# Quadratic Equations | Speed Time

Ex 4.3 Q10. CBSE 10th Maths NCERT exercise solution

#### Video Explanation

Step 1: Frame equations based on the information about the speed and time taken by the express and passenger trains.
Let the speed of the express train be s km/hr.
Statement: Speed of the express train is 11km/hr more than that of passenger train.
∴ speed of the passenger train = (s – 11)km/hr

Time taken by express train to travel 132 km = $$frac{132}{s}$ hours Time taken by passenger train to travel 132 km = $\frac{132}{s-11}$ hours Statement: Express train takes 1 hour lesser time than the passenger train ∴ $\frac{132}{s}$ = $\frac{132}{s-11}$ - 1 $\frac{132}{s-11}$ - $\frac{132}{s}$ = 1 Because s ≠ 0 and s ≠ 11$if s = 11, speed of passenger train will be 0. Then the passenger train would not have travelled at all), multiply the equation by s(s-11)
132s – 132(s – 11) = s(s – 11)
132s – 132s + 1452 = s2 - 11s
s2 – 11s - 1452 = 0

##### Step 2: Factorize the equation to find the value of ‘s’

s2 – 44s + 33s – 1452 = 0
s(s – 44) + 33(s – 44) = 0
(s – 44) (s + 33) = 0
s = 44 or s = -33
Speed cannot be negative
∴ speed of the express train = 44km/hr
Speed of the passenger train = s - 11 = 44 - 11 = 33 km/hr