SOLUTION: 110 members of a sports club play at least one of the games, football, basketball and volleyball. If 20 play football and basketball only, 15 play football and volleyball only, 26

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Question 1206774: 110 members of a sports club play at least one of the games, football, basketball and volleyball. If 20 play football and basketball only, 15 play football and volleyball only, 26 play basketball and volleyball only, x play all the three games, 2x each play only one game, how many play basketball altogether?
Found 2 solutions by MathLover1, ikleyn:
Answer by MathLover1(20849)   (Show Source): You can put this solution on YOUR website!

here is link with answer to your question:
note: only difference is "24 play football and volleyball only", and you have "15 play football and volleyball only"

https://myschool.ng/questions/view/academic-questions/238536
Answer by ikleyn(52776)   (Show Source): You can put this solution on YOUR website!
.
110 members of a sports club play at least one of the games, football, basketball and volleyball.
If 20 play football and basketball only, 15 play football and volleyball only, 26 play basketball and volleyball only,
x play all the three games, 2x each play only one game, how many play basketball altogether?
~~~~~~~~~~~~~~~~~~~~~~~~


As it is presented in the post, wording is incorrect.
A correct wording should go this way: 


    110 members of a sports club play at least one of the games, football, basketball and volleyball. 
    If 20 play football and basketball only, 15 play football and volleyball only, 26 play basketball and volleyball only, 
    x play all the three games, 2x play football only, 2x play basketball only, 2x play volleyball only, 
    how many play basketball altogether?

Then the solution is as follows: 

We have 7 separate disjoint subsets:

    - playng F only (2x);
    - playng B only (2x);
    - playng V only (2x);

    - playing FB only (20);
    - playing FV only (15);
    - playing BV only (26);

    - playing all three games FBV (x).


Then we can write the equation for the total as the disjoint sum of its parts

      110 = 20 + 15 + 26 + 2x + 2x + 2x + x,

or

      110 - 20 - 15 - 26 = 7x

             49          = 7x

              x          = 49/7 = 7.


Then the number of those playing basketball is  

    B_only + (FB)_only + (BV)_only + FBV = 2x + 20 + 26 + x = 2*7 + 20 + 26 + 7 = 67.


ANSWER.  67 play basketball.

Solved, with corrected wording.


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It is very soft form from my side to say that wording in this problem is incorrect.

It is simply   .




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