Topic Question Set


Q 1 :    

An organization awarded 48 medals in event 'A', 25 in event 'B' and 18 in event 'C'. If these medals went to total 60 men and only five men got medals in all the three events, then, how many received medals in exactly two of three events?       [2023]

                                                                                       

  • 10 

     

  •  

  • 21 

     

  • 15

     

 



Q 2 :    

Let Ω be the sample space and AΩ be an event. Given below are two statements

(S1) : If P(A) = 0, then A = ϕ

(S2) : If P(A) = 1, then A = Ω

Then                                                                                                                                   [2023]

  • both (S1) and (S2) are true

     

  • only (S1) is true

     

  • only (S2) is true

     

  • both (S1) and (S2) are false

     

 



Q 3 :    

The number of elements in the set {n:|n2-10n+19|<6} is ___________ .                  [2023]

 



 



Q 4 :    

The number of elements in the set {n:10n100 and 3n-3 is a multiple of 7} is __________ .                      [2023]

 



 



Q 5 :    

Let λR and let the equation E be |x2|-2|x|+|λ-3|=0. Then the largest element in the set S={x+λ:x is an integer solution of E} is _________ .    [2023]

 



 



Q 6 :    

Let A={n[100,700]N:n is neither a multiple of 3 nor a multiple of 4}. Then the number of elements in A is          [2024]

  • 280

     

  • 300

     

  • 310

     

  • 290

     

(B)

Multiples of 3 are 102,105,,699

an=699=102+(n-1)·3

597=3(n-1)n=200

Now, multiples of 4 are 100,104,,700

     am=700=100+(m-1)·4

m=151

Multiples of 3 and 4 are 108,120,,696

     ap=696=108+(p-1)·12

p=50

     n(34)=n(3)+n(4)-n(34)

     =n+m-p=200+151-50=301

The number of elements in A=601-301=300

 



Q 7 :    

Let S={xR:(3+2)x+(3-2)x=10}. Then the number of elements in S is           [2024]

  • 4

     

  • 2

     

  • 0

     

  • 1

     

(B)

(3+2)x+(3-2)x=10

Let 3+2=t

tx+1tx=10

Let tx=y, then y+1y=10

y2-10y+1=0

y=10±100-42=10±962=10±462=5±26

(3+2)x=5±26

xln(3+2)=ln(5+26) or ln (5-26)

x=ln(5+26)ln(3+2)  or  x=ln(5-26)ln(3+2)

So, two real values of x.

    The number of elements in S is 2.

 



Q 8 :    

Let A and B be two finite sets with m and n elements respectively. The total number of subsets of the set A is 56 more than the total number of subsets of B. Then the distance of the point P(m, n) from the point Q(−2,−3) is                [2024]

  • 6

     

  • 8

     

  • 10

     

  • 4

     

(C)

Total number of subsets of set A=2m

Total number of subsets of set B=2n

According to question, 2m-2n=56

2n(2m-n-1)=23×7

2n=23 and 2m-n=8=23

n=3 and m-n=3m=3+n=3+3=6

m=6 and n=3

Now, distance between the points P(6, 3) and Q(-2, -3) is given by PQ=(6+2)2+(3+3)2=64+36=10 units

 



Q 9 :    

In a survey of 220 students of a higher secondary school, it was found that at least 125 and at most 130 students studied Mathematics; at least 85 and at most 95 studied Physics; at least 75 and at most 90 studied Chemistry; 30 studied both Physics and Chemistry; 50 studied both Chemistry and Mathematics; 40 studied both Mathematics and Physics and 10 studied none of these subjects. Let m and n respectively be the least and the most number of students who studied all the three subjects. Then m+n is equal to _______.                    [2024]



(45)

Total number of students=n(S)=220

n(M)[125,130], n(P)[85,95], n(C)[75,90]

n(MPC)=220-10=210

n(MP)=40, n(PC)=30, n(MC)=50

n(MPC)=n(M)-n(MP)+n(MPC)

n(MPC)=210+(40+30+50)-n(M)

   (n(MPC))max=n=min{n(MP),n(PC),n(MC)}=30

   (n(M))max=130+95+90=315

(n(MPC))min=m=330-315=15

n+m=45



Q 10 :    

If S={aR:|2a-1|=3[a]+2{a}}, where [t] denotes the greatest integer less than or equal to t and {t} represents the fractional part of t, then 72aSa is equal to ________ .                     [2024]



(18)

We have, S={a:|2a-1|=3[a]+2{a}}

Since for x, x=[x]+{x}

  3[a]+2{a}=2a+[a]

  |2a-1|=2a+[a]

For 2a-1<0, we have

        -2a+1=2a+[a]

4a=1-[a]4a+[a]-1=0

a=14

Now, if 2a-1>0a>12

 2a-1=2a+[a]

[a]=-1 which is not possible as a>12

So, a=14S

  72aSa=72×14=18