Question 11 please solve - Maths - Pair of Linear Equations in Two Variables

Question

Question 11 please solve

Lovina Kansal · Accepted Answer

let the original rate of walking be x km/hr and the distance covered be y km.

therefore time taken to cover y km = distance/speed = y / x hr

therefore

if he walks 1/2 km/hr faster , he takes 1 hour less, therefore eq is:

$\frac{y}{x + \frac{1}{2}} = \frac{y}{x} - 1 \frac{2 y}{2 x + 1} = \frac{y}{x} - 1 \frac{2 y}{2 x + 1} = \frac{y - x}{x} 2 x y = 2 x y + y - 2 x^{2} - x y = 2 x^{2} + x ............... (1)$

if he walks 1 km / hr slower, he takes 3 more hours:

$\frac{y}{x - 1} = \frac{y}{x} + 3 \frac{y}{x - 1} = \frac{y + 3 x}{x} x y = (x - 1) (y + 3 x) x y = x y - y + 3 x^{2} - 3 x y = 3 x^{2} - 3 x ............. (2)$

from eq(1) and eq(2):

$2 x^{2} + x = 3 x^{2} - 3 x x^{2} - 4 x = 0 x (x - 4) = 0 x = 0 o r x = 4$

since x is the speed , it can not be zero, therefore rate of walking = 4 km/hr

and
$y = 3 \times 4^{2} - 3 \times 4 y = 3 \times 16 - 13$

y = 48-12 = 36 km

thus the distance covered is 36 km.