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

Question

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 .

A. 8
B. 6
C. 4
D. 2
E. 0

KarishmaB · Accepted Answer

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.

fluke · Answer

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)

gmat1220 · Answer

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

Posted from my mobile device

subhashghosh · Answer

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

dina98 · Answer

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.

ScottTargetTestPrep · Answer

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

abalodi · Answer

Hi Scott, I solved this using a formula i learned in TTP as follows... Can you let me know if my approach is correct. I ended up with the same answer.

33= 63 (A,B, and C) - 37(double overlaps) - 2x(all three overlaps) + N (0 in this case since everyone plays something)

Upon solving x =3.5 and then i just worked my way inside out to figure out only cricket and ended up with 2

ScottTargetTestPrep · Answer

There's a mistake in your calculations; if you solve 33 = 63 - 37 - 2x, you'll find x = -3.5, which is not correct because the number of people who play all of the sports can be neither negative nor fractional.

The "double overlaps" and "all three overlaps" in the formula you mention refer to the number of people who play exactly two of the sports and exactly three of the sports, respectively. The numbers 15, 12 and 10 in this problem, on the other hand, are the number of people who play at least two of the three sports, so they all include the number of people who play all three sports as well.

In order to use that formula with this question, you need to determine the number of people who play exactly two of the sports, for which you need the number of people who play all three of the sports. That's why it's not practical to use that formula for this question.

abalodi · Answer

thank you for the explanation, understood. I guess i got lucky.

Could you explain when to use the formula you used Total = #C + #F + #H - #(C and F) - #(F and H) - #(C and H) + #(C and F and H) + None vs the formula that i used?

ScottTargetTestPrep · Answer

The biggest difference between the formula I used earlier and the formula you used is that in my formula, the double overlaps include the triple overlap as well; whereas, in your formula, the double overlaps are the number of elements which belong to two sets but do not belong to three sets.

When you add the number of elements of sets A, B and C; the elements which belong to exactly two sets are counted twice and the elements which belong to all three sets are counted three times. One way of obtaining the number of elements that belong to A or B or C is subtracting from A + B + C the number of elements that belong to exactly two sets and twice the number of elements that belong to all three sets. That's what's done in your formula. The other way of obtaining the same number is subtracting from A + B + C the number of elements that belong to at least two sets; in which case you will have subtracted the number of elements that belong to all three sets three times. To get the correct number, one should add the number of elements that belong to all three sets and that's what's done in my formula.

Kinshook · Answer

Given: 1. In a class of 33, 20 play cricket, 25 play football and 18 play hockey. 2. 15 play both cricket and football, 12 play football and hockey and 10 play cricket and hockey. Asked: If each student plays at least one game, how many students play only cricket? Screenshot 2020-04-29 at 10.04.53 AM.png 33 = 20 + 25 + 18 - (15+12+10) + x x = 7 Number of students playing only cricket = 20 - (8+7+3) = 2 IMO D

1616 · Answer

Hello  ScottTargetTestPrep ,

In these kind of questions, How can we determine whether the question is talking about exactly two or at least two?
Can you please help, I took me 4+ minutes and still an incorrect answer.

Bunuel · Answer

In this question, "15 play both cricket and football" implies that 15 individuals play both sports, but they might also participate in other sports. If the statement were "15 play exactly two sports: cricket and football," it would mean that these 15 individuals play only those two sports and no other. Therefore, without additional restrictions, if some individuals are grouped into two categories, it does not necessarily exclude them from being part of other categories.

Kinshook · Answer

@AnkitKGiven: 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.
Asked: If each student plays at least one game, how many students play only cricket?
33 = 20 + 25 + 18 - 15 - 12 - 10 + All
All = 7

Number of students that play only cricket = 20 - 7 - (15-7) - (10-7) = 20 - 7 - 8 - 3 = 2

IMO D

Madhavsinghyadav · Answer

GMATNinja  
how do i solve this one with your method of three overlapping sets?

sherifY · Answer

I dont understand why we minus AuB but we add the AuBuC

GMATNinja · Answer

The other methods in the thread might be more efficient in this case, and it's not the easiest method to describe in writing. But in case it would help, here's what it would look like: tally method.PNG Here's what's going on: you'd start by letting "x" be the number of students who play all three sports. Those students account for 3x "tallies", where a "tally" represents a unique student-sport combination (e.g. if Tim plays cricket, football, AND hockey, that would be 3 distinct "tallies"). So the total number of tallies is 20+25+18 -- the sum of the tallies for each individual sport. That makes the number who play only two sports: (15-x)+(12-x)+(10-x) = 37-3x. Each of those students represent TWO distinct tallies, so the total tallies for that group is 2*(37-3x) = 74-6x. Next, calculate the number who play just one sport: 33-x-(37-3x) = 2x-4. Each of those students only represents one tally, so the total tallies for that group is also 2x-4. The sum for each group has to equal the total number of tallies, so 20+25+18 = 63 = (3x)+(74-6x)+(2x-4) = 70-x. Since 63 = 70-x, x is 7. We're almost done! Since 7 play all three sports, we know that 8 play just cricket and football (15-7) and 3 play just cricket and hockey (10-7). That's 11 students who play cricket plus one other sport, along with the 7 who play all three sports (including cricket).... so 18 who play cricket plus AT LEAST one other sport, leaving only 2 who play JUST cricket (20-18). So... yeah. Not sure if that helps at all. This method is a bit more elegant in a video than in a forum post.

eeshajain · Answer

Can someone please explain why there is a difference in the formula used in this question vs the formula used here  https://gmatclub.com/forum/of-the-300-subjects-who-participated-in-an-experiment-using-virtual-re-134147.html?

In the cricket/football/hockey question, we are adding the count of people doing all 3 sports, 
Total = #C + #F + #H - #(C and F) - #(F and H) - #(C and H) + #(C and F and H) + None,

whereas in the other question we subtracted twice the count of inter-section of all three groups
Total = # in (Group A) + # in (Group B) + # in (Group C) – # in (Groups of Exactly Two) – 2   + # in (Neither)