In a class of 33, 20 play cricket, 25 play football and 18 play hockey

I suggest using venn diagram than formula in this one. So we need fluke to illustrate this with diagram :-) I can explain my reasoning.
Let x be the all three. So we have three " both" overlaps. They are
15-x c and f
12-x f and h
10-x c and h
The three "singular" regions
C only = cricket - overlaps
=> 20 - (15-x + x + 10-x) = x-5

Similarly, if you calculate f only and h only we have:
F only : x-2
H only : x-4

None= 0 it's given.

Add all the pieces of the puzzle-
Total=33 = x-5 + 15-x+ x + 10-x+ x-2+ 12-x+ x-4
X = 7

Hence C only is x - 5 = 7-2 = 2

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C + CF + CH - CFH = 20

F + CF + FH - CFH = 25

H + FH + CH - CFH = 18

C + F + H + 2(CF + CH + FH) - 3CFH = 63

=> C + F +H + 2(15 + 12 + 10) - 3CFH = 63

=> c+F+H - 3CFH = 63 - 74 = -11


C + F + H + CF + CH + FH - 3CFH + CFH = 33

C + F + H + 37 - 2CFH = 33

=> C + F + H - 2CFH = -4

=> CFH = 7

=> C + 15 + 10 - 7 = 20

=> C = 20 - 18 = 2

This took me longer than ten minutes.

My approach:

I drew a venn diagram with variables C for those who only play cricket, F for those who only play football, H for those who play Hockey only.

Those who play two sports - I defined these in terms of X, which I took as the number of players who played all three sports.

So, 33 = c + f + h + x + (15-x) + (10-x) + (12-x) (call this Eq 1)
20 = c + (10 - x) + (15 - x) + x
25 = f + x + (15 - x) + (12 - x)
18 = h + x + (10-x) + (12-x)

Find out that c = x - 5, f = x - 2, h = x-4.

Substitute these variables in eq 1 and you will get x + 26 = 33.
X = 7. C = 2.

So, answer is D.

We can let the number of students who play all 3 sports = x and we can use the following equation:

Total = #C + #F + #H - #(C and F) - #(F and H) - #(C and H) + #(C and F and H) + None

33 = 20 + 25 + 18 - 15 - 12 - 10 + x + 0

33 = 26 + x

x = 7

Thus, the number of students who play only cricket is:

#(C only) = #C - #(C and F) - #(C and H) + #(C and F and H)

#(C only) = 20 - 15 - 10 + 7

#(C only) = 2

Answer: D

Hi Scott, I solved this using a formula i learned in TTP as follows... Can you let me know if my approach is correct. I ended up with the same answer.

33= 63 (A,B, and C) - 37(double overlaps) - 2x(all three overlaps) + N (0 in this case since everyone plays something)

Upon solving x =3.5 and then i just worked my way inside out to figure out only cricket and ended up with 2

There's a mistake in your calculations; if you solve 33 = 63 - 37 - 2x, you'll find x = -3.5, which is not correct because the number of people who play all of the sports can be neither negative nor fractional.

The "double overlaps" and "all three overlaps" in the formula you mention refer to the number of people who play exactly two of the sports and exactly three of the sports, respectively. The numbers 15, 12 and 10 in this problem, on the other hand, are the number of people who play at least two of the three sports, so they all include the number of people who play all three sports as well.

In order to use that formula with this question, you need to determine the number of people who play exactly two of the sports, for which you need the number of people who play all three of the sports. That's why it's not practical to use that formula for this question.

thank you for the explanation, understood. I guess i got lucky.

Could you explain when to use the formula you used Total = #C + #F + #H - #(C and F) - #(F and H) - #(C and H) + #(C and F and H) + None vs the formula that i used?

The biggest difference between the formula I used earlier and the formula you used is that in my formula, the double overlaps include the triple overlap as well; whereas, in your formula, the double overlaps are the number of elements which belong to two sets but do not belong to three sets.

When you add the number of elements of sets A, B and C; the elements which belong to exactly two sets are counted twice and the elements which belong to all three sets are counted three times. One way of obtaining the number of elements that belong to A or B or C is subtracting from A + B + C the number of elements that belong to exactly two sets and twice the number of elements that belong to all three sets. That's what's done in your formula. The other way of obtaining the same number is subtracting from A + B + C the number of elements that belong to at least two sets; in which case you will have subtracted the number of elements that belong to all three sets three times. To get the correct number, one should add the number of elements that belong to all three sets and that's what's done in my formula.

Given:
1. In a class of 33, 20 play cricket, 25 play football and 18 play hockey.
2. 15 play both cricket and football, 12 play football and hockey and 10 play cricket and hockey.

Asked: If each student plays at least one game, how many students play only cricket?



33 = 20 + 25 + 18 - (15+12+10) +x
x = 7

Number of students playing only cricket = 20 - (8+7+3) = 2

IMO D

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