Differentiation (Part-38) | S N De

 

18. Using the set operations find the H.C.F. of the numbers $\,\,12,15 \,\text{and}\, 18.$

Sol. Let $\,\,A,B,C\,\,$ denotes the set of factors of $\,\,12,15 \,\text{and}\, 18\,\,$ respectively.  $\\ \therefore A=\{1,2,3,4,6,12\},\,\,B=\{1,3,5,15\},\,\,C=\{1,2,3,6,9,18\}.\\ A \cap B \cap C= \{1,3\}.\\ \text{Clearly,}\,\,A \cap B \cap C$ contains the H.C.F. of the numbers $\,\,12,15 \,\text{and}\, 18.$

19(i) Applying the set operations find the L.C.M. of the numbers $\,\,15,25,30.$ 

Sol. Let $\,\,A=\text{multiples of 15}\,=\{15,30,45,60,75,90,105,120,135,150,....\} \,\\B=\text{multiples of 25}\,=\{25,50,75,100,125,150,....\}    \,\\C=\text{multiples of 30}\,=\{30,60,90,120,150,.....\}\\ \therefore A \cap B \cap C=\{150,300,.....\} \\ \text{So, the L.C.M. of }\,\, 15,25,30= 150.$ 

19(ii) If $\,\,r,s,t\,\,$ are prime numbers and $\,p,q\,$ are the positive integers such that  LCM of $\,p,q\,$ is $\,\,r^2t^4s^2,\,\,$ then find the number of ordered pair $\,(p,q).$

Sol. It is given that LCM of $\,(p,q)=r^2t^4s^2$ 

That is, at least one of $\,p\,$ and $\,q\,$ must have $\,r^2,\,t^4\,\,\,\text{and}\,\,s^2\,\,$ in their prime factorizations.

Now, consider the cases for power of $\,r\,$ as follows:

Case 1:  $\,p\,$ contains $\,r^2\,$ 

then $\,q\,$ has $\,r^k\,$ with $k=(0,1)$.

That is, number of ways $=2.$

Case 2:   $\,q\,$ contains $\,r^2\,$ then $\,p\,$ has $\,r^k\,$  with $\,k=(0,1).$

That is, number of ways $=2.$

Case 3:  Both $\,p\,$ and $\,q\,$ contains $\,\,r^2\,\,$ then, number of ways $=1.$

Therefore, exponent of $\,r\,$ may be chosen in $\,2+2+1=5\,$ ways.

Similarly, exponent of $\,t\,$ may be chosen in $\,4+4+1=9\,$ ways and exponent of $\,s\,$ may be chosen in $\,\,2+2+1=5\,$ ways.

Thus, the total number of ways is:

$\,\,5×9×5=225.$

Hence, the number of the ordered pair $\,(p,q)\,$ is $\,225.$

20. Using a Venn diagram or otherwise, solve the following problem:

In a class of $\,70\,$ students, each student has taken either English or Hindi or both. $\,45\,$ students have taken English and $\,30\,$ students have taken Hindi. How many students have taken both English and Hindi?

Alternatively, we know, $\,\, n(E \cup H)= n(E)+n(H)-n(E \cap H)\\ \implies 70=45+30-n(E \cap H) \\ \implies n(E \cap H)=45+30-70=5$  

21. Use a Venn diagram to solve the following problem:

In a statistical investigation of 1003 families of Kolkata it was found that 63 families had neither a radio nor a T.V., 794 families had a radio and 187 a television. How many families in that group had both a radio and a T.V.?

From the Venn Diagram, we see  that $ \,63\, $ families had neither a radio nor a T.V.,  and so Out of}  $\,1003\, \,$ only $\,\,1003-63=940\,\,$  families use either TV or Radio.

 So, In the Venn diagram the required result will be the Yellow Portion which indicates the families that use both TV and Radio. 

Their number is : $=(187+794)-940=41$

22. A market research group conducted a survey of $\,1000\,$ consumers and reported that $\,720\,$ consumers liked product $\,A\,$ and $\,450\,$ consumers liked product $\,B\,$. What is the least number that must have liked both products?

Sol. We know, $\,\, n(A \cup B)= n(A)+n(B)-n(A \cap B)\\ \implies 1000=720+450-n(A \cap B) \\ \implies n(A \cap B)=720+450-1000=170 \\ \therefore 170 \,\,\text{is the least number that must have liked both products}$

23. In a town $\,60 \%\,$ of the people read magazine A whereas $\,25 \%\,$ do not read magazine A but they read magazine B. Now calculate the percentage of those who do not read any of magazine A or B. Also find the highest and lowest possible figures of those who read magazine B.

Sol. The percentage of people who neither read magazine A nor B is :   

$=100-(60+25)=15.$   

Detailed Explanation:

This can be interpreted as follows:  

According to the problem, $\, n(A)=60,\,\, n(B)-n(A \cap B)=25$  

Let $\,N\,$  denotes the no. of people who don't read any magazine.   

Clearly, the sets $\,N\,$ and $\,A \cup B\,$ are disjoint set.   

Now, we know, $\,n(A \cup B)=n(A)+n(B)-n(A \cap B) \\ \implies n(A \cup B)=60+25=85$

Also, $\,n(A \cup B \cup N)= n(A \cup B) +n(N)\\ \implies 100=85+n(N) \\ \implies n(N)=100-85=15$    

So, the percentage of those who do not read any of magazine A or B is : $\,\,15 \%$

2nd part : Maximum  no. of people who read B is possible only when all apart from $\,\,n(N)\,$ read B i.e. $=100-n(N)=100-15=85$  

Lowest possible figures of those who read magazine B, follows from the question that says:  

$\,25 \%\,$ do not read magazine A but they read magazine B. So, the min. no. of people who read B is possible by those $\,\,25 \% \,$ who only read B but not A.

24. (i) Two finite sets  $\,A\,$  and  $\,B\,$  have respectively $\,p\,$ and $\,q\,$ elements. If the total number of subsets of $\,A\,$ is  $\,56\,$  more than the total number of subsets of  $\,B\,$ , then find the values of  $\,p\,$  and  $\,q.\,$ 

Sol. Since two finite sets  $\,A\,$  and  $\,B\,$  have respectively $\,p\,$ and $\,q\,$ elements, the total number of subsets of $\,A\,$ and $\,B\,$  are : $\,2^p\,$ and $\,2^q\,$ respectively.

$\therefore  2^p-2^q=56 \\ \implies 2^q(2^{p-q}-1)=2^3 \times 7 \cdots (1) \\ \text{Comparing both sides of (1), we get}\,\, 2^q=2^3 \implies q=3 \\  \text{and,}\,\,\,2^{p-q}-1=7 \\ \implies 2^{p-q}=7+1=8=2^3\\ \implies p-q=3 \\ \implies p-3=3 \\ \implies p=3+3=6.  $

24(ii) Two finite sets  $\,A\,$  and  $\,B\,$  have m and n elements respectively. Find the maximum and minimum elements are in $\,\, A \cup B$.

Sol. We know, $\,\,\,n(A \cup B)=n(A)+n(B)-n(A \cap B) \\ \implies n(A \cup B)=m+n-n(A \cap B) \\ \text{Now, the value of }\,\, n(A \cup B) \,\,\text{is maximum when the value of}\,\, n(A \cap B)\,\,\text{is minimum that is ZERO.}\\ \text{So, the max. value of } n(A \cup B)=m+n.\\ \text{Similarly, the value of } \,\,n(A \cup B) \text{is min. whenever the value of } n(A \cap B)\,\, \text{is max. i.e. }\,\, \{m,n\} \\ \text{So, the min. value of}\,\, n(A \cup B)=m+n-\{m,n\}$

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