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Directions: Study the following information carefully and answer the questions given beside:
 
A survey of a group of 1250 students is conducted to know about their likeness of sports. 68% students like Football, 69.6% like Cricket and 64% like Badminton. 14.4% of them like only Football, 16.8% like only Cricket and 8.8% like only Badminton. Now, answer the following questions based on this information.
1
What per cent of the students like, at most, one sport from the given sports?
» Explain it
C
Following the common explanation, we get

The total number of people who like only one sport = 180 + 210 + 110 = 500

∴   Reqd% =  500  × 100 = 40%
1250

Common Explanation:

As per the given information, we get

No. of students who like football = 68% of 1250 = 850

No. of students who like Cricket = 69.6% of 1250 = 870

And, No. of students who like Badminton = 64% of 1250 =  800

From the further information, we get

No. of students who like only Football = 14.4% of 1250 = 180

No. of students who like only Cricket = 16.8% of 1250 = 210

No. of students who like only Badminton = 8.8% of 1250 = 110

Suppose, x students like both Football and Cricket, y students like both Football and Badminton, z students like Badminton and Cricket and k students like all three games.


  x + y + k + 180 = 850

or, x + y + k = 670       ...(i)

Similarly, 

x + z + k = 660      ...(ii)

y + z + k = 690       ...(iii)

And, x + y + z + k = (1250 – 180 + 210 + 110)

x + y + z + k = 750    ....(iv)

By solving the equations (i), (ii), (iii) and (iv), we get

k = 520 or, x = 60, y = 90, z = 80


 
2
How many students are there who like any two sports from the given three sports.
» Explain it
B
Following the common explanation, we get 
 
The total no. of students who like any two sports = x + y + z = 60 + 90 + 80 = 230 

Common Explanation:

As per the given information, we get

No. of students who like football = 68% of 1250 = 850

No. of students who like Cricket = 69.6% of 1250 = 870

And, No. of students who like Badminton = 64% of 1250 =  800

From the further information, we get

No. of students who like only Football = 14.4% of 1250 = 180

No. of students who like only Cricket = 16.8% of 1250 = 210
No. of students who like only Badminton = 8.8% of 1250 = 110

Suppose, x students like both Football and Cricket, y students like both Football and Badminton, z students like Badminton and Cricket and k students like all three games.


  x + y + k + 180 = 850

or, x + y + k = 670       ...(i)

Similarly, 

x + z + k = 660      ...(ii)

y + z + k = 690       ...(iii)

And, x + y + z + k = (1250 – 180 + 210 + 110)

x + y + z + k = 750    ....(iv)

By solving the equations (i), (ii), (iii) and (iv), we get

k = 520 or, x = 60, y = 90, z = 80


 

3
What per cent of the students like Football and Badminton both but not Cricket?
» Explain it
D
Following the common explanation, we get

The total number of students who like Footaball and Badminton = 90

And the total number of students = 1250

Reqd % =  90  × 100 = 7.2%
1250

Common Explanation:

As per the given information, we get

No. of students who like football = 68% of 1250 = 850

No. of students who like Cricket = 69.6% of 1250 = 870

And, No. of students who like Badminton = 64% of 1250 =  800

From the further information, we get

No. of students who like only Football = 14.4% of 1250 = 180

No. of students who like only Cricket = 16.8% of 1250 = 210

No. of students who like only Badminton = 8.8% of 1250 = 110

Suppose, x students like both Football and Cricket, y students like both Football and Badminton, z students like Badminton and Cricket and k students like all three games.


  x + y + k + 180 = 850

or, x + y + k = 670       ...(i)

Similarly, 

x + z + k = 660      ...(ii)

y + z + k = 690       ...(iii)

And, x + y + z + k = (1250 – 180 + 210 + 110)

x + y + z + k = 750    ....(iv)

By solving the equations (i), (ii), (iii) and (iv), we get

k = 520 or, x = 60, y = 90, z = 80


 
4
How many students are there who like all three sports?
» Explain it
D
We can clearly get from the common explanation that, there are 520 students (representing k) who like all the three sports.

Common Explanation:

As per the given information, we get

No. of students who like football = 68% of 1250 = 850

No. of students who like Cricket = 69.6% of 1250 = 870

And, No. of students who like Badminton = 64% of 1250 =  800

From the further information, we get

No. of students who like only Football = 14.4% of 1250 = 180

No. of students who like only Cricket = 16.8% of 1250 = 210

No. of students who like only Badminton = 8.8% of 1250 = 110

Suppose, x students like both Football and Cricket, y students like both Football and Badminton, z students like Badminton and Cricket and k students like all three games.


  x + y + k + 180 = 850

or, x + y + k = 670       ...(i)

Similarly, 

x + z + k = 660      ...(ii)

y + z + k = 690       ...(iii)

And, x + y + z + k = (1250 – 180 + 210 + 110)

x + y + z + k = 750    ....(iv)

By solving the equations (i), (ii), (iii) and (iv), we get

k = 520 or, x = 60, y = 90, z = 80


 
5
What per cent of students like both Football and Cricket but not Badminton?
» Explain it
B
Following the common explanation, we get

Total number of students who like only Football and Cricket = 60

And the total number students = 1250

 Reqd% =  60  × 100 = 4.8%
1250
  Common Explanation:

As per the given information, we get

No. of students who like football = 68% of 1250 = 850

No. of students who like Cricket = 69.6% of 1250 = 870

And, No. of students who like Badminton = 64% of 1250 =  800

From the further information, we get

No. of students who like only Football = 14.4% of 1250 = 180

No. of students who like only Cricket = 16.8% of 1250 = 210

No. of students who like only Badminton = 8.8% of 1250 = 110

Suppose, x students like both Football and Cricket, y students like both Football and Badminton, z students like Badminton and Cricket and k students like all three games.


  x + y + k + 180 = 850

or, x + y + k = 670       ...(i)

Similarly, 

x + z + k = 660      ...(ii)

y + z + k = 690       ...(iii)

And, x + y + z + k = (1250 – 180 + 210 + 110)

x + y + z + k = 750    ....(iv)

By solving the equations (i), (ii), (iii) and (iv), we get

k = 520 or, x = 60, y = 90, z = 80