Question
Download Solution PDFएक माणूस A ते B पर्यंतचा प्रवास 36 किमी/तास वेगाने 74 मिनिटांमध्ये करतो आणि B ते C पर्यंतचा प्रवास 45 किमी/तास वेगाने 111 मिनिटांमध्ये पूर्ण करतो. संपूर्ण प्रवासाचा सरासरी वेग शोधा.
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFदिल्याप्रमाणे:
एक माणूस A ते B पर्यंतचा प्रवास 36 किमी/तास वेगाने 74 मिनिटांमध्ये करतो आणि B ते C पर्यंतचा प्रवास 45 किमी/तास वेगाने 111 मिनिटांमध्ये पूर्ण करतो.
वापरलेले सूत्र:
सरासरी वेग = एकूण अंतर/एकूण घेतलेला वेळ
गणना:
घेतलेल्या वेळाचे गुणोत्तर = 74 : 111
वेळ = 2 : 3
सरासरी वेग = \(\frac{{36\ \times\ 2\ +\ 45\ \times\ 3}}{{2\ +\ 3}}\)
सरासरी वेग = 207/5
सरासरी वेग = 41.4 किमी/तास
∴ संपूर्ण प्रवासाचा सरासरी वेग 41.4 किमी/तास आहे.
Last updated on Jun 12, 2025
-> OSSC CGL Final Merit List has been released on the official website.
->The merit list consist the names of the candidates found suitable after the CV round.
->Earlier,The OSSC CGL Call Letter had been released for the Document Verification (Advt No.1249/OSSC).
-> Interested candidates registered from 29th November 2024 to 28th December 2024.
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