In a class of 50 students, 20 play Hockey, 15 play Cricket and 11 play Football.7 play both Hockey and Cricket, 4 play Cricket and Football and 5 play Hockey and football.
If 18 students do not play any of these given sports, how many students play exactly two of these sports?
A 12
B 10
C 11
D 15
E 14
how many students play exactly two of these sports
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- niketdoshi123
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Total students = 50
Students who play Hockey
a+d+g+e = 20 ---- (1)
Students who play Cricket
b+f+g+d = 15 ---(2)
Students who play Football
c+e+g+f = 11 ----- (3)
Both Hockey & Cricket
d+g = 7 ------ (4)
Both Cricket & Football
g+f = 4 ------(5)
Both Football & Hockey
g+e = 5 ------ (6)
18 students do not play any given sports,
then a+b+c+d+e+f+g = 50-18=32 -----(7)
We need to find number of students playing exactly two of these sports
i.e d+e+f
adding (1),(2) &(3)
a+b+c+2d+2e+2f+3g = 20+15+11=46 ------ (8)
subtracting (7) from (8)
d+e+f+2g=14 ---- (9)
adding (4),(5) &(6)
d+e+f+3g = 5+4+7 = 16 ----(10)
subtracting (9) from (10)
g=2
substituting value of g in (9)
d+e+f =10
hence ans is B
- aneesh.kg
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In problems such as these, using the equations is easier than using Venn-Diagrams.
Equation 1:
Total - neither = H + C + F - [H&C + C&F + F&H] + H&C&F
50 - 18 = 20 + 15 + 11 - [7 + 4 + 5] + H&C&F
32 = 30 + H&C&F
H&C&F = 2
Equation2:
Total - neither = H + C + F - [people in only two groups] - 2*H&C&F
(using the value of H&C&F from above)
50 - 18 = 20 + 15 + 11 - [people in exactly two groups] - 2*2
people in exactly two groups = 10
[spoiler](B)[/spoiler] is the answer.
Note:
One should have the three methods - (i) drawing Venn Diagrams, (ii) using the equations above, and (iii) the Table approach- in your repository so that you can choose one of the depending on what will be easier for the given problem.
I solved a similar problem a few weeks back here:
https://www.beatthegmat.com/the-fastest- ... tml#465320
where I also explained the derivation of the Equation 2.
Equation 1:
Total - neither = H + C + F - [H&C + C&F + F&H] + H&C&F
50 - 18 = 20 + 15 + 11 - [7 + 4 + 5] + H&C&F
32 = 30 + H&C&F
H&C&F = 2
Equation2:
Total - neither = H + C + F - [people in only two groups] - 2*H&C&F
(using the value of H&C&F from above)
50 - 18 = 20 + 15 + 11 - [people in exactly two groups] - 2*2
people in exactly two groups = 10
[spoiler](B)[/spoiler] is the answer.
Note:
One should have the three methods - (i) drawing Venn Diagrams, (ii) using the equations above, and (iii) the Table approach- in your repository so that you can choose one of the depending on what will be easier for the given problem.
I solved a similar problem a few weeks back here:
https://www.beatthegmat.com/the-fastest- ... tml#465320
where I also explained the derivation of the Equation 2.
Aneesh Bangia
GMAT Math Coach
[email protected]
GMATPad:
Facebook Page: https://www.facebook.com/GMATPad
GMAT Math Coach
[email protected]
GMATPad:
Facebook Page: https://www.facebook.com/GMATPad
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