Question

A man walks for x hours at a speed of

Answer 1

Given:

A man walks for x hours at a speed of  (x + 1) km/h and cycles for (x - 1) hours at  a speed of (2x + 5) km/h.

Total distance  travelled is 90 km.

To find:

The value of x.

Solution:

We know that,

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

$Speed\times Time=Distance$

A man walks for x hours at a speed of  (x + 1) km/h, so walking distance is

$D_1=(x+1)(x)$ km

The man cycles for (x - 1) hours at  a speed of (2x + 5) km/h, so the cycling distance is

$D_2=(2x+5)(x-1)$ km

Now,

Total distance = 90 km

$D_1+D_2=90$

$(x+1)x+(2x+5)(x-1)=90$

$x^2+x+2x^2-2x+5x-5-90=0$

$3x^2+4x-95=0$

$3x^2+19x-15x-95=0$

$x(3x+19)-5(3x+19)=0$

$(3x+19)(x-5)=0$

$3x+19=0\text{ and }x-5=0$

$x=\dfrac{-19}{3}\text{ and }x=5$

Time cannot be negative. So, the only possible value of x is 5.