GRE Prep Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
Track
Your Progress
Practice
Pays
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
At a convention of monsters, 2/5 have no horns,
[#permalink]
10 Jul 2019, 14:29
10 Jul 2019, 14:29
Expert Reply
6
Bookmarks
00:00
Question Stats:
86% (02:07) correct
13% (02:05) wrong based on 81 sessions
Hide Show timer Statistics
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,
[#permalink]
17 Sep 2019, 07:08
17 Sep 2019, 07:08
4
Expert Reply
1
Bookmarks
\(x- (\frac{2x}{5}+\frac{x}{7}+\frac{x}{3}) = 26\)
\(\frac{13x}{105} = 26\)
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,
[#permalink]
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
.
