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02 Dec 2019, 00:43
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A
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
45%
(medium)
Question Stats:
69% (01:59) correct
31%
(02:13)
wrong
based on 196
sessions
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Aaron, Ben, and Cliff all make resolutions to exercise more in the new year. Aaron resolves to exercise on every fourth day, Ben resolves to exercise on every seventh day, and Cliff resolves to exercise on every tenth day. If they follow their resolutions exactly, what is the probability that they all exercised on a randomly selected day, assuming this is not a leap year?
A. \(\frac{2}{365}\)
B. \(\frac{1}{280}\)
C. \(\frac{1}{73}\)
D. \(\frac{13}{65}\)
E. \(\frac{28}{73}\)
Are You Up For the Challenge: 700 Level Questions
A. \(\frac{2}{365}\)
B. \(\frac{1}{280}\)
C. \(\frac{1}{73}\)
D. \(\frac{13}{65}\)
E. \(\frac{28}{73}\)
Are You Up For the Challenge: 700 Level Questions
02 Dec 2019, 01:18
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My way of reasoning
\(P=\frac{{Favorable outcomes}}{{Possible outcomes}}\)
1) Favorable outcomes: we need the LCM for \(2^2\), \(7\) and \(2*5\) LCM=140; \(\frac{365}{140}=2.\) SO we have got only two days when they all exersice
2) Posible outcomes: normal year does have 365 days
\(\frac{2}{365}\)
IMO
Ans: A
\(P=\frac{{Favorable outcomes}}{{Possible outcomes}}\)
1) Favorable outcomes: we need the LCM for \(2^2\), \(7\) and \(2*5\) LCM=140; \(\frac{365}{140}=2.\) SO we have got only two days when they all exersice
2) Posible outcomes: normal year does have 365 days
\(\frac{2}{365}\)
IMO
Ans: A
02 Dec 2019, 04:18
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Explanation:
Only 2 Days are there on which all 3 will exercise.
i.e. 140 th & 280th Day.
Probability : Favourable outcome/ total no. of outcomes
= 2/365
IMO-A
Please give kudos, if you find my explanation good enough
Only 2 Days are there on which all 3 will exercise.
i.e. 140 th & 280th Day.
Probability : Favourable outcome/ total no. of outcomes
= 2/365
IMO-A
Please give kudos, if you find my explanation good enough 
