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h-n-h/3=36
3h-(n-h)=108
h=n+108/4

n+108 should be div by 4
n+108/4<=n
108<=3n
n>=36
n<=180

Poss value----36,40,44.........180

180-36/4+1
144/4+1
36+1=37
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C is the correct answer.

To score exactly 36 points, an archer must have shot the target at least 36 times.

All the archers can still end up with exactly 36 points as long as their missed shots are 3 times more than their hit shot. For xample 3 and 1, 6 and 2, 9 and 3, etc. 180 - 36 = 144 spare targets for this hit and miss with the ratio 3 - 1. So that means 144/4 = 36 other archers that get to hit and miss to score exactly 36 points.
Plus 1 archer who shoots 36/36 with 0 miss to score 36 points, we have 36 + 1 = 37 archers in total.
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Let a be the no. of targets hit and b be the no. of targets missed by the archer.
Given a-b/3 = 36. we need to find the pairs of a and b satisfying these values.

1) a=36, b=0
2) a= 37, b=3.

If you see the values of a are increasing by 1 and values of b are increasing by 3. this will continue till a+b=180.
also the value of a+b is increasing by 4, starting from 36.

Hence from 36 ( 4X9) to 180 ( 4X45) there are a total of 37 pairs of a and b values including both.

Hence the answer is 37.
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i am going with option c.
we are given, winning points - h, points lost - 1/3m, total score will be h - 1/3m = 36
h -1/3(n-h) = 36
4/3h - 1/3n = 36
4h - n = 108
constraint, max n<180
4h - 108 < 180 = h < 72
min, m>0, h=36, m=0 considering 36 points, 36 targets, h>36
as no 2 archers shot at same no. of targets, unique hit and miss for all
total archers = 72-36+1 = 37.
option a and b are incorrect as they undercount
option d and e are incorrect as they overcount
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let total number of targets be x
A hit=+1 point
miss=-(1/3) point
to get 36, miss a multiple of 3
minimum target
to get 36 points, archer must shoot 36 and miss 0
Min x=36
Max x=180
multiples of 4 in 36 to180
(last-fist)/spacing+1
(180-36)/4+1
144/4+1 =
36+1=37
ans: C 37
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Let, number of targets hit =H & number of targets missed= M
H-M/3 =36
3H-M=108 or M=3H-108
Total number of targets shot= N=H+M
N= H+3H-108 or N= 4H-108

M>= 0
3H-108 >= 0 or H>=36

Atmost 180 targets can be shot,
N=< 180
4H-108 =<180 or H=<72
We get, 36=<H=<72
Maximum number of archers in the group = 72-36+1= 37.

C
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This question took me some time to understand. So there are 180 targets, each archer (we dont know how many are there) can choose to hit as many as they want (30, 40, 50, 180 etc) but the catch is all of them totalled 36 each and no one shot at same # of targets. Hit the target +1; miss -1/3.

So someone could choose 36, hit them all and score 36. Someone could choose 37 hit 36 miss one but then score doesnt match 36. we have to give 3 multiples to targets missed so we can get whole number score. Trying 40, 37 hit, 3 missed we get 36. Trying 44, 38 hit 6 missed we get 36, so this is increasing by multiple of 4. we can keep going 36, 40, 44, 48, .... 180, counting we get 37.
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I selected answer C.
Let's begin by beginning with the smallest possible case.

That is, one archer hits 36/36 shots and scores 36 points.

Logically, the next smallest case would be the archer who hits 37 shots but misses 3, bringing the total points earned to 36, and total attempted shots to 40.

Our goal is to find out the maximum number of this sequence of numbers which does not exceed 180.

Notice to calculate the number of missed shots, we can subtract one from the archer's number and multiply by 3. I.e. Archer #1 missed 0 shots, Archer #2 missed 3 shots (2-1 = 1 ) * 3, and Archer #3 missed (3-1=2) *3 = 6 shots, etc.

So now we can use the answer choices to backtrack our work. Let's start with D. If the 45th archer missed 44*3 = 132 shots, how many points would he have had to score in order to make sure his score is 36? Well, he would have had to score 44+36=80 points which exceeds 180, so this is too high.

Let's try 37. 37-1= 36*3 = 108 missed shots, so he would have had to score 36+36=72 shots to get 180. That's exactly 180, so this is our answer.
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Let \(w\) be the number of targets hit.
Let \(l\) be the number of targets missed.

Constraints:
\(w - \frac{1}{3}l = 36\)
\(w + l \le 180\) (where \(w + l\) cannot be greater than 180)

Listing out the possibilities
\(l = 0, w = 36\)
\(l = 3, w = 37\)
\(l = 6, w = 38\)
\(l = 9, w = 39\)
...
\(l = 102, w = 70\)
\(l = 105, w = 71\)
\(l = 108, w = 72\)

The number of possibilities = \(72 - 35 = 37\).
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Answer: C) 37 Archers
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ok so h+m=180 and h-(m/3)=36

lets say m=0 then h=36 is one possibility

when T has to remain 36 but each player had different targets..we see everytime h increases by 1, 3 of m needs to be in place to get 36... ie, player will play 4 shots, 1 good, 3 bad to nullify..this keep mutiplying in the same ratio

so possibility is 36,40,44,48....180

\( ((180-36)/4))+1 =37 is the answer (C)\)
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let x be the successful hits
y be the number of misses, and T be to total shot a person took

so T = x + y
given for every successful hit, 1 point and for every miss -1/3 and each archer scored 36

so x(1) - 1/3(y) = 36
we know that y = T - x
x - 1/3T +1/3x = 36
4/3x - 1/3T = 36
x = (T + 108)/4

now X cannot be a fraction or a decimal, and as 108 is divisible by 4. T should also be divisible by 4

so T can be 0,4,8,12,16...
but to find out the exact range of T
we know that X is less than T meaning

(T+108)/4 <= T
T+108 <= 4T
T >=36

similarly, T itself cannot be more than the number of targets, so T <=180

range of T then would be as follows -> 36<= T <= 180
and we derived T is always divisible by 4, so possible values include 36, 40, 44, ... 180

to find count... 36 is first term, 180 is last term, and difference between each is 4

(180-36)/4 + 1 = 36 + 1 = 37
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Let the number of shots taken be N, number of hits be H and number of misses be M

N = H + M; M = N-H
H - M/3 = H - (N-H)/3 = 36
3H - (N-H) = 108
4H - N = 108; N = 4H - 108

N <= 180
4H - 108 <= 180
H <= 288/4 = 72

H <= N = 4H - 108
3H >= 108
H >= 108/3 = 36

There is unique number of shots for each number of hit since N = 4H - 108

36 <= H <=72

Maximum number of archers in the group = 72 - 36 + 1 = 37

IMO C
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This was slightly difficult for me and I'm not sure of the answer.

Got 36 (B) as the answer here
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Tough question... here are my thoughts

If every archer finished with a score of 36 and each target hit is one point, then the minimum shots each person could have taken is 36.
We need to find other combinations of wins and losses that could end with a score of 36 since no two archers shot at the same number of targets
Combination of wins and lossesScoreShots taken/ number of targets
36 wins,0 losses3636
37 wins, 3 losses3640
38 wins, 6 losses3644
39 wins, 9 losses3648
40 wins, 12 losses3652

There is a pattern of multiples of 4 to have a score of 36 under the number of shots taken. Assume that an archer shot as many as 180 times, and an archer shot as low as 36 times, how many different combinations could have amounted to 36, by different people.

Since the minimum score is 36, we need to find how many multiples of are between 36 and 180 to find the maximum number of archers in the group.
Max term = 180
Min term = 36
Common difference = 4
(180-36)/ 4 = 144/4 = 36

Bunuel
In an archery contest with 180 different targets, an archer may shoot at any number of these targets. For each target hit, the archer earns 1 point, and for each target missed, the archer loses 1/3 point. A group of archers participated in the contest, and no two archers in the group shot at the same number of targets. If every archer in the group finished with a score of exactly 36 points, what is the maximum possible number of archers in the group?

A. 35
B. 36
C. 37
D. 38
E. 45


 


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N number of target archer shoots at
0 <= n <= 180

H number of hits
M number of misses

H+M = N

score is H - 1/3M = 36

since M=N-H

H - 1/3 (N-H) =36

H -1/3 N + 1/3H = 36
4/3 H = 36 + 1/3N

4H + 108 +N
H = 108 +N/4

H is integer so 108+N must be multiple of 4

108 + N <= 4N
108 <= 3N
n>= 36, 40, 44 .... 180

180 -36 / 4 +1 = 36 + 1 = 37

Answer C 37
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let
t= number of targets the archer attempted
h= number of target hits
misses=t-h


The score is
h-1/3(t-h)=36

3h-(t-h)=108
4h-t=108

t=4h-108
h=t+108/4

h must be an integer, so t+108 must be divisible by 4
Since 108 is divisible by 4, t must be a multiple of 4
h=<t

t+108/4=<t
t+108<=4t
108<=3t
t>=36

As per arithmetic sequence
180-36/4+1
144/4+1
36+1
=37

Bunuel
In an archery contest with 180 different targets, an archer may shoot at any number of these targets. For each target hit, the archer earns 1 point, and for each target missed, the archer loses 1/3 point. A group of archers participated in the contest, and no two archers in the group shot at the same number of targets. If every archer in the group finished with a score of exactly 36 points, what is the maximum possible number of archers in the group?

A. 35
B. 36
C. 37
D. 38
E. 45


 


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