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A certain kickball team named "The good guys" won 1/3 of their first 2
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19 Dec 2021, 13:05
19 Dec 2021, 13:05
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Question Stats:
60% (02:45) correct
40% (01:48) wrong based on 5 sessions
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A certain kickball team named "The good guys" won 1/3 of their first 24 games. If the teams lose no more than 1/9 of their remaining games, what is the least number of games they must play to ensure they win more than they loose?
a) 9
b) 10
c) 11
d) 12
e) 13
May someone tell me where is my mistake?
I tried solving backward by starting with a
\((1/9)* 9 + 16 = 17 \)
and the winning will be \((8/9) * 9 + 8 = 16 \) option a wrong
no for the rest, all of the options will give us the first option 1.something .. I round it to 2.
Until I reach 13, where the winning games is more than the lost games.
a) 9
b) 10
c) 11
d) 12
e) 13
May someone tell me where is my mistake?
I tried solving backward by starting with a
\((1/9)* 9 + 16 = 17 \)
and the winning will be \((8/9) * 9 + 8 = 16 \) option a wrong
no for the rest, all of the options will give us the first option 1.something .. I round it to 2.
Until I reach 13, where the winning games is more than the lost games.
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Re: A certain kickball team named "The good guys" won 1/3 of their first 2
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04 Mar 2022, 00:58
04 Mar 2022, 00:58
1
Asmakan wrote:
A certain kickball team named "The good guys" won 1/3 of their first 24 games. If the teams lose no more than 1/9 of their remaining games, what is the least number of games they must play to ensure they win more than they loose?
a) 9
b) 10
c) 11
d) 12
e) 13
May someone tell me where is my mistake?
I tried solving backward by starting with a
\((1/9)* 9 + 16 = 17 \)
and the winning will be \((8/9) * 9 + 8 = 16 \) option a wrong
no for the rest, all of the options will give us the first option 1.something .. I round it to 2.
Until I reach 13, where the winning games is more than the lost games.
a) 9
b) 10
c) 11
d) 12
e) 13
May someone tell me where is my mistake?
I tried solving backward by starting with a
\((1/9)* 9 + 16 = 17 \)
and the winning will be \((8/9) * 9 + 8 = 16 \) option a wrong
no for the rest, all of the options will give us the first option 1.something .. I round it to 2.
Until I reach 13, where the winning games is more than the lost games.
Games Won = \(\frac{1}{3}(24) = 8\)
Games Lost = \(24 - 8 = 16\)
Let, the total number of remaning games be \(x\)
Games Lost of the remaining = \(\frac{x}{9}\), and
Games Won of the remaining = \(x - \frac{x}{9} = \frac{8x}{9}\)
As per the question;
Total games Won ≥ Total games Lost
\(8 + \frac{8x}{9} ≥ 16 + \frac{x}{9}\)
\(\frac{8x}{9} - \frac{x}{9} ≥ 8\)
\(7x ≥ 72\)
i.e. \(x ≥ 10.28\)
Hence, option C
gmatclubot
Re: A certain kickball team named "The good guys" won 1/3 of their first 2 [#permalink]
04 Mar 2022, 00:58
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