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At a convention of monsters, 2/5 have no horns,
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10 Jul 2019, 14:29
10 Jul 2019, 14:29
Expert Reply
6
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00:00
Question Stats:
87% (02:10) correct
12% (01:52) wrong based on 70 sessions
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At a convention of monsters, \(\frac{2}{5}\) have no horns, \(\frac{1}{7}\) have one horn, \(\frac{1}{3}\) have two horns, and the remaining 26 have three or more horns. How many monsters are attending the convention?
(A) 100
(B) 130
(C) 180
(D) 210
(E) 260
(A) 100
(B) 130
(C) 180
(D) 210
(E) 260
Re: At a convention of monsters, 2/5 have no horns,
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17 Sep 2019, 07:08
17 Sep 2019, 07:08
4
Expert Reply
1
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\(x- (\frac{2x}{5}+\frac{x}{7}+\frac{x}{3}) = 26\)
\(\frac{13x}{105} = 26\)
x=210
OR
The total has to be a value that would be divisible by 3, 5 & 7.
D
\(\frac{13x}{105} = 26\)
x=210
OR
The total has to be a value that would be divisible by 3, 5 & 7.
D
General Discussion
Re: At a convention of monsters, 2/5 have no horns,
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17 Sep 2019, 03:15
17 Sep 2019, 03:15
1
Carcass wrote:
At a convention of monsters, \(\frac{2}{5}\) have no horns, \(\frac{1}{7}\) have one horn, \(\frac{1}{3}\) have two horns, and the remaining 26 have three or more horns. How many monsters are attending the convention?
(A) 100
(B) 130
(C) 180
(D) 210
(E) 260
(A) 100
(B) 130
(C) 180
(D) 210
(E) 260
any explanation will be appreciated
.
