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

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

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.

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

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

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

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

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:

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.

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

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

Answer is B.

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)

Let 'O', 'I', 'II', 'III' be the number of students who play zero, one, two and three games respectively.
Now, O = 18
The total number of students playing at least one game is given as follows:
I + II + III = 50 - 18 = 32
Now, those who play two games contribute to the total number of games played by 2 units, similarly, those who play three games contribute by 3. Hence,
I + 2 II + 3 III = 20 + 15 + 11 = 46
Subtracting these equations we get
II + 2 III = 14
Consider the second statement.
7 play Hockey and Cricket, 4 play Cricket and Football and 5 play Hockey and football.
Adding them up, we get the entire segment of II(people playing two sports) once and the entire segment of III thrice.
II + 3 III = 7 + 4 + 5 = 16
Subtracting this from the last equation, we get III = 2
Hence, II = 14 - 2*2 = 10.

Label the overlaps as follows:


Complete the diagram;


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

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

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.

\(\{Total\} = \{Hockey\} + \{Cricket\} + \{Football\} - \{HC + CH + HF\} + \{All \ three\} + \{Neither\}\)
(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.[/quote

The only thing that I dont get is why in the formula we add the number of those who play ALL THREE. Just it does not sit in my head. The question that confuses me is that why we add ALL THREE as additional players? Cannot their number be hidden in preceding part of the formula? Hope could explain what I meant.

Have you checked this topic: ADVANCED OVERLAPPING SETS PROBLEMS ? It explains the logic behind the formula in detail.

As per your calculation and the same i got it comes to around -2x=2.
But you just considered it as 'X'

Can u explain why