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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
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
10 Jul 2026, 02:19
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Bunuel
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
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
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
