LEARN MATH STEP BY STEP THROUGH VERY EASY PROCESS

SETS - PROBLEM & SOLUTION

**PROBLEM & SOLUTION -**

**Example.1) If n(ξ) = 750, n(A) = 450, n(B) = 400, and n(A ∩ B) = 325, draw
a Venn Diagram to find –**

**(i) n(A ∪ B) (ii) n(A ∪ B)’ (iii) n(A – B)**

**Ans.) **

**(i) n(A ∪ B)**

**Given, n(ξ) = 750, n(A) = 450,
n(B) = 400, and n(A ∩ B) = 325**

**n(A ∪ B) = n(A) + n(B) – n(A ∩ B) = 450 +
400 – 325 = 525**

**The given sets are intersecting
sets. The Venn Diagram is as shown.**

**(ii) n(A ∪ B)’ = n(ξ) – n(A ∪ B)
= 750 – 525 = 225**

**(iii) n(A – B) = n(A) – n(A ∩ B) = 450 –
325 = 125 (Ans.)**

**Example.2) Out of 100 candidates who
appeared in an entrance exam test in English and Mathematics, 80 passed in at
least one subject. If 60 passed in English and 65 in Mathematics, find –**

**(i) How many passed in both the
subjects **

**(ii) How many passed in English only**

**(iii) How many failed in Mathematics**

**Ans.) Let, ξ be the set of all the candidates who
took the test, A the set of candidates who passed in English and B of those who
passed in Mathematics. Then –**

**Given, n(ξ) = 100, n(A ∪ B) =
80, n(A) = 60, n(B) = 65,**

**(i) n(A ∩ B) = n(A)
+ n(B) – n(A ∪ B) = 60 + 65 – 80 = 45**

**So, Number of candidates who
passed in both the subjects = 45**

**(ii) Number of candidates who
passed only in English **

** = n(A – B) **

** = n(A) – n(A ∩ B) = 60 –
45 = 15**

**(iii) Number of candidates who
failed in Mathematics **

** = n(B’) = n(ξ) – n(B)**

** =
100 – 65 = 35 (Ans.)**

**Example.3) 100 of the students at a
language learning school take at least one of the languages French, German, and
Russian, suppose 65 study French, 45 study German, 42 Study Russian, 20 study
French and German, 25 study French and Russian and 15 Study German and Russian.
Find the number of Students who study all the three languages.**

**Ans.) Let F, G, and R denotes
the sets of studying French, German, and Russian respectively. Then –**

**n(F ∪ G ∪ R) = n(F) + n(G) +
n(R) – n(F ∩ G) – n(F ∩ R) – n(G ∩ R) + n(F ∩ G ∩ R)**

**now, n(F ∪ G ∪ R) =
100, because 100 of the students study at least one of the languages in this
school. n(F) = 65, n(G) = 45, n(R) = 42, **

**French + German = n(F ∩ G) = 20,**

**French + Russian = n(F ∩ R) = 25, and**

**German + Russian = n(G ∩ R) = 15.**

**So, n(F ∪ G ∪
R) = n(F) + n(G) + n(R) – n(F ∩ G) – n(F ∩ R) – n(G ∩ R) + n(F ∩ G ∩ R)**

**=>
100 = 65 + 45 + 42 – 20 – 25 – 15 + n(F ∩ G ∩ R)**

**=>
n(F ∩ G ∩ R) = 100 – 92 = 8**

**So, 8 students are there in these 100 students
who are studying all three languages.**

**We may use this result to fill in the Venn
Diagrams, we have –**

**8 study all three languages.**

**So,**

**=> 20 – 8 = 12 study French & German but
not Russian,**

**=> 25 – 8 = 17 study French and Russian but
not German,**

**=> 15 – 8 = 7 study German and Russian but
not French.**

**Now, **

**=> 65 – 12 – 8 – 17 = 28 study only French,**

**=> 45 – 12 – 8 – 7 = 18 study only German,**

**=> 42 – 17 – 8 – 7 = 10 study only Russian.**

**120 – 100 = 20 do not study any of the
languages**

**Thus, from the completed diagram, we can
observe that 28 + 18 + 10 = 56 students study exactly one of the three
languages. (Ans.)**

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