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20 Apr 2020, 05:16
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C
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In a tournament, there are two groups of teams. Group A has 20 teams and Group B has 25 teams. The average (arithmetic mean) scores of teams in Group A and B are 20 and 25, respectively. The highest and lowest scores in Group A are 25 and 15, respectively and the highest and lowest scores in Group B are 32 and 24, respectively. What can be the minimum and maximum value of Group A’s average if 5 teams are transferred from Group B to Group A?
A. 20.8 and 22.8
B. 20.8 and 22.4
C. 20.8 and 21.8
D. 20.8 and 21.6
E. 20.6 and 22.6
Are You Up For the Challenge: 700 Level Questions
A. 20.8 and 22.8
B. 20.8 and 22.4
C. 20.8 and 21.8
D. 20.8 and 21.6
E. 20.6 and 22.6
Are You Up For the Challenge: 700 Level Questions
yashikaaggarwal
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20 Apr 2020, 05:43
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Answer should be B
Average of Group A is 400 [A=20*20]
The lowerst score of group B team is 24.
Let's suppose there were 5 members in team B with 24 (minimum) score and they are transferred.
So 400+24*5=400+120=520
New Average=520/25=20.8
Now suppose there were 5 members in team B with 32 (maximum) score and they are transferred.
So 400+32*5=400+160=560
New Average=560/25=22.4
Even if the score Exceeds 24 or is short from 32.
The range will nearly lie between 20.8-22.4
So answer should be B
Posted from my mobile device
Average of Group A is 400 [A=20*20]
The lowerst score of group B team is 24.
Let's suppose there were 5 members in team B with 24 (minimum) score and they are transferred.
So 400+24*5=400+120=520
New Average=520/25=20.8
Now suppose there were 5 members in team B with 32 (maximum) score and they are transferred.
So 400+32*5=400+160=560
New Average=560/25=22.4
Even if the score Exceeds 24 or is short from 32.
The range will nearly lie between 20.8-22.4
So answer should be B
Posted from my mobile device
23 Apr 2020, 06:48
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Bunuel
...................A......B
# of teams....20...25
Average........20...25
5 teams are to be transferred from B to A..
Let Group B score be such that 5 teams have highest possible score 32, 5 teams have lowest score 24 and remaining 15 hare adjusted accordingly..
32, 32, 32, 32, 32, .
MAX average of A after transfer..
All 5 have maximum possible, so take remaining 20 as minimum possible = 24. Total of these 20 teams = 20*24=480.
Thus, maximum possible score of these 5 teams = 25*25-24*20=145, and average= \(\frac{20*20+145}{25}=\frac{545}{25}=21.8\)
MIN average of A after transfer..
All 5 have minimum possible, that is 24, so NEW average of A= \(\frac{20*20+24*5}{25}=\frac{520}{25}=20.8\)
So Min and Max = 20.8 and 21.8
C
