Question
Download Solution PDF96 किमीचा प्रवास जलद ट्रेनने (A) स्लो ट्रेनने (B) पेक्षा एक तास कमी लागतो. जर B चा सरासरी वेग A पेक्षा 16 किमी/ता कमी असेल, तर A चा सरासरी वेग (किमी/तास मध्ये) आहे:
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFतपशीलवार उपाय:
ट्रेन A चा वेग x किमी/तास असावा.
ट्रेनचा वेग B = (x - 16) किमी/ता
प्रश्नानुसार
\(\Rightarrow \frac{{96}}{{x - 16}} - \frac{{96}}{x}{\rm{}} = {\rm{}}1\)
\(\Rightarrow \frac{{96{\rm{\;}} \times {\rm{\;}}16}}{{x\left( {x - 16} \right)}}{\rm{ }} = {\rm{}}१\)
⇒ x 2 - 16x = 96 x 16
⇒ x 2 - 16x - 1536 = 0
⇒ (x - 48) (x + 32) = 0
⇒ x = 48 आणि x = -32 (शक्य नाही)
∴ ट्रेनचा वेग A = 48 किमी/ता
शॉर्टकट युक्ती पर्यायांमधून जा
ट्रेनचा वेग 48 किमी/तास असू द्या.
आणि ट्रेनचा वेग B = 48 - 16 = 32
⇒ (96/32) - (96/48) = 1
⇒ 3 - 2 = 1
⇒ 1 = 1 (समाधानी)
Last updated on Jul 7, 2025
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