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At a convention of monsters, 2th has no horns, th has one horn, Ard has two horns, and 5 7 3 remaining 26 have three or more horns. How many monsters are attending the convention?

(C) 180

(E) 260

(B) 130

(A) 100

(D) 210

Can anyone solve this question
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Can anyone solve this question
at-a-convention-of-monsters-2-5-have-no-horns-1-7-have-one-horn-300364.html#p2316091

the answer would be 210
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You can use the options only to figure out the answer.
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Not necessary to use options only. The clue here is "remaining 26". Assume total monsters are X. What the problem statement essentially says is" 1 - ( sum of all monsters with fractions) = 26
1-(92/105)X= 26
X= 210
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deaddumbledore
Not necessary to use options only. The clue here is "remaining 26". Assume total monsters are X. What the problem statement essentially says is" 1 - ( sum of all monsters with fractions) = 26
1-(92/105)X= 26
X= 210
Can u tell me how u got 92
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anybody knows how to solve this ?
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Ayush1725
At a convention of monsters, 2th has no horns, th has one horn, Ard has two horns, and 5 7 3 remaining 26 have three or more horns. How many monsters are attending the convention?

(C) 180

(E) 260

(B) 130

(A) 100

(D) 210
(1-sum of all monsters)=26, which is 13/105X=26 ----> x=210

Ayush1725
Can u tell me how u got 92
2/5+1/7/1/3

*2/5+1/7+1/3 =92/105
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How to solve this
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Ayush1725
How to solve this
251. the seq will be like 500, 499.5, 499, 498 ..... So, S(499) = 500-498*0.5 = 251
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Ayush1725
At a convention of monsters, 2th has no horns, th has one horn, Ard has two horns, and 5 7 3 remaining 26 have three or more horns. How many monsters are attending the convention?

(C) 180

(E) 260

(B) 130

(A) 100

(D) 210
B
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Ultimately deduce the problem till
Sn=S1- (n-1)/2
Then S499=500-(499-1)/2=251

Actually Sn=Sk- (n-k)/2
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agnee
Ultimately deduce the problem till
Sn=S1- (n-1)/2
Then S499=500-(499-1)/2=251
Can you write the answer in a paper and send i still didn’t understand
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