If a student runs at 1.5 times his usual speed, he reaches his school 20 minutes early. If he runs at 0.5 times
his usual speed, how late will he reach his school?
(1) 60 min
(2) 30 min
(3)45 min
(4) 15 min
🧠 Problem Overview
A student runs at 1.5 times his usual speed and reaches school 20 minutes early.
If he runs at 0.5 times his usual speed, how late will he reach school?
Options:
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60 minutes
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30 minutes
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45 minutes
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15 minutes
📘 Concepts Used
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Speed and time are inversely proportional:
Time∝1Speed\text{Time} \propto \frac{1}{\text{Speed}}
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When speed increases, time taken decreases.
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When speed decreases, time taken increases.
🔍 Step-by-Step Solution
Let the usual time to school = T minutes
✅ Case 1: Speed = 1.5 × usual speed
New time=T1.5=23T\text{New time} = \frac{T}{1.5} = \frac{2}{3}T
This new time is 20 minutes less than the usual time:
T−23T=20⇒13T=20⇒T=60 minutesT – \frac{2}{3}T = 20 \Rightarrow \frac{1}{3}T = 20 \Rightarrow T = 60 \text{ minutes}
So, the usual time to reach school is 60 minutes.
✅ Case 2: Speed = 0.5 × usual speed
New time=T0.5=2T=2×60=120 minutes\text{New time} = \frac{T}{0.5} = 2T = 2 \times 60 = 120 \text{ minutes}
So, he takes 120 – 60 = 60 minutes more, which means he is 60 minutes late.
🎯 Final Answer: (1) 60 minutes
🧮 Summary Table
| Speed Variation | Arrival Status | Time Difference |
|---|---|---|
| 1.5× usual speed | 20 minutes early | -20 min |
| 0.5× usual speed | 60 minutes late | +60 min |
💡 Why It Matters
This type of speed-time question is a favorite in exams like:
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SSC
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Banking
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Railway exams
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CAT, GMAT, and other aptitude-based tests
It tests your understanding of ratios, inverse proportion, and time management.
✅ Key Takeaways
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When solving speed-time problems, assume a variable for the usual time or distance.
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Use inverse proportion when speed changes.
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Always compare new time with the original time to find early or late arrival.
