In a class of 65 students 40 like cricket, 25 like football and 20 lik

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In a class of 65 students 40 like cricket, 25 like football and 20 like hockey. 10 students like both cricket and football, 8 students like football and hockey and 5 students like all three sports. If all the students like at least one sport, then the number of students who like both cricket and hockey is

A. 6
B. 7
C. 8
D. 10
E. 12

A. 6
B. 7
C. 8
D. 10
E. 12

ArunSharma12 · Answer

In a class of 65 students 40 like cricket, 25 like football and 20 like hockey. 10 students like both cricket and football, 8 students like football and hockey and 5 students like all three sports. If all the students like at least one sport, then the number of students who like both cricket and hockey is

A. 6
B. 7
C. 8
D. 10
E. 12

Attachment:

img1.png [ 11.16 KiB | Viewed 17161 times ]

If all the students like at least one sport then
a+b+c+x+y+z+m =65 --1
a+x+y+m = 40
b+x+z+m= 25
c+y+z+m = 20
a+b+c+2(x+y+z)+3m = 85 --2
x+m = 10
z+m = 8
m = 5
x = 5 & z = 3

subtracting equation 1 & 2,
x+y+z+2m = 20
x+y+z+m = 15
y+m = 15 - (5+3) = 7

Ans: B

lacktutor · Answer

Total —65 
Cricket —40 
Football — 25 
Hockey —20 
Cricket and Football —10
Football and hockey —8 
Hockey and Cricket —x 
All three sports —5 
Neither —0 
——> 40 +25+20 —( (10–5)+ (x—5)+(8–5)) —2*5 + 0 = 65 
85–(x+ 3) —10 = 65
x = 85 —13–65 = 7

Answer ( B)

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GMATWhizTeam · Answer

Solution: 
 • Number of students who like cricket and football only = 10 -5 = 5
• Number of students who like football and hockey only = 8-5 = 3
• a +b + c + 5 + 3 + 5 +d =65
 o a + b + c +d = 52…(i) 
 https://www.dropbox.com/s/h4wxgsynexbcjtu/49.jpg?dl=1 
• The total number of students who like cricket = 40
 o a + d + 5 + 5 = 40
o a + d = 30 
• Total number of students who like football =25
 o b + 5 + 5 + 3 = 25
o b = 12 
• The total number of students who like hockey = 40
 o c + d + 5 + 3 = 20
o c + d = 12
  a + b + c +2d = 54…(ii)  
• Now, on subtracting equation (i) from (ii), we get
 o d = 2
 Total number of students who like hockey and cricket = d + 5 = 2 + 5 = 7  
Hence, the correct answer is  Option B.

exc4libur · Answer

t=a+b+c-both-2mid+neither
65=40+25+20-both-2(10)+0
-10=-both, both=10

both=cf-mid+fh-mid+ch-mid
10=10-5+8-5+x-5
-8=-15+x, x=7

Ans (B)

MrSengupta · Answer

The answer is (B) . We can solve this sum using set theory of circles having Cricket , Football, Hockey . Let us say the the people playing only Cricket, Football and hockey be denoted as a, b and c.

Therefore a+b+c+5+5+3+x=65 (let x denote the no of people playing Cricket and Hockey both only)
=> a+b+c+x=52

Again a+b+c+2(8+x)+3(5)=85
=> x=2

Therefore total ans = 5+2 =7

DeusTacitus22 · Answer

For three sets A, B and C, n(AᴜBᴜC) = n(A) + n(B) + n(C) – n(A∩B) – n(B∩C) – n(C∩A) + n(A∩B∩C)
Let(C∩H)=x
=> 65 = 40+25+20-10-8-x+5
=>x = 20-18+5 =>x = 7

People who only like Cricket and Hockey = 7

Therefore, 
Answer = 7, Option B

GMATinsight · Answer

Total = 65
Cricket = 40
Football = 25
Hockey = 20
Cricket and Football = 10
Football and Hockey = 8
Cricket and Hockey = x
All the games = 5

65 = 40+25+20 - (10+8+x) + 5

i.e. x = 7

Answer: Option B

freedom128 · Answer

Nett Total = 65 (nobody does no sport). 
P(Cricket) = P(C) = 40
P(Football) = P(F) = 25
P(Hockey) = P(H) = 20
P(C n F n H) = 5
P(C n F) = 10
P(F n H) = 8

Q. P(C n H) ?

Nett total = P(C) + P(F) + P(H) – P(C n F) – P(F n H) – P(C n H) + P(C n F n H)

65 = 40 + 25 + 20 - 10 - 8 - P(C n H) + 5

P(C n H) = 7
--> the number of students who like both cricket and hockey is 7

FINAL ANSWER IS (B)

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unraveled · Answer

In a class of 65 students 40 like cricket, 25 like football and 20 like hockey. 10 students like both cricket and football, 8 students like football and hockey and 5 students like all three sports. If all the students like at least one sport, then the number of students who like both cricket and hockey is

A. 6
B. 7
C. 8
D. 10
E. 12

Attachment:

img1.png [ 11.16 KiB | Viewed 17161 times ]

If all the students like at least one sport then
a+b+c+x+y+z+m =65 --1
a+x+y+m = 40
b+x+z+m= 25
c+y+z+m = 20
a+b+c+2(x+y+z)+3m = 85 --2
x+m = 10
z+m = 8
m = 5
x = 5 & z = 3

subtracting equation 1 & 2,
x+y+z+2m = 20
x+y+z+m = 15
y+m = 15 - (5+3) = 7

Ans: B

monikakumar · Answer

In a class of 65 students 40 like cricket, 25 like football and 20 like hockey. 10 students like both cricket and football, 8 students like football and hockey and 5 students like all three sports. If all the students like at least one sport, then the number of students who like both cricket and hockey is

A. 6
B. 7
C. 8
D. 10
E. 12

A. 6
B. 7
C. 8
D. 10
E. 12

total=play hockey+cricket+football -(play only hockey and football+only footbal and cricket+only hockey and cricket)-2(all three)
65=85-(5+3+x)-2(5)
x=2.
so play hockey and cricket is 7
Ans B

madgmat2019 · Answer

let number of students who like both cricket and hockey be x

65 = 40+25+20 - 10 - 8 - x + 5 + 0
x = 7

OA:B

Upami · Answer

Ans _ B

All Combined - 40+25+20=85 but there are only 65 students hence 20 are in common category

C+F+H=5
So, C+F= 5 and F+H=3 Total (5+5+3)=13 hence 7 most be common between Cricket and Hockey.

Kinshook · Answer

Given: 1. In a class of 65 students 40 like cricket, 25 like football and 20 like hockey. 2. 10 students like both cricket and football, 8 students like football and hockey and 5 students like all three sports. Asked: If all the students like at least one sport, then the number of students who like both cricket and hockey is Screenshot 2020-04-29 at 2.47.42 PM.png 40 + 17 + 8- x = 65 x = 2 The number of students who like both cricket and hockey is = 5 + x = 7 IMO B

bumpbot · Answer

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In a class of 65 students 40 like cricket, 25 like football and 20 lik

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