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

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alamin8

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09 Jul 2026, 09:26

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If exactly 36 points earned, meaning out of 180, 72 needs to be targeted.
72 x 1 = 72
108 x (1/3) = 36
Total 72 - 36 = 36
Therefore, 36 + 1 = 37 (because there is one endpoint, which needs to be added).

Answer: C

gaurikhanna17

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09 Jul 2026, 09:34

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IMO : 37
well i am not very sure of the answer here though, but this is how i got the answer :
equation 1 : a-1/3b =36
3a-b =108-->i
equation 2 : total shots (t)= a+b
equation 3 : b>=0
so using equation 3 in equation 1 ==>3a-108>=0
3a>=108
a<=36---->iv

also maximum shots an archer can shoot is 180
therefore, a+b<=180-->v
using eqn i in eqn v ==> a+3a-108<=180
4a<=288
a<=72----->vi
using vi and iv
36<=a<=72
therefore to get all the possible values of t ===> 72-36+1 = 37

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09 Jul 2026, 09:50

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Hits - Miss = Total Score
H - M/3 = 36
3H - M = 108 (This is equation 1)

Now for total shots
H + M = N
H + 3H - 108 = N
4H - 108 = N

Next, we find range,
Since, misses cant be 0
3H - 108 >= 0
H >= 36

N can not exceed 180
4H - 108 <= 180
H <= 72

Total possible archers = 72 - 36 + 1 = 37 (c)

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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?

Hits=x ; Misses= y ; Total targets=z
z=x+y

Each hit gives +1 and miss -1/3.
Total: x-(1/3)h= 36
y=3x-108 ___(y=z-x)
z-x=3x-108 : z=4x-108_____(1)

One cannot hit more targets one shoot. So, x≤z ___(z=4x-108)
x≤4x-108 ; x≥36
Minimum number of hits= 36

Archer can shoot atmost 180 targets.
z≤180
4x-108≤180 ; x≤72
Maximum number of hits= 72

x can take any value between 36 and 72 inclusive.
Total= 72-36+1= 37

C

Saurav1408

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09 Jul 2026, 10:18

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ALWAYS CHECK FOR WHAT IS GIVEN:
We have been told total 180 targets are to hit.
1 point awarded for hitting a target
-1/3 for target missed.

No archers with same number of targets hit
Every archer scores exactly 36 points.

To start this question we need to know that the must hit at least 36 targets to get 36 points.
So let us now write the ways to achieve 36 points
(we shall know that 1/3 + 1/3 + 1/3 = 1, we could say 1/3 + 1/3 + 1/3 = x

36, 37-x, 38-2x, 39-3x, 40-4x the individual total targets hit would be 36, 37+3, 38+6, and so on.
So the number of targets would be in sequence; 36, 40, 44, 48,..., 180
So this is an Arithmetic Progression with
Initial term(a) = 36, Final term(l) = 180 and Common Difference(d) = 4
So to find the total number of terms, we could use formula of 'n'th term in Arithmetic Progression: a+(n-1)d = 'n'th term
Our 'n'th term is the last term(l), put the values
36+(n-1)4=180
n= 37
Hence 37 is the answer

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09 Jul 2026, 10:19

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if the archer never miss one shot , the minimum shots needed to reach 36 point is 36 hits.
the number of shots required to keep the score same , while adding more shot is = 1 hit + 3 miss = 4 shots ; as 3 miss equals to the reduction of 1 point the total of 1 hit and 3 miss iz zero , keeping the score 36 same as before.
so , the sequence can be : 36 ,40,44,48,52,56,60,64................180

Here , a =36
d =4
n=?

so , a+(n-1)d=180
or, 36+ (n-1)4 =180
or, (n-1)4= 144
or , n-1 = 144/4
or, n=36+1
or, n=37
so ans is 37 .

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09 Jul 2026, 10:43

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Let the number of targets hit be x and number of targets missed be y. Noy for each archer, there is a different combination of x and y but 2 constraints remain the same.

1. Since each x adds 1 point, each y reduces 1/3 points and the total of these for every archer is 36, we have equation 1x-(y/3) = 36 or 3x-y=108 (eq1). The value of x and y will be different for each archer will be different, but the sum of the points cannot exceed 108 in the above equation
2. There were 180 targets and hence x+y have to be <=180 (ineq2).

With this info, we need to identify the minimum of value of x and y which satisfies both the conditions.

If I take x<36, y will turn negative. Let's take an example. x=35, then 3(35)-y = 108, -y = 108-105, -y=3 or y=-3 which is not possible
Hence, the minimum value of x, where y is at least 0, is 36 (in effect an archer who has hit 36 targets and then kind of stopped)

With each increment of 1 in x, for the eq1 to hold true, y will increase by 3.
However, the maximum value of x can be 72, where y will be 108 (thus x+y = 180) and 3(72)-108 = 36

If x increases beyond 72, ineq2 will be breached. So, count of all possible values of x = 72-36+1 = 37

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We are given that every hit is +1 point and every miss is -1/3 points. Also total number of targets is 180.
Let number of hits by an archer be h and number of misses be m.
Thus eqn for scoring exactly 36 points will be:

h*1 - m*(1/3) = 36
h - m/3 = 36

h = 36+m/3 _____ (1)

Now since it is given to us that every pair of (h,m) is unique, we need to find the total number of such pairs that can exist.
Also, h+m <= 180 since it can never exceed 180.

From here I started substituting different values of m in eqn (1) which can give me h+m <=180. Also since h has to be a whole number, m can only take values of multiples of 3
at m = 0, h = 36
m = 3, h = 37
m = 6, h = 38
m = 9, h = 39

I started seeing a pattern here where for every increase of 3 in m, h would increase by 1.

I then looked at the options. For 35 pairs of (h,m) to exist, the 35th such m would be 3*34 = 102 and the corresponding 35th h would be 36+34 = 70. (3*34 because in the first pair, m = 0). The sum of h+m in this case would be 172. Thus From here I just tried more combinations till I reached h+m = 180
For, m = 105, h = 71
m = 108, h = 72

So adding these 2 pairs to the previous 35 I got the answer to be 37.

I marked (C)

PS: I dont really think this was a 555-605 level question. It seemed like atleast a 655-705 question if not more. Just curious as to how are competition questions difficulties marked?

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Bunuel

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OPTION C. Not sure about my soln but here's my approach - there should be minimum 36 correct , if 37 then 3 mistakes, 38 , 6 mistakes , 39, 9 mistakes and so on ........ till 72 -> 108mistakes , total = 180. 72-36+1 = 37.

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09 Jul 2026, 11:24

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Let A1 , A2.... etc denote the archers

A1 = hit 36, 0 missses --> Points = 36; Total targets = 36

Similarly,
A2 = hit 37, missed 3 ---> points 36, Targets = 40
A3 = hit 38, missed 6 ---> points = 36, Targets = 44
A4 = hit 39, missed 9 ---> Points = 36, Targets = 48
So every increase increase in a point would lead to an increase in overall 4 targets hits (1 point and 3 for negetives) to keep the points at 36

Hence this becomes an AP : 36 ----> 180 with a common difference of 4
Hence number of terms = (180 - 36) /4 + 1 = 37 Answer is C

Bunuel

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09 Jul 2026, 11:36

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As the number of shots cannot be equal. Then only one archer could shoot 36 targets correct. Other will have shot more than 36. Considering 37 hits.

37*1-3*(1/3)=36. thus targets = 40,

similarly, for 38 correct hits-

38*1-3*(1/3)=36, thus targets = 44.

We can notice a pattern here. The targets are increasing by 4, each time you increase the number of correct hits. So as the max targets is 180 and 36*4=180, then only 36+1 = 37 archers are there.

Bunuel

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09 Jul 2026, 11:38

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My answer is C.) 37

Let number of hits= x; number of misses=y
Also, x-y/3=36 and x+y<=180 (total targets)
x=36+y/3. (For x to be integer (number of hits), y is a multiple of 3)

Lets check options-
A. 35- If 35 archers then we can find total targets-
Every archer has hit different number of targets so 35th archer must have missed 35*3= 105 targets.
When y=105, then x=36+105/3= 71. x+y=176 which is less than maximum targets (180)
B. 36- following the same as above-
when y=36*3=108, then x=72. x+y=180 which is the maximum number of targets hence number of archers who missed atleast 1 target are 36.

Also if an archer does not miss any target and hits 36 targets then his score is 36. Therefore max archers are 36+1= 37. Which is option C.

Bunuel

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09 Jul 2026, 11:52

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Bunuel

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Solved this using Arithmetic progression
+1 for every hit, -1/3 for every miss
Number of hits by an archer = x
number of misses = y
we know that all archers end up at 36 points
thus x- y/3 = 36
hence 3x - y = 108

For archers to be maxed, the number of targets hit per archer should be minimised
Minimum Targets an archer can hit to get 36 points = 36 ( all hits, 0 misses => x = 36, y=0)
Next number of targets = 40 ( 37 hits, 3 misses)
Next = 44 ( 38 hits, 6 missess) and so on

We find an arithmetic series - 36, 40, 44......
Last term of the series = The number of targets = 180
Using series formula,
180 = 36 + (n-1)4
we get n = 37

Hence max archers in the group = 37 (C)

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09 Jul 2026, 11:55

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I found it very simple... just count

Like for bang on 36 points one of the archer will shoot first 36 arrows correct

Then to loose one point 3 target has to be missed and to compensate for this point, extra 1 arrow to be shoot correct which will nullify lost point ... so next archer will shoot 4 more arrows to secure 36 and his total will be 40...

similarly it will increase by 4 to finish at 36 points for other archer Theron. Every 4 arrows shot will have 3 wrong and 1 right to nullify each other and remains at 36 points as asked in question. this way it will give maximum possible number of archers in the list

so starting from 36, it will be multiple of 4 and last being 180.

Total final number= (180-36)/4+1=37

Ans- C

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Let's assume no of target hits = X and Target Missed = Y ; X and Y both will be integer
There will be 2 equations,
1: X+Y<=180
2: X-(Y/3)=36 == X=36 +(Y/3)
Since X and Y both are integer so from equation 2, Y can be 0 and multiple of 3, then only X can be integer.
Thus, the group of unique (X,Y) created will be ( 36,0) , (37,3)....(72,108) ; beyond this first equation won't hold true so a maximum of 37 combinations can exist which will sum up as 36. So right answer is 37.

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09 Jul 2026, 12:06

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IMO C
Let X be the no of target hits & Y be the number of misses
Then x-y/3= 36 -----(1)
Also Min no of X+Y =36 (where Y=0 & X=36) -----(2)
Also X+Y<=180
Solving Y<=108, then X<=72
There max group = [36,72] both inclusive, => 72-36+1 = 37. Ans C

Bunuel

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09 Jul 2026, 12:12

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Based on logic, 36 is the maximum point any archer received, or rather all archers received same point. Note that they earn 1 point, hence 36 archers+1=37 archers can be maximum. Hence our answer is C.

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09 Jul 2026, 12:19

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h = hits
m = misses
total: n=h+m
h-(1/3)m = 36
3h-m=108
m=3h-108
n=h+(3h-108) = 4h-108 -----(I)

since misses is non-negative integer:
3h-108 >= 0
h>=36

total maximum targets for each archer is 180
n <= 180
4h-108 <= 180
h <= 72
36<= h <= 72

n is different for each archer. in eq I, each different value of h gives unique value of n and each value of n represents unique number of targets, each value of n also represents an archer.

we find all the possible values of h and hence n in the above range to find the maximum possible number of archers:
(72-36)+1 = 37 archers.

Answer C

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09 Jul 2026, 13:21

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Let the number of target attempts be x and h be the number of hits.
So number of misses will be x-h.
From the question stem we get :
h - (1/3)(a-h) = 36
and we get => h = (108 + x)/4

h must be integer so the number of attempts x must be a multiple of 4.
Since max value of x = 180 we can cay :
h = (x + 108)/4 <= x
x+ 108 <= 4x
3x>=108
Or overall we can say
36<=x<=180

Now the number of multiples from 36, 40,....180
I tried AP formula to calculate the number of terms with first term 36 and Common difference 4
n = 37

IMO Ans is C.

Bunuel

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