A man completed a trip of 136 km in 8 hours. Some parts of the trip was covered at 15 km/hr and the remaining at 18 km/hr. Find the part of the trip covered at 18 km/hr.

Total distance = 136 km

Total time = 8 hours

Let the part of the trip that was covered at 15 km/h be x km.

So, the remaining part, (136 - x)km, was covered at 18km/hr.

As we know, Time = $\dfrac{Distance}{Speed}$

And

Total time = time taken to cover x km at 15 km/h + time taken to cover (136−x) km at 18 km/h.

$⇒ \dfrac{x}{15} + \dfrac{136 - x}{18} = 8\\[1em]$

Since L.C.M. of 15 and 18 = 90, multiply each term with 90 to get:

$⇒ \dfrac{x \times 90}{15} + \dfrac{(136 - x) \times 90}{18} = 8 \times 90$

⇒ x $\times$ 6 + (136 - x) $\times$ 5 = 720

⇒ 6x + 680 - 5x = 720

⇒ x + 680 = 720

⇒ x = 720 - 680

⇒ x = 40

Other part = 136 - x km

= 136 - 40 km

= 96 km

Hence, the man covered 96 km at the speed of 18 km/hr.