CBSE 11th Standard Maths Subject Sets HOT Questions 2 Mark Questions 2021
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CBSE 11th Standard Maths Subject Sets HOT Questions 2 Mark Questions 2021
11th Standard CBSE
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Reg.No. :
Mathematics
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In a class of 60 students, 25 students play cricket, 20 students play Tennis and 10 students play both the gmes. Then, find the number of students who play neither games.
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Let F1 be the set of parallelograms, F2 be the set of rectangles, F3 be the set of rhombus and F4 be the set of squares. Then, show that F1 is equal to the union of all sets.
(a) -
In a town of 840 persons read Hindi, 300 read English and 200 read both newspapers. Then, find the number of persons who read neither of the newspapers
(a) -
Each students in a class of 40 students study atlest one of the subjects English,Mathematics and Economics. 16 students study English, 22 Economics and 26 Mathematics, 5 study English and Economics, 14 Mathematics and Economics and 2 English, Economics and Mathematics. Find the number of students who study English and Mathematics
(a)
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CBSE 11th Standard Maths Subject Sets HOT Questions 2 Mark Questions 2021 Answer Keys
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Let C and T respectively donote the set of students who play Cricket and Tennis, and set U donotes the set of all students in a class.
Then n(C) = 25, n(T) = 20, n(C\(\)\(\cap \)T) = 10 and n (U)= 60
We know that, n(C\(\)\(\cup \)T) = n(C) + n(T) - n(C\(\)\(\cap \)T)
\( \Rightarrow \) n(C\(\)\(\cup \) T) = 25 + 20 - 10 = 45 - 10 = 35
Now the Number of students who play neither game
n(C'\(\cap \)T' ) = n(C'\(\cup \)T' )
= n (U) - n(C\(\cup \)T) = 60 - 35= 25
Hence, 25 students play neither games. -
All rectangles, Rhombus and square are parallelograms because its opposite sides are equal and parallel.
Therefore \({ F }_{ 2 }\subset { F }_{ 1 },{ F }_{ 3 }\subset { F }_{ 1 }\) and \({ F }_{ 4 }\subset { F }_{ 1 }\)
\({ F }_{ 1 }={ F }_{ 2 }\cup { F }_{ 3 }\subset { F }_{ 4 }\) -
Let H and E respectively doneto the set of persons who read Hindi and English newspapers and let U set of all person in a town.
Then, we have n (U) = 840,
n (H) =450,
n (E) = 300
and n(H \(\cap \) E) = 200
Clearly, n(H \(\cup \) E) = n(H) + n(E) - n(H \(\cap \) E)
n(H \(\cup \) E) = 450 + 300 - 200
= 750 - 200 = 550
Now, the number of persons who read neither of the newspaper is given by
n(H' \(\cap \) E' ) = n(H' \(\cup \) E') [by De Morgan's law]
= n (U) - n(H\(\cup \)E)
= 840 - 550 = 290
Hence, 290 Persons read neither of the newspapers -
Let A, B and C denote the set of students who study English, Economics and Mathematics, respectively.
Then, we have,
Total number of students, n (A \(\cup \) B \(\cup \) C) = 40
Number of students who study English, n (A) = 16
Number of students who study Eoconomics, n (B) = 22
Number of students who study Mathematics, n (C) = 26
Number of students who study English and Economics, n (A \(\cap \) B) = 5
Number of students who study Mathematics and Economics, n( B \(\cap \) C) = 14
and number students who study all subjects,
n (A \(\cup \) B \(\cup \) C) = 2
Clearly, n (A \(\cup \) B \(\cup \) C) = n(A) + n(B) - n(C) - n (A \(\cap \) B) - n( B \(\cap \) C) - n( A \(\cap \) C) - n (A \(\cap \) B \(\cap \) C)
40 = 16 +22+ 26 - 5- 14 n ( C \(\cap \) A) + 2
40 = 66 - 19 - n ( C \(\cap \) A)
n ( C \(\cap \) A) = 47 - 70
=7
Hence the number of students who study English and Mathematics are 7.
