He needs to purchase some work clothes and notices that J.C. Penney is having a 25% off of your total purchase sale. He decides
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For this type of problem, it is best to use Venn diagrams as shown in the picture. The areas where the circles intersect are the mutual events that occur together. The area where all circles intersect is denoted as x. These are the students who play all sports. Assuming all of the students play sports in the school, all of the numbers in the circles should add up to 405. The remaining area would be the difference. The solution is as follows:
Students who play tennis and hockey: 45 - x
Students who play hockey and softball: 60 - x
Students who play only tennis and softball: 39 - x
Students who only play tennis:
251 - 45 + x - 39 +x -x = x + 167
Students who only play hockey:
157 - x - 45 + x - 60 + x = x + 52
Students who only play softball:
111 - x - 60 + x - 39 + x = x + 12
The sum of all of these should be 405:
45-x+60-x+39-x+x+167+x+52+x+12 = 405
Solving for x,
x = 30
Therefore, there are 30 pupils who play all sports; 15 pupils who play tennis and hockey; 30 pupils who play hockey and softball; 9 pupils who play tennis and softball; 197 pupil who only play tennis; 82 pupils who only play hockey; and 42 pupils who only play softball.
Students who play tennis and hockey: 45 - x
Students who play hockey and softball: 60 - x
Students who play only tennis and softball: 39 - x
Students who only play tennis:
251 - 45 + x - 39 +x -x = x + 167
Students who only play hockey:
157 - x - 45 + x - 60 + x = x + 52
Students who only play softball:
111 - x - 60 + x - 39 + x = x + 12
The sum of all of these should be 405:
45-x+60-x+39-x+x+167+x+52+x+12 = 405
Solving for x,
x = 30
Therefore, there are 30 pupils who play all sports; 15 pupils who play tennis and hockey; 30 pupils who play hockey and softball; 9 pupils who play tennis and softball; 197 pupil who only play tennis; 82 pupils who only play hockey; and 42 pupils who only play softball.