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A train covers a distance of (300\text{ km}). If its speed were (10\text{ km/h}) more, it would take (1\text{ h}) less. What was the actual speed of the train?एक ट्रेन (300\text{ km}) दूरी तय करती है। यदि उसकी चाल (10\text{ km/h}) अधिक होती, तो समय (1\text{ h}) कम लगता। ट्रेन की वास्तविक चाल क्या थी?
Answer and explanationउत्तर और व्याख्या
Correct answer:सही उत्तर: (50\text{ km/h})(50\text{ km/h})
Let the actual speed be (x\text{ km/h}), then (\frac{300}{x}-\frac{300}{x+10}=1). This gives (x^2+10x-3000=0), so the positive root is (x=50).वास्तविक चाल (x\text{ km/h}) मानें, तब (\frac{300}{x}-\frac{300}{x+10}=1)। इससे (x^2+10x-3000=0), इसलिए धनात्मक मूल (x=50) है।
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