Answer

Verified

448.8k+ views

Hint: Represent the football, basketball, cricket as the sets F, B, C. Then write down the values of

Complete step-by-step answer:

n(F), n(B), n(C)…….. $n\left( F\cup B\cup C \right)$and $n\left( F\cap B\cap C \right)$ .

Then further use the formula

$n\left( F\cup B\cup C \right)=n\left( F \right)+n\left( B \right)+n\left( C \right)-\left\{ n\left( F\cap B \right)+n\left( B\cap C \right)+n\left( F\cap C \right) \right\}+n\left( F\cap B\cap C \right)$ and get the desired results.

In the question it is said that a college awarded 38 medals in football, 15 medals in basketball and 20 in cricket. Then it is said that there medals went to a total of 58 men and only 3 men got the medals in all three sports and hence we have to find the number of men who received exactly in two of the three sports.

Let the number of men who got medals in football be represented as n(F), similarly for basketball let it be represented as n(B) and in cricket as n(C).

For number of people who won in all the three sports will be $n\left( F\cap B\cap C \right)$, and the number of people who got medals in two of three is $\left\{ n\left( F\cap C \right)+n\left( F\cap B \right)+n\left( B\cap C \right) \right\}$and for total number men be represented as $n\left( F\cup B\cup C \right).$

So now we will represent the data as,

n(F) = 38

n(B) = 15

n(C) = 20

$n\left( F\cap B\cap C \right)=3$

$n\left( F\cup B\cup C \right)=58$

So now we will use the formula which is

$n\left( A\cup B\cup C \right)=n\left( A \right)+n\left( B \right)+n\left( C \right)-\left\{ n\left( A\cap B \right)+n\left( B\cap C \right)+n\left( A\cap C \right) \right\}+n\left( A\cap B\cap C \right)$

Now instead of sets A, B, C we will put sets of football (F), Basketball (B) and cricket (C).

We will write as,

$n\left( F\cup B\cup C \right)=n\left( F \right)+n\left( B \right)+n\left( C \right)-\left\{ n\left( F\cap B \right)+n\left( B\cap C \right)+n\left( F\cap C \right) \right\}+n\left( F\cap B\cap C \right)$

So now putting the values of n(F) = 38, n(B) = 15, n(C) = 20, $n\left( F\cap B\cap C \right)=3$and$n\left( F\cup B\cup C \right)$ we get,

$58=38+15+20-\left\{ n\left( F\cap B \right)+n\left( F\cap C \right)+n\left( B\cap C \right) \right\}+3$

So,

$n\left( F\cap B \right)+n\left( F\cap C \right)+n\left( B\cap C \right)=\left( 38+15+20+3 \right)58=18$

Hence the number of men who got two out of three medals are 18.

Therefore the correct answer is option (a).

Note: Students are generally confused between the signs $\cup $ and $\bigcap $ . Actually $\left( A\cup B \right)$ means that we have to include both the elements of A and B. \[\left( A\cap B \right)\] means that we have to consider only that element which is common to both A and B.

Complete step-by-step answer:

n(F), n(B), n(C)…….. $n\left( F\cup B\cup C \right)$and $n\left( F\cap B\cap C \right)$ .

Then further use the formula

$n\left( F\cup B\cup C \right)=n\left( F \right)+n\left( B \right)+n\left( C \right)-\left\{ n\left( F\cap B \right)+n\left( B\cap C \right)+n\left( F\cap C \right) \right\}+n\left( F\cap B\cap C \right)$ and get the desired results.

In the question it is said that a college awarded 38 medals in football, 15 medals in basketball and 20 in cricket. Then it is said that there medals went to a total of 58 men and only 3 men got the medals in all three sports and hence we have to find the number of men who received exactly in two of the three sports.

Let the number of men who got medals in football be represented as n(F), similarly for basketball let it be represented as n(B) and in cricket as n(C).

For number of people who won in all the three sports will be $n\left( F\cap B\cap C \right)$, and the number of people who got medals in two of three is $\left\{ n\left( F\cap C \right)+n\left( F\cap B \right)+n\left( B\cap C \right) \right\}$and for total number men be represented as $n\left( F\cup B\cup C \right).$

So now we will represent the data as,

n(F) = 38

n(B) = 15

n(C) = 20

$n\left( F\cap B\cap C \right)=3$

$n\left( F\cup B\cup C \right)=58$

So now we will use the formula which is

$n\left( A\cup B\cup C \right)=n\left( A \right)+n\left( B \right)+n\left( C \right)-\left\{ n\left( A\cap B \right)+n\left( B\cap C \right)+n\left( A\cap C \right) \right\}+n\left( A\cap B\cap C \right)$

Now instead of sets A, B, C we will put sets of football (F), Basketball (B) and cricket (C).

We will write as,

$n\left( F\cup B\cup C \right)=n\left( F \right)+n\left( B \right)+n\left( C \right)-\left\{ n\left( F\cap B \right)+n\left( B\cap C \right)+n\left( F\cap C \right) \right\}+n\left( F\cap B\cap C \right)$

So now putting the values of n(F) = 38, n(B) = 15, n(C) = 20, $n\left( F\cap B\cap C \right)=3$and$n\left( F\cup B\cup C \right)$ we get,

$58=38+15+20-\left\{ n\left( F\cap B \right)+n\left( F\cap C \right)+n\left( B\cap C \right) \right\}+3$

So,

$n\left( F\cap B \right)+n\left( F\cap C \right)+n\left( B\cap C \right)=\left( 38+15+20+3 \right)58=18$

Hence the number of men who got two out of three medals are 18.

Therefore the correct answer is option (a).

Note: Students are generally confused between the signs $\cup $ and $\bigcap $ . Actually $\left( A\cup B \right)$ means that we have to include both the elements of A and B. \[\left( A\cap B \right)\] means that we have to consider only that element which is common to both A and B.

Recently Updated Pages

Let X and Y be the sets of all positive divisors of class 11 maths CBSE

Let x and y be 2 real numbers which satisfy the equations class 11 maths CBSE

Let x 4log 2sqrt 9k 1 + 7 and y dfrac132log 2sqrt5 class 11 maths CBSE

Let x22ax+b20 and x22bx+a20 be two equations Then the class 11 maths CBSE

Let x1x2xn be in an AP of x1 + x4 + x9 + x11 + x20-class-11-maths-CBSE

Let x1x2x3 and x4 be four nonzero real numbers satisfying class 11 maths CBSE

Trending doubts

How many crores make 10 million class 7 maths CBSE

Change the following sentences into negative and interrogative class 10 english CBSE

Fill the blanks with proper collective nouns 1 A of class 10 english CBSE

Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE

The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths

Write a letter to the principal requesting him to grant class 10 english CBSE

Give 10 examples of Material nouns Abstract nouns Common class 10 english CBSE

What are the public facilities provided by the government? Also explain each facility

Write an application to the principal requesting five class 10 english CBSE