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Aaron, Ben, and Cliff all make resolutions to exercise more in the new

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Math Expert
Joined: 02 Sep 2009
Posts: 60555
Aaron, Ben, and Cliff all make resolutions to exercise more in the new  [#permalink]

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02 Dec 2019, 00:43
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Difficulty:

35% (medium)

Question Stats:

87% (02:13) correct 13% (02:26) wrong based on 15 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. $$\frac{28}{73}$$

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Senior Manager
Joined: 13 Feb 2018
Posts: 495
GMAT 1: 640 Q48 V28
Aaron, Ben, and Cliff all make resolutions to exercise more in the new  [#permalink]

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02 Dec 2019, 01:18
1
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
Senior Manager
Joined: 16 Feb 2015
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Location: United States
Concentration: Finance, Operations
Re: Aaron, Ben, and Cliff all make resolutions to exercise more in the new  [#permalink]

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02 Dec 2019, 04:18
1
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
Re: Aaron, Ben, and Cliff all make resolutions to exercise more in the new   [#permalink] 02 Dec 2019, 04:18
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