MBOSE Class 10 Question Bank 2027 Mathematics

FOR MEGHALAYA CLASS 10 PREPARATION

MBOSE Class 10
Question Bank 2027
Mathematics
EXAM YEAR TYPE SUBJECT

Meghalaya Class 10 2027 Question Bank Maths

Notes · Sample Papers · Previous Year Papers · Mock Tests

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CM IMPACT Guidebook for Students
(With Important Questions and Answers)

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Mathematics
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Class-X

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Published by
Education Department
Government of Meghalaya

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An Initiative under

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Chief Minister’s Initiative to Maximize Pass Achievement and Classroom Triumph (CM

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IMPACT)

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REAL NUMBERS
Important points/Summary
1. Prime Number: A number which has exactly two factors i. e. 1 and the number itself is called a
prime number. 2 is the smallest prime number.

2. Composite number: A number which is not prime is called a composite number. A composite
number has more than two factors. The smallest composite number is 4.
3. Co-primes: Two numbers p and q are called co-primes if their HCF is unity i. e.
HCF (p, q) = 1.
𝒑
4. Rational number: A number which is in the form 𝒒 , where p, q are positive integers and q ≠ 0 is
called a rational number. e.g. 2, -3, ½, ¾ , .. are all rational numbers
5. Irrational number: A number which is not a rational is called an irrational number.
E. g ∛2, √3, π, ….. are all irrational numbers.
6. Fundamental theorem of Arithmetic: - Every composite number can be expressed (factorized) as
a product of primes, and this factorisation is unique, apart from the order in which the prime factors
occur.
7. For any two positive integers a and b then
i. HCF (a, b) X LCM (a, b) = a X b (product of two numbers)
𝒑𝒓𝒐𝒅𝒖𝒄𝒕 𝒐𝒇 𝒕𝒘𝒐 𝒏𝒖𝒎𝒃𝒆𝒓𝒔
ii. HCF (a, b) = 𝑳𝑪𝑴 (𝒂,𝒃)
𝒑𝒓𝒐𝒅𝒖𝒄𝒕 𝒐𝒇 𝒕𝒘𝒐 𝒏𝒖𝒎𝒃𝒆𝒓𝒔
iii. LCM (a, b) = 𝑯𝑪𝑭(𝒂,𝒃)
𝑯𝑪𝑭 𝑿 𝑳𝑪𝑴
iv. One number =
𝒂𝒏𝒐𝒕𝒉𝒆𝒓 𝒏𝒖𝒎𝒃𝒆𝒓
𝑯𝑪𝑭 𝑿 𝑳𝑪𝑴
v. Other number = 𝒐𝒏𝒆 𝒏𝒖𝒎𝒃𝒆𝒓

Section-A Ans.(B)
Multiples Choice Questions (MCQ) - 1 Mark
4. For any two positive integers a and b, HCF
1. The exponent of 2 in the prime factorisation (a, b) X LCM (a, b) is equal to:
of 144 is: (A) a+ b (B) a – b (C) a x b (D) a/b
(A) 5 (B) 4 (C) 3 (D) 2
Ans.(C)
Ans: (B)
5. Every composite number can be expressed
2. The product of a non-zero rational number
as a product of:
and an irrational number is:
(A) Co-primes (B) primes
(A) An irrational number
(C) twin primes (D) none of these
(B)a rational number(C) one
Ans. (B)
(D) zero
(MBOSE Suppl.)
Ans.(A)
6. Every composite number has:
(A) At least two factors
3. The sum or difference of a rational number (B) at least three factors
and an irrational number is: (C)Only one factor
(D)None of these
(A) A rational number
(B) An irrational number Ans:(B)
(C) One
(D) Zero

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Ans. (B)
is always:
a
(A) Non-terminating 14. The LCM of 6 and 20 is
(B) non-terminating and non- repeating (A) 40 (B) 60 (C)80 (D)100
Ans (B)
(C)terminating or non- terminating repeated
(D) none of these 15. If n is any natural numbers then 6n
Ans. (C) always ends with

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(A) 1 (B) 0 (C) 3 (D) 6

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Ans. (D) 6
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8. The HCF of p and q which are relatively
primes is:
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16. A natural number which has exactly two
(A) 1
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(B) p (C) q (D) p x q
(A) Rational number (B) whole a
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factors i. e., 1 and the number itself is called a:
Ans. (A) number
(C) composite number (D) prime number
9. The LCM of p and q which are relatively Ans. (D)
primes is:
(A) 1 (B) p (C) q (D) p q 17. A natural number which is not prime and
Ans. (D) has more than two factors is called a/an:

om (C) Odd natural number (D) prime number
(A) Composite number (B) whole number
10. The prime factorization of 96 is
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(A)2 x35 2 4
(B) 2 x3 5
(C) 2 x3 (D)2 6 Ans. (A)

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Ans (C)
18. For some even integer q, every odd
integer is of the form:
11. Product of two co-prime numbers is 117. (A) q (B) q + 2 (C) 2q (D) q – 1
Their then LCM is Ans. (D)

(A)13 (B) 39 (C)117 (D)9 19. HCF of the smallest composite number
Ans=(C)117 and smallest prime number is:

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(A) 1 (B) 2 (C) 4 (D) 8

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12. . HCF of two different numbers is 24, Ans. (B)

em18 is 36 and HCF of P
which one of the following can be their
20. . If LCM of P s
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and 18 is 2.lThen
LCM.

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a (B) 3 (C)1 (D) 4
(A)12 (B) 18 (C) 36 (D) 48

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Ans (D) 48
(A)2
13. A number which cannot be expressed in the
Ans= (D) 4)
form a/b, where ‘a’ and ‘b’ are both integers
and b ≠ 0 is called a/an:
(A) Rational number
(B) Irrational number

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(C) Composite number

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(D) Prime number

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s POLYNOMIALS g
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IMPORTANT POINTS/SUMMARY:-
1. Polynomial: An algebraic expression of the form ,
P (x) = an xn + …….+ a1 x1 + a0 , an ≠0 is called a polynomial in x whose degree is n.
2. Degree of the polynomial: The highest power of the variable x in a polynomial p(x) is called its
degree.
3. A Polynomial of degree 1 is called a linear polynomial.
4. A polynomial of degree 2 is called a quadratic polynomial.

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5. A polynomial of degree 3 is called a cubic polynomial.

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6. A polynomial of degree 4 is called a bi-quadratic polynomial.

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7. Constant polynomial: a polynomial of degree 0 is called a constant polynomial.

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8. Zero of the polynomial: A Real number ‘k’ is called the zero of the polynomial P (x) if P (k) = 0.

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9. A linear polynomial has at most one zero.

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10. A quadratic polynomial has at most two zeroes.
11. A cubic polynomial has at most three zeroes.
12. A bi-quadratic polynomial has at most four zeroes.
13. If any graph representing the polynomial P(x)cuts the x-axis at one point then it has one zero; if it
cuts the x-axis at two points then it has two zeroes and if it cuts the x-axis at three points then it
has three zeroes.
𝒃 𝒄𝒐𝒏𝒔𝒕𝒂𝒏𝒕 𝒕𝒆𝒓𝒎
14. The zero of a linear polynomial P (x) = ax + b, a ≠0 is - 𝒂 = - 𝒄𝒐−𝒆𝒇𝒇𝒊𝒄𝒊𝒆𝒏𝒕 𝒐𝒇 𝒙

15. Relationship between zeroes and co-efficient of quadratic polynomial

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Let α and β be the two zeroes of a quadratic polynomial P (x) = ax2 + bx + c, a ≠0

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𝒃
16. Sum of the zeroes = α + β = - 𝒂

a s
𝒄𝒐−𝒆𝒇𝒇𝒊𝒄𝒊𝒆𝒏𝒕 𝒐𝒇 𝒙

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= - 𝒄𝒐−𝒆𝒇𝒇𝒊𝒄𝒊𝒆𝒏𝒕 𝒐𝒇𝒙𝟐

17. Product of the zeroes = α β =
𝒄
𝒂 a g
𝒄𝒐𝒏𝒔𝒕𝒂𝒏𝒕 𝒕𝒆𝒓𝒎
=
𝒄𝒐−𝒆𝒇𝒇𝒊𝒄𝒊𝒆𝒏𝒕 𝒐𝒇𝒙𝟐

Sl. Pair of Lines Compare the ratios Graphical Algebraic
No. Representati Interpretation

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1 x − 2y = 0 a₁/a₂ ≠ b₁/b₂ Intersecting Exactly one solution

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3x + 4y − 20 = 0 lines (unique)

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4x + 6y − 18 = 0 lines solutions

3 x + 2y − 4 = 0 a₁/a₂ = b₁/b₂ ≠ c₁/c₂ Parallel lines No solutions
2x + 4y − 12 = 0

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28. Product of zeroes of the polynomial p(x) =
21. A quadratic polynomial in variable x is of x2 – 3 is:
the form ax2 + bx + c where a, b and c are (A)– 3 (B) 3 (C) √3 (D) - √3
real numbers and
(A) a=0 (B) a ≠ 0 Ans. (A)

(C) a and b are both 0 (D) none of these.
29. The number of zeroes of a polynomial of
Ans (B)
degree n is: (MBOSE 2025 Suppl)
(A) Equal to n (B) greater than n
22. A polynomial of degree 3 is called a:
(C) less than n (D) less than or equal to n
(A) Linear polynomial
Ans. (D)less than or equal to n
(B) Quadratic polynomial
(C) Cubic polynomial
30. If the graph of y = p(x), where p(x) is a
(D) Biquadratic polynomial
polynomial, does not intersect the x- axis
Ans. (C)
then the number of zeros is:
(A) 1 (B) 2 (C) 3 (D) No Zeros
23. A quadratic polynomial can have at most:
Ans. (D)
(A) 1 zero (B) 2 zeroes
(C) 3 zeroes (D) 4 zeroes
31. If the graph of y = p(x) where p(x) is a
Ans. (B) (MBOSE 2025)
polynomial, intersects the x- axis at one
point only then the number of zeros is
24. The degree of a constant polynomial is:
(A) 1 (B) 2 (C) 3 (D) 4
(A) 2 (B) 1 (C) -1 (D) 0
Ans. (A)
Ans. (D) (MBOSE 2025 Suppl)

32. The number of zeroes of a polynomial
25. The degree of a zero polynomial is: p(x) is the number of points where the
(A) Always zero (B) never zero graph of y=p(x) cuts the:
(C) Negative (D) undefined (A) y-axis (B) x-axis

Ans. (D) (C) origin (D) none of these

Ans (B)
26. if 3 is a zero of the polynomial p(x)=
4x²+kx+12,
then the value of k is: 33. The zero of a linear polynomial P (x) = a x +
(A) 16 (B) –16 (C) 48 (D) –48
Ans (B) b, where a, b are real numbers, is:
(A) - a/b (B) – b/a (C) – (ab) (D) a/b
27. Sum of zeroes of the polynomial p(x) = x – 2
Ans. (B) – b/a
3x + 2 is:
(A) 2 (B) 3 (C) -2 (D) -3 34. For any polynomial p (x), if p (a) =0, then
Ans. (B) ‘a’ is called: (MBOSE 2025)
(A) Constant of the polynomial
(B) Zero of the polynomial

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(C) Degree of the polynomial 42. The quadratic polynomial with 0 and – 1/7
(D) Coefficient of the polynomial
a
as its two zeroes is:
Ans.(B) zero of the polynomial (A) 7x2 + x (B) x2 – 7 x
(C) x2 + 7x (D) 7x2 – x
35. Value of k for which x=2 is a solution of Ans. (A)
the equation 5x²–4x+(2+k)=0
(A) 10 (B) –10 (C) 14 (D) –14 43. If the graph of y = x2 – 2x – 35 cuts the x-

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axis at (7, 0) and (- 5, 0) then the zeroes of

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Ans (D)

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the polynomial x2 – 2x – 35 are:

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36. A bi-quadratic polynomial is a polynomial (A) 7, - 5 (B) 0, – 5 (C) 0, 7 (D) 0, 0
of degree:
l a s Ans. (A)
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(A) 1
g (B) 2 (C) 3 (D) 4
Ans. (D) 4 44. A Polynomial of degree 1 is called:
(A) bi-quadratic (B) cubic
(C) quadratic (D) linear
37. . If the product of zeroes of ax2 - 6x - 6
is 4 then a is : Ans. (D)
−3 3 2 −2
(A) 2 (B) 2 (C) 3 (D) 3 45. If the graph of a polynomial lies above or

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Ans (A) below the X - axis without touching or

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intersecting it then the number of its zeroes is

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38. Which of the following is a polynomial?

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−1
(A) 0 (B) 1 (C) 2 (D) 3

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(A) x2+ 𝑥 2 (B) √𝑥 + 2

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a
(C) x2+ 3x (D) x-3+ 1
Ans (C)

39. The zeroes of a quadratic polynomial 3x2 –
x – 4 are:
(A) 1, - 4/3 (B) – 1, 3/4
(C) – 1, 4/3 (D) -1, - 4/3

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Ans. (C)

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40. The zeroes of a quadratic polynomial x2 –

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(B) √15, – √15
(C) 3√5, - 3 √5 (D) 5√3, - 5 √3
Ans. (B)

41. The quadratic polynomial with 2 as sum and
– 8 as product of its zeroes is:
(A) x2 – 2x – 8 (B) x2 + 2x – 8

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(C) x2 – 2x + 8 (D) x2 + 2x + 8
Ans. (A)
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Pair of Linear Equations in two variables
Important points/summary: a
i. A pair of linear equation in two variables x and y is of the form :
a1 x + b1 x + c1 = 0 ………… (1)
a2 x + b2 x + c2 = 0 …………. (2),
where a1, a2 , b1 b2,c1, c2 are real numbers and
a12 + b1 2 ≠0 ; a22 + b2 2 ≠0

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ii. The graphs represented by a pair of linear equations

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a1 x + b1 x + c1 = 0 ………… (1)

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a2 x + b2 x + c2 = 0 …………. (2)
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e pair of equations has unique solution and the graphs representing the systemgoflas
Case 1: if 𝑎 ≠ 𝑏 , them
are straight line graphs and three cases arise:

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𝑎 𝑏

a 𝑐 lines. The system of equation is consistent (dependent).
1 1

𝑎 g 𝑏
l a
2 2

Case 2: if a = = , the pair of equations has infinitely many solutions and the graphs representing
equations are intersecting
1 1 1
𝑎 𝑏2 𝑐2 2
the system of equations are coincident lines. The system of equations is dependent (consistent).
𝑎 𝑏 𝑐
Case 3: if 𝑎1 = 𝑏1 ≠ 𝑐1 , the pair of equations has no solution and the graphs representing the system of
2 2 2
equations are parallel lines. The system of equations is inconsistent.

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49. The system of equations a1x +b1y + c1 = 0

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46. If the graphs of two lines pass through the 2 2 2

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same points, then the system of equations if:

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a
representing these lines is:
(A) Consistent (C) a1/a2 = b1/b2 = c1/c2 (D) none of these
(B) inconsistent Ans. (A) (MBOSE 2025 Suppl)
(C) Consistent and dependent
(D) Not dependent 50. The system of equations a1x +b1y + c1 = 0
Ans. (C) consistent and dependent and a2x +b2 y + c2 = 0 has no solution if:
(A) a1/a2 ≠ b1/b2 (B) a1/a2 = b1/b2 ≠ c1/c2

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47. The pair of equations x = 3 and y = 5 has: (C) a1/a2 = b1/b2 = c1/c2 (D) none of these

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(B) one solution Ans. (B)

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(D) infinitely many

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a represents lines which are:
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51. The pair of equations x = a and y = b

a Ans. (B) graphically
(A) Parallel
48. If the ten’s and unit’s digit of a two digit (B) coincident
number are 𝑝 𝑎𝑛𝑑 𝑞 respectively, then the (C) Intersecting at (a, b)
number will be –
(A) 𝑝𝑞 (B) 10𝑝 − 𝑞 (D) Intersecting at (b, a)
Ans. (C)
(C) 10𝑝 + 𝑞 (D) 10𝑞 + 𝑝

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Ans: (C)

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52. The system of equations -3x + 4y = 5 and
9 15
2
x – 6y + 2 = 0 has:

(A) Unique solution (B) infinite solutions
(C) No solutions (D) none of these
Ans. (B)

53. If x = a and y = b is the solution of the
equations x – y = 2 and x + y = 4, then the
values of a and b are respectively:
(A) 3 and 5 (B) 3 and 1
(C) 5 and 3 (D) -1 and -3
Ans. (B) 3 and 1

54. If (4, k) is a solution of the equation 3x + 2y
– 22 = 0 then the value of k is:
(A) - 4 (B) - 5 (C) 4 (D) 5
Ans. (D)

55. If the sum of zeroes of the polynomial p(x) =
2x2 –k √2 x + 1 is √2 then k is equal to:
(A) √2 (B) 2 (C) 2√2 (D) 1/2
Ans. (B) 2

56. A linear equation in two variables has
infinitely many solutions which can be
represented on –
(A) a circle (B) a triangle
(C) a number line (D) the Cartesian plane
Ans: (D)
57. If the line y = mx − 3 passes through the
point (5, 3), then the value of m is –
5 6
(A) 6
(B) 5
(B) 5 (D) 6
Ans: (B)

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QUADRATIC EQUATIONS
Important points/summary:
1. The standard form of a quadratic equation in the variable x is of the form :
ax2 + bx + c = 0, where a, b, c are real numbers and a ≠0.

2. A quadratic equation ax2 + bx + c = 0, has at most two roots
−𝒃±√𝒃𝟐 −𝟒𝒂𝒄
3. The solution of the quadratic equation ax2 + bx + c = 0 is given by x = 𝟐𝒂
, which is known
as the quadratic formula.
4. D = b2 – 4ac, is called the Discriminant of the quadratic equation ax2 + bx + c = 0, where a, b, c are
real numbers and a ≠0.
5. In D = b2 – 4ac, three cases arise:
If D ˃ 0, ax2 + bx + c = 0 has real and unequal/distinct roots
If D = 0, ax2 + bx + c = 0 has real and equal roots
If D < 0, ax2 + bx + c = 0 has no real roots

(A) b2 – 4ac = 0 (B) b2 – 4ac > 0
58. Which of the following is a quadratic (C) b2 – 4ac < 0 (D) b= 0
equation? Ans. (C) (MBOSE 2025 Suppl)
(A) x2 – 2x = (-2)(3 – x) 62. If x = 1is the common root of ax2 + bx + 2 =
(B) x2 + 3√x + 2 = 5 0 and x2 + x + b = 0, then a equals:
(C) (x – 2)(x + 1) = (x – 1) (x + 3) (A) 1 (B) 0 (C) 3 (D) 4
(D) (x + 2)3 = 2 x (x2 – 1) Ans.(B)
Ans. (A)
63. If the polynomial 𝑎𝑥 2 + 𝑏𝑥 + 𝑐 is a perfect
59. If the roots of the equation ax2 + bx + c = 0 square, then
(A) 𝑏 2 = 4𝑎𝑐 (B) 𝑏 = −4𝑎𝑐
are equal, then b equals to:
(C) 𝑎2 = 4𝑏𝑐 (D) 𝑎2 = −4𝑏𝑐
(A)√2𝑎𝑐 (B) ±√4𝑎𝑐
Ans: (A)
(C) −√2𝑎𝑐 (𝐷) − √4𝑎𝑐
Ans. (B)
64. The roots of the quadratic equation 100 x2 –
60. If x = 3 is a solution of the quadratic 20 x + 1 = 0 are:
equation 3x2 + (k – 1)x + 9 = 0, then k (A) 10, 10 (B) –10, 10
equals to: (C) 1/10, 1/10 (D)1/10, –1/10
(A) 11 (B) – 11 (C) 13 (D) -13 Ans. (C)
Ans.(B)
65. 𝑎𝑥 2 + 𝑏𝑥 + 𝑐 = 0, 𝑎 > 0, 𝑏 = 0, 𝑐 > 0 has
(A) One real roots
61. The roots of the equation ax2 + bx + c = 0
(B) no real roots
are not-real if: (C) two equal roots

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(D) two distinct real roots (A) 𝑎, 𝑎 (B)−𝑎2 , 𝑎2
Ans: (B) a
(C) −𝑎, −𝑎 (D) 𝑎, −𝑎
Ans: (D)
66. If the roots of a quadratic equation ax2 + bx
+ c = 0, a ≠ 0 are real and equal, then which
of the following relation is true?

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(A) a = b2/c (B) b2 = ac

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(C) ac = b2/4 (D) c = b2/a
Ans. (C)
.co s e m
em la
l a s a g
ag
67. The Discriminant of the quadratic equation
x2 + 8 x + 16 = 0 is:
(A) 3 (B) 2 (C) 1 (D) 0
Ans. (D)

68. The solutions of (x – 3) (x + 6) = 0 are:
(A) -3, 6 (B) 3, -6 (C) -3, -6 (D) 3,6

m
co
Ans. (B)

.
s em
l a
69. If b2 – 4ac ˃ 0, then the quadratic equation
ax2 + bx + c = 0 has:
(A) Real and equal roots a g
(B) Real and unequal roots
(C) No real roots (D) none of the above
Ans. (B)

m
70. If b2 – 4ac = 0, then the roots of the

m - b/2a, - b /2a (B) –b/2a, b/2a .co
c. o (A)
quadratic equation ax2 + bx + c = 0 are:

m
em (C) 2b/a, – 2b/a (D) a/2b, - a/2b la se
l as a g
ag
Ans. (A)

71. If 2x2 – 2kx + 1 = 0 has real and equal roots,
then the value of k is:
(A) ± 1 (B) 2 (C) ± 2 (D) 1
Ans. (C)

c om m .
𝑥 𝑎

m. e
72. The roots of the quadratic equation 𝑎 = 𝑥 , x,
a ≠ 0, are –
e l as
l a s a g
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ARITHMETIC PROGRESSION
Important points/summary a
a. The sequences or set of terms arranged according to some definite law, whose general term can be
expressed as a formula are called progressions.

b. A.P: A sequence is called an Arithmetic Progression if the difference of any of its terms and the
preceding term is always the same.
That is, t n + 1 – t n = constant, which is called the common difference of the A.P.
c. Common difference, d = t n + 1 – t n , where n= 1,2,3,……

m
c. o
Like, d = t2 – t1 = t3 – t2 = t4 – t3

m
co
d. nth term of an AP:
an = a1 + (n – 1)d
. e m
e m
or, tn = a + (n – 1)d
la s
l a s a g
ag
e. sum of n terms of an AP:
n
Sn = 2 {2a + (n – 1) d}

n
or, Sn = 2 {a + l}

77. The first three terms of an AP when the first
73. In an AP, if d= 2 and fourth term, a4= 13

m are:
term, a= 10 and common difference, d = 10

o
c (A) 10, 20, 30
then a is:

.
em
(A) 6 (B) 7 (C) 20 (D) 28
(B) 0, 10, 100

las
Ans. (B) 7
(C) 10, - 10, - 20 (D) 10, 5, 0

ag
Ans. (A)
74. In an AP, if a= 2, d = 0, n = 1000 then an, is:
(A) 2 (B) 10 (C) 1000 (D) 2000
78. If common difference of an AP is 5, then
Ans. (A)
a16 – a15 is:
(A) 1 (B) 31 (C) 5 (D) 15
75. In an AP: 3, 1, -1, -3, …… the first term ‘a’
Ans. (C)

m
and common difference ‘d’ are respectively:

m .co
(A) 3, 2 (B) 1, -2 (C) 2, 3 (D) 3, -2

c. o Ans. (D) m
79. Which of the following is not an AP ?

se
(A) 4, 10, 16, 22, . . . . ..

em 76. Which of the following is an AP? la
(B) 1, −1, −3, −5, . . . . . ..

las g
a 9, 27, . . . . . . ..
(C) 0, −4, −8, −12, . . . . . ..

ag (A) 2, 4, 8, 16, …….
(D) 1, 3,

(B) -1.2, -3.2, -5.2, -7.2, …… Ans. (D)

(C) 1,3, 9, 27, …….
(D) 0.2, 0.22, 0.222, 0.2222,….. 80. The missing term in the box of the AP: 2, □,
Ans. (B) 26,……. Is:
(A) 6 (B) 12 (C) 13 (D) 14

m .
c. o m
Ans. (D)

s e
sem g l a
la a
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81. The 4th term of the AP
1 5 9
, , ,. . . . . . is – 88. In an A.P: a, a+ d, a + 2d, a + 3d, ………., its
3 3 3
4 11 general term an equals to:
(A) 3 (B) 3
13 12 (A) a + n d (B) {a + (n - 1)d}
(C) 3 (D) 3
Ans. (C) (C) (a + n)/2 (D) {n (a + n)d}
Ans. (B)
82. If k, 2k – 1, k + 4 are three consecutive terms
of an AP, then the value of k is: 89. The list of numbers −10, −6, −2,
2, . . .. . . . is –
(A) – 2 (B) 3 (C) – 3 (D) 6
(A) An AP with 𝑑 = −16
Ans. (B) (B) an AP with 𝑑 = 4
(C) An AP with 𝑑 = −4

83. If 2n + 3 is the nth term of an AP, then the (D) not an AP
common difference is:
(A) 2 (B) 4 (C) 5 (D) 6 Ans. (B)
Ans. (A)

90. In an AP, the difference between t n + 1 and t n
84. The first three terms of an AP if an = 2n + 5
is called the:
are:
(A) First term (B) last term
(A) 1, 2, 3 (B) 7, 10, 13
(C) Common difference (D) next term
(C) 7, 9, 11 (D) 11, 13, 17
Ans. (C)
Ans. (C)

91. If a = - 2, d = 5 then the value of t10 is equal
85. The number of odd numbers between
to:
0 𝑎𝑛𝑑 50 is –
(A) 24 (B) 25 (A) 23 (B) 33 (C) 43 (D) 53

(B) 23 (D) 26 Ans. (C)

Ans. (B)
92. The common difference of the AP: 0, ¼, ½,
¾, ……… is:
86. The sum of first n terms of an AP, whose
(A) ¼ (B) 1/2 (C) 3/4 (D) 0
first term ‘a’ and last term ‘l’ is:
Ans. (A)
(A) n/2 (a + 2l) (B) n /2 (a + l)
(C) (a + l) (D) n/2
Ans. (B)

87. The next term of the series √2, √8, √18, √32,
. . . . . . is –
(A) √48 (B) √54

(C) √50 (D) √60
Ans. (C)

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c. o

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s a g

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.co s e m
em la
l a s a g
ag

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. co
s em
g l a
a

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c. o se
m
s em g la
gla a
a

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.co s e m
e m l a
l a s a g
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TRIANGLES
Important points/summary:
1. Triangle: A Triangle is a plane figure bounded by three line segments. It has three sides,
three vertices and three angles.
2. Scalene triangle: A triangle in which no two of its sides are equal is called a scalene triangle.
3. Isosceles triangle: A triangle in which two of its sides are equal is called an isosceles triangle.
4. Similar figures: Two geometric figures having the same shape but different sizes are said to
be similar figures.
5. Congruent figures: Two geometric figures having the same shape but same sizes are said to
be congruent figures.
6. Similar triangles: Two triangles are said to be similar if
i. Their corresponding angles are equal and
ii. their corresponding sides are proportional
7. Equilateral triangle: An equilateral triangle is a triangle in which all the three sides are
equal.
8. Mid- point theorem: The straight line joining the middle points of two sides of a triangle is
parallel to the third side.
9. AAA Similarity: If in two triangles, corresponding angles are equal then the triangles are
similar.
10. AA Similarity: If in two angles of one triangle are respectively equal to two angles of another
triangle then the two triangles are similar.
11. SSS Similarity: If the corresponding sides of two triangles are proportional then they are
similar.
12. SAS Similarity: If in two triangles, one pair of corresponding sides is proportional and the
included angle is equal then the two triangles are similar.
13. In an equilateral triangle having side ‘a’ units, then
√𝟑
i. Altitude = 𝟐 a units
√𝟑
ii. Area = 𝟒 a2 sq. units

94. The ratio of any two corresponding sides in
93. All geometrical congruent figures are: two equiangular triangles is always:
(A)Not similar (B) similar (A)The same (B) different
(C) Unequal (D) none of the above (C) Similar (D) None of the above
Ans. (B) Ans. (A)

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m
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c. o 95. If two angles of one triangle are respectively

e m
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s g
100. In a triangle, if the perpendicular from the
equal to two angles of another triangles then
a
vertex to the base bisects the base. The
the two triangles are similar. This is referred triangle is:
to as the: (A) Scalene (B) isosceles
(A) AA Similarity Criterion for two (C) Obtuse –angled (D) right- angled
triangles Ans. (B)
(B) SAS Similarity Criterion for two

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triangles 101. Given ∆𝐴𝐵𝐶 ~ ∆𝑃𝑄𝑅, ∠ A = 300 and ∠ Q =

m
co m
900, then the value of ∠ R + ∠ B is

. e
(C) AAA Similarity Criterion for two

m s
(A) 1200

e la
triangles

a s g
0
(B) 150

l a
(D) SSS Similarity Criterion for two

ag
(C) 1800
triangles
(D) 900
Ans. (A)
Ans. ( C)
102. In a triangle, the line segment joining from
96. If ∆ABC and ∆DEF are two triangles in
𝐴𝐵 𝐵𝐶
one vertex to the mid-point of the opposite
which = then the two triangles are
𝐷𝐸 𝐷𝐹 side is called its:
similar if: (A) Median (B) perpendicular

o m (C) Hypotenuse (D) angle bisector
c Ans. (A)
(A) ∠ A = ∠ F (B) ∠B = ∠D

.
em
(C) ∠A = ∠D (D) ∠B =∠ E (MBOSE 2025 Suppl)
Ans. (B)
l a s
a g
103. In the ∆ABC with XY ‖ BC, if AX = 3 cm,
XB = 5 cm, AY = 6 cm, then AC is equal to:
97. The area of an equilateral triangle of side ’a’ (A) 10 cm (B) 13 cm
𝟐√𝟑 (C) 14cm (D) 16 cm
is (A) 3a2 (B) 𝒂 Ans. (D) 16 cm
√𝟑 √𝟑
(C) 𝟒 a (D) a2
𝟒

Ans. (D) (MBOSE 2025) 104. If ∆ABC ̴ ∆DEF, such that 2AB = DE and
BC = 8 cm, then EF equals:

m
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(A) 4cm (B) 8 cm

m .
98. The length of the altitude of an equilateral

c. o (A) 3 (B) √3 (C) √3/2 (D) 2√3
(C) 12 cm (D) 16 cm

m
se
triangle of side 2 cm is: Ans. (D) 16 cm

s em Ans. (B) g la
l a a
ag
(MBOSE 2025 Suppl) 105. The perimeter of an equilateral triangle with
side 9 cm is: (MBOSE 2025) suppl.
(A) 9 cm (B) 18 cm
𝐴𝐵 𝐵𝐶 𝐶𝐴
99. In a ∆ ABC and ∆ DEF, = = 𝐷𝐸 𝐸𝐹 𝐹𝐷
(C) 27 cm (D) 36 cm
Ans. (C)
Which of the following is true
(A) ∆ ABC ~ ∆ DEF by SSS Criterion
(B) ∆ ABC ~ ∆ DEF by SAS Criterion
(C) ∆ ABC ~ ∆ DEF by AA Criterion

m .
co m
(D) ∆ ABC is not similar to ∆ DEF

m . s e
a
Ans. (A) ∆ ABC ~ ∆ DEF by SSS Criterion

se g l
l a a
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COORDINATE GEOMETRY
Important points/summary: a

m
m c. o
c. o e m
s em two mutually perpendicular axes intersecting each other at a point O, calledgthelas
la are O (0,0).
XOX’ and YOY’ are

g a
a in the ordered pair (x, y), x is called the abscissa or x-coordinate.
Origin, whose coordinates
1. Abscissa:
2. Ordinate: in the ordered pair (x, y), y is called the ordinate or y-coordinate.
3. Collinearity of three points: three points are collinear if the sum of any two distances is equal to
the third distance.
4. The coordinates of any point on the X-Axis are (x,0) (as y = 0)

m
5. The coordinates of any point on the Y-Axis are (0,y) (as x = 0)

. c o
6. Distance formula: the distance between the two points A (x1 , y1 ) and B (x2 , y2 ) is given by

m
e O (0, 0) is given by OP = √(x + y )
𝟐 𝟐

7. The distance of any point P (x , y ) from thesorigin
AB = √(𝒙 – 𝒙 ) + (𝒚 – 𝒚 )
𝟐 𝟏 𝟐 𝟏

g la
2 2

8. Section formula:
a
A (x1 , y1 ) P (x, y) B (x2 , y2 )

m n
The coordinates of the point P (x, y) which divides internally the line segment joining the points
A (x1 , y1 ) and B (x2 , y2 ) in the ratio m : n are given by

o m
𝑚𝑥2 +𝑛𝑥1 𝑚𝑦2 +𝑛𝑦1
(x, y) = ( . )

m c
c. o Let M (x, y) be the mid point of the line segment joining the points A (x ,sye)mand. B (x , y ) in
𝑚+𝑛 𝑚+𝑛

9. Mid- point formula:

s em the ratio 1 : 1 are given by g l a
1 1 2 2

gla a
a
𝑥1 +𝑥2 𝑦1 +𝑦2
(x, y) = ( . ) 2 2

10. Centroid: the point of concurrency of the three medians of a triangle is called the centroid of a
triangle.

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e m l a
l a s a g
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106. The perimeter of a circle is called its: (C) (a/2, b/2) (D) (a, b)
(A) Area (B) Diameter Ans. (C)
(C) Radius (D) Circumference
Ans. (D) circumference 113. The coordinates of a point on X –axis are of
107. The coordinates of a point on X-axis is of the form:
(A) (0, x) (B) (x, 0)
the form:
(C) (0, y) (D) (y, 0)
(A) (x, y) (B) (x, 0) Ans. (B)
(C) (y, 0) (D) (0, y)
114. The mid-point of the line segment joining
Ans. (B)
the points P (- 4, 5) and Q(4, 6) lies on:
(A) Origin (B) x - axis
108. The distance of any point P (x, y) from
Origin is: (MBOSE 2025) (C) y- axis (D) neither x-axis or y-axis

(A)√𝑥 2 − 𝑦 2 (B)√𝑦 2 − 𝑥 2 Ans. ( C)

(C)√𝑥 2 + 𝑦 2 (D)√𝑥 3 + 𝑦 3
115. The perimeter of a triangle with vertices
Ans. (C) (0,4), (0, 0) and (3, 0) is:
(A) 12 units (B) 14 units

109. In which quadrant does the point (-3, 5) (C) 16 units (D) 8 units

lie? Ans. (A)
(A) first quadrant (B) second quadrant
116. The point (3, - 4) lies in the:
(C) third quadrant (D) fourth quadrant
(A) first quadrant (B) second quadrant
Ans. (B)
(C) third quadrant (D) fourth quadrant
Ans. (D)
110. The distance of the point P (3, 4) from the
origin is: (MBOSE 2025 Suppl)
117. The coordinates of the point P (x, y) which
(A) 4 unit (B) 3 units
divides the line segment joining the points
(C) 5 units (D) 7 units
A(x1, y1) and B(x2, y2)in the ratio m:n is:
Ans. (C)
𝑚𝑥2 +𝑛𝑥1 𝑚𝑦2 +𝑛𝑦1
(A)( 𝑚−𝑛
. 𝑚−𝑛 )
𝑚𝑥2 −𝑛𝑥1 𝑚𝑦2 −𝑛𝑦1
111. The coordinates of the midpoint of the line (B)( 𝑚+𝑛 . 𝑚+𝑛 )
𝑚𝑥2 +𝑛𝑥1 𝑚𝑦2 +𝑛𝑦1
segment joining the points A (- 2, 8) and B (C)( 𝑚+𝑛 . 𝑚+𝑛 )
𝑚𝑥2 −𝑛𝑥1 𝑚𝑦2 −𝑛𝑦1
(- 6, - 4) is: (D)( . )
𝑚−𝑛 𝑚−𝑛
(A) (4, 2) (B) (- 4, 2)
Ans. (C)
(C) (4, - 2) (D) (- 4, - 2)
Ans. (B) 118. The coordinates of a mid- point of the line
segment AB with end points A (x 1, y1) and B (x 2, y 2)
is:
112. The coordinates of the midpoint of the line
𝑥 −𝑥 𝑦 −𝑦 𝑥 +𝑥 𝑦 +𝑦
segment joining the points A (a, b) and B (0, (A) ( 1 2 2 . 1 2 2 ) (B) ( 1 3 2 . 1 3 2 )
𝑥 +𝑥 𝑦 +𝑦 𝑥 +𝑥 𝑦 +𝑦
0) is: (C) ( 1 4 2 . 1 4 2 ) (D) ( 1 2 2 . 1 2 2 )
Ans. (D)
(A) (a + b/2, a) (B) (a + b, b)

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c. o 119. The perpendicular distance of a point from

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126. Centroid of a triangle is the point of
Y- axis called:
a
concurrency of its three:
(A) Ordinate (B) abscissa (A) Angle bisectors (B) medians
(C) altitude (D) none of the above (C) Altitudes (D) perpendicular bisectors
Ans. (B) Ans. (B)
120. The ordinates of all points on a horizontal 127. The ordinates of all points on a horizontal
line are: line are:

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(A) Parallel (B) perpendicular (A) Parallel (B) perpendicular

m
.co e m
(C) Equal (D) coincident

m s
(C) equal (D) coincident

e la
Ans. (C)

a s g
Ans. (C)

l a
ag (B) – 1
121. The abscissa of any point on the Y – axis is:
(A) 1 (C) 0 (D) 2
Ans. (C)

𝑎
122. If P (4, ) is the mid- point of the line
2

segment joining the points A (- 6, 5) and B

m
co
(- 2, 3) then ‘a’ equals:

.
em
(A) - 6 (B) – 8 (C) – 12 (D) 12

a s
Ans. (B)

g l
123. The distance between the points A (4, k) and a
B (1, 0) is 5 units then k equals:
(A) 4 (B) – 4 (C) 0 (D) ± 4
Ans. (D)

m
124. The distance between the points P (0, 5) and

m 5 units (B) 5 √2 units (C) 2√ 5 units .co
Q (- 5, 0) is:

c. o (A) se
m
em (D) la
s
√10 units

l a a g
ag
Ans. (B)

125. If the end points of a diameter of a circle are
(1, 2) and (3, 4) then the coordinates of the
Centre are:
(A) (2, 4) (B) (2, 3)
(C) (1, 2) (D) (4, 6)

m .
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Ans. (B)

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em
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s g
Introduction to Trigonometry
Important points/summary: a
Trigonometry is a branch of Mathematics which deals with the measurements of sides and angles of
a triangle and their relationships.

m
m c. o
c. o e m
e m ∆ABC right angled at B and let θ< 90 at C be an acute angle then thelasixs
trigonometric a s g
Consider a right 0

l a
agθ =
ratios are defined as :
𝑃𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 𝑃 1
1. sin 𝐻𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒
= 𝐻
(sin θ= ) (reciprocals)
𝑐𝑜𝑠𝑒𝑐𝜃
𝑏𝑎𝑠𝑒 𝐵 1
2. cos θ = 𝐻𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒 =𝐻 ( cos θ= 𝑠𝑒𝑐𝜃 )
𝑃𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 𝑃 1
3. tan θ = 𝐵𝑎𝑠𝑒
=𝐵 (tan θ= 𝑐𝑜𝑡𝜃 )
𝐵𝑎𝑠𝑒 𝐵 1
4. cot θ = = (cot θ= )
𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 𝑃 tan 𝜃

m
co
𝐻𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒 𝐻 1
5. sec θ = = ( sec θ= )

.
𝐵𝑎𝑠𝑒 𝐵 𝑐𝑜𝑠𝜃

em
𝐻𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒 𝐻 1
cosec θ= (cosec θ= )

s tan θ =
6. =

a
𝑃𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 𝑃 𝑠𝑖𝑛𝜃

g l
sin 𝜃 cos 𝜃
7. Relation between Sin θ, cos θ and tanθ: cot θ = 𝑠𝑖𝑛𝜃
a
𝑐𝑜𝑠𝜃
;

8. Relation between Sin θ, cos θ and cot θ is
9. Some Identities/formulae:
a. sin2θ + cos2θ = 1
b. 1 + tan2θ = Sec2θ
c. 1 + cot2θ = cosec2 θ
Table showing trigonometric ratios of some angles:

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Ratios Some specific Angles ↓

m .
c. o
↓

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se
00 300 450 600 900

em la
sin 0 1 1 √3 1

las g
2 √2

a
2

g
1 1

a
cos 1 √3 0
2 √2 2

tan 0 1 1 √3 Undefined
√3

cot undefined √3 1 1 0
√3

sec 1 2 √2 2 Undefined
√3

cosec undefined 2 √2 2 1

m .
√3

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128. The value of 1 + tan2 450 is: (D) tan θ/cos θ = sin θ
(A) - 1 (B) 0 (C) 1 (D) 2 Ans. (B)
Ans. (D) 138. The relation between sin θ, cos θ and cot θ
129. If cos θ =1, then the value of θ is: is:
(A) 0 0
(B) 30 0
(C) 60 0
(D) 90 0
(A)cos θ/sin θ = cot θ (B) sin θ/cos θ = tan θ
Ans. (A) (C) tan θ/sin θ = cos θ (D) tan θ/cos θ = sin θ
130. The value of 3 cot2 A - 3 cosec2 A is equal Ans. (A)
to: (MBOSE 2025 Suppl) 139. If θ = 300 then the value of cos2θ – sin2θ is:
(A) - 3 (B) 0 (C) 3 (D) 3/2 (A) 1 (B) 1/2 (C) -1/2 (D) -1
Ans. (A) Ans. (B) (MBOSE 2025)
131. In ∆ABC right angle at B, if AC = 13cm, 140. The reciprocal of cosine θ is:
BC = 5 cm and AB = 12 cm then sin A is (A) tan θ (B) sec θ (C) cosec θ (D) sin θ
equal to: Ans. (B)
(A) 13/5 (B) 5/13 (C) 12/13 (D) 13/12 141. tan 2 θ + 1 is equal to:
Ans. (B) (A) cot 2 θ (B) sec 2 θ
132. The value of 9 sec2θ - 9 tan2θ is: (C) cosec 2 θ (D) cos2 θ
(A) 0 (B) 1 (C) 9 (D) 10 Ans. (B)
Ans. (C) 142. 1 – sin 2 θ is equal to:
2 tan 30𝑜 (A) cot 2 θ (B) sec 2 θ
133. The value of is equal to:
1−tan2 30𝑜
(C) cosec 2 θ (D) cos2 θ
(A) cos 600 (B) sin 600
Ans. (D)
(C) tan 600 (D) sin 300
Ans. (C)
134. Sin 2A = 2 sin A is true when A equals to:
(A) 00 (B) 300 (C) 450 (D) 600
Ans. (A) (MBOSE 2025 Suppl)
135. If θ is an acute angle of a right-angled
triangle, then which of the following equations is
not true?

(A) sin θ cot θ = cos θ (B) cos θ tan θ = sin θ
(C) cosec2θ – cot2θ = 1 (D) tan2θ – sec2 θ = 1

Ans. (D) tan2 θ – sec2 θ = 1

136. The value of sec 600 is:
(A) √3 /2 (B) 1/2 (C) 2 (D) 1
Ans. (C)
137. The relation between sin θ, cos θ and tan θ
is:
(A) cos θ/sin θ = tan θ
(B) sin θ/cos θ = tan θ
(C) tan θ/sin θ = cos θ

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Some Applications in the Trigonometric Ratios
Important points/ summary

Definitions
1. Line of Sight
The line drawn from the eye of an observer to a point in the object where the person is viewing is
called the line of sight.

2. Angle of Elevation
If you look upwards at an object, say the top of a tree, the angle formed
between the horizontal and your line of sight is called angle of elevation.

Note: Sun’s altitude or Sun’s elevation is nothing but angle of elevation.

3. Angle of Depression
If you look downwards at an object, the angle formed between the
horizontal and your line of sight is called angle of depression.

Note: Both the angle of elevation and depression are measured from the horizontal.
Note: To solve the sum of height & distance, diagram is mandatory.

with the wall, then the height of the wall is
(A) 15√3 m (B) 15√3/2 m
143. The angle formed between the line of sight (C) 15 m/2 (D) 15 m
and the horizontal when you look towards the
height of a tower is Ans: (C) 15 /2 m
(A) Angle of elevation
146. If the height of a vertical pole is √3 times
(B) Angle of depression
the length of its shadow on the ground, then the
(C) Corresponding angles
angle of elevation of the Sun at that time is
(D) Alt. int. angles
(A) 30° (B) 60°
Ans: (A) Angle of elevation (C) 45° (D) 75°

144. The angle formed between the line of sight Ans: (B) 60°
and the horizontal line when you look from the
top of a high rise building towards a car on a 147. The ratio of the length of a rod and its shadow
street is is 1:√3. The angle elevation of the rod is
(A) Angle of elevation (A) 30° (B) 45°
(B) Angle of depression (C) 60° (D) 90°
(C) Alt. int. angles Ans: (A) 30°
(D) Straight angles
148. The ___________ of an object is the angle
Ans: (B) Angle of depression formed by the line of sight with the horizontal when
145. A ladder 15 m long just reaches the top of a the object is below the horizontal level.
vertical wall. If the ladder makes an angle of 60° (a) line of sight

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c. o (b) angle of elevation

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em (c) angle of depression l a s
s g
152. From the top of a cliff 25 m high, the angle

(d) none of these a
of elevation of a tower is found to be equal to the
angle of depression of the foot of the tower. The
height of the tower is
Ans: (c) angle of depression
(A) 25 m (B) 50 m
149. A pole is being broken by the wind at the point (C) 75 m (D) 100 m
of trisection of the pole, then the top strikes the
Ans: (B) 50 m
ground at an angle of
(A) 30° (B) 60° 153. If the angle of elevation of the top of a tower

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c. o
(C) 90° (D) 45° from two points distant a and b from the base in

m
the same straight line with it are complementary,

co m
Ans: (A) 30°

. e
then the height of the tower is

m s
(B) √ab

la
(A) ab

e
150. The shadow of a tower is equal to its height at

s g
(D) √(a/b)

a
10:45 a.m. The sun’s altitude is (C) a/b
(A) 30°
g l
(B) 45°
a
a
Ans: (B) √ab
(C) 60° (D) 90°
154. A pole 6 m high casts a shadow 2√3 m long
Ans: (D) 45°
on the ground, the Sun’s elevation is
151. If a pole of height 20 m makes an angle of (A) 60° (B) 75°
elevation 45° with the ground, then the distance (C) 30° (D) 90°
between the point on the ground and the foot of the
Ans: (A) 60°
pole is
(A) 10 m (B) 15 m

m
(C) 20 m (D) 40 m

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Ans: (C) 20 m

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s Circles
Important points/summary a g
1. Circle: A circle is a collection of all points in a plane which is equidistant from a fixed point in the
plane. The fixed point is called the Centre of a circle and the constant distance is called its radius.
O is the Centre of a circle and OP= r, is its radius

r P
O

m
m c. o
.co e m
2. The perimeter/length/path of a circle is called its circumference.

em
3. A chord passing through the Centre of a circle is called its diameter. Or, the longest chord of a

la s
a s a g
circle is called its diameter.

l
ag
4. Tangent: A line which intersects the circle at one point only is called a tangent line.
5. The point which the tangent intersects the circle is called the point of contact.
6. A line which intersects the circle in two distinct points is called secant of the circle.
7. There can be only two tangents from a point outside the circle.
8. One and only one tangent can be drawn through a point lying on the circle.
9. The lengths of tangents drawn from an external point to a circle are equal.
10. A tangent to a circle is perpendicular to the radius through the point of contact.

o m
. c point of a diameter is called a:
159. A part of the circle whose end points are end

em
155. If tangents PA and PB from a point P to a

a s
circle with Centre O are inclined to each other at

l
(A) Circumference (B) segment

a g
an angle of 800 then <POA is equal to:
(C) Semicircle (D) perimeter
(A) 500 (B) 600 (C) 700 (D) 800
Ans. (C) (MBOSE 2025)
Ans. (A)
160. A tangent PQ at a point P of a circle of
156. From a point Q, the length of the tangent to
radius 5 cm meets a line through the Centre O at
a circle is 4 cm and the distance of Q from the
a point Q so that OQ = 12cm, then length PQ
Centre is 5 cm, then the radius of a circle is:
is:

m
(A) 1 cm (B) 5 cm (C) 3 cm (D ) 4 cm
(A) 12 cm (B) 13 cm

m .co
Ans. (C) (C) 8.5 cm (D) √119 cm

co
. of a circle are: m
se
157. The tangents drawn at the end points of a Ans. (D) √119 cm

em la
161. A line which intersects a circle at two

s
diameter

a a g
distinct points is called:

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(A) Equal (B) parallel (A) Tangent (B) secant
(C) Perpendicular (D) intersecting (C) Diameter (D) chord of a circle
Ans. (B) Secant
Ans. (B) (MBOSE 2025 Suppl)
162. If the angle between two tangents drawn
158. The distance between two parallel tangents
from a point P to a circle of radius r and centre O
of a circle of radius 8 cm is: is 90°, the OP is equal to
(A) 8 cm (B) 12 cm (A) 2r (B) r/2
(C) 14 cm (D) 16 cm (C) √2 r (D) √2 / r

om m .
Ans. (D) (MBOSE 2025)

c
Ans: (C) √2 r

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163.If two tangents inclined at an angle 60° are
drawn to a circle of radius 3 cm, then length of
each tangent is equal to
(A) √ 3/2 cm (B) 3 cm
(C) 3√3 cm (D) 6 cm
Ans: (C) 3√3 cm
164. Two circles touch each other externally at P.
AB is a common tangent to the circles touching
them at A and B. The value of ∠APB is
(A) 30° (B) 45°
(C) 60° (D) 90°
Ans: (D) 90°

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AREA RELATED TO CIRCLES
Important points/summary:
𝑐 𝑐𝑖𝑟𝑐𝑢𝑚𝑓𝑒𝑟𝑒𝑛𝑐𝑒
1. π = ( )
𝑑 𝑑𝑖𝑎𝑚𝑒𝑡𝑒𝑟

2. 𝑐 = πd = 2πr ( As d = 2r)
3. Area of a circle = π r 2
4. Area of a semicircle = ½ π r 2
5. Area of a ring = π ( R2 - r 2)
6. Angle described by a minute hand in 1 hour ( 60 minutes) = 3600

(that is, Angle described by a minute hand in 1 minute = 60 )

7. The portion of a circular region enclosed by two radii and the corresponding arc is called the
sector of the circle.
8. The portion of a circular region enclosed between a chord and the corresponding arc is called
the segment of the circle.
πr2 θ
9. 𝐴𝑟𝑒𝑎 𝑜𝑓 𝑡ℎ𝑒 𝑠𝑒𝑐𝑡𝑜𝑟 =
360

(θ is the sector angle)
πθ 1
Area of the segment = 𝑟 2 ( 360 - 2
sin θ) [ when θ > 900 𝑢𝑠𝑒 , sin(1800 − θ) = sin θ ]
πr θ
10. Length of an arc of a circle =
180

11. Area of the major segment = π r2 – area of minor segment
πr θ
12. Perimeter of a sector = 2r + 180

165. The portion (or part) of a circular region 166. The portion (or part) of the circular region
enclosed between a chord and the corresponding arc enclosed by two radii and the corresponding
is called a/an: arc is called the/an:
(A)Arc of the circle (B) Perimeter of a circle (A) Arc of the circle
(C) Sector of a circle (D) segment of a circle. (B) Perimeter of a circle
Ans. (D) (C) Sector of a circle
(D) Segment of a circle
Ans. (C) (MBOSE 2025 Suppl)

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173. The total surface area of a hemispherical
167. If r is the radius of a circle and θ is the
aobject of radius r is:
angle subtended by an arc of length l , then (A) π r 2 (B) 2π r 2
area of the sector of a circle is: (C) 3π r 2 (D) 4π r 2
1 Ans. (C) (MBOSE 2025)
(A) 2 l r (B) 2π r l
2 1
(C) rl (D) l r θ
3 2
174. The length of a diagonal of a cube of side

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c. o
Ans. (A)
‘a’ is:

m
c. o m
(A) a √3 (B) 3 √a (C) √3a (D) a/√3

circumference m
168. The Area of a circle is 49π cm2. Its
s e
e la
Ans. (C)

la s g
175. The area of square is the same asathe area of
is:

(C) a
g
(A) 7π cm (B) 14π cm
21π cm (D) 28π cm
circle. Their perimeters are in the ratio of:
Ans. (B)
(A) 1:1 (B) 2:π (C) π: 2 (D) 2: √π
Ans. (D)
169. In a circle of radius 21cm, an arc subtends
an angle of 600 at the centre. The length of
176. The perimeter of a circle is equal to that of a
an arc is:(take π = 22/7)

m (A) 22 : 7
square, then the ratio of their areas is

. co
(A) 22 cm (B) 44 cm
(B) 14 : 11

em
(C) 132 cm (D) 231 cm

s
(C) 1 : 22 (D) 11 : 14
Ans. (A) (MBOSE 2025 Suppl)

g l a Ans: (B) 14 : 11

170. The angle made by the minute hand of a
a
177. A garden roller has circumference of 4 m. the
clock at its centre in 15 minutes duration
number of revolutions it makes in moving 40 m is:
is:
(A) 8 (B) 10 (C) 12 (D) 16
(A) 600 (B) 800 (C) 900 (D) 1800
Ans. (B)
Ans. (C)

o m
171. If the area and circumference of a circle are

m .c
178. The angle described by a minute hand in half

c. o (A) 2 units m
numerically equal, then its diameter is:

s e (C) 180
an hour is:

em (C) 4 units a
(B) 3 units

s Ans. (C) gl
0 0 0
(A) 60 (B) 120 (D) 3600

la a
(D) 6units

ag Ans. (C) (MBOSE 2025)
179. The circumference of a circle of diameter D
units is:
172. If the circumference of a circle increases
(A) π D ( B) 2πD (C) 4π/2 (D) πD/2
from 2π to 4π, then its area is:
Ans. (A) π D
(A) Four times (B) tripled
(C) doubled (D) halved
180. If ‘r’ and ‘h’ represent the radius of the base

om .
Ans. (A)

c m
and height of a right circular cone respectively

e m. as e
then its curved surface area is:

a s g l
l a
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(A) π r h (B) π r2 h 186. The difference between circumference and
(C) π r (√ r2 + h2) (D) π r h2 diameter of a circle is 90 cm, then the radius
Ans. (C) of a circle is:
(A) 24. 5 cm (B) 23 cm
181. The area of a sector whose perimeter is four (C) 22. 5cm (D) 21 cm
times its radius ’r’ units is: Ans. (D)
(A) r2/4 (B) 2r2 (C) r2 (D) r2/2
Ans. (C) 187. If the radius of a circle is 7 cm, then its
circumference is equal to

182. Distance covered by a wheel in one (A) 22 cm (B) 44 cm
revolution is equal to: (C) 77 cm (D) 88 cm
(A) Diameter of a wheel (B ) radius of a wheel
Ans: 44 cm
(C) Circumference of a wheel (D) None of these
Ans. (C)
188. If the diameter of the circle is 7 cm, then its
area is equal to
183. The circumference of a bicycle wheel that
(A) 7/2 cm² (B) 77/2 cm²
makes 5000 revolutions in moving 11 km is:
(Take π = 22/7) (C) 154/2 cm² (D) 308/3 cm²

(A) 500 cm (B)250 cm Ans: (B) 77/2
(C) 220 cm (D) 150 cm
Ans. (C)

184. The area of a circle when area and
circumference are numerically equal is:
(A) 2π (B) 4π (C) 6π (D) 8π
Ans. (B)
185. When area and circumference are
numerically equal then the radius of the
circle is equal to:
(A) 1 unit (B) 2 units
(C) 3 units (D) 4 units
Ans. (B)

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SURFACE AREA AND VOLUMES
Important points/summary:
CUBOID: If l, b, h represents length, breadth and height of a cuboid, then
i. Total Surface Area = 2 (lb + bh + hl) sq. units.
ii. Lateral Surface Area = 2 (l + b) sq. units
= Area of four walls of a room
iii. Volume = (l x b x h) cu. Units

CUBE: If ‘a’ represents side of a cube, then
iv. Total Surface Area = 6 a2 sq. units.
v. Lateral Surface Area = 4 a2 sq. units
vi. Volume = a3 cu. Units

RIGHT CIRCULAR CYLINDER: If ‘r’ and ‘h’ be the radius and height of the right circular cylinder
then
vii. Area of each circular base = π r2 sq. units
viii. Curved Surface Area = 2 π r h sq. units
ix. Total Surface Area = 2 π r (h + r) sq. units.

Volume = π r 2 h cu. Units
CONE: If ‘r’, ‘h’ and ‘l’ represent the radius, height and slant height of the cone then
i. Area of each circular base = π r2 sq. units
ii. Curved Surface Area = π r l sq. units
iii. Total Surface Area = π r (l + r) sq. units.
iv. l = √( r2 + h2 ) units
v. Volume = 1/3 π r 2 h cu. Units

SPHERE: If ‘r’ is the radius of the sphere then
vi. Curved Surface Area = 4 π r 2 sq. units
vii. Volume = 4/3 π r 3 cu. Units

HEMISPHERE: If ‘r’ is the radius of the hemisphere then
viii. Curved Surface Area = 2 π r 2 sq. units
ix. Total Surface Area = 3 π r 2 sq. units.
x. Volume = 2/3 π r 3 cu. Units

SPHERICAL SHELL: Let ‘R’ and ‘r’ be the outer and inner radii of the spherical shell then
xi. Outer Surface Area = 4 π R 2 sq. units
xii. Inner Surface Area = 4 π r 2 sq. units
xiii. Volume = 4/3 π ( R 3 - r3) cu. Units

Note: 1 cm³ = 1 ml
1 litre = 1000 ml
= 1000 cm³

189. The total surface area of a hemisphere with 190. The volume of a sphere of radius R is:
radius ‘R’ is: (A) 2/3 π R3 (B) 4/3π R3
(A) π R2 (B) 2π R2 (C) 3π R2 (D) 4π R2 (C) 3π R3 (D) 4π R3
Ans. (C) Ans. (B)

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m
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c. o 191. During the conversion of a solid from one

e m
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(C) 2π r(r - h) (D) 3π r (r +h)
shape to another, the volume of the new shape
a
Ans. (A)
will: 200. If the curved surface area of a cone is 308
(A) Increase (B) decrease cm2 and its slant height is 14 cm, then radius
(C) Remain unaltered (D) Depends on the shape of the base is: (MBOSE 2025)
Ans. (C) (A) 14 cm (B) 7 cm (C) 21 cm (D) 28 cm
192. If the radius of a sphere becomes 3 times, Ans. (B)

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then its volume will becomes: 201. If the radius of a sphere is doubled then the

m
.co e m
(A) 3 times (B) 6 times (C) 9 times (D) 27 times ratio of the volumes of the first sphere to the
Ans. (D)
e m
new sphere is: (MBOSE 2025)
s
la 1:8
l a s
193. Thee surface areas of two spheres are in the (A) 1:2 (B) 1:4 (C) 1:6
a g(D)

ag (B) 16:9 (C) 4:3 (D) 16 :9
ratio 16:9. The ratio of their volume is: Ans. (D)
(A) 64:27 3 3
202. Let ‘r’ and ‘h’ be the radius and height
Ans. (A) (MBOSE 2025 Suppl) respectively of a right circular cylinder, then
194. If the length of each side of a cube is 4 m, its curved surface area is:
then diagonal of the cube is (A) 2π r(r + h) (B) 2πrh2
(A) 3√3 m (B) 4√3 m (C) 5√3 m (D) 6√3 m (C) 2πr h (D) 2πr (r - h)

m Let ‘r’ and ‘h’ be the radius and height
c. o 203. respectively
Ans. (B) 4√3 m Ans. (C)

em
195. The radius of a hemisphere is 7 m. The total

s
surface area of the hemisphere is of a cone, then its volume is

g l a equal to:

a
(A) 264 m² (B) 624 m² (C) 462 m² (D) 642 m²
(A) π r2 h (B) πr2
Ans. (C) 462 m²
(C) 1/3 πr2h (D) 2/3π r h
196. The volume of a cube of an edge of 4 cm is:
Ans. (C)
(A) 16 cm3 (B) 64 cm3 (C) 128 cm3
204. What is the maximum length of a pencil
(D) 256 cm3
that can be placed in a rectangular box
Ans.(B) (MBOSE 2025 Suppl) of dimensions 10 cm × 6 cm × 5 cm?

m
197. Eight metallic sphere each of radius 2mm

c. o(D) √201 cm
(A) √161 cm (B) √173 cm

m the radius of the new sphere is:
c. (A)o Then
are melted and recast into a single sphere.

Ans. (A) √161 cm em
(C) √191 cm

em 4 mm (B) 4.5 mm (C) 5.5 mm (D) 6 mm l a s
l as 205. The radiusg(in cm) of the largest right
a that can be cut out from a cube
ag
Ans. (A)
circular cone
198. The total surface area of a cube of length ‘a’
of side 4.2 cm is
units is:
(A) 2.1 cm (B) 4.2 cm (C) 8.4 cm (D) 1.05 cm
(A) 3a2 sq. units (B) 4a2 sq. units
Ans. (A) 2.1 cm
(C) 6a2 sq. units (D) 8a2 sq. units
206. Vertical cross-section of a right circular
Ans. (C) cylinder is always a
199. The total surface area of a right circular

m .
(A) square (B) rhombus

co m
cylinder of radius ‘r’ and height ‘h’ is:

. e
(C) rectangle (D) trapezium
(A) 2π r (r +h)
e m
(B) 2π r h Ans. (C) rectangle
l as
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STATISTICS
Important points/summary: a
1. MEAN OF GROUPED DATA: Let x1, x2, x3,…..xn be n observations whose frequencies are f1, f2,
f3, …. …fn respectively i. e. x1 occurs f1 times, x2 occurs f2 times and so on, then the mean ẋ (average)
f1x1 + f2 x2 + ….+ fn xn ∑ fixi ∑ fixi
of the data is given by ẋ = f1 + f2 + ….+ fn
= ∑fi
= 𝑛
(i= 1,2,3,..n)

𝐮𝐩𝐩𝐞𝐫 𝐋𝐢𝐦𝐢𝐭+𝐋𝐨𝐰𝐞𝐫 𝐋𝐢𝐦𝐢𝐭

m
2. Class mark of a class interval is defined as class mark =

c. o
𝟐

m
.co e m
3. Mode of grouped data: The item which occurs more frequently in a data is called its mode.

em la s
l a s a g
ag
4. Modal class: in a grouped frequency distribution, the class which has the maximum frequency is
called Modal class.

5. The data which have only one mode is called unimodal, if it contains two modes, it is called bimodal,
if it contains three modes, it is called trimodal. In general, if it contains more than one mode, it is
called multimodal.

6. Median of grouped data: Median is the value of the middlemost item when the data are arranged in
ascending or descending order of magnitude.

m
. co
em
n
7. Median class: the class which corresponds to 2 the observation is called the median class.

l a s
g
8. Median of ungrouped data: Let the total number of items in the data be n, then
i. If n is odd,
n+1
a
Median = value of ( 2 )th item

ii. If n is even,
1 n n
Median= 2[ 𝑣𝑎𝑙𝑢𝑒 𝑜𝑓 2 th item + value of (2 + 1)th item]

m
co
9. The empirical relationship between mean, mode and median is given by 3 Median = Mode + 2 Mean

m .
c. o
10. Cumulative frequency of a particular value of the variable (or class) is the sum total of all the

m
se
frequencies up to that value (or the class).

em la
11. Ogive or cumulative frequency curve: The graphical representation of a cumulative frequency

las distribution is called an Ogive or cumulative frequency curve.
a g
ag
IMPORTANT FORMULAE:
∑ fixi
i. Mean = , (Direct method)
∑fi
where x = class marks and f = frequencies
i i

∑ fidi
ii. Mean = a + ∑fi
,
(Assumed mean method)
where di = xi – a, a = assumed mean
∑ fiui
iii. Mean = a + h ∑fi ,

m .
co m
(Step deviation method)

. s e
xi – a

m a
where ui = ℎ , h = class size

se g l
l a a
ag For more Question Papers, Sample Papers, Notes & Syllabus visit Page 41 of 113

f1 −f0
iv. Mode = l + 2f1 −f0−f2 x h,
Where l = lower limit of the modal class
h= class size
f1 = frequency of modal class
f0 = frequency of the class preceding the modal class
f1 = frequency of the class succeeding the modal class
n
−c
v. Median = l + 2 f x h,
Where l = lower limit of the median class
f = frequency of median class
𝒏
𝟐
= half of the cumulative frequency
c = cumulative frequency preceding the median class

207. The empirical relationship between mean, (A) Mean
mode and median in asymmetrical distribution is: (B) Mode
(C) Median
(A) Mode = 3 Median – 2 Mean
(D) Standard deviation
(B) Mode = 3 Median + 2 Mean
Ans: (D) Standard deviation
(C) Mode = 2 Median – 3 Mean
(D) Mode = Median – 2 Mean
212. The sum of the values of all the
Ans. (A) (MBOSE 2025 Suppl)
observations divided by the total number of
observations is called:
208. Mode is:
(A) Mean (B) mode
(A) Least frequent value
(C) median (D) Frequency
(B) Middle most value
Ans. (A)
(C) Most frequent value
(D) None of the above
213. In a frequency distribution, the class having
Ans. (C)
the maximum frequency is called:
(A) Class mark (B) class size
209. The cumulative frequency table is useful in
(C) Median class (D) modal class
determining the:
Ans. (D) (MBOSE 2025)
(A) Mean (B) Median
(C) Mode (D) All of the above
214. The wickets taken by a bowler in 10 cricket
Ans. (B)
matches are as follows: 2, 6, 4, 5, 0, 2, 1, 3,
2, 3
210. The difference between the highest and
lowest values of the observations is called Then the mode of the data is:
(A) Frequency (A) 3 (B) 2 (C) 4 (D) 5
(B) Range Ans. (B)
(C) Mean
(D) Class interval
215. In the formula of mode, f₁ stands for
Ans: (B) Range
(A) Lowest frequency
211. Which of the following is not a measure of
(B) Frequency of the class succeeding
central tendency?
the modal class
(C) Frequency of the class preceding the

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m
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c. o modal class

e m
em l a s
s g
median class. Then the formula for finding
(D) Frequency of the modal class

Ans: (D) Frequency of the modal class
a
median is median equals:
(A) l + (N/2 - c)h/f (B) l - (N/2 - c)h/f
(C) l + (N/2 + c)h/f (D) l + (N/2 )h/f
Ans. (C)
216. The middlemost observation of every data
arranged in order is called:
222. The median of the following data:

m
(A) Mode (B) Median

c. o
6, 8, 15, 16, 9, 22, 21, 25, 18 is:

m
(C) Mean (D) Deviation

.co e m
(A) 21 (B) 18 (C) 16 (D) 9

s
Ans. (B)

em la
Ans. (C)

l a s a g
g (B) 5
217. The mean of first 10 natural numbers is:

a
223. Mode is
(A) 5.5 (C)6 (D)10
Ans. (A) (A) Average value
(B) Most frequent value
(C) Middle value
218. The difference between the minimum and (D) Difference of values
maximum values of the data is called: Ans. (B) Most frequent value
(A) Class limits (B) class interval

o m
c central tendency?
(C) Class size (D) range of data

.
224. Which of the following is not the measure of

em
Ans.(D) (MBOSE 2025 Suppl)

l a s (A) Mean (B) mode
219. The value of ½ (Lower Limit + Upper
Limit) of any class interval is called: a g (C) median (D) Standard deviation
(A) Class size (B) class limits Ans. (D)
(C) class mark (D) class interval 225. The formula for finding mean by direct
Ans. (C) (MBOSE 2025) method is equal to:
(A) ∑fi xi/∑fi (B) ∑fi xi/fi
220. The median class is (C) ∑ xi/∑fi (D) ∑fi xi

m
co
(A) Class with highest frequency Ans. (A)

m .
c. o m
(B) Class containing median

se
(C) Class whose cumulative frequency just

s em la
exceeds N/2

la
(D) First class
a g
ag Ans: (C) Class whose cumulative frequency
just exceeds N/2

221. If l = lower limit of the median class,
N/2 = half of the cumulative frequency
c = cumulative frequency of the preceding class,
h= height of each class, f = frequency of the

m .
.co s e m
e m l a
l a s a g
ag For more Question Papers, Sample Papers, Notes & Syllabus visit Page 43 of 113

For more Question Papers, Sample Papers, Notes & Syllabus visit Page 44 of 113

m
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c. o

e m
em l a s
s g
PROBABILITY
Important points / Summary:
1. OUTCOMES: The observations of any experiment are called its outcomes.
a
2. Sample space: The set of all possible
outcomes of an experiment is called a sample space.
e.g. {1, 2, 3, 4, 5, 6 } is the sample space for an unbiased die.
3. RANDOM EXPERIMENT: An experiment whose outcomes cannot be predicted in advance is
called a random experiment.

m
c. o
4. EVENT: The possible outcomes of a trial are called Events.

m
co
5. Probability: The theoretical probability of an event ‘E’ , written as P(E) is defined as

. e m
m la s
Num,ber of outcomes favourable to E

e
P (E) =

s g
𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑎𝑙𝑙 𝑝𝑜𝑠𝑠𝑖𝑏𝑙𝑒 𝑜𝑢𝑡𝑐𝑜𝑚𝑒𝑠

l a a
6. The probability that the event E will not happen, denoted by E’ is given by

ag P (E ) = ’ Num,ber of unfavourable outcomes to E
𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑎𝑙𝑙 𝑝𝑜𝑠𝑠𝑖𝑏𝑙𝑒 𝑜𝑢𝑡𝑐𝑜𝑚𝑒𝑠
’
7. Sum of both P (E) and P (E ) is always unity. i. e. P (E) + P (E’) = 1
 P (E’) = 1 - P (E)
8. The probability of a sure event (or certain event) is 1
i.e. if P (E) = 1, E is a sure event
9. The probability of an impossible event is 0
i.e. if P (E) = 0, E is an impossible event

m (A) 5/14 (B) 11/14 (C) 3/14 (D)1/3
. co
em
226. A die is thrown once. The probability of

a s
getting a number less than 3 is: Ans. (C)
(A)1/2
Ans. (B)
(B) 1/3
(MBOSE 2025) ag
(C)1/4
l(D) 1/6 232. If the probability of a player winning a game
is 0. 79, then the probability of his losing the
227. A die is thrown once. The probability of same game is:
getting a prime number is: (A) 1. 79 (B) 0. 31
(A) 1/2 (B) 1/3 (C) 2/3 (D) 1/6 (C) 0. 21% (D) 0. 21
Ans. (A) 1/2 Ans. (D) 0. 21
228. The probability of a sure Event is: 233. A die is thrown once, then the probability of

o m
mAns. (C) (MBOSE 2025 Suppl) c
(A) 1/6 (B) 1/2 .(C) 1/3
(A) 0 (B) 1/2 (C) 1 (D) non-existent getting an even prime number is:

c.229.o The probability of an impossible Event is: e m
(D) 2/3

em (A) 0 (B) 1/2 (C) 1 (D) non-existent s
la at random from the
s
Ans. (A) (MBOSE 2025 Suppl)

la g
g letters ofathe words ‘MATHEMATICS’ then
234. A letter is selected

a Ans. (A)
230. If P (E) = 0.05, then P (E’) equals: the probability of getting the letter M is:
(A) 0.05 (B) 0.5 (C) 0.9 (D) 0.95 (A) 2/11 (B) 6/11 (C) 4/11 (D) 5/11
Ans. (D) Ans. (A)
231. A jar contains 6 red, 5 black, and 3 green 235. In a musical chair game, the person playing
marbles of equal size. The probability that a the music has been advised to stop playing

m .
co
randomly drawn marble would be green in the music at any time within 2 minutes after

. e m
m s
colour is: the play starts. Then the probability that the

se g l a
l a a
ag For more Question Papers, Sample Papers, Notes & Syllabus visit Page 45 of 113

music will stop within one and half minutes 238. Which of the following numbers best
after starting the game is: expresses the probability of a man dying before
completing 180 years of age?
(A) 1/2 (B) 1/4 (C) 3/4 (D) 1/3
(A) 0.5 (B) 0.6
Ans. (C)
236. If two coins are tossed at the same time, the (C) 0.2 (D) 1
probability of getting tail on the top face of both the Ans: (D) 1
coins is
(A) ¾ (B) 1/2
239. A card is drawn at random from a well
(C) ¼ (D) 1 shuffled deck of 52 playing cards. What is the
Ans: (C) 1/4 probability of getting a card of club or a queen?

237. The probability that a number selected at (A) 1/13 (B) 4/13
(C) 2/13 (D) 1/52
random from the numbers 3, 4,5,6,7, 8, 9 is a
multiple of 4 is: Ans. (B) 4/13

(A) 1/7 (B) 2/7 (C) 5/7 (D) 6/7
Ans. (B) (MBOSE 2025 Suppl)

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m
m .co
c. o

e m
em l a s
s a g

m
m c. o
.co s e m
em la
l a s a g
ag

m
. co
s em
g l a
a

m
m .co
c. o se
m
s em g la
gla a
a

m .
.co s e m
e m l a
l a s a g
ag For more Question Papers, Sample Papers, Notes & Syllabus visit Page 47 of 113

Section-B

Very Short Answer Questions (2 Marks)

1. Express 3825 as a product of its prime 2|510 2|92
factors 3|255 2|46
5 |85 |23
17

510 = 2 x 3 x 5 x 17
92 = 22 x 23
∴HCF =2
𝑃𝑟𝑜𝑑𝑢𝑐𝑡 𝑜𝑓 𝑡𝑤𝑜 𝑛𝑢𝑚𝑏𝑒𝑟𝑠
LCM= 𝐻𝐶𝐹
510 𝑥 92
= 2

= 23460
5.Given that HCF (306, 657) = 9, find the
2. Find the least Number which is exactly
LCM (306, 657). (MBOSE 2025)
divisible by 18, 24 and 36.
We know,
HCF X LCM = Product of two numbers
2|18 2 |24 2|36
 9 x L CM = 306 X 657
3| 9 2|12 2|18 306 𝑋 657
3 2 |6 3 |9  LCM= 9
|3 3  LCM = 34 X 657 = 22338
2
15 = 2 x 3
24 = 23 x 3 6. Prove that 7√5 is irrational.
30 = 22 x 32
Ans Let us assume, to the contrary, that 7√5 is
∴ L C M = 23 x 32 = 72
rational
∴ The required number = 72
𝑎
7√5 = 𝑏 , where a and b are
3. Find the LCM of the smallest two digit
integers and b ≠ 0
number and the smallest composite
number. 7√5 b = a
Ans The smallest two digit number = 10 𝑎
√5 = 7𝑏
The smallest composite number = 4
𝑎
But 7𝑏 is rational and so √5 is rational
 10 = 2 × 5

4 = 22 But this contradicts the fact that √5 is
irrational
 LCM (10, 4) = 2 × 2 × 5
Hence, 7√5 is irrational
= 20
Hence Proved. (MBOSE 2025)
4. Find the HCF and LCM of 510 and 92 by
applying the fundamental theorem of 7. Find the HCF and LCM of 21 and 315 by
Arithmetic prime factorisation method.

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m
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c. o

em
em l a s
s g
3 |21 3|315 Soln:- Here, to find the required smallest
|7 3|105
5 |35 a
number, we will find the LCM of 306
and 657.
|7
So, 306 = 2 × 32 × 17
21 = 3 x 7
315 = 32 x 5 x 7 657 = 32 × 73
H C F = 3 x 7 = 21
∴ L C M = 32 x 5 x 7 = 315 ∴ 𝐿𝐶𝑀 = 2 × 32 × 17 × 73 = 22338

m
c. o
Hence, the required smallest number is

m
c. o
22338. (Ans)

m
𝑛
8. Check weather 6 can end with the digit 0

m the digit zero, then it s e
for any natural number 𝑛.

e la
12. There is a circular path around a sports

a s g
𝑛 field. Sonia takes 18 minutes to drive one

a
Ans. If 6 ends with

l
g prime factorization of 6 must
round of the field, while Ravi takes 12

athe
should be divisible by 10
minutes for the same. Suppose they both
𝑛 start from the same point and at the same
So,
contain the primes 2 and 5. This is not time, and go in the same direction. After
possible as the only primes in the factorisation how many minutes will they meet again at
of 6𝑛 are 2 and 3. Thus, there is no value of 𝑛 the starting point?
in natural numbers for which 6𝑛 ends with the Soln: To find the time after which they
digit zero. meet again at the starting point, we have to

m We have,
find the LCM of 18 and 12 minutes.
9. Find the HCF and LCM of 12, 15 and 21

. co
by applying the prime factorisation

em
method.

a s
18 = 2 × 32

l
Soln:-
2|12
2 |6
3|15
|5
3|21
|7 a g 12 = 22 × 3

|3 ∴ 𝐿𝐶𝑀 = 22 × 32

12 = 22 x 3 = 4 × 9 = 36
15 = 3 x 5
Hence, they will meet again at the starting
21 = 3 x 7
point after 36 minutes. (Ans)

m
∴ HCF=3

co
13. Show that 5 − √3 is an irrational number.

m bells ring at intervals of 4, 7, 12, and .
And LCM = 22 x 3 x 5 x 7 = 420

c. o 10. Four m
Soln:- Let us assume that 5 − √3 is a rational

m se
gla
number.

s e 14 seconds respectively. If they begin

l a a
𝑝
Now, 5 − √3 = 𝑞 , where 𝑝 & 𝑞 are

ag
ringing together at 12 o’clock, when will
they next ring together? integers, having no common factor except 1
n
and 𝑞 ≠ 0.
Sol :- 𝑝
 5 − = √3
𝑞
LCM of 4, 7, 12, and 14 = 84 5𝑞−𝑝
 √3 =
𝑞
5𝑞−𝑝
∴ The bell will ring together after 84 seconds or Since 𝑞 is rational number as 𝑝 & 𝑞 are
1min 24seconds or at 12:01:24secs integers.
11. Write the smallest number which is ∴ √3 is also a rational number which is a

m .
contradiction.

co
divisible by both 306 and 657.

. e m
Thus, our assumption is wrong.

e m l as
l a s a g
ag For more Question Papers, Sample Papers, Notes & Syllabus visit Page 49 of 113

Hence, 5 − √3 is an irrational number. Hence, the required angles are
(Ans) 99° 𝑎𝑛𝑑 81°.

14. Is the following pair of linear equations 17. Check whether the system of linear
consistent? Justify your answer. equations 5𝑥 − 4𝑦 + 8 = 0, 7𝑥 + 6𝑦 −
2𝑎𝑥 + 𝑏𝑦 = 𝑎, 4𝑎𝑥 + 2𝑏𝑦 = 2𝑎; 𝑎, 𝑏 ≠ 0. 9 = 0 intersect at a point or not.
Soln: Yes.
𝑎 2𝑎 1 Soln:- Here,
Here, 𝑎1 = 4𝑎 = 2
2
𝑏1𝑏 1
𝑏2
= 2𝑏 = 2 𝑎1 = 5, 𝑏1 = −4, 𝑐1 = 8
𝑐1 −𝑎 1
𝑐2
= −2𝑎 = 2 𝑎2 = 7, 𝑏2 = 6, 𝑐2 = −9
𝑎 𝑏 𝑐
Since, 𝑎1 = 𝑏1 = 𝑐1
2 2 2 𝑎1 5
=
∴ The given system of linear equations is 𝑎2 7
consistent
𝑎 𝑏 𝑐 𝑏1 −4 −2
15. On comparing the ratios 𝑎1 , 𝑏1 𝑎𝑛𝑑 𝑐1 , 𝑏2
= 6 = 3
2 2 2
find out whether the lines representing the
𝑐1 −8
following pair of linear equations intersect =
𝑐2 9
at a point, are parallel or coincident
5𝑥 − 4𝑦 + 8 = 0 𝑎 𝑏
∵ 𝑎1 ≠ 𝑏1
2 2
7𝑥 + 6𝑦 − 9 = 0
 The given system of linear equations
Soln:- We have, intersects at a point.
5𝑥 − 4𝑦 + 8 = 0 ----------(i)
7𝑥 + 6𝑦 − 9 = 0 ---------(ii) 18. Find out whether the lines representing a
Here, 𝑎1 = 5, 𝑏1 = −4, 𝑐1 = 8 pair of linear equations 2𝑥 − 3𝑦 − 8 = 0,
𝑎2 = 7, 𝑏2 = 6, 𝑐2 = −9 4𝑥 − 6𝑦 − 9 = 0 has a unique solution or
𝑎1 5 𝑏 −4 −2
So, 𝑎 = 7 𝑎𝑛𝑑 𝑏1 = 6 = 3 not.
2 2
𝑎 𝑏
Since, 𝑎1 ≠ 𝑏1 Soln:- Here,
2 2
∴ Equations (i) and (ii) represent
intersecting lines. (Ans) 𝑎1 = 2, 𝑏1 = −3, 𝑐1 = −8

16. The larger of two supplementary angles 𝑎2 = 4, 𝑏2 = −6, 𝑐2 = −9
exceeds the smaller by 18 degrees. Find
𝑎1 2 1
them 𝑎2
= 4= 2

Soln:- Let the supplementary angles be 𝑏1 −3 1
= −6 = 2
𝑥 𝑎𝑛𝑑 𝑦 ( 𝑥 > 𝑦). 𝑏2
Now, we are given that
𝑐1 −8 8
𝑥 + 𝑦 = 180° --------- (i) = =
𝑐2 −9 9
& 𝑥 − 𝑦 = 18°
 𝑥 − 18° = 𝑦 ----------- (ii) 𝑎 𝑏 𝑐
Substituting 𝑦 = 𝑥 − 18° in equation (i), Here, 𝑎1 = 𝑏1 ≠ 𝑐1
2 2 2
we have
𝑥 + 𝑦 = 180°  The given system of linear equations does not
 𝑥 + 𝑥 − 18° = 180° have a unique solution.
 2𝑥 = 180° + 18°
 2𝑥 = 198° 19. Determine whether the following system
198° of linear equations is consistent or
 𝑥=
2
 𝑥 = 99° inconsistent.
Then, from equation (ii), we have 2𝑥 − 3𝑦 = 8
𝑦 = 𝑥 − 18°
 𝑦 = 99° − 18° 4𝑥 − 6𝑦 = −5
 𝑦 = 81°

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m
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c. o Soln:-

e m
em l a s
s g
Here, ⇨ 𝑦 = 35 − 24

𝑎1 = 2, 𝑏1 = −3, 𝑐1 = −8 aHence, the required numbers are
⇨ 𝑦 = 11

𝑎2 = 4, 𝑏2 = −6, 𝑐2 = 5
𝑎1 2 1 24 𝑎𝑛𝑑 11.
= =
𝑎2 4 2
𝑏1 −3 1 22. Solve the system of linear equations 𝑥 +
= −6 = 2
𝑏2 𝑦 = 14, 𝑥 − 𝑦 = 4 by substitution
𝑐1 −8
= 5 method.
𝑐2

m
Soln :- 𝑥 + 𝑦 = 14 -------------(i)

c. o
𝑎 𝑏 𝑐
Here, 𝑎1 = 𝑏1 ≠ 𝑐1

m
co
2 2 2
𝑥−𝑦 =4

m
-------------(ii)

m .
 The given system of linear equations is
s e
e la
 𝑥 = 4+𝑦 ………….(iii)

s g
inconsistent.

a
20. Checklwhether the following system of a
ag equations 9𝑥 + 3𝑦 + 12 =
Substituting the value of 𝑥 in eqn (i),
we get
linear
0, 18𝑥 + 6𝑦 + 24 = 0 represent co- 𝑥 + 𝑦 = 14
incident lines or not.
 4 + 𝑦 + 𝑦 = 14 (from (iii))
Soln:-:- Here,
 4 + 2𝑦 = 14
𝑎1 = 9, 𝑏1 = 3, 𝑐1 = 12
 2𝑦 = 14 − 4
𝑎2 = 18, 𝑏2 = 6, 𝑐2 = 24

om
𝑎1 9 1
 2𝑦 = 10

c
= 18 = 2

.
𝑎2

em
𝑏1 3 1 10
= 6= 2  𝑦= 2

s
𝑏2

l a
𝑐1 12 1
= 24 = 2

g
𝑐2  𝑦=5
𝑎
Here, 1 = 1 =
𝑎2
𝑏
𝑏2
𝑐1
𝑐2
a Putting 𝑦 = 5 in eqn (i), we get

𝑥 + 𝑦 = 14
 The given system of linear equations represents
co-incident lines.  𝑥 + 5 = 14

21. Sum of the two numbers is 35 and their  𝑥 = 14 − 5
difference is 13. Find the numbers.  𝑥=9

o m
c
Soln.:-Let the two numbers be 𝑥 𝑎𝑛𝑑 𝑦, where

m 23. Total number of legs .in a herd of goats and
𝑥 = 9 and𝑦 = 5

c. o em this situation in the
𝑥 > 𝑦.

em s
la equation in two variables.
Now, According to the question,

s g
hens is 40. Represent

la
we have

Soln.: Let a
form of a linear

ag 𝑥 + 𝑦 = 35
𝑥 − 𝑦 = 13
(i)
(ii)
the number of goats and hens be
𝑥 𝑎𝑛𝑑 𝑦 respectively.

Adding equations (i) & (ii), we get Since, Number of legs of 1 goat = 4

2𝑥 = 48 ∴ Number of legs of 𝑥 goats = 4𝑥
48 Since, Number of legs of 1 hen = 2
⇨ 𝑥= 2
⇨ 𝑥 = 24 ∴ Number of legs of 𝑦 hens = 2𝑦

o m m .
24 + 𝑦 = 35.c e
Substituting 𝑥 = 24 in equation (i), we get Now, according to the question, we have

em l as
s
4𝑥 + 2𝑦 = 40

la a g
a g For more Question Papers, Sample Papers, Notes & Syllabus visit Page 51 of 113

⇨ 2𝑥 + 𝑦 = 20 (Dividing both sides by The coach of a cricket team buys 7 bats
2) and 6 balls for ₹ 3800. Later, she buys 3
bats and 5 balls for ₹ 1750.
Which is the required linear equation in
two variables. Soln:- Let the cost of each bat be ₹ 𝑥 and the cost
24. The difference between two numbers is 26 of ball be ₹ 𝑦
and one number is three times the other.
Find them.  The cost of 7 bats is ₹ 7𝑥 and the cost of 6
balls is ₹ 6𝑦 and their total cost is ₹ 3800
Soln : Let the two number be 𝑥 and 𝑦 (𝑥 > 𝑦)
7𝑥 + 6𝑦 = 3800
 According to the question,
Again, the cost of 3 bats is ₹ 3𝑥 and the cost of 5
𝑥 − 𝑦 = 26 ------------(i) balls is ₹ 5𝑦 and their cost is ₹ 1750
𝑥 = 3𝑦 ------------(ii)  3𝑥 + 5𝑦 = 1750
Putting 𝑥 = 3𝑦 in eqn (i), we get 27. Find the value of 𝑘 for which the equation
𝑥 2 + 𝑘(2𝑥 + 𝑘 − 1) + 2 = 0 has real and
𝑥 − 𝑦 = 26 equal roots.
3𝑦 − 𝑦 = 26 Soln: Given quadratic equation is –
𝑥 2 + 𝑘(2𝑥 + 𝑘 − 1) + 2 = 0
 2𝑦 = 26
 𝑥 + 2𝑘𝑥 + (𝑘 2 − 𝑘 + 2) = 0
2

Here,𝑎 = 1, 𝑏 = 2𝑘 𝑎𝑛𝑑 𝑐 = (𝑘 2 − 𝑘 + 2)
2613
𝑦 = For equal roots, we have \
21 𝐷𝑖𝑠𝑐𝑟𝑖𝑚𝑖𝑛𝑎𝑛𝑡 = 0
 𝑏 2 − 4𝑎𝑐 = 0
𝑦 = 13  (2𝑘)2 − 4 × 1 × (𝑘 2 − 𝑘 + 2) = 0
 4𝑘 2 − 4𝑘 2 + 4𝑘 − 8 = 0
Putting 𝑦 = 13 in eqn (ii), we get  4𝑘 − 8 = 0
 4𝑘 = 8
𝑥 = 3𝑦 8
 𝑘=4
 𝑘=2
𝑥 = 3 × 13
Hence, the required value of 𝑘 𝑖𝑠 2.
𝑥 = 39 2
28. If 𝑥 = 3 𝑎𝑛𝑑 𝑥 = −3 are roots of the
 The two numbers are 39 and 13 quadratic equation 𝑎𝑥 2 + 7𝑥 + 𝑏 = 0,
find the values of 𝑎 𝑎𝑛𝑑 𝑏.
25. The cost of a note book is thrice the cost of Soln:-Let us assume the quadratic equation be
a pen. Write this situation as a linear 𝑎𝑥 2 + 𝑏𝑥 + 𝑐 = 0
equation in two variables. −𝑏
Now, Sum of the roots = 𝑎
2 −7
Soln.:- Let the cost of one notebook and one  3
+ (−3) = 𝑎
pen be 𝑅𝑠 𝑥 𝑎𝑛𝑑 𝑅𝑠 𝑦. 
2−9
= 𝑎
−7
3
−7 −7
Now, by the question, we have  =
3 𝑎
(−7)×3
𝑥 = 3𝑦  𝑎 = (−7)

⇨ 𝑥 − 3𝑦 = 0  𝑎=3
𝑐
And, Product of the roots = 𝑎
Which is the required linear equation in 2 𝑏
 × (−3) = 𝑎
two variables. 3
𝑏
 −2 =
26. From a pair of linear equations in the 𝑎
 −2𝑎 = 𝑏
given problem: (MBOSE 2025)
 −2 × 3 = 𝑏

For more Question Papers, Sample Papers, Notes & Syllabus visit Page 52 of 113

m
m .co
c. o

e m
em l a s
s g
 −6 = 𝑏 9
𝑥 = 0, 2
a
 𝑏 = −6
Hence, 𝑎 = 3 𝑎𝑛𝑑 𝑏 = −6.
32. Check whether (𝑥 + 4)(𝑥 − 4) =
29. Find the roots of the quadratic equation 𝑥(𝑥 + 2) + 8 is a quadratic equation or
4𝑥 2 + 4√3𝑥 + 3 = 0 by applying the not.
quadratic formula. Soln:- (𝑥 + 4)(𝑥 − 4) = 𝑥(𝑥 + 2) + 8
Soln: We have, 4𝑥 2 + 4√3𝑥 + 3 = 0
Here, 𝑎 = 4, 𝑏 = 4√3 𝑎𝑛𝑑 𝑐 = 3  (𝑥)2 − (4)2 = 𝑥 2 + 2𝑥 + 8
∴ 𝐷𝑖𝑐𝑟𝑖𝑚𝑖𝑛𝑎𝑛𝑡, 𝐷 = 𝑏 2 − 4𝑎𝑐
m
c. o
= (4√3)2 − 4 × 4 × 3  𝑥 2 − 16 − 𝑥 2 − 2𝑥 − 8 = 0

m
co m
= 48 − 48

. e


It is not in the form 𝑎𝑥 + 𝑏𝑥 + 𝑐 = 0as
=0 −2𝑥 − 24 = 0

e m
Since, 𝐷 = 0, roots exists and are equal.
l
a s g
2

𝑥= l a
Thus, using quadratic formula, we have

ag
−𝑏±√D −4√3±0 −4√3 −√3
= = =  The given equation is not a quadratic
2𝑎 2×4 8 2
equation
Hence, the roots of the given equation are
−√3 −√3 33. Find the roots of the quadratic equation
𝑎𝑛𝑑 .
2 2 2𝑥 2 + 𝑥 − 6 = 0 by factorization.
30. Find the discriminant and nature of roots Soln:- 2𝑥 2 + 𝑥 − 6 = 0
of the quadratic equation 2𝑥 2 − 3𝑥 − 2 =
0.  2𝑥 2 + 4𝑥 − 3𝑥 − 6 = 0

o m 2𝑥(𝑥 + 2) − 3(𝑥 + 2) = 0
. c  (𝑥 + 2)(2𝑥 − 3) = 0
Soln:- Here,

s em
𝑎 = 2, 𝑏 = −3, 𝑐 = −2

g l a
a
 Discriminant= 𝑏 2 − 4𝑎𝑐  Either, or,

= (−3)2 − 4 × 2 × (−2) 𝑥+2=0 2𝑥 − 3 = 0

= 9 + 16 𝑥 = −2 2𝑥 = 3
3
= 25˃ 0 𝑥 = 2

Thus, the nature of roots are real and unequal Hence the roots are −2, 2
3

m
m .co
31. Solve the quadratic equation2𝑥 2 =

o
34. Represent the given situations in the form

. c m
9𝑥.

se
of a quadratic equations.

emSoln:-
la
2 The area of a rectangular plot is 528m2.

s
2𝑥 = 9𝑥

l a a g
The length of the plot (in metres) is one

ag
 2𝑥 − 9𝑥 = 0 2
more than twice its breadth.
Soln:- Let the length (in metres) of the plot be
𝑥 (2𝑥 − 9) = 0
𝑥 and the breadth (in metres) of the plot be
 Either, or, 𝑦

𝑥=0 2𝑥 − 9 = 0  Area of the plot= (𝑥𝑦)𝑚2

 2𝑥 = 9 𝑥𝑦 = 528 ………(i)

m .
9
 Also, 𝑥 = 1 + 2𝑦 ……… (ii)

co
𝑥= 2

. e m
m as
Using (ii) in (i), we get

se g l
l a a
ag For more Question Papers, Sample Papers, Notes & Syllabus visit Page 53 of 113

(1 + 2y) y = 528  x = 9000/5
 2y2 + y – 528 = 0, is a quadratic equation
 x = 1800
35. Find the nature of the quadratic
Thus, cost of one bat = ₹ 1800
equation𝑥 2 + 7𝑥 + 10 = 0.

Soln:-The quadratic equation is 𝑥 2 + 7𝑥 + 10 = 0 38. Find two consecutive positive numbers
whose square have the sum 85.
Here,
Soln:-Let the two consecutive positive numbers
𝑎 = 1, 𝑏 = 7, 𝑐 = 10 be 𝑥 𝑎𝑛𝑑 𝑥 + 1

Discriminant (𝐷) = 𝑏 2 − 4𝑎𝑐  According to the question,

= (7)2 − 4 × 1 × 10 𝑥 2 + (𝑥 + 1)2 = 85

= 49 − 40  𝑥 2 + 𝑥 2 + 2𝑥 + 1 = 85

=9  2𝑥 2 + 2𝑥 + 1 − 85 = 0

∵𝐷 > 0  2𝑥 2 + 2𝑥 − 84 = 0

 The given equation has two real and  2(𝑥 2 + 𝑥 − 42) = 0
distinct roots
 𝑥 2 + 𝑥 − 42 = 0
36. The cost of a note book is thrice the cost of
a pen. Write this situation as a linear  𝑥 2 + 7𝑥 − 6𝑥 − 42 = 0
equation in two variables.
 𝑥(𝑥 + 7) − 6(𝑥 + 7) = 0
Soln.:- Let the cost of one notebook and one
pen be 𝑅𝑠 𝑥 𝑎𝑛𝑑 𝑅𝑠 𝑦. (𝑥 + 7)(𝑥 − 6) = 0

Now, by the question, we have  Either, or,

𝑥 = 3𝑦 𝑥+7=0 𝑥−6=0
⇨ 𝑥 − 3𝑦 = 0 𝑥 = −7 𝑥 = 6
37. Which is the required linear equation in
two variables. The coach of a cricket team (Rejected)
buys 3 bats and 2 balls for ₹ 5700. Later
 The two positive numbers are 6 and 7
he buys 2 bats and 3 balls for ₹ 4050. Find
the cost of one bat? 39. In the adjoining figure, 𝐷𝐸 ∥ 𝐵𝐶. Find 𝐸𝐶
Soln:- Let ₹x and ₹y be the cost of each where 𝐴𝐷 = 1.5𝑐𝑚, 𝐷𝐵 =
bat and each ball respectively 3𝑐𝑚 𝑎𝑛𝑑 𝐴𝐸 = 1𝑐𝑚.
(MBOSE 2025 Suppl)
 By question,

3𝑥 + 2𝑦 = 5700 ------------(i) x 3

Also, 2𝑥 + 3 𝑦 = 4050 … … … … . . (𝑖𝑖) x 2

Multiplying (i) by 2 and (ii) by 3 we get

9 x + 6 y = 17100

4 x + 6y = 8100

Subtracting, 5x = 9000

For more Question Papers, Sample Papers, Notes & Syllabus visit Page 54 of 113

m
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c. o 40. In ∆𝐴𝐵𝐶, 𝐷 𝑎𝑛𝑑 𝐸 are points on the

e m
em l a s
s sides AB and AC respectively such that
𝐴𝐷 2 a g
Soln:- ∆𝐴𝐵𝐶, in
𝐷𝐸 ∥ 𝐵𝐶. If = and 𝐴𝐸 = 7.2𝑐𝑚, find
𝐷𝐵 3 which D is the mid-
AC.
point of AB and
Soln:- ∵ 𝐷𝐸 ∥ 𝐵𝐶 𝐷𝐸 ∥ 𝐵𝐶

 By Basic To Prove:- 𝐴𝐸 = 𝐸𝐶
Proportionality
Proof: In ∆𝐴𝐵𝐶,

m
Theorem,

m c. o
co
∵ 𝐷𝐸 ∥ 𝐵𝐶

m
𝐴𝐷 𝐴𝐸

. e
=

s
𝐷𝐵 𝐸𝐶

m la
 By Basic Proportionality Theorem

s e g
2 7.2 𝑐𝑚
 3 = 𝐸𝐶

l a a
𝐴𝐷 𝐴𝐸

ag
3 × 7.23.6 = --------------(i)
𝐸𝐶 = 21
𝑐𝑚 𝐷𝐵 𝐸𝐶

𝐸𝐶 = 10.8 𝑐𝑚 ∵ D is the mid-point of AB
𝐴𝐶 = 𝐴𝐸 + 𝐸𝐶
 AD = DB -------------(ii)
= 7.2 𝑐𝑚 + 10.8 𝑐𝑚
= 18 𝑐𝑚  From eqn (i) and (ii), we get
41. In the adjoining figure, 𝐷𝐸 ∥ 𝐴𝐶 and 𝐷𝐸 ∥
𝐵𝐹 𝐵𝐸 𝐴𝐷 1 𝐴𝐸
𝐴𝐸. Prove that 𝐹𝐸 = 𝐸𝐶 . =
𝐴𝐷1 𝐸𝐶

. c om
𝐴𝐸
1 = 𝐸𝐶

em
las
𝐴𝐸 = 𝐸𝐶

ag
𝐷𝐸 bisects 𝐴𝐶

Hence Proved

43. S and T are points on side PR and QR of
Soln:- In ∆𝐴𝐵𝐶,
∆𝑃𝑄𝑅 such that ∠𝑃 = ∠𝑅𝑇𝑆. Show that
𝐷𝐸 ∥ 𝐴𝐶 ∆𝑅𝑃𝑄~∆𝑅𝑆𝑇.
Soln:- In ∆𝑅𝑃𝑄 𝑎𝑛𝑑 ∆𝑅𝑇𝑆
 By Basic Proportionality Theorem

o m
m c
. 𝑎𝑛𝑔𝑙𝑒)
𝐵𝐷 𝐵𝐸 ∠𝑃 = ∠𝑅𝑇𝑆 (𝐺𝑖𝑣𝑒𝑛)

c. o Again, in ∆𝐴𝐵𝐸,
= 𝐸𝐶 ------------(i)

m
e[𝐴𝐴 𝑆𝑖𝑚𝑖𝑙𝑎𝑟𝑖𝑡𝑦]
𝐷𝐴

s
∠𝑃𝑅𝑄 = ∠𝑆𝑅𝑇 (𝐶𝑜𝑚𝑚𝑜𝑛

s em g l a
a
∆𝑅𝑃𝑄~∆𝑅𝑇𝑆
l Hencea
ag
𝐷𝐹 ∥ 𝐴𝐶
Showed
 By Basic Proportionality Theorem
𝐵𝐷 𝐵𝐹 44. In the adjoining figure, ∆𝑂𝐴𝐵~∆𝑂𝐶𝐷.
= 𝐹𝐸 ------------(ii)
𝐷𝐴 When 𝐴𝐵 = 8𝑐𝑚, 𝐵𝑂 = 6.4𝑐𝑚, 𝑂𝐶 =
 From eqn (i) and (ii), we get 3.5𝑐𝑚 𝑎𝑛𝑑 𝐶𝐷 = 5𝑐𝑚, find 𝑂𝐴 𝑎𝑛𝑑 𝐷𝑂.

𝐵𝐹 𝐵𝐸
𝐹𝐸
= 𝐸𝐶 Hence Proved

m .
co
42. Prove that the line drawn from the mid-point

. e m
of one side of a triangle parallel to another

e m as
side bisects the third side.

a s g l
l a
ag For more Question Papers, Sample Papers, Notes & Syllabus visit Page 55 of 113

Soln:- ∵∆𝑂𝐴𝐵~∆𝑂𝐶𝐷
𝐴𝐵 𝑂𝐴 𝑂𝐵 Soln:- A D
 = =
𝐶𝐷 𝑂𝐶 𝑂𝐷

8 𝑂𝐴 6.4 𝑐𝑚
5 = 3.5 𝑐𝑚 = 𝑂𝐷

8 𝑂𝐴
Taking, =
5 3.5 𝑐𝑚

5 × 𝑂𝐴 = 8 × 3.5 𝑐𝑚

8 × 3.50.7 𝑐𝑚 In ∆𝑠 𝑃𝐴𝐵 𝑎𝑛𝑑 𝑃𝐷𝐶
 𝑂𝐴 = 51

∠𝐴 = ∠𝐷 (𝐸𝑎𝑐ℎ 900 )
 𝑂𝐴 = 5.6 𝑐𝑚
∠𝐴𝑃𝐵 = ∠𝐷𝑃𝐶 (𝑉𝑒𝑟𝑡𝑖𝑐𝑎𝑙𝑙𝑦 𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒 𝑎𝑛𝑔𝑙𝑒𝑠)
8 6.4 𝑐𝑚
Taking, 5 = 𝑂𝐷
∆𝑃𝐴𝐵~∆𝑃𝐷𝐶 (𝐴𝐴 𝑆𝑖𝑚𝑖𝑙𝑎𝑟𝑖𝑡𝑦)
8 × 𝑂𝐷 = 5 × 6.4 𝑐𝑚 𝐴𝑃 𝑃𝐵
 =
𝐷𝑃 𝑃𝐶
5 × 6.40.8 𝑐𝑚
 𝑂𝐷 = 81
𝐴𝑃 × 𝑃𝐶 = 𝐷𝑃 × 𝑃𝐵
 𝑂𝐷 = 4 𝑐𝑚 Hence Proved
45. In the adjoining figure, ∠𝐴𝑃𝑄 = ∠𝐵. Prove 47. X and Y are points on side PQ and PR of a
that ∆𝐴𝑃𝑄~∆𝐴𝐵𝐶. If 𝐴𝑃 = 3.8𝑐𝑚, 𝐴𝑄 = ∆𝑃𝑄𝑅. If PX = 2cm, XQ = 4cm and 𝑋𝑌 ∥
3.6𝑐𝑚, 𝐵𝑄 = 2.1𝑐𝑚 𝑎𝑛𝑑 𝐵𝐶 = 𝑄𝑅, prove that 3𝑋𝑌 = 𝑄𝑅.
4.2𝑐𝑚 𝑓𝑖𝑛𝑑 𝑃𝑄.
Soln:- In ∆𝑃𝑋𝑌 𝑎𝑛𝑑 ∆𝑃𝑄𝑅
Soln:- In ∆𝐴𝐵𝐶 𝑎𝑛𝑑 ∆𝐴𝑃𝑄,
∠𝑃 = ∠𝑃 (𝐶𝑜𝑚𝑚𝑜𝑛 𝑎𝑛𝑔𝑙𝑒)
∠𝐴𝑃𝑄 = ∠𝐵 (𝐺𝑖𝑣𝑒𝑛)
∠𝑄 = ∠𝑃𝑋𝑌 (∵ 𝑋𝑌 ∥ 𝐵𝐶)
∠𝐴 = ∠𝐴 (𝐶𝑜𝑚𝑚𝑜𝑛 𝑎𝑛𝑔𝑙𝑒)
∆𝑃𝑋𝑌~∆𝑃𝑄𝑅 (𝐴𝐴 𝑆𝑖𝑚𝑖𝑙𝑎𝑟𝑖𝑡𝑦)
∆𝐴𝐵𝐶~∆𝐴𝑃𝑄 (𝐴𝐴 𝑆𝑖𝑚𝑖𝑙𝑎𝑟𝑖𝑡𝑦)
𝑃𝑋 𝑋𝑌
𝐴𝑃 𝑃𝑄 𝑃𝑄 = 𝑄𝑅
𝐴𝐵 = 𝐵𝐶
𝑃𝑋 𝑋𝑌
3.8 𝑃𝑄  𝑃𝑋 +𝑋𝑄 = 𝑄𝑅
3.6 +2.1 = 4.2
2 𝑋𝑌
3.8 𝑃𝑄  = 𝑄𝑅
 = 2+4
5.7 4.2
2 𝑋𝑌
5.7 × 𝑃𝑄 = 3.8 × 4.2 𝑐𝑚  6
= 𝑄𝑅

3.80.2 × 4.21.4 1 𝑋𝑌
 𝑃𝑄 = cm  3
= 𝑄𝑅
5.70.30.1

 𝑃𝑄 = 2.8 𝑐𝑚  3𝑋𝑌 = 𝑄𝑅

46. Two ∆𝑠 𝐵𝐴𝐶 𝑎𝑛𝑑 𝐵𝐷𝐶, right angled at A Hence Proved
and D respectively are drawn on the same
48. A vertical tower of length 6m casts a
base BC and on the same side of BC. If
shadow 4cm long on the ground and at the
AC and DB intersect at P, prove that AP ×
same time a tower casts a shadow 28m
PC = DP × PB.
long. Find the height of the tower.

For more Question Papers, Sample Papers, Notes & Syllabus visit Page 56 of 113

m
m .co
c. o Soln:-

e m
em l a s
s g
50. In the adjoining figure, 𝐷𝐸//𝐵𝐶. If 𝐴𝐷 =

a
𝑥, 𝐷𝐵 = 𝑥 − 2, 𝐴𝐸 = 𝑥 + 2 & 𝐸𝐶 = 𝑥 −
Let AC be the height of the tower and BC = 1, find the value of 𝑥.
28 m be the length of its shadow.
Soln: In ∆ 𝐴𝐵𝐶,
∵ 𝐷𝐸//𝐵𝐶
Using Basic Proportionality Theorem, we
have
𝐴𝐷 𝐴𝐸
𝐷𝐵
= 𝐸𝐶

m
𝑥 𝑥+2
 = 𝑥−1

c. o
𝑥−2

m
 𝑥(𝑥 − 1) = (𝑥 −

.co e m
2)(𝑥 + 2)

m s
 𝑥2 − 𝑥 = 𝑥2 − 4

la
Let DF = 6 m be height of the pole and EF = 4

s e  −𝑥 = −4
g
a
m be the length of its shadow.

l 𝑎𝑛𝑑 ∆𝐷𝐸𝐹, a
 𝑥 = 4 (Ans)

a g
In ∆𝐴𝐵𝐶
51. In the adjoining figure,
𝐴𝑂
𝑂𝐶
=
𝐵𝑂
𝑂𝐷
=
∠𝐵 = ∠𝐸 (𝑠𝑎𝑚𝑒 𝑖𝑛𝑐𝑙𝑖𝑛𝑎𝑡𝑖𝑜𝑛 𝑎𝑡 𝑡ℎ𝑒 𝑠𝑎𝑚𝑒 𝑡𝑖𝑚𝑒) 1
𝑎𝑛𝑑 𝐴𝐵 = 5 𝑐𝑚. Find the value of 𝐷𝐶.
2
0
∠𝐶 = ∠𝐹 (𝑒𝑎𝑐ℎ 90 ) Soln: In ∆ 𝐴𝑂𝐵 𝑎𝑛𝑑 ∆ 𝐶𝑂𝐷,
𝐴𝑂 𝐵𝑂
= (Given)
∆𝐴𝐵𝐶~∆𝐷𝐸𝐹 (𝐴𝐴 𝑆𝑖𝑙𝑖𝑙𝑎𝑟𝑖𝑡𝑦) 𝑂𝐶 𝑂𝐷
∠𝐴𝑂𝐵 = ∠𝐶𝑂𝐷 (Vertically
𝐴𝐶 𝐵𝐶
opposite angles)
𝐷𝐹 = 𝐸𝐹 So, by SAS Similarity Criterion, we have

m
co
∆ 𝐴𝑂𝐵~∆ 𝐶𝑂𝐷

.
𝐴𝐶 28 𝐴𝑂 𝐵𝑂 𝐴𝐵
6 𝑚 = 4  = =

em
𝑂𝐶 𝑂𝐷 𝐷𝐶

s
1 5
 = (∵ 𝐴𝐵 = 5 𝑐𝑚)

l a
287 × 6𝑚 2 𝐷𝐶
𝐴𝐶 =

g
 1 × 𝐷𝐶 = 2 × 5

a
41
 𝐷𝐶 = 10 𝑐𝑚. (Ans)
𝐴𝐶 = 42𝑚
52. In ∆𝐴𝐵𝐶, right angled at B, AB = 24cm,
 The height of the tower is 42 𝑚 BC = 7cm, find Sin A.
Soln :- By Pythagoras Theorem in ∆𝐴𝐵𝐶, we get

49. If ∆ 𝐴𝐵𝐶~∆ 𝐷𝐸𝐹 such that 2𝐴𝐵 = 𝐴𝐶 2 = 𝐴𝐵2 + 𝐵𝐶 2
𝐷𝐸 𝑎𝑛𝑑 𝐵𝐶 = 8 𝑐𝑚. Then find the length
𝐴𝐶 2 = (24𝑐𝑚)2 + (7𝑐𝑚)2

m
of 𝐸𝐹.

m .c o
𝐴𝐶 2 = 576𝑐𝑚2 + 49𝑐𝑚2

c. o 𝐴𝐶 = 625𝑐𝑚 m
s e
em
2 2

s g l a
la a
𝐴𝐶 = √625𝑐𝑚
2 2

ag Soln: Since ∆ 𝐴𝐵𝐶~∆ 𝐷𝐸𝐹 (Given)
𝐴𝐵 𝐵𝐶
𝐴𝐶 = 25𝑐𝑚
2

∴ 𝐷𝐸 = 𝐸𝐹 𝐵𝐶 7
𝐴𝐵 8
sin 𝐴 = 𝐴𝐶 = 25
 2𝐴𝐵
= 𝐸𝐹 (∵ 𝐷𝐸 = 2𝐴𝐵)
1 8
 = 𝐸𝐹 53. Find the value of 𝑠𝑖𝑛600 𝑐𝑜𝑠300 +
2
 1 × 𝐸𝐹 = 8 × 2 𝑠𝑖𝑛300 𝑐𝑜𝑠600
 𝐸𝐹 = 16 𝑐𝑚. (Ans)
Soln:- 𝑠𝑖𝑛600 𝑐𝑜𝑠300 + 𝑠𝑖𝑛300 𝑐𝑜𝑠600

c om m .
√3 √3 1 1

m. e
= × 2 +2×2

s
2

se g l a
l a a
ag For more Question Papers, Sample Papers, Notes & Syllabus visit Page 57 of 113

3
= 4+ 4
1
L H S = R H S

3+1 Hence Verified. (MBOSE 2025 Suppl)
= 4
56. Prove that (1 − 𝑠𝑖𝑛2 𝜃)𝑠𝑒𝑐 2 𝜃 = 1
4
=4
Soln:- L H S = (1 − 𝑠𝑖𝑛2 𝜃)𝑠𝑒𝑐 2 𝜃

=1 = 𝑐𝑜𝑠 2 𝜃 × 𝑠𝑒𝑐 2 𝜃
54. Prove that 2𝑐𝑜𝑠 2 300 − 1 = 𝑐𝑜𝑠600 1
= 𝑐𝑜𝑠 2 𝜃 × 𝐶𝑜𝑠2 𝜃
Soln:- L H S = 2𝑐𝑜𝑠 2 300 − 1
=1
2
√3
= 2×( ) −1 =RHS
2

3 Hence Proved
= 2×4−1

3 57. Prove that (𝑠𝑒𝑐 2 𝐴 − 1)(𝑐𝑜𝑠𝑒𝑐 2 𝐴 − 1) = 1
= 2−1
Ans L H S = (𝑠𝑒𝑐 2 𝐴 − 1)(𝑐𝑜𝑠𝑒𝑐 2 𝐴 − 1)
3−2
= 2 = 𝑡𝑎𝑛2 𝐴 × 𝑐𝑜𝑡 2 𝐴
1
= = 𝑡𝑎𝑛2 𝐴 ×
1
2
𝑡𝑎𝑛2 𝐴

= 𝑐𝑜𝑠600 =1
=RHS
=RHS
Hence Proved
Hence Proved
0 0
55. If 𝐴 = 60 , 𝐵 = 30 , verify that 3
𝑐𝑜𝑠(𝐴 + 𝐵) = 𝑐𝑜𝑠𝐴𝑐𝑜𝑠𝐵 − 𝑠𝑖𝑛𝐴𝑠𝑖𝑛𝐵 58. If 𝑠𝑖𝑛𝐴 = , calculate 𝑡𝑎𝑛𝐴
4

Soln:- L H S = 𝑐𝑜𝑠(𝐴 + 𝐵) Soln:- Let ABC be a right ∆, right angled at C
𝐵𝐶 3
= 𝑐𝑜𝑠(600 + 300 ) 𝑠𝑖𝑛𝐴 = 𝐵𝐴 = 4

= 𝑐𝑜𝑠900  BC = 3𝑘 and BA = 4𝑘, where 𝑘 is a
=0 positive number

R H S = 𝑐𝑜𝑠𝐴𝑐𝑜𝑠𝐵 − 𝑠𝑖𝑛𝐴𝑠𝑖𝑛𝐵  By using Pythagoras Theorem, we get

= 𝑐𝑜𝑠600 𝑐𝑜𝑠300 − 𝑠𝑖𝑛600 𝑠𝑖𝑛300 𝐴𝐵2 = 𝐵𝐶 2 + 𝐴𝐶 2

1 √3 √3 1  (4𝑘)2 = (3𝑘)2 + 𝐴𝐶 2
=2× 2 − 2 ×2
 16𝑘 2 = 9𝑘 2 + 𝐴𝐶 2
√3 √3
= − 4
4
16𝑘 2 − 9𝑘 2 = 𝐴𝐶 2
√3− √3
= 4  7𝑘 2 = 𝐴𝐶 2

= 4
0
 √7𝑘 2 = 𝐴𝐶

=0  √7 𝑘 = 𝐴𝐶

For more Question Papers, Sample Papers, Notes & Syllabus visit Page 58 of 113

m
m .co
c. o

e m
em l a s
s g
𝐵𝐶 62. If 𝑠𝑖𝑛𝜃 − 𝑐𝑜𝑠𝜃 = 0, then find the value of
𝑡𝑎𝑛𝐴 = 𝐴𝐶

a
𝑠𝑖𝑛4 𝜃 + 𝑐𝑜𝑠 4 𝜃.
3𝑘 Soln:- Given: 𝑠𝑖𝑛𝜃 − 𝑐𝑜𝑠𝜃 = 0
=  𝑠𝑖𝑛𝜃 = 𝑐𝑜𝑠𝜃
√7 𝑘
𝑠𝑖𝑛𝜃
3  𝑐𝑜𝑠𝜃 = 1 (Dividing both side by 𝑐𝑜𝑠𝜃)
=
√7  𝑡𝑎𝑛𝜃 = 1
 𝑡𝑎𝑛𝜃 = 𝑡𝑎𝑛45°
 𝜃 = 45°
Now, 𝑠𝑖𝑛4 𝜃 + 𝑐𝑜𝑠 4 𝜃

m
= 𝑠𝑖𝑛4 45° + 𝑐𝑜𝑠 4 45°

c. o
4 4

m
1 1

co
1−𝑡𝑎𝑛2 450 = ( 2) + ( 2 )

m
59. Find the value of √ √

. e
1+𝑡𝑎𝑛2 450
1 1

m la s
= +

e
1−𝑡𝑎𝑛2 450 4 4

s g
Soln:-

a
1+1

a
1+𝑡𝑎𝑛2 450

l
= 4

ag
2
1 – 12 =4
= 1 + 12
1
= 2 (Ans)
1 −1
= 1+1
1−𝑠𝑖𝑛𝜃
=
0 63. Prove that 1+𝑠𝑖𝑛𝜃 = (𝑠𝑒𝑐𝜃 − 𝑡𝑎𝑛𝜃)2 .
2
1−𝑠𝑖𝑛𝜃
Soln:- 𝐿𝐻𝑆 = 1+𝑠𝑖𝑛𝜃
=0 1−𝑠𝑖𝑛𝜃 1−𝑠𝑖𝑛𝜃
= ×
1+𝑠𝑖𝑛𝜃 1−𝑠𝑖𝑛𝜃

m
1 1 (1−𝑠𝑖𝑛𝜃)2

co
60. Prove that 1+𝑡𝑎𝑛2 𝜃 + 1+𝑐𝑜𝑡 2 𝜃 = 1 = 1−𝑠𝑖𝑛2 𝜃

.
(1−𝑠𝑖𝑛𝜃)2

em
1 1 = 𝑐𝑜𝑠2 𝜃
= +

s
Soln:- LHS
1+𝑡𝑎𝑛2 𝜃 1+𝑐𝑜𝑡 2 𝜃

a
1−𝑠𝑖𝑛𝜃 2

l
= ( 𝑐𝑜𝑠𝜃 )

g
1 1

a
= + [∵ 1 + 1 𝑠𝑖𝑛𝜃 2
𝑠𝑒𝑐 2 𝜃 𝑐𝑜𝑠𝑒𝑐 2 𝜃 =( − )
𝑐𝑜𝑠𝜃 𝑐𝑜𝑠𝜃
𝑡𝑎𝑛2 𝜃 = 𝑠𝑒𝑐 𝜃 𝑎𝑛𝑑 1 + 𝑐𝑜𝑡 𝜃 = 𝑐𝑜𝑠𝑒𝑐 2 𝜃]
2 2
= (𝑠𝑒𝑐𝜃 − 𝑡𝑎𝑛𝜃)2
= 𝑅𝐻𝑆 (Proved)
= 𝑐𝑜𝑠 2 𝜃 + 𝑠𝑖𝑛2 𝜃 𝑠𝑖𝑛60°+𝑐𝑜𝑠30° √3
64. Prove that 1+𝑠𝑖𝑛30°+𝑐𝑜𝑠60° = 2 .
=1 𝑠𝑖𝑛60°+𝑐𝑜𝑠30°
Soln:- 𝐿𝐻𝑆 = 1+𝑠𝑖𝑛30°+𝑐𝑜𝑠60°
√3 √3
=RHS 2
+
2
= 1 1
1+ +

m
2 2
Hence Proved

co
√3+√3

m .
2
=

c. o
2+1+1

=m
1+𝑡𝑎𝑛2 𝐴

se
2
61. Express 1+𝑐𝑜𝑡 2 𝐴 in terms of 𝑡𝑎𝑛𝐴 2√3

em Soln:- la =
2

s
4
1+𝑡𝑎𝑛2 𝐴

la g
2
(MBOSE 2025)

a
1+𝑐𝑜𝑡 2 𝐴 2√3 2

g
×4

a
2
𝑠𝑒𝑐 2 𝐴
= √3
𝑐𝑜𝑠𝑒𝑐 2 𝐴 =
2
1
= 𝑅𝐻𝑆 (Proved
𝑐𝑜𝑠2 𝐴
= 1
𝑠𝑖𝑛2 𝐴
65. A die is thrown once. What is the
probability of getting a number less than
1 𝑠𝑖𝑛2 𝐴 3?
= 2 ×
𝑐𝑜𝑠 𝐴 1
Soln:- Total number of possible outcomes
𝑠𝑖𝑛2 𝐴

m .
= 𝑐𝑜𝑠2 𝐴 = 6 {𝑖𝑒 1, 2, 3, 4, 5, 6}

.co s e m
m
= 𝑡𝑎𝑛2 𝐴 Number of favourable outcomes less than 3

se g l a
l a a
ag For more Question Papers, Sample Papers, Notes & Syllabus visit Page 59 of 113