Title Mathematics ÔÇó Final Step-A ÔÇó SOL.pdf ✅

Vidyamandir Classes VMC | Final Step-A 1 Class XI | Mathematics Solutions to Final Step-A | Mathematics Sets, Relations, Statistics & Mathematical Reasoning 1.(A) Mean         100 2 100 1 1 .... 1 100 2 1 50 101 101 d d d x d           M.D.   50 49 ...... 0 .... 50 2 2 3 .... 50   101 101 i x x d d d d d d d d d x n                 2 1 2 3 .... 50   50 51 101 101 d     d     M.D.  x   255 (given)  50 51 255 101 d     101 10.1 10 d   2.(C) Here, P is False and Q is True  V V F T 1 2 , , .     3.(D) Let 5 observations be 1, 2, 6, and . x y  Mean   1 2 6 5 x y x       5 5 9    x y  x y   16 Variance   1 2 2 5 i   x x    1 2 2 124 1 4 36 25 5       x y  2 2 x y   704    2 2 x x    16 704  x y   16  2 x x    16 224 0  x  8 12 2  x  8 12 2 and y  8 12 2 M.D.   1 1 5 x x x    1 1 1 5 2 5 6 5 8 12 2 5 8 12 2 5 4 3 1 3 12 2 3 12 2 5 5                            1 1 8 3 12 2 12 2 3 8 24 2 8.38 5 5                 4.(D) Contra-positive of p q  is   q p  . Given statement is in the form of p q  , so its contra-positive is If the area of a square does not increase four times, then its side is not doubled. 5.(D) Question based on Probability A  family owning a phone ; B  family owning a car     25 15 , 100 100 P A P B     65 100 P A B      65 100 P A B      35 100 P A B          35 100 P A P B P A B          5 and 2000 100 P A B n A B          5 100 n A B n S    n S   40,000
Vidyamandir Classes VMC | Final Step-A 2 Class XI | Mathematics  40,000 families live in the town    5 100 P A B    5% families own both a car and a phone.    35 100 P A B    35% families own either a car and a phone. 6.(C) Combined mean 1 1 2 2 1 2 n x n x x n n    Given n n x x 1 2 1     70, 30, 54, 60  2 70 54 30 60 70 30    x    x 2  74  Mean wage of the night shift workers is Rs. 74. 7.(A) Given statement is p q  p : It is raining Its contra-positive is     q p   i.e. q p  q : I will come Thus, contra-positive of given statement is "If I will come, then it is not raining". 8.(D) Given statement is P R Q     Q P R      Negation of this statement is   Q P R     9.(C) Not reflection : For a   / 2 (real number) a a P ,   as 2 2 2 sec tan tan 1 a a a    Symmetric : Let a b P ,    2 2 sec tan 1 a b     2 2 sec 1 tan a b    2 2 tan sec 1 a b    2 2 sec tan 1 b a    b a P ,   Transitive : Let a b P b c P , and ,       2 2 2 2 sec tan 1 and sec tan 1 a b b c      2 2 2 2 sec tan sec tan 1 1 a b b c       2 2 sec tan 1 a c    a c P ,   10.(A) Let the observations in ascending order be 1 2 3 1 2 , , ,...., , , n n n x x x x x x  Mean 1 2 2 .... 2 n x x x x n      ...... (i) Median (for even) 1 2 n n x x    which lies between n x and 1 . n x  So, the observations below the median are 1 2 , ,...., n x x x and the remaining observations are 1 2 2 , ,...., . n n n x x x   New mean  1 2 1 2 2 5 5 .... 5 3 3 .... 3            2 n n n n x x x x x x n                 1 2 3 2 .... 5 3 1 2 n x x x x n n x n          [From (i)] Thus, new mean increases by 1.
Vidyamandir Classes VMC | Final Step-A 3 Class XI | Mathematics 11.(B) p q  means if p then q. Contra-positive of p q  is   q p  . Let p : It rains ; q : I go to school. So, the given statement in the form of  p q  . Thus, its contrapositive is  q p  . Hence, contapositive is “ If I do not go to school, it rains”. 12.(C)  p q p q p q p q                               p q q p f           q q f   fallacy   p 13.(D) M.D. i x X n    (By defn.) Obviously, new mean   X 5 New M.D.  x X i 5 5    n       M.D. 14.(B) p q r     is equivalent to  p q p r       by truth table p q r q r  p q r     p q  p r   p q p r       T T T T T T T T T T F T T T F T T F T T T F T T T F F F F F F F F T T T T T T T F T F T T T T T F F T T T T T T F F F F T T T T Equivalent 15.(C) A     2, 1, 0, 1, 2 B   0, 0 , 1, 1 , 2, 2 , 2, 2         Number of elements in the power set of 4 R  2 . 16.(B) Mean 1 2 .... n x x x x n      Variance   i 2 x x n      New mean   1 2 1 2 .... n n d d d x x x na d x a n n              New variance          2 2 2 2 2 i i i i d d x a x a x x x x n n n n                     Mode of 1 2 , ,...., n x x x is M  Mode of 1 2 , ,...., n      x a x a x a is   M a Mode has the highest frequency  Both the statements are true. 17.(C) Contrapositive of p q  is   q p  . Given statement is in the form of  p q  . So, its contrapositive form is  q p  . Hence option (C).
Vidyamandir Classes VMC | Final Step-A 4 Class XI | Mathematics 18.(A) i x x n   (mean)  . i x n x  Sum of all observations (incorrect)    40 20 800 . Sum of all observations (correct)     800 33 53 820  Correct mean sum correct   820 41 20 20    19.(C) Energy element of set A has distinct image in A. So R is a one-one and onto function and hence it has an inverse. 20.(A) i x x n   (mean)  i x n x  Sum of observations    7 5 35  5 th observation       35 6 7 8 10 4   Variance     2 2 7 1 0 1 9 9 4 5 i i x x x n n             21.(D) Clearly, reflective as a a R a A , .     5, 12  R but, 12, 5  R So not symmetric 3, 5  R and 5, 12  R  3, 12 ,   R so transitive. 22.(B) p q q p  p q p     p q  p p q     p q  p p q     p q  p p q     T T T T T T T T T T T F T T T T F F F F F T F T T T F T T T F F T T F T F T T T Clearly, p q p p p q       and   are equivalent. 23.(A) (B) P : 4 divides 8 Q : 4 divides 2 8 But 4 is not a prime  P Q R   is not True. (C) Q :12 divides 2 6 But 12 is not a prime  Q R  is not True. (D) Q :12 divides 2 6 But 12 is not a prime  Q P  is not divide 6.  Q P  is not true (A) Q m: divides 2 n R m: is a prime then m divides n  Q R P   is True. e.g. 3 divides 2 12 and 3 is a prime then 3 also divides 12.