Each of 90 students participated at least 1 of the track tryouts: High jump, long jump, 100 meter dash. If 20 students participated in high jump tryout, 40 students participated in the long jump tryout and 60 students participated in the 100 meter dash tryout, and if 5 students participated in all 3 tryouts, how many students participated in only two of these tryouts?
(A) 25
(B) 20
(C) 15
(D) 10
(E) 5
[spoiler]OA: B[/spoiler]
3 Overlapping sets
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Total Students: 90mgm wrote:Each of 90 students participated at least 1 of the track tryouts: High jump, long jump, 100 meter dash. If 20 students participated in high jump tryout, 40 students participated in the long jump tryout and 60 students participated in the 100 meter dash tryout, and if 5 students participated in all 3 tryouts, how many students participated in only two of these tryouts?
(A) 25
(B) 20
(C) 15
(D) 10
(E) 5
[spoiler]OA: B[/spoiler]
Total Participations: 120
5 students participated in all three: 15 participations
Remaining students: 85
Remaining Participations: 105
As every student participated in atleast 1,
students who participated in 2 of the three compensate for the difference in no. of students and no. of participations: 105-85 = 20
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Here is the formula for 3 overlapping groups:mgm wrote:Each of 90 students participated at least 1 of the track tryouts: High jump, long jump, 100 meter dash. If 20 students participated in high jump tryout, 40 students participated in the long jump tryout and 60 students participated in the 100 meter dash tryout, and if 5 students participated in all 3 tryouts, how many students participated in only two of these tryouts?
(A) 25
(B) 20
(C) 15
(D) 10
(E) 5
[spoiler]OA: B[/spoiler]
T = A + B + C - (AB + AC + BC) - 2(ABC)
The big idea with overlapping group problems is to SUBTRACT THE OVERLAPS.
When we add together everyone in A, everyone in B, and everyone in C:
Those in exactly 2 of the groups (AB+AC+BC) are counted twice, so they need to be subtracted from the total ONCE.
Those in all 3 groups (ABC) are counted 3 times, so they need to be subtracted from the total TWICE.
By subtracting the overlaps, we ensure that no one is overcounted.
In the problem above:
T = 90.
A = high jump = 20.
B = long jump = 40.
C = dash = 60.
The number participating in exactly 2 events = AB + AC + BC = x.
Since 5 students participate in all 3 events, ABC = 5.
Plugging these values into the formula, we get:
90 = 20 + 40 + 60 - x - 2(5)
x = 20.
The correct answer is B.
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Explanation without any Formulamgm wrote:Each of 90 students participated at least 1 of the track tryouts: High jump, long jump, 100 meter dash. If 20 students participated in high jump tryout, 40 students participated in the long jump tryout and 60 students participated in the 100 meter dash tryout, and if 5 students participated in all 3 tryouts, how many students participated in only two of these tryouts?
(A) 25
(B) 20
(C) 15
(D) 10
(E) 5
Answer: Option B
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We can use the following formula:mgm wrote:Each of 90 students participated at least 1 of the track tryouts: High jump, long jump, 100 meter dash. If 20 students participated in high jump tryout, 40 students participated in the long jump tryout and 60 students participated in the 100 meter dash tryout, and if 5 students participated in all 3 tryouts, how many students participated in only two of these tryouts?
(A) 25
(B) 20
(C) 15
(D) 10
(E) 5
Total = number of high jump + number of long jump + number of 100-meter dash - number who did two events - 2(number who did all 3 events) + number who did zero events
Since we see that each student participated in at least 1 event, the number who did zero events is zero.
Filling in the rest of the equation, we have:
90 = 20 + 40 + 60 - D - 2(5) + 0
90 = 110 - D
D = 20
Answer: B
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