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GMAT Club World Cup 2026 (Day 4): In an archery contest with 180

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twinkle2311

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10 Jul 2026, 00:54

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Let H = hits and M = misses
Score = 36, so
H - (M/3) = 36
-> 3H - M = 108

Targets attempted = H+M
Substitute M = 3H -108
Targets attempted = 4H-108

So the no. of targets attempted must be a multiple of 4.
Also, it must lie between 36(all hits) and 180(all targets).
Possible values are 36,40,44,......., 180
No. of values = (180-36)/4 + 1 = 37
Since no 2 archers attempted the same no. of targets, the maximum no. of archers is 37.

Ans : C

FlushLoop

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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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h - m/3 = 36

3h - m = 108 -- Equation 1

The number of targets shot at is: h + m

Using Equation 1

h + m = h + (3h - 108)

= 4h - 108

4h - 108 ≤ 180

4h ≤ 288 h ≤ 72

Since m ≥ 0

3h - 108 ≥ 0

h ≥ 36

So h can be any integer from 36 to 72, inclusive.

That's: 72 - 36 + 1 = 37

Option C

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10 Jul 2026, 02:44

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Let's break down the stem first:
Total targets- 180
Points for hitting the target- +1
Points for missing the target- -(1/3)

Let archers in the team= A

Points for all archers= 36
(no two archers hit the same amount of targets)

Let targets hit: X
Let targets Missed: Y
Let total shots taken: T

T=X+Y ----(1)
Points-
X(1) - Y(1/3)=36 ----(2)
X = 36 + (Y/3) ----(3)

Since the archer scores 36 points, best case scenario is that they take 36 shots and hit targets every time.
After that, for every target they miss, they lose points, so they will have to hit targets to make up for the deficit in the 36.

Therefore-
X >= 36 ----(4)

Using (3)-
X = 36 + (Y/3)
3X = 108 + Y
3X - 108 = Y

Now using (1) again:
X+Y=T
X + (3X-108)=T
4X-108=T ----(5)

Since, T can be 180 at the most-
4X-108<=180
4X<=288
X<=72 ----(6)

Combining (4) and (6)-
36<=X<=72
Possible values of X: 37

Now since no two archers hit the same number of targets, Using equation (5), every integer value of X will give a new value of T.
Therefore, possible values of T=37

Ans: C

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10 Jul 2026, 02:47

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IMO, answer should be (C) 37

Let N = Total targets shot and H = Total targets hit.
Also, H <= N

Since each Hit earns +1, and each miss loses -1/3 point:

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

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

Given max targets could be 180 only so, N <=180

So, N must be a multiple of 4 and between 36 and 180 (inclusive).

Number = (180-36)/4 +1 = 37

Since archer shoots a different no of targets, the max number of archers is the number of valid n values.

Answer: C: 37

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10 Jul 2026, 02:57

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Total no of targets =180
No of hits = a and misses =b
a*1-b/3=36
3a-b=36*3=108
3a-108=b
b>=0 (misses cannot be less than 0)
3a-108>=0 that means a>=36
total points < = 180
max point a person can score without any miss is 180*1 =180
therefore whatever is total score of all participants
a+b<180
a+3a-108<180
4a<=288
a<=72

therefore 36<=a<=72

a could be 72-36+1= 37

Bunuel

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10 Jul 2026, 03:00

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s = Shots
h = hits

h- s-h/3 = 36
4h = s + 108
s + 108 will be divisble by 4

Min possible s will be 36
max is 180

Value of s can be from 36, 40, 44, 48 to 108

180-36/4 + 1 = 37

Ans is 37 c

Bunuel

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10 Jul 2026, 03:11

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Let n be the number of targets shot. H-hits; M-misses

Hit=+1
Miss=-1/3
H - M/3 = 36
Putting M = N-H
H-(N-H)/3 = 36

solving, H= (N+108)/4
So N can be 4,8, etc upto 180
But hits cannot exceed shots.
H <= N
N + 108 / 4 <=N
Therefore N>=36

Possible numbers are multiples of 4 from 36 to 180
180-36 / 4 +1 = 37 numbers
So 37 maximum archers are present

Bunuel

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The answer is C.37
For an archer shooting n targets with h hits score = h -(n-h)/3=36, so 4h-n=108
Each archer needs a distinct n, and h = (n+108)/4 must be a whole number with h no more than n, and n is no more than 180. For h to be whole, n must be a multiple of 4. The condition is that h is at most n means n must be at least 36. And n is at most 180.
So valid values of n are 36,40, 44 and all the way up to 180. That gives (180-36)/4+1=37 archers.

Bunuel

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10 Jul 2026, 03:36

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Each person can hit any no. of targets, let be n.
n can be maximum of 180

by points if h be the misses,
(1*h)-(1/3 * (n-h)) = 36
h=(n+108) / 4 = (n/4)+ 27..n has to a multiple of 4

misses = n-h = n- ((n+108) / 4) = (3/4)*(n-36)
for misses >= 0 , n has to be minimum 36

so 36 <= n <=180 & multiples of 4:

No. of terms = (180-36)/4 + 1 = 37

Ans 37

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Total Targets = 180
Targets that can be hit ≤ 180

Points per archer = $a - \frac{b}{3} = 36$

$a = 36 + \frac{b}{3}$ ----------(I)

a= targets hit, b = targets lost

also, a + b ≤ 180-------(II)

we have see for how many values of b is a+b ≤180

so,
substituting I in II

$36 + \frac{b}{3 }+ b ≤ 180$

b ≤ 108

so b can have$ \frac{108 - 0}{3} + 1 = 37$ values

C

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The minimum number of shots is 36 (all targets hit)
The next option would be that teh archer would get another target hit and three more targets missed to compensate (4 more shots in total).

We can continue until we get 180 shots. To calculate the number of targets hit in that case:

36 + (180-36)/4 = 72

36 original targets hit + 36 additional targets hit + 108 targets missed = 180 shots

So there are 72-36+1 = 37 possible number of archers in the group

Answer C

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10 Jul 2026, 05:26

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let k is the number of targets shot and a is the number of shots hit.

a - (k-a)/3 = 36
4a-k = 108
0=<k=<180
so 27<a=<72

ans : e

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10 Jul 2026, 05:30

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Option C - 37

Total points = N =36

Hit + miss = Total points
H + M = N - 1
H - 1/3 M = 36

Solving

3H - M = 108
M = 3H-108 - 2

Putting the value back in 1

H +(3H -108) = N
4H - 108 = N - 3

So M and H >= 0 ,they cannot be negative as they targets
So putting value in 2

3H - 108 >=0
H >= 36

total targets <= 180
so putting in 3
4H - 108 <=180
H <= 72

So 72 < H> 36
Number of archers = 72 - 36 + 1 = 37
1 is included as when counting btw 72 and 36 we miss out on actual 36

Bunuel

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I got option choice C as an answer!!
--- here h= Number of target hit , M = to number of target miss ,, n = Total number of target shots ,,,N = H+ M
---- Here the equation is 1 h - 1/3 m = 36 ,, 3h - m = 108 ,, m = 3h - 108 ,, N = H + M after putting m value n = 4h - 108
------ Now I try to find the maximum and the minimum value ,, m > = 0 always so ,, m = > 3h - 108 > = then h > = 36
and we know as given in statement n < = 180 so ,, 4h - 108 < = 180 so h < = 72
--- Now that we know the value of h should be In between 36 and 72 inclusive and n = 4h - 108 For every separate value of H we will get a different value of N so the number of values we can get is 72 - 36 + 1 = 37 so We get 37 different Maximum values

--- That's how I got my answer if anybody has any better methods or shortcut feel free to tag : )

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10 Jul 2026, 05:47

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archer1: hits=36, missed=0, 36+0=36

For every extra successful shot, three missed shots must be added to compensate:
archer2: hits=37, missed=3, 37+3=40
archer3: hits=38, missed=6, 38+6=44
archer4: hits=39, missed=9, 39+9=48
archer5: hits=40, missed=12, 40+12=52

It will continue until the total numder of shots reaches 180.

Serie is [36, 40, 44, 48, 52,... 180] that has (180-36)/4 + 1 = 37 elements

The correct answer is C

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Let total hits be h and total shots shot at be t and misses be t-h
So we get h-1/3(t-h)=36. Simplify that and you have 4h-t= 108
Therefore t= 4h-108
t is less/equal 180 4h-108</= 180
h</=72
So 36<= h<=72
We need to find possible values using the equation t=4h-108. So we get 36, 40, 44-----180 which forms an Arithmetic Sequence with a common difference d. So no. of terms 180-36/4+1= 37
Ans C

Bunuel

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10 Jul 2026, 06:03

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total(T), Point won(p), Point deducted(d)
T=p+d

and given p-d/3=36
or 3p-d=108

add both equations- 4p=108+T
p=(108+T)/4

p is an integer, hence T must be divisible by 4
And T<=180

point won- will have to be amongst 4,8,12...180

so all multiples of 4 in 180- 180/4=45

45 different archers can satisfy the conditions.

Answer E

Bunuel

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10 Jul 2026, 06:14

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h=hits
m=missed

constraints:
m >= 0
h+m <= 180

h-m/3 = 36
3h-m = 108

m = 3h - 108
as m >= 0 then h >= 36

adding h:
h+m = 4h - 108 <= 180
h <= 72

So: 36 <= h <= 72 -> 72-36+1 = 37 possible number of archers

IMO C

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Possible cases:
targets hit = 36, total = 36
targets hit = 37 and targets missed = 3, total = 40
targets hit = 38 and targets missed = 6, total = 44
targets hit = 39 and targets missed = 9, total = 48
...

secuence is 36, 40, 44, 48, ... 180

number of elements: (180-36)/4 + 1 = 36 + 1 = 37

The answer is C

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Question is interesting.

My first thought is that I need a combination all possible distinct values of x such that if archer hits x targets and misses y then sum of x - y/3 = 36
and given x + y <= 180
and we can also say since we want total score 36 => x > 36

Not sure how can I get the exact count. choose ans D as an instinct

Bunuel

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