Abus takes 2 hrs less for a journey of 300 km, if its speed is increased by 5km/hr from its usual speed, then the usual speed in km/hr is

Let the usual speed of bus be x km/hr.

Time = Distance / Speed = 300/x hrs

According to the question,

=> (300/x - 2) (x + 5) = 300

=> (300 - 2x)/x (x + 5) = 300

=> 300x + 1500 - 2x² - 10x = 300x

=> 2x² + 10x - 1500 = 0

=> 2(x² + 5x - 750) = 0

=> x² + 30x - 25x - 750 = 0

=> x(x + 30) - 25(x + 30) = 0

=> (x - 25) (x + 30) = 0

=> x = 25 or x = - 30

∴ x ≠ - 30 (Because speed can't be negative)

Hence,

The usual speed of bus = 25 km/hr

12*c

step-by-step explanation:

x + y = 90 5x – 10y = 90 90 – 10y = 5x 90 + 10y = 5x

step-by-step explanation:

last month maria hiked the 5-mile mountain trail a number of times and she hiked the 10-mile canal trail several times. let x represent the number of times she hiked the 5-mile trail, and let y represent the number of times she hiked the 10-mile trail. if she hiked a total of 90 miles, which equation can be used to find the number of times maria hiked each trail? x + y = 90 5x – 10y = 90 90 – 10y = 5x 90 + 10y = 5x