Surbhi was travelling to point B from point A with speed of 45 km/hr.

1.	Surbhi was travelling to point B from point A with speed of 45 km/hr. After 1 hour, Pragati also started to travel from point A to point B. Pragati reached the point B 30 minutes before Surbhi. If Pragati would have decreased his speed by 6 km/hr then both would have reached point B at the same time. Find the original speed of Pragati. 1. 55 km/hr2. 54 km/hr3. 56 km/hr4. 64 km/hr5. 60 km/hr
Answer» Correct Answer - Option 5 : 60 km/hr Let the original time taken by Surbhi to reach point B from point A = t hours Then, original time taken by Pragati to reach point B from point A = (t – 1.5) hours Also, let original speed of Pragati = v km/hr So, the distance between point A and B = v × (t – 1.5) = (v – 6) × (t – 1) = 45t ---- (1) \Rightarrow v/(v – 6) = (t – 1)/(t – 1.5) \Rightarrow v/6 = (t – 1)/0.5 {applying a/b = c/d \Rightarrow a/(a – b) = c/(c – d)} \Rightarrow v = 12(t – 1) ---- (2) From eq. (1) and eq. (2) \Rightarrow 12(t – 1)(t – 1.5) = 45t \Rightarrow 4(t²– 2.5t +1.5) = 15t \Rightarrow 4t²– 10t + 6 = 15t \Rightarrow 4t²– 25t + 6 = 0 \Rightarrow 4t²– 24t - t + 6 = 0 \Rightarrow 4t(t – 6) – (t – 6) = 0 \Rightarrow (4t – 1)(t – 6) = 0 \Rightarrow t = ¼ or t = 6 Since t > 1, so t = ¼ not possible, thus t = 6 \Rightarrow v = 12(6 – 1) = 60 km/hr

Surbhi was travelling to point B from point A with speed of 45 km/hr. After 1 hour, Pragati also started to travel from point A to point B. Pragati reached the point B 30 minutes before Surbhi. If Pragati would have decreased his speed by 6 km/hr then both would have reached point B at the same time. Find the original speed of Pragati. 1. 55 km/hr2. 54 km/hr3. 56 km/hr4. 64 km/hr5. 60 km/hr

Answer» Correct Answer - Option 5 : 60 km/hr

Let the original time taken by Surbhi to reach point B from point A = t hours

Then, original time taken by Pragati to reach point B from point A = (t – 1.5) hours

Also, let original speed of Pragati = v km/hr

So, the distance between point A and B = v × (t – 1.5) = (v – 6) × (t – 1) = 45t ---- (1)

\Rightarrow v/(v – 6) = (t – 1)/(t – 1.5)

\Rightarrow v/6 = (t – 1)/0.5 {applying a/b = c/d \Rightarrow a/(a – b) = c/(c – d)}

\Rightarrow v = 12(t – 1) ---- (2)

From eq. (1) and eq. (2)

\Rightarrow 12(t – 1)(t – 1.5) = 45t

\Rightarrow 4(t²– 2.5t +1.5) = 15t

\Rightarrow 4t²– 10t + 6 = 15t

\Rightarrow 4t²– 25t + 6 = 0

\Rightarrow 4t²– 24t - t + 6 = 0

\Rightarrow 4t(t – 6) – (t – 6) = 0

\Rightarrow (4t – 1)(t – 6) = 0

\Rightarrow t = ¼ or t = 6

Since t > 1, so t = ¼ not possible, thus t = 6

\Rightarrow v = 12(6 – 1) = 60 km/hr