - 2400 = 0$

$x(x + 60) - 40(x + 60) = 0$

$(x + 60)(x - 40) = 0$

so,

x=-60. or. x=40

speed cannot be negative,

so,

the original speed of car is 40 km/hr