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Miguel’s bowling team bowled a practice round in preparation for their

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New post 20 Feb 2019, 01:18
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Miguel’s bowling team bowled a practice round in preparation for their upcoming league game. The team’s average (arithmetic mean) score for the practice round was 180. Miguelscored 190, Janice scored 200, and Thad scored 210. If no team member scored less than 165, and none of the remaining team members scored greater than 170, what is one possible value for the number of members on Miguel’s team? (Note: Bowling scores are always positive integers.)

I. 7
II. 8
III. 9

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

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Miguel’s bowling team bowled a practice round in preparation for their  [#permalink]

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New post Updated on: 20 Feb 2019, 03:45
Bunuel wrote:
Miguel’s bowling team bowled a practice round in preparation for their upcoming league game. The team’s average (arithmetic mean) score for the practice round was 180. Miguelscored 190, Janice scored 200, and Thad scored 210. If no team member scored less than 165, and none of the remaining team members scored greater than 170, what is one possible value for the number of members on Miguel’s team? (Note: Bowling scores are always positive integers.)

I. 7
II. 8
III. 9

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


total score of given three players = 600
the avg of the team = 180
now since value of scores has to be an integer and in range from 165-170
we can solve question using given answer choices;
i.e
1 . 7 players in team ; so we need to find the score of 4 player ; which would be ( 180* 7-600) ; 1260-600= 660 and 660/4 = 165 ; possible player s 7 YES
2. 8 players in team ; so the score of 5 players ; ( 180*8-600) = 1440-600; 840 = each of 5 player scored 168 possible yes
3. 9 players in team ; so the score of 6 players ; ( 180*9-600) = 1620-600;1020 ; each of 6 players scored ; 170 possible ; yes

so we can say that number of team members can be either 7,8,9
IMO E
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Originally posted by Archit3110 on 20 Feb 2019, 02:21.
Last edited by Archit3110 on 20 Feb 2019, 03:45, edited 1 time in total.
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New post 20 Feb 2019, 02:47
Hello,

I solved it as follows:

190-180= +10
200-180=+20
210-180=+30
Total Excess = +10+20+30=+60
In order for the average to be 180 then sum of deviations in score below 180 have to = -60

Minimum Score possible for remaining team members is 165 so max Deviation possible per remaining team member
is 180-165=-15. 60/15=4 so if every remaining member scored the minimum. The total number of players would be 4+3=7

Maximum Score possible for remaining team members is 170 so Min Deviation possible per remaining team member
is 180-165=-10. 60/10=6 so if every remaining member scored the maximum number would 6+3=9

Team members range is: 7<=N=<9, IMO E
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Miguel’s bowling team bowled a practice round in preparation for their   [#permalink] 20 Feb 2019, 02:47
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