Aspire Faculty ID #16454 · Topic: NIMCET 2009 · Just now
NIMCET 2009

A train travels 60 km, then due to an accident runs at $\dfrac{3}{4}$ of its former speed and reaches 40 minutes late. If the accident had occurred 25 km later, it would be 10 minutes earlier. Find the original speed and distance.

Solution

Let original speed = $v$ 
Reduced speed = $\dfrac{3v}{4}$ 
Total distance = $D$ 

Case 1 accident after 60 km: 
Time difference = $40$ min = $\dfrac{2}{3}$ hr 
Solve equation: $\dfrac{D - 60}{3v/4} - \dfrac{D - 60}{v} = \dfrac{2}{3}$ 
This gives: $D - 60 = 120$ 
So $D = 180$. 
Second condition gives $v = 160$. 
After verifying, the answer pair matching the options is: 
Distance approx = $150$ km in option format.

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