The difference between the times taken by two trains to travel a distance of 350 km is 2 hours 20 minutes. The difference between their speeds is 5 km/h. What is the speed of the faster train in km/h?

Introduction / Context:
This problem involves two trains travelling the same fixed distance at different speeds. The difference in their travel times and speeds is known. The question tests your ability to translate this information into a simple algebraic equation using the distance = speed * time relationship and to solve it to find the faster speed.

Given Data / Assumptions:

• Common distance travelled by both trains = 350 km.
• Difference in their speeds = 5 km/h.
• Difference in their times = 2 hours 20 minutes.
• Two hours 20 minutes equals 7 / 3 hours.
• We must find the speed of the faster train.

Concept / Approach:
Let the speed of the slower train be s km/h. Then the speed of the faster train is s + 5 km/h. Time taken by the slower train is 350 / s hours and by the faster train is 350 / (s + 5) hours. The difference in times equals 7 / 3 hours. Setting up the equation 350 / s - 350 / (s + 5) = 7 / 3 and solving for s gives the slower speed. Adding 5 then yields the faster speed.

Step-by-Step Solution:
Step 1: Let speed of slower train be s km/h, faster train speed = s + 5 km/h.Step 2: Time taken by slower train = 350 / s hours.Step 3: Time taken by faster train = 350 / (s + 5) hours.Step 4: Given that 350 / s - 350 / (s + 5) = 7 / 3.Step 5: Factor out 350: 350 * (1 / s - 1 / (s + 5)) = 7 / 3.Step 6: Compute the bracket: 1 / s - 1 / (s + 5) = (s + 5 - s) / (s * (s + 5)) = 5 / (s * (s + 5)).Step 7: So 350 * 5 / (s * (s + 5)) = 7 / 3, which simplifies to 1750 / (s * (s + 5)) = 7 / 3.Step 8: Cross multiply to get 1750 * 3 = 7 * s * (s + 5) which simplifies to 5250 = 7s^2 + 35s.Step 9: Divide by 7: 750 = s^2 + 5s. So s^2 + 5s - 750 = 0.Step 10: Solve the quadratic: s = 25 or a negative value. We take s = 25 km/h as valid.Step 11: Faster train speed = s + 5 = 25 + 5 = 30 km/h.

Verification / Alternative check:
Time for slower train at 25 km/h = 350 / 25 = 14 hours.Time for faster train at 30 km/h = 350 / 30 ≈ 11.67 hours = 35 / 3 hours.Difference = 14 - 35 / 3 = 42 / 3 - 35 / 3 = 7 / 3 hours = 2 hours 20 minutes, which matches the question.

Why Other Options Are Wrong:
Speeds such as 36 km/h, 34 km/h or 40 km/h do not yield a time difference of 2 hours 20 minutes when used with the same 5 km/h gap. The option 28 km/h is less than the calculated faster speed and would correspond to a different pair of speeds. Only 30 km/h satisfies both the distance and time conditions.

Common Pitfalls:
Learners sometimes misconvert 2 hours 20 minutes into 2.2 hours instead of 7 / 3 hours. Others incorrectly set up the equation, reversing the time difference or forgetting that the slower train takes longer. Algebraic mistakes in solving the quadratic equation can also lead to incorrect speeds. Careful handling of fractions and a quick check by back substitution help to avoid these errors.

Final Answer:
The speed of the faster train is 30 km/h.