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# If r, s, t, and u are distinct, consecutive prime integers less than 3

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Math Expert
Joined: 02 Sep 2009
Posts: 58340
If r, s, t, and u are distinct, consecutive prime integers less than 3  [#permalink]

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08 Feb 2019, 04:43
00:00

Difficulty:

65% (hard)

Question Stats:

46% (02:03) correct 54% (02:25) wrong based on 35 sessions

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If r, s, t, and u are distinct, consecutive prime integers less than 31, then which of the following could be the average (arithmetic mean) of r, s, t, and u ?

I. 4.25
II. 9
III. 22

A. I only
B. I and II only
C. I and III only
D. II and III only
E. I, II and III

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Re: If r, s, t, and u are distinct, consecutive prime integers less than 3  [#permalink]

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08 Feb 2019, 04:54
Bunuel wrote:
If r, s, t, and u are distinct, consecutive prime integers less than 31, then which of the following could be the average (arithmetic mean) of r, s, t, and u ?

I. 4.25
II. 9
III. 22

A. I only
B. I and II only
C. I and III only
D. II and III only
E. I, II and III

Key word: r, s, t, and u are distinct, consecutive prime integers less than 31

S1 = 17/4 = 4.25
2,3,5,7

S2 = 36/4 = 9
5,7,11,13

S3 = 88/4 = 22
17,19,23,29

E
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If r, s, t, and u are distinct, consecutive prime integers less than 3  [#permalink]

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08 Feb 2019, 05:05
Bunuel wrote:
If r, s, t, and u are distinct, consecutive prime integers less than 31, then which of the following could be the average (arithmetic mean) of r, s, t, and u ?

I. 4.25
II. 9
III. 22

A. I only
B. I and II only
C. I and III only
D. II and III only
E. I, II and III

Nice question

primes less than 31.

2-3-5-7-11-13-17-19-23-29.

1) 2 + 3 + 5 + 7 = 17.

17 / 4 = 4.26.

2) 5 + 7 + 11 + 13 =36

36 / 4 = 9.

3) 17 + 19 + 23 + 29 =88

88/4 = 22.

Intern
Joined: 25 Jan 2019
Posts: 4
Re: If r, s, t, and u are distinct, consecutive prime integers less than 3  [#permalink]

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09 Feb 2019, 06:34
Bunuel Is there any other better approach to solve this question?
Re: If r, s, t, and u are distinct, consecutive prime integers less than 3   [#permalink] 09 Feb 2019, 06:34
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