Question : A fast local of Mumbai takes 45 minutes less than a slow local for a journey of 150 km. If the speed of the fast local is 10 km/hr more than that of the slow local, find the speed (in km/hr) of the slow local train.
Option 1: 30
Option 2: 35
Option 3: 45
Option 4: 40
Correct Answer: 40
Solution : Let's denote the speed of the slow local train as S km/hr. ⇒ The fast local train's speed is (S + 10) km/hr, as given in the problem. The time taken by the slow local train to cover a distance of 150 km: We know, $\text{Time} = \frac{\text{Distance}}{\text{Speed}}$ ⇒ The time taken by the slow local is $\frac{150}{S}$ hours. The time taken by the fast local (which is 45 minutes or $\frac{45}{60}$ hours less than the slow local) is $\frac{150}{S+10}$ hours. Now, according to the question, the fast local takes 45 minutes less than the slow local, ⇒ $\frac{150}{S}−\frac{45}{60}=\frac{150}{S+10}$ ⇒ $\frac{150}{S}−\frac{3}{4}=\frac{150}{S+10}$ ⇒ $4\times150(S+10)−3S(S+10)=4\times150S$ ⇒ $600(S+10)−3S(S+10)=600S$ ⇒ $600S+6000−3S^2−30S=600S$ ⇒ $3S^2+30S−6000=0$ ⇒ $S^2+10S−2000=0$ ⇒ $(S+50)(S−40)=0$ Speed cannot be negative, so $S=40$ km/hr. Hence, the correct answer is 40.
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