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.

Important for :

1

C

Following the common explanation, we getThe total number of people who like only one sport = 180 + 210 + 110 = 500

∴ Reqd% = |
500 | × 100 = 40% |

1250 |

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

Similarly,

x + z + k = 660

y + z + k = 690

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

x + y + z + k = 750

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

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

2

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

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

Similarly,

x + z + k = 660

y + z + k = 690

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

x + y + z + k = 750

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

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

3

D

Following the common explanation, we getThe total number of students who like Footaball and Badminton = 90

And the total number of students = 1250

Reqd % = | 90 | × 100 = 7.2% |

1250 |

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

Similarly,

x + z + k = 660

y + z + k = 690

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

x + y + z + k = 750

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

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

4

D

We can clearly get from the common explanation that, there are 520 students (representing k) who like all the three sports.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

Similarly,

x + z + k = 660

y + z + k = 690

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

x + y + z + k = 750

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

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

5

B

Following the common explanation, we getTotal number of students who like only Football and Cricket = 60

And the total number students = 1250

∴ Reqd% = |
60 | × 100 = 4.8% |

1250 |

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

Similarly,

x + z + k = 660

y + z + k = 690

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

x + y + z + k = 750

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

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