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Every member of a local chess club is assigned one of two skill levels

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Every member of a local chess club is assigned one of two skill levels  [#permalink]

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New post 09 Dec 2019, 00:06
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Question Stats:

15% (04:27) correct 85% (02:54) wrong based on 26 sessions

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Every member of a local chess club is assigned one of two skill levels: A or B. At the beginning of the month, there were 'a' level A members and 'b' level B members. The ratio of a to b was 5:4. At the end of the month, the ratio of level A members to level B members has changed to 7:8. If nobody left the club and no new members joined during the month, how many total members are in the club?

(1) 49 < a < 79
(2) 29 < b <59


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Every member of a local chess club is assigned one of two skill levels  [#permalink]

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New post 09 Dec 2019, 08:57
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Bunuel wrote:
Every member of a local chess club is assigned one of two skill levels: A or B. At the beginning of the month, there were 'a' level A members and 'b' level B members. The ratio of a to b was 5:4. At the end of the month, the ratio of level A members to level B members has changed to 7:8. If nobody left the club and no new members joined during the month, how many total members are in the club?

(1) 49 < a < 79
(2) 29 < b <59


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Analyzing the question:
Since the ratios are 5:4 and 7:8, we know that the total amount of club members must be able to be divided by 5 + 4 and 7 + 8. Hence the total amount must have a factor of 9 and 15. The LCM of 9 and 15 is 9 * 15 / 3 = 45. Therefore the total amount of members is a multiple of 45. 'a' is equal to 5/9 of this multiple of 45, and 'b' is equal to 4/9 of this multiple of 45. Hence 'a' must be a multiple of 25 and 'b' must be a multiple of 20, and they must have the same multiplier.

Statement 1:
'a' must be a multiple of 25, here we can have either 50 or 75 so insufficient.

Statement 2:
'b' must be a multiple of 20, we can only have b = 40 here so this is sufficient.

Ans: B

Note: If we change statement 2 to 19 < b < 59. Then we can select either b = 20 or b = 40 so it would be insufficient on its own. However combining with the original statement 1, we can only choose a = 50 paired with b = 40 since they must have the 5:4 ratio (or the same case of total of 90). In that case, we would choose C.
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Re: Every member of a local chess club is assigned one of two skill levels  [#permalink]

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New post 13 Dec 2019, 08:53
Bunuel wrote:
Every member of a local chess club is assigned one of two skill levels: A or B. At the beginning of the month, there were 'a' level A members and 'b' level B members. The ratio of a to b was 5:4. At the end of the month, the ratio of level A members to level B members has changed to 7:8. If nobody left the club and no new members joined during the month, how many total members are in the club?

(1) 49 < a < 79
(2) 29 < b <59


If no members joined or left during the period and the ratio decreased from 5/4 to 7/8;
Then some members of 'a' must have advanced to 'b'; a/b=5/4 became a-x/b+x=7/8;

Total members at the start was 5x+4x=9x
Total members at the end was 7y+8y=15y
Total members lcm(9x,15y)=(3^2*5xy)=45k

a=5/9*45k=25k
b=4/9*45k=20k

(1) 49 < a < 79 insufic

a=(50,75)

(2) 29 < b <59 sufic

b=40; a=5b/4=50; a+b=90.

Ans (B)
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Re: Every member of a local chess club is assigned one of two skill levels   [#permalink] 13 Dec 2019, 08:53
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