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In a class of 33, 20 play cricket, 25 play football and 18 play hockey [#permalink]

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25 Mar 2011, 22:23

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In a class of 33, 20 play cricket, 25 play football and 18 play hockey. 15 play both cricket and football, 12 play football and hockey and 10 play cricket and hockey. If each student plays at least one game, how many students play only cricket?

Re: In a class of 33, 20 play cricket, 25 play football and 18 play hockey [#permalink]

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26 Mar 2011, 00:05

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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

Re: In a class of 33, 20 play cricket, 25 play football and 18 play hockey [#permalink]

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26 Mar 2011, 00:41

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AnkitK wrote:

In a class of 33, 20 play cricket, 25 play football and 18 play hockey. 15 play both cricket and football, 12 play football and hockey and 10 play cricket and hockey. If each student plays at least one game, how many students play only cricket? 1. 8 2. 6 3. 4 4. 2 Explain with a diagram .

Re: In a class of 33, 20 play cricket, 25 play football and 18 play hockey [#permalink]

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26 Mar 2011, 00:52

Yeah last equation is horrible. I doubt if algebraic equation can be dealt faster since its based on rules. It's like a cop waiting to get u a speeding ticket. Great !

AnkitK wrote:

Thnkx gmat1220 .Dear Fluke pls explain your approach to the question..Also is there any short cut to such questions?

Re: In a class of 33, 20 play cricket, 25 play football and 18 play hockey [#permalink]

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17 Feb 2016, 23:44

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Re: In a class of 33, 20 play cricket, 25 play football and 18 play hockey [#permalink]

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23 Mar 2016, 12:48

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.

Re: In a class of 33, 20 play cricket, 25 play football and 18 play hockey [#permalink]

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13 Sep 2016, 18:29

Step 1: Apply the formula TOTAL = Cricket + Football + Hockey + Neither + All - (Sum of 2 overlaps). This gives ALL = 7

Step 2: Reduce 7 from all two overlaps relevant to cricket. This gives only cricket and Football and only cricket and hockey. Formula method : (Cricket and Football)+ (Cricket and Hockey) - 2(All). Let's call it R

In a class of 33, 20 play cricket, 25 play football and 18 play hockey. 15 play both cricket and football, 12 play football and hockey and 10 play cricket and hockey. If each student plays at least one game, how many students play only cricket?

Now think of the cricket circle. 20 play cricket. 15 play cricket and football (including 7 who play all three). 10 play cricket and hockey (including 7 who play all three). So 15 + 10 - 7 (because 7 is double counted, we subtract it once) = 18 play at least one sport other than cricket too. Therefore, only 20 - 18 = 2 play cricket only.
_________________

In a class of 33, 20 play cricket, 25 play football and 18 play hockey. 15 play both cricket and football, 12 play football and hockey and 10 play cricket and hockey. If each student plays at least one game, how many students play only cricket?

1. 8 2. 6 3. 4 4. 2 5. 0

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
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