In a class of 40 students, 16 take maths, 21 take chem., 18 take phy., 9 take math & chem., 5 take math & phy., # of students who take... maths only & (chem. + phy.) only are equal & 3 students take all three subjects. Find the number of students who take maths only.

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16/40 students take math21/40 students take chemistry18/40 students take physics9/40 take math and chemistry5/40 take math and physics3/40 take all three subjects This doesn't add up because this would imply that 9+5+3 students take math, but not only math; however, that equals 17...

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16/40 students take math
21/40 students take chemistry
18/40 students take physics

9/40 take math and chemistry
5/40 take math and physics
3/40 take all three subjects

This doesn't add up because this would imply that 9+5+3 students take math, but not only math; however, that equals 17 students?

Edit: because I assumed that the three groups were mutually exclusive.  ie. the 3 taking all three subjects were not accounted for in the math/chemisty or math/physics group.

If they are not mutually exclusive then it would be be 16 - (9+5-3) = 5 students taking math only.

To solve for the number of students taking physics and chemisty sum the number of students taking physics and the num of students taking chemistry first: 21+18 = 39

Subtract from it all other possible combinations: 39 - 5 (math only) - 9 (math and chemistry) - 5 (math and physics) - 7 (physics only) - 8 (chemistry only) + 3 (all three) = 8

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