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

     

(3)

n(A)=48,  n(B)=25,  n(C)=18

n(ABC)=60  [Total]

n(ABC)=5

(ABC)=|A|-|AB|+|ABC|

|AB|=48+25+18+5-60=36

Number of men who received exactly 2 medals

=|AB|-3|ABC|=36-15=21



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

     

(1)

Ω be the sample space and A be an event. If P(A)=0A=ϕ If P(A)=1A=Ω

Then both statements are true.

 



Q 3 :

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

 



(6)

We have, |n2-10n+19|<6, n

 |(n-5)2-6|<6 

 0<(n-5)2<12 

 (n-5)2=1,4,9 

 n-5=±1,±2,±3 

 6 values of n exist.



Q 4 :

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

 



(15)

3n-3=7k, k

 3n=7k+3, k               ...(i)

Now, 33 mod 7

           33-1 mod 7

           361 mod 7

           373 mod 7

           3133 mod 7 

  n=1,7,13, will satisfy equation (i) 

This forms an A.P. with common difference 6.

Now, N[10,100] 

 n=13,19, 

Now, 13+(n-1)·610013+6n-6100 

6n100-76n93n15.5n=15

So, there are 15 such numbers.



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]

 



(5)

|x|2-2|x|+|λ-3|=0 

(|x|-1)2+|λ-3|-1=0                       ...(i) 

Given that x is an integer |x|-1 is an integer

(|x|-1)2-1 is also an integer. Thus |λ-3| is also an integer. Therefore λ is also an integer, (|x|-1)21

Case I: |x|-1=0x=±1 

|λ-3|=1λ=4,2 

Possible values of λ+x=5,3,1. 

Case II: |x|-1=1x=±2 

 |λ-3|=0λ=3

Possible values of x+λ=5,1. 

Considering all possible cases, the maximum value of λ+x=5.



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

     

(2)

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

     

(2)

(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

     

(3)

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

 



Q 11 :

The number of elements in the set S={(x,y,z):x,y,zZ,x+2y+3z=42,x,y,z0} equals _______ .               [2024]



(169)

x+2y+3z=42

At z=0,x+2y=42

y can take all value from 0 to 21

Hence 22 solutions.

Similarly,

at z=1, x+2y=39    20 solutions

at z=2, x+2y=36    19 solutions

at z=3, x+2y=33    17 solutions

at z=4, x+2y=30    16 solutions

at z=5, x+2y=27    14 solutions

at z=6, x+2y=24    13 solutions

at z=7, x+2y=21    11 solutions

at z=8, x+2y=18    10 solutions

at z=9, x+2y=15      8 solutions

at z=10, x+2y=12    7 solutions

at z=11, x+2y=9        5 solutions

at z=12, x+2y=6        4 solutions

at z=13, x+2y=3        2 solutions

at z=14, x+2y=0        1 solutions

Total cases = 169 solutions.

 



Q 12 :

Let the set C={(x,y)x2-2y=2023, x,yN}. Then (x,y)C(x+y) is equal to ________ .              [2024]



(46)

We have, x2-2y=20232y=x2-2023

  Possible values of (x,y) are (45,1)

So, (x,y)C(x+y)=(45+1)=46

 



Q 13 :

A group of 40 students appeared in an examination of 3 subjects - Mathematics, Physics, and Chemistry. It was found that all students passed in at least one of the subjects, 20 students passed in Mathematics, 25 students passed in Physics, 16 students passed in Chemistry, at most 11 students passed in both Mathematics and Physics, atmost 15 students passed in both Physics and Chemistry, atmost 15 students passed in both Mathematics and Chemistry. The maximum number of students passed in all the three subjects is __________ .    [2024]



(10)

 



Q 14 :

Let A={(α,β)R×R:|α1|4 and |β5|6} and B={(α,β)R×R:16(α2)2+9(β6)2144}. Then          [2025]

  • AB

     

  • neither AB nor BA

     

  • BA

     

  • AB={(x,y):4x4,1y11}

     

(3)

A:|α1|4 and |β5|6

  4α14 and 6β56

  3α5 and 1β11

B:16(α2)2+9(β6)2144

B:(α2)29+(β6)2161

From Diagram, BA.



Q 15 :

Let A = {1, 2, 3, ..., 10} and B={mn:m,nA, m<n and gcd(m,n)=1}.

Then n(B) is equal to :          [2025]

  • 36

     

  • 31

     

  • 37

     

  • 29

     

(2)

We have, A = {1, 2, 3, ..., 10}

B={mn:m,nA, m<n and gcd(m,n)=1}

For  m = 1, n = 2, 3, ..., 10    9 cases

       m = 2, n = 3, 5, 7, 9       4 cases

       m = 3, n = 4, 5, 7, 8, 10 5 cases

       m = 4, n = 5, 7, 9          3 cases

       m = 5, n = 6, 7, 8, 9       4 cases

       m = 6, n = 7                  1 cases

       m = 7, n = 8, 9, 10        3 cases

       m = 8, n = 9                  1 cases

       m = 9, n = 10                1 cases

    n(B) = 31.



Q 16 :

Let S={(m,n):m,n{1,2,3,.....,50}}. If the number of elements (m,n) in S such that 6m+9n is a multiple of 5 is p and the number of elements (m,n) in S such that m+n is a square of a prime number is q, then p+q is equal to _______.          [2026]



(1333)

S={1,2,3,,50}

p=(6m+9n) is divisible by 5

No. of ways

6m=(5λ+1)m=5k+1

9n=(10-1)n=10μ-1 if n is odd

n must be odd

10μ+1 if n is even

No. of ways=50×25=1250

q(m+n) is square of a prime

  m+n=4 m+n=9 m+n=25 m+n=49
No. of
ways 3 8 24 48

 

q=3+8+24+48=83

=p+q=1250+83=1333



Q 17 :

The number of elements in the set

S={x:x[0,100] and 0xt2sin(x-t)dt=x2} is_____.  [2026]



(16)

0xt2sin(x-t)dt=x2

Use by parts

-x2cosx+x2+2x2cosx-2xsinx

-x2cosx+2xsinx+2cosx-2=x

cosx=1

x=0,2π,4π,,30π

Total Elements=16



Q 18 :

Let A={x:|x2-10|6}  and  B={x:|x-2|>1}. Then    [2026]

  • AB=(-,1](2,)

     

  • AB=[-4,-2][3,4]

     

  • A-B=[2, 3)

     

  • B-A=(-,-4)(-2,1)(4,)

     

(4)

|x2-10|6

-6x2-106

4x216

A=[-4,-2][2,4]

|x-2|>1

B=(-,1)(3,)

AB=(-,1)[2,)

AB=[-4,-2](3,4]

A-B=[2,3]

B-A=(-,-4)(-2,1)(4,)



Q 19 :

Let S be the set of the first 11 natural numbers. Then the number of elements in 

A={BS:n(B)2 and the product of all elements of B is even} is _______. [2026]



(1979)

A={1,2,3,,11}

 n(B)2 and product of all elements in B is even

Case (i):

n(B)=2C211-C26

n(B)=3C311-C36

n(B)=4C411-C46

n(B)=5C511-C56

n(B)=6C611-C66

n(B)=7C711

n(B)=11C1111

 number of set B=r=211Cr11-r=26Cr6

=211-(1+11)-(26-7)

=2048-64-5

=1979

Alternate Solution:

Total subsets=211

No. of subsets having odd terms only=26

No. of subsets having one term only & also having even terms=5

Required ways=211-26-5=1979