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

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chasing725

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09 Jul 2026, 15:06

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The minimum number of shots required = 36. All hits an no miss.

After that for every one hit, 3 miss are allowed. Hence, next after 36 shots, the next number of targets = 36 + 4 = 40

Post that it will be 36 + (2*4) = 44, then 36 + (3*4) = 48 and so on.

36, 40, 44, 48 ,.. until 180

Hence, we have a AP.

Number of terms = (180-36)/4 + 1 = 144/4 + 1 = 36+1= 37

Option C

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09 Jul 2026, 15:38

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n = number of targets
h = number of targets hit
m = targets missed

n=h+m
score = h - 1/3m = 36
multuply by 3: 3h - m =108
Since m=n-h
3h -(n-h) = 108
n = 4h-108
reason this^^ n must be divisible by 4

every archer whos scores 36 must have shot a number of targets that is a mutliple of 4

possible number of targets shot start at around 36 as thats the bare minimum need and going up to 180

number of values = 180-36/4 + 1 = 37

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09 Jul 2026, 16:49

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Lets say archer hit total x targets and misses y out total targets n = x+y
so total score = x - y/3 now we can say y= n-x
So Total score= x - (n-x)/3
Now it has mentioned that archer scores exactly 36
36 = x -(n-x)/3 => 108 = 3x - n +x = 4x-n
=> 4x-n= 108 when score is exactly 36.

let's find value for x = (108 + n)/4
here we can say that x atleast be equal to n or less than hit<= total. And n has to be divisible of 4
lets consider x=n

so 4x= 108 + x so x= 36. So we can clearly say that n or x has the minimum value of 36.
now we can see the pattern of different number of targets are, 36, 40, 44, .....180
Now we can count the total number of member from AP concept.
(180-36)/4 +1 = 144/4 +1 = 36+1 = 37.
So the maximum number of archers with score 36 would be 37.
Answer C
(This is kind of 700+ difficulty level. almost took me 5 mins to understand solve this)

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09 Jul 2026, 18:03

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hits be H and misses be M
H+M=180

H+1/3 M= 36
3H-M=108

M can be 0 or above

H+M <=180
H+3H-108<= 180
H<=72

For minimum hits. 3H-Y = 108, H=36 Y=0, These are the minimum values to satisfy the equation
so, H>=36

so if archers score all the possible points between 36 and 72 (inclusive) then total 37 archers are possible.

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To solve this problem I set up the score equation $(H-\frac{1}{3M}=36)$ alongside the total shots equation$ (T= H + M$). Being: $T$ - Total targets shot at, $H$ - Hits, $M$ - misses. Isolating the hits ($H$) and subsituting it into the total shot formula gives you direct relationship: $T= 36 +\frac{ 4}{3M}$. Since the total number of shots cannot exceed the 180 available targets: $T≤180$, you find that the maximum number of misses ($M$) is 108. Because both $T$ and $M$ must be whole numbers, $M$ has to be a multiple of 3; therefore, you simply count how many multiples of 3 exist from 0 to 108 with: $\frac{108}{3}+1=37$. $+1$ because $0$ is missing. Each of these 37 unique miss values generates a unique total number of shots ($T$), giving you a maximum limit of 37 archers. CORRECT ANSWER: C

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09 Jul 2026, 20:38

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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?

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

1 points for 1 shot on target
-1/3 points for 1 missed shot
i.e. every 3 missed shots nullify 1 target shot

To find out max possible number of archers we need to consider to two extreme scenarios

1. One archer hits all shots on target and misses nothing still gets 36. Which gives the number of shots taken = 36.

2. One archer utilizes all 180 shots but still gets 36. i.e. apart from 36 target shots, all other target shots got nullified because of missed shot. Say no. of target shots = k, so number of missed shots = 3k. Which gives total number of shots = 36+k+3k
Or, 36+4k =180
Or, 4k = 144
Or, k = 36

Since every archer in the group taken different number of shots. So, the number of archers will be = no. of elements of the set {0,1,2,...,k} = 36+1 = 37.. Ans (C).

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09 Jul 2026, 20:59

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C - 37

Set h = targets hit, n m = targets missed. The score condition is h - m/3 = 36, so m = 3h - 108

Since misses cannot be negative it forces h >= 36.

Now for the shots, an archer shoots h + m = 4h - 108 targets, and there are only 180 targets available, so 4h - 108 <= 180 gives h <= 72.

Therefore h runs from 36 to 72.

Counting the range between 72-36 + 1 = 37.

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09 Jul 2026, 21:02

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The correct answer is 37. I set up the following equation where x is targets and y is the number hit

y - (1/3)*(x-y)=36 then multiplied it by 3 to get rid of the fraction
4y - x = 108 -> y = (x + 108)/4

Since y (targets hit) cannot exceed x (number of targets, (x+108)/4 <= x, so n + 108 <= 4x, then x>= 36.

x is also <= 180 as stated in the problem.

Since y must be a whole number, x + 108 must also be divisible by 4

The multiples of 4 between 36 and 180 equals (180 - 36)/4 +1 = 37

The answer is 37.

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09 Jul 2026, 21:26

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Points for a hit=+1, and Points for a miss=-1/3
Therefore, the number of misses will be in multiples of 3.

Least misses=0
The archer with 0 misses will hit 36 targets.

To maximize misses, we have to minimize hits and maximize shots.
Let us try with the highest possible number first, i.e. 180.
Assuming x misses and 180-x hits, we have 180-x-x/3=36.
Thus, x=108 misses.

Since misses will be in multiples of 3 starting from 0, i.e. 0,3,6,...,108.
Each possible value of misses will denote one archer, therefore there are 37 archers [{(108-0)/3}+1]

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09 Jul 2026, 21:44

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So over here , we are given the following information:
Let the total number of shoots that an archer participate be x, total shoot that an archer hits be y, hence total shoot an archer loses be x-y
Score if target achieved = y*1
Score if taget missed = -1/3 (x-y)
Total score = 36 points
Hence y - 1/3 (x-y) = 36
3y -x+y = 108
4y -x =108
4y = 108 + x
y = 108/4 + x/4
y = 27 + x/4

Now since the total number of shoots of an archer has to be a whole number, x should be a multiple of 4
It is also given that an archer cannot shoot more than 180 targets, and that the total score is 36 if y = 36, x = (36-27)*4 = 36
Accordingly, we have the range of range of x from 36 to 180 in multiple of 4 for 1 archer

We are required to find the maximum total archers in the group and 2 archers can't have the same number of targets (x needs to be different)
Accordingly, the maximum possible archers in the group = number of terms in the series -> 36, 40, 44, ..., 176, 180 (Arithmetic Progression)
Number of terms in a series = last term - first term / common difference + 1
= 180-36/4 + 1
= 144/4 + 1
= 36 + 1
= 37
Hence the answer is Option C = 37

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09 Jul 2026, 22:08

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h = +1, m = -1/3
h+m<=180
score = h-m/3 = 36
now, min value of m = 0, therefore min h = 36(if m increases, h should also increase)
as score is integer, only multiple of 3 can be values of m = 0,3,6...; h for these values will be 36,37,...
for max,
3h-108+h <=180, so h<=72; so h = 36, 37 .....72
so total number of unique value/archers= 72-36+1 = 37.

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09 Jul 2026, 23:12

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Bunuel

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let h = number of targets hit
m = no of targets missed
n = h+m
given, $n<=180$
$h+m <=180$
also given , 1 point for hit, -1/3 for miss, given archer finishes with 36 points , so
$h-m/3 = 36$
$m = 3h-108$
we know, $m>=0, so 3h-108>=0 \\
h>=36$
then, $n= h+3h-108 $
$n = 4h-108$
we also know that max no of targets is 180
$4h-108 <=180$
$4h<=288 so h<=72$
so, $36<=h<=72$
from the above equation for n = 4h-108, we know that every unique value of h will have a unique value for n. so the maximum possible no of archers is the number of possible integer values of h
so max possible number of archers = range of h = 72-36 +1 = 37
C

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09 Jul 2026, 23:19

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T = total targets shot
H = targets hit
M = targets missed

Since every shot is either a hit or a miss we can say T = H + M

Equation from given: H - M/3 = 36

Multiply by 3 -----> 3H - M = 108

Substitute M = T - H:

-----> 3H - (T - H) = 108

-----> 4H - T = 108

----->4H = T + 108

H = (T + 108)/4

For H to be an integer (T + 108)/4 must be an integer.

Since 108 is divisible by 4 T must be divisible by 4 (number property)

Possible values of T could be 4, 8, 12, ..., 180

------> There are 180/4 = 45 multiples of 4.

Now check that H <= T (equivalently, M >= 0)

(T + 108)/4 <= T

T + 108 <= 4T

108 <= 3T

T >= 36

So valid values of T are:

36, 40, 44, ..., 180

Number of valid multiples can be found by: Number of terms = ((Last - First)/Common Difference) + 1

((180 - 36)/4) + 1

= 144/4 + 1

= 36 + 1

= 37

Answer: C

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09 Jul 2026, 23:45

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let x1 (shots on target), x2 (shots missed), x3 (shots not attempted)
x3= 180-x1-x2

points = x1*1 - 1/3*x2 + (180-x1-x2)*0 =36
==> x1-x2/3=36
min x1 == 36, x2=0
for every additional x1 we need 3 no of x2 to keep the score same
for max x2==> when x3=0 or x1+x2=180
==> x1- (180-x1)/3 =36
==> x1max =72
so total no of possibilities =72-36+1= 37

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09 Jul 2026, 23:56

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Let,

number of targets an archer shoots = n
number of target hit = h

If misses = n - h

Score = h - (n-h)/3 = 36

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

n = 4h - 108 <= 180
h <= 72

Also,

n = 4h -108 >= h
h>= 36

So, h varies from 36 to 72, having (72-36 = 37 ) 37 values,
Corresponsingly, n are all distinct, so there are 37 distinct number of targets shot.

Answer: C (37)

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We can start by exploring the problem. The easiest case is when it's all hits and no misses, which is 36 hits. The next possibility is one more hit making it 37 hits, being balanced by 3 misses, because $-\frac{1}{3} * 3 = -1$
So, total 40 shots.
Following is two extra hits, now being balanced by six misses. $36+2+6= 44$
We can see we have here, an arithmetic progression, with
$a1 = 36, and An = 36+(n-1)*4=36-4 +4n=32+4n$
The last possible hit amount is 180 if the max amount of players gets 36 points, therefore: $32+4n=180,4n=180-30-2=150-2=148/:4->n= 37$

Option (C)

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10 Jul 2026, 00:33

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let x be no of hit target, y be no of missed target -> x-y/3=36 -> 3x = 108+y -> y=3x-108
we know x+y <=180 -> x+3x-108<=180 -> x<=72 (1)
we also know y>=0 -> y/3>=0 -> -y/3 <=0 -> x-y/3 <= x -> 36 <=x (2)
from 1 and 2 we have: 36<=x<=72
x can also represent number of archers as an archer can end up target from 36-72 hit points
--> the maximum number of archer is all the valid values of x = 72-36+1=37 -> Choose C

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OK so this question can be solved using algebraic equations so the equation is
X + Y/3 = 36
now by the trial and error method it's obvious that the first value of X would be 36 and Y would be 0 now if we will go forward X would be 37 and Y would be 3 so as you can see in the first case the total sum is 36 and in the second case the total sum is 40 and if we'll go forward it will come to come out to be as X equal to 38 and Y equal to 6 that is total sum come out to be 44 cancel and so on. So where we will stop?

We will stop when we'll get the sum to be 180 and if you have noticed no sum is s same as maintaining the constraint mentioned in the question. So moving forward what we have is 36, 40, 44, 48...... .180. This is an AP and number of terms in it is 37. [ 180 = 36 +(n-1)*4; n = 37]

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10 Jul 2026, 00:46

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total targets =180
each hit = earn 1 point.
each miss= lose 1/3 point.

no two people had same num of targets.
each person scored 36.

find: max archers in group.

assume total hits for one person =x
total mis= y
total targets for one person = N
so total target shot, N= x+y

points earned,

N = x- (1/3)y
we also know each has scored 36 points.

36 = x - (1/3) y
108 = 3x- y

y= 3x-108

put this value of y into first eqn,
N= x + 3x-108
N = 4x -108

so here if we can find the range of x, we can find the range of N(archers)

we know total targets are 180.
N<= 180
4x-108 <= 180
4x <=288
x<= 72

there are maxim 72 hits.
now we need to know min hits as well.

num of misses at least zero since it cant be in negative side.

y >= 0
3x - 108 >= 0
3x >= 108
x>=36

36 <= x <= 72

the reason why we needed range of x becuase we were told that no two people had same target shots. so with the num of hit we can find the N using N= 4x-108

total x = 72-36+1 = 37

choice C

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