In a class of 50 students, 20 play Hockey, 15 play Cricket and 11 play

Question

In a class of 50 students, 20 play Hockey, 15 play Cricket and 11 play Football. 7 play both Hockey and Cricket, 4 play Cricket and Football and 5 play Hockey and football. If 18 students do not play any of these given sports, how many students play exactly two of these sports?

A. 12
B. 10
C. 11
D. 15
E. 14

A. 12 
B. 10
C. 11 
D. 15
E. 14

Bunuel · Accepted Answer

Notice that "7 play both Hockey and Cricket" does not mean that out of those 7, some does not play Football too. The same for Cricket/Football and Hockey/Football.

\{Total\} = \{Hockey\} + \{Cricket\} + \{Football\} - \{HC + HF + CF\} + \{All \ three\} + \{Neither\}  
(For more check  ADVANCED OVERLAPPING SETS PROBLEMS )

50 = 20 + 15 + 11 -(7 + 4 + 5) + \{All \ three\} + 18 ;
 \{All \ three\}=2 ;

Those who play ONLY Hockey and Cricket are 7 - 2 = 5;
Those who play ONLY Cricket and Football are 4 - 2 = 2;
Those who play ONLY Hockey and Football are 5 - 2 = 3;

Hence, 5 + 2 + 3 = 10 students play exactly two of these sports.

Answer: B.

PareshGmat · Answer

Answer = 10
Using Venn Diagram
Bunuel, can you please tell if this method is correct?. Got x -ve in this case

josemnz83 · Answer

This question asks for the number of students who played exactly two sports? Why does the second formula not work here?

Bunuel · Answer

Notice that "7 play both Hockey and Cricket..." does NOT mean that these 7 students play ONLY Hockey and Cricket, some might play Football too. The same for "4 play Cricket and Football and 5 play Hockey and football". So, we cannot use the second formula directly. Also notice that we don't know the number of students who play all three sports.

But we CAN use the first formula, find the number of students who play all three and then find the number of students who play exactly two of the sports.

Hope it's clear.

MJ23 · Answer

Hi All,

That's exactly what I did, started with formula 1 and then used #2 once had "g" (all three):

Formula #1: 50=20+15+11-(7+4+5)+18 --> 2 (all three or "g")
Formula #2: 50=20+15+11-x-(2*2)+18 which leads to 50=60-x ; x=10

Hope this helps.

Cheers,
MJ

BrainLab · Answer

I've solved this one with the second formula
Let X be the area where all 3 overlap and Exactly 2-Groups overlaps = 2-Group ovelaps - 3*X --> 
50=20+15+11 -(7+4+5-3X) - 2X+18
X=2 and 16-3X=10

ScottTargetTestPrep · Answer

Since 18 students do not play any of the three sports, 50 - 18 = 32 students must play at least one of the 3 sports. This total can be formulated as follows:

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

Thus, we have:

32 = 20 + 15 + 11 - 7 - 4 - 5 + #(H and C and F)

32 = 30 + #(H and C and F)

2 = #(H and C and F)

Since #(H and C) = 7 (which also include those who play Football), but we’ve found that #(H and C and F) = 2, there must be 7 - 2 = 5 students who play Hockey and Cricket only. Similarly, there must be 4 - 2 = 2 students who play Cricket and Football only, and 5 - 2 = 3 students who play Hockey and Football only. Thus, there must be 5 + 2 + 3 = 10 students who play exactly 2 sports.

Answer: B

ankitprad · Answer

Hi Bunuel Could you tell why are we considering, from the given statement ' 7 people who play both hockey and cricket', there are possibly people also playing football. In your post https://gmatclub.com/forum/advanced-ove ... 44260.html in eg 6 , which, I think, is similar to this problem - it is given that the roster had 9 names in common between E & M. Why then are we not considering the possibility of the common names on the E&M roster include people who have also taken statistics ? Thanks

Bunuel · Answer

1. Let me ask you back: how does the phrase "7 play both Hockey and Cricket..." could mean that those 7 does not play any other sports? 2. In the example, you give, it's the same: "E and M had 9 names in common" does not mean that any from those 9 do not belong to S. If you check solutions here: https://gmatclub.com/forum/when-profess ... 43149.html, you'll see that it can be a case. For example: https://gmatclub.com/forum/download/file.php?id=22009

avinash07 · Answer

As No of students who dont play one of these games is 18.
So,No. of students who plays at least one of these fame is 50-18=32.
Hence, 8+x+7-x+x+5-x+4+x+4-x+2+x+5-x=32

30-x=32 or x=2.

No. of students who play exactly 2 game=7-x+5-x+4-x=16-3x =16-6=10.Ans.

Kinshook · Answer

Given: In a class of 50 students, 20 play Hockey, 15 play Cricket and 11 play Football. 7 play both Hockey and Cricket, 4 play Cricket and Football and 5 play Hockey and football. 
Asked: If 18 students do not play any of these given sports, how many students play exactly two of these sports?

Total = A + B + C - Two + Three + None
50 = 20 + 15 + 11 - 7 - 4 - 5 + Three + 18
Three = 50 - 46 + 16 - 18 = 2

Total = A + B + C - Exactly Two - 2*Three + Non
50 = 20 + 15 + 11 - Exactly Two - 4 + 18
Exactly Two = 46 +14 - 50 = 10

IMO B

Basshead · Answer

I find in these types of questions it's an absolute must to draw a Venn diagram:

Answer is B.

CrackverbalGMAT · Answer

Solution:

Let Hockey = H ; Football= F ; Cricket = C

We know Total = H +C+ F - [(H∩C) + (C∩H) + (H∩F)] + (H∩C∩F) + Neither

=> 50 = 20+15+11 -[ 7+4+5 ] + (H∩C∩F) + 18

=> H∩C∩F = 2

Exactly two of these sports = Only HC + Only HC + Only HF

Number of students playing ONLY Hockey and Cricket

= (H∩C)-(H∩C∩F ) = 7 - 2 = 5

Number of students playing ONLY Cricket and Football

= (C∩F) - (H∩C∩F )= 4-2 = 2

Number of students playing ONLY Hockey and Football

= (H∩F) - (H∩C∩F ) = 5-2 = 3

Thus, number of students play exactly two of these sports. = 5+2+3=10 (Option B)

Hope this helps

Devmitra Sen(Quants GMAT expert)

BrentGMATPrepNow · Answer

Label the overlaps as follows:
 https://i.imgur.com/YJKP9Eg.png

Complete the diagram;
 https://i.imgur.com/b2Po0TK.png

If 18 students do not play any of these given sports, then the sum of the parts inside the circles = 50 - 18 =  32

So we can write the following equation: (20 - a - b - x) + (15 - b - c - x) + (11 - a - c - x) + a + b + c + x =  32 
Simplify: 46 - a - b - c - 2x = 32
Rearrange to get:   14 - 2x = a + b + c

If 7 play both Hockey and Cricket, we can write: b + x = 7
If 4 play Cricket and Football, we can write: c + x = 4
If 5 play Hockey and football, we can write: a + x = 5
Add these three equations to get:  3x +  a + b + c  = 16  
Substitute to get:  3x +  14 - 2x  = 16  
Simplify to get:  x + 14 = 16  
Solve:  x = 2

Plug  x = 2  into the 3 equations above to get: a = 2, b = 5 and c = 3, which means  a + b + c = 10

Answer: B

GMATGuruNY · Answer

T = H + C + F - (HC + CF + HF) - 2(HCF) + N 
In this equation:
HC = only H and C
CF = only C and F
HF = only H and F
HCF = all 3 sports
N = none of the 3 sports

In a class of 50 students, 20 play Hockey, 15 play Cricket and 11 play Football.
18 students do not play any of these given sports. 
Plugging these values into the blue equation, we get:
50 = 20 + 15 + 11 - (HC + CF + HF) - 2(HCF) + 18
 (HC + CF + HF) + 2(HCF) = 14

7 play Hockey and Cricket: 
HC + HCF = 7
 4 play Cricket and Football: 
CF + HCF = 4
 5 play Hockey and football: 
HF + HCF = 5
Adding these 3 equations, we get:
 (HC + CF + HF) + 3(HCF) = 16

Subtracting the red equation from the green equation, we get:
HCF = 2
Plugging HCF=2 into the red equation, we get:
HC + CF + HF + 2(2) = 14
HC + CF + HF = 10

B

GMATGuruNY · Answer

An alternate approach -- one that does not require advance knowledge of a formula -- is to GUEES AND CHECK.

7 play both Hockey and Cricket, 4 play Cricket and Football and 5 play Hockey and Football. 
Meaning:
7 students play AT LEAST H AND C
4 students play AT LEAST C AND F
5 students play AT LEAST H AND F

Case 1: 1 student plays all 3 sports 
Subtracting this 1 student from each of the three groups above, we get:
Number who play only H and C = (number who play at least H and C) - (number who play all 3) = 7-1 = 6
Number who play only C and F = (number who play at least C and F) - (number who play all 3) = 4-1 = 3
Number who play only H and F = (number who play at least H and F) - (number who play all 3) = 5-1 = 4
Adding together the resulting values, we get:
Number who play exactly two sports = 6+3+4 = 13
Since 13 is not among the answer choices, Case 1 is not viable.

Case 2: 2 students play all 3 sports 
Subtracting these 2 students from each of the three groups above, we get:
Number who play only H and C = (number who play at least H and C) - (number who play all 3) = 7-2 = 5
Number who play only C and F = (number who play at least C and F) - (number who play all 3) = 4-2 = 2
Number who play only H and F = (number who play at least H and F) - (number who play all 3) = 5-2 = 3
Adding together the resulting values, we get:
Number who play exactly two sports = 5+2+3 = 10
Since 10 is among the answer choices, Case 2 is viable.

If the number who play all 3 sports INCREASES, the number who play exactly two sports will DECREASE below 10.
Since no answer choice is less than 10, only Case 2 is viable, implying that the number who play exactly two sports = 10

B

Vibhatu · Answer

Although it is not written that Any student play all three sports. Is it not a mistake ? Please help

Posted from my mobile device

GMATGuruNY · Answer

7 play both Hockey and Cricket, 4 play Cricket and Football, and 5 play Hockey and Football . 
 How many students play exactly two of these sports? 
If no one plays all 3 sports. then the question can be answered -- far too easily -- simply by adding the values in the blue sentence above.
The result would be a silly problem that every test-taker would solve correctly.

Further, if no student plays all 3 sports, then the following formula would apply:
Total = (total who play H) + (total who play C) + (total who play F) - (number who play two sports) + (number who play none of the sports) = 20 + 15 + 11 - (7 + 4 + 5) + 18 = 48
Since the actual total is not 48 but 50, there must be some students not included in the equation above: the number who play all 3 sports.

DanTheGMATMan · Answer

Standard 3 group overlapping sets with a little twist:

https://www.youtube.com/watch?v=p7nTztjM-Dw

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