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

Question

Competition Mode Question

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 16957 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)

Posted from my mobile device

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)

Posted from my mobile device

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 16957 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

Automated notice from GMAT Club BumpBot: 

A member just gave Kudos to this thread, showing it’s still useful. I’ve bumped it to the top so more people can benefit. Feel free to add your own questions or solutions.

 This post was generated automatically.

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

Prep Toolkit

Top 5 GMAT Debrief Videos

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards

GMAT Problem Solving (PS) Questions

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

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards