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

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

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New post Updated on: 10 Feb 2017, 23:31
3
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D
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Question Stats:

63% (02:17) correct 37% (02:48) wrong based on 308 sessions

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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?

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

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Originally posted by AnkitK on 25 Mar 2011, 22:23.
Last edited by abhimahna on 10 Feb 2017, 23:31, edited 1 time in total.
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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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New post 26 Mar 2011, 00:41
3
2
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 .


With the formula;

\(n(ALL) = n(C)+n(F)+n(H)-n(C \cap F)-n(F \cap H)-n(C \cap H)+n(C \cap F \cap H)+n(Neither)\)

\(n(ALL) = 33\)
\(n(C) = 20\)
\(n(F) = 25\)
\(n(H) = 18\)
\(n(C \cap F) = 15\)
\(n(F \cap H) = 12\)
\(n(C \cap H) = 10\)
\(n(Neither)=0\) As each student plays at least one game.

Putting them in the formula;
\(33 = 20+25+18-15-12-10+n(C \cap F \cap H)+0\)
\(33 = 26+n(C \cap F \cap H)\)
\(n(C \cap F \cap H)=33-26=7\)

\(n(C) = n(Only Cricket) + n(C \cap H) + n(C \cap F) - n(C \cap F \cap H)\)
\(n(Only Cricket) = n(C) - n(C \cap H) - n(C \cap F) + n(C \cap F \cap H) = 20-10-15+7 = 2\)

Ans: "D" (Well, I see no E here. Not a GMAT question)
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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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New post 25 Mar 2011, 23:33
After a very lengthy calculation I got 2. What is the source?

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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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New post 25 Mar 2011, 23:42
Well it was asked in one of the public examinations.yes the answer is 2 only.But how?
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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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New post 26 Mar 2011, 00:05
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1
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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Re: In a class of 33, 20 play cricket, 25 play football and 18 play hockey [#permalink]

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New post 26 Mar 2011, 00:16
Thnkx gmat1220 .Dear Fluke pls explain your approach to the question..Also is there any short cut to such questions?
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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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New post 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?


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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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New post 26 Mar 2011, 01:03
AnkitK wrote:
Thnkx gmat1220 .Dear Fluke pls explain your approach to the question..Also is there any short cut to such questions?


Please go through this;

http://gmatclub.com/forum/formulae-for-3-overlapping-sets-69014.html#p729340

You will never have to memorize the formula(s) if you understand this fully.
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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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New post 26 Mar 2011, 07:10
1
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
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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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New post 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.

So, answer is D.
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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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New post 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

Step 3: Total Cricket - R = Only cricket

Option D
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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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New post 13 Sep 2016, 22:47
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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



Total = 33 = 20 + 25 + 18 - 15 - 12 - 10 + n(All three) + 0
n (All three) = 7

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.
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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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New post 08 Sep 2017, 01:08
TOTAL - NEITHER = ΣA - Σ(ONLY2)+Σ(ALL 3)
ΣA=20+25+18=63
Σ ONLY2 = 37
NEITHER =0
TOTAL 33
33-0=63-37+ALL
ALL=7
ONLY CRICKET= A - (A&B-K) -(A&C-K)-K
= 20- (15- 7)-(10-7)-7
= 20 -8-3-7 =2
OPTION D
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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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New post 11 Sep 2017, 16:01
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
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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Re: In a class of 33, 20 play cricket, 25 play football and 18 play hockey   [#permalink] 11 Sep 2017, 16:01
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