Bihar Board Class 10 · Maths · Matric exam 2027
Bihar Board Class 10 Maths Question Paper 2023
The board’s 2023 question paper, set J — all 138 questions as printed; 100 objective answers from the board’s own key; full model answers for 3 written questions, how the answer begins for 35 more. Each question names its chapter, where that chapter’s most-asked questions are.
Time 3 h 15 min
Section A · Objective questions (100)
sin^2 37° - sin^2 53° =
- (A) 0
- (B) 1 ✓
- (C) -1
- (D) 2
Answer (the board’s key): (B)
The value of sin^2 48° - cos^2 42° is
- (A) 0 ✓
- (B) 1
- (C) -1
- (D) 2
Answer (the board’s key): (A)
If tan theta = sqrt(3), then the value of sec theta is
- (A) 2/sqrt(3)
- (B) sqrt(3)/2
- (C) 1/2
- (D) 2 ✓
Answer (the board’s key): (D)
If cos theta = sec theta, then the value of cos^2 theta + sec^2 theta is
- (A) 21/4
- (B) 17/4 ✓
- (C) 29/4
- (D) 33/4
Answer (the board’s key): (B)
cosec^2 theta - tan^2(90° - theta) =
- (A) 0
- (B) 1 ✓
- (C) -1
- (D) 2
Answer (the board’s key): (B)
cos 20° sin 70° + sin 20° cos 70° =
- (A) 1 ✓
- (B) 2
- (C) 0
- (D) -1
Answer (the board’s key): (A)
sin 38° - cos 52° =
- (A) 0 ✓
- (B) 1
- (C) sqrt(3)/2
- (D) -1
Answer (the board’s key): (A)
If three coins are tossed simultaneously then the probability of getting the same outcomes in all three coins is
- (A) 1/2
- (B) 1/4 ✓
- (C) 1/3
- (D) 3/8
Answer (the board’s key): (B)
If P(E) = 0.05, then P(E') is equal to
- (A) 0.94
- (B) 0.95 ✓
- (C) 0.03
- (D) 0.05
Answer (the board’s key): (B)
(sin^2 theta - 1) / cos^2 theta =
- (A) sec^2 theta
- (B) tan^2 theta
- (C) -1 ✓
- (D) 1
Answer (the board’s key): (C)
The ratio of the volumes of two cylinders of equal height is 4:9. The ratio of the areas of their curved surfaces will be
- (A) 2:3 ✓
- (B) 4:9
- (C) 16:81
- (D) 1:3
Answer (the board’s key): (A)
If the diameter and the height of a cone and a cylinder be equal then what will be the ratio of their volumes?
- (A) 2:3
- (B) 1:3 ✓
- (C) 3:4
- (D) 1:2
Answer (the board’s key): (B)
(tan 45° - tan 30°) / (1 + tan 45°.tan 30°) =
- (A) (sqrt(3)-1)/(sqrt(3)+1) ✓
- (B) (sqrt(3)+1)/(sqrt(3)-1)
- (C) 1/sqrt(3)
- (D) 1
Answer (the board’s key): (A)
If the points A(x, 2), B(-3, -4) and C(7, -5) are collinear, then the value of x is
- (A) - 63 ✓
- (B) - 60
- (C) 63
- (D) none of these
Answer (the board’s key): (A)
The area of triangle OAB with vertices A(a, 0), O(0, 0) and B(0, b) (in square units) is
- (A) ab
- (B) 1/2 ab ✓
- (C) 1/2 a^2 b^2
- (D) 1/2 b^2
Answer (the board’s key): (B)
If the point (d, a) lies on the line 2x + 3y = 5, then the value of a is
- (A) 1 ✓
- (B) 1/2
- (C) 3
- (D) 1/3
Answer (the board’s key): (A)
Common root in x^2 - 3 = 0 and x^2 - 8x + 15 = 0 is
- (A) 5
- (B) 3 ✓
- (C) - 3
- (D) - 5
Answer (the board’s key): (B)
From an external point P to a circle, two tangents PA and PB are drawn. If PA = 5 cm, then the length of PB will be
- (A) 10 cm
- (B) 5 cm ✓
- (C) 15 cm
- (D) none of these
Answer (the board’s key): (B)
The second term of an A.P. is 13 and its 5th term is 25. Its common difference is
- (A) 2
- (B) 3
- (C) 4 ✓
- (D) 5
Answer (the board’s key): (C)
The arithmetic mean of 13 and 17 is
- (A) 13
- (B) 15 ✓
- (C) 17
- (D) None of these
Answer (the board’s key): (B)
The sum of the circumference and radius of a circle is 51 cm. Then the radius of the circle is
- (A) 7 cm ✓
- (B) 14 cm
- (C) 21 cm
- (D) 28 cm
Answer (the board’s key): (A)
A sphere of radius 8 cm is melted and recast into a cone of height 32 cm. The radius of the base of the cone is
- (A) 8 cm ✓
- (B) 9 cm
- (C) 10 cm
- (D) none of these
Answer (the board’s key): (A)
If the area of a circle is equal to the sum of the areas of circles of radii 3 cm and 4 cm respectively, then the radius of that circle is
- (A) 7 cm
- (B) 12 cm
- (C) 5 cm ✓
- (D) none of these
Answer (the board’s key): (C)
A side of an equilateral triangle ABC is a. Its area is
- (A) sqrt(3)/4 a
- (B) sqrt(3)/4 a^2 ✓
- (C) 4/sqrt(3) a^2
- (D) 1/sqrt(3) a^2
Answer (the board’s key): (B)
ABC is an isosceles triangle in which angle C = 90 degree. If AC = 6 cm, then AB =
- (A) 6 cm
- (B) 6 sqrt(2) cm ✓
- (C) 2 sqrt(6) cm
- (D) 4 sqrt(2) cm
Answer (the board’s key): (B)
tan(theta)/(sec(theta)-1) + tan(theta)/(sec(theta)+1) =
- (A) 2tan(theta)
- (B) 2sec(theta)
- (C) 2cosec(theta) ✓
- (D) 2tan(theta)sec(theta)
Answer (the board’s key): (C)
Tangent PQ at a point P of a circle of radius 5 cm meets a line through the centre O at a point Q such that OQ = 12 cm. Length of PQ is
- (A) 12 cm
- (B) 13 cm
- (C) 8.5 cm
- (D) sqrt(119) cm ✓
Answer (the board’s key): (D)
The probability of getting a tail when a coin is tossed is
- (A) 0
- (B) 1/2 ✓
- (C) 1
- (D) 1/4
Answer (the board’s key): (B)
Two dice are thrown simultaneously. The probability that the sum of the numbers appearing on these dice is 10, is
- (A) 1/4
- (B) 1/6
- (C) 1/12 ✓
- (D) 1/36
Answer (the board’s key): (C)
sin 90° * tan 60° * cos 30° =
- (A) 1/2
- (B) sqrt(3)/2 ✓
- (C) 2/3
- (D) 0
Answer (the board’s key): (B)
In the form of p / (2^n * 5^m), 0.67 can be written as
- (A) 67 / (2^0 * 5)
- (B) 67 / (2^2 * 5^2) ✓
- (C) 67 / (2 * 5)
- (D) none of these
Answer (the board’s key): (B)
sin^2(theta) =
- (A) 1 - cos^2(theta) ✓
- (B) cos^2(theta) - 1
- (C) sqrt(1 - cos^2(theta))
- (D) sqrt(cos^2(theta) - 1)
Answer (the board’s key): (A)
The solution of the pair of equations 3x - 2y + 3 = 0, 4x + 3y - 47 = 0 is
- (A) x = 1, y = 3
- (B) x = 4, y = 5
- (C) x = 5, y = 9 ✓
- (D) x = 8, y = 5
Answer (the board’s key): (C)
The first three terms of an A.P. are (3y - 1), (3y + 5) and (5y + 1) respectively. The value of y is
- (A) -3
- (B) 4
- (C) 5 ✓
- (D) 2
Answer (the board’s key): (C)
The common difference of the A.P. 6, 9, 12, ... is
- (A) 3 ✓
- (B) -3
- (C) 6
- (D) -6
Answer (the board’s key): (A)
sin 10° =
- (A) cos 10°
- (B) cos 80° ✓
- (C) sec 80°
- (D) cosec 80°
Answer (the board’s key): (B)
The 40th term of 8, 5, 2, ... is
- (A) -107
- (B) -109 ✓
- (C) -111
- (D) -113
Answer (the board’s key): (B)
Which term of A.P. 11, 15, 19, ... is 51 ?
- (A) 9th
- (B) 10th
- (C) 11th ✓
- (D) 12th
Answer (the board’s key): (C)
Which of the following is equal to 1/sqrt(2) ?
- (A) sin 30°
- (B) sin 60°
- (C) cos 45° ✓
- (D) cos 60°
Answer (the board’s key): (C)
If tan theta = sqrt(3), then what will be the value of theta ?
- (A) 90°
- (B) 45°
- (C) 30°
- (D) 60° ✓
Answer (the board’s key): (D)
The median of 2, 7, 5, 3, 8 is
- (A) 2
- (B) 3
- (C) 5 ✓
- (D) 7
Answer (the board’s key): (C)
The mode of 7, 2, 5, 7, 7, 3, 7 is
- (A) 2
- (B) 3
- (C) 5
- (D) 7 ✓
Answer (the board’s key): (D)
The class mark of the class interval 15 - 25 is
- (A) 10
- (B) 20 ✓
- (C) 30
- (D) 40
Answer (the board’s key): (B)
The length of the class intervals of the classes 2 - 5, 5 - 8, 8 - 11, 11 - 14 is
- (A) 2
- (B) 3 ✓
- (C) 4
- (D) 3.5
Answer (the board’s key): (B)
(sec A - cos A)(cot A + tan A) =
- (A) tan A . sin A
- (B) tan A . sec A ✓
- (C) sin A . cot A
- (D) sec A . sin^2 A
Answer (the board’s key): (B)
If the mean of four consecutive odd numbers is 6, then the largest number is
- (A) 5
- (B) 9 ✓
- (C) 21
- (D) 15
Answer (the board’s key): (B)
What should be the length of the classes to calculate the mode from the formula ?
- (A) Equal ✓
- (B) Unequal
- (C) Very large
- (D) None of these
Answer (the board’s key): (A)
The mean and mode of a frequency distribution are 28 and 16 respectively. What will be its median ?
- (A) 22
- (B) 23.5
- (C) 24 ✓
- (D) 24.5
Answer (the board’s key): (C)
sin 45° . tan 45° . sec 45° =
- (A) 0
- (B) 1 ✓
- (C) 2
- (D) -2
Answer (the board’s key): (B)
The total surface area of a cone is
- (A) pi r l
- (B) pi r^2 h
- (C) pi r(r + l) ✓
- (D) 2 pi r h
Answer (the board’s key): (C)
If alpha, beta are the zeroes of any quadratic polynomial p(x) = 6x^2 - 7x - 3, then alpha + beta =
- (A) 7/6 ✓
- (B) -7/6
- (C) 7/3
- (D) 1/2
Answer (the board’s key): (A)
If zeroes of the polynomial x^2 + ax + b are equal but of opposite sign, then a =
- (A) 1
- (B) -1
- (C) 0 ✓
- (D) b
Answer (the board’s key): (C)
Which of the following is a quadratic polynomial whose zeroes are 0 and 5?
- (A) x^2 + 5x + 5
- (B) x^2 - 5x ✓
- (C) x^2 + 5x
- (D) x^2 - 5
Answer (the board’s key): (B)
The value of a for which the system of equations ax + 10y = 9, 2x + 5y = 11 has no solution is
- (A) 2
- (B) -2
- (C) -4
- (D) 4 ✓
Answer (the board’s key): (D)
One root of 2x^2 + 5x - 12 = 0 is
- (A) 0
- (B) 1
- (C) -4 ✓
- (D) 4
Answer (the board’s key): (C)
The discriminant of the equation 3sqrt(3)x^2 + 10x + sqrt(3) = 0 is equal to
- (A) 44
- (B) 54
- (C) 64 ✓
- (D) 74
Answer (the board’s key): (C)
What is the power of 3 in the prime factorisation of 234 ?
- (A) 3
- (B) 4
- (C) 2 ✓
- (D) 1
Answer (the board’s key): (C)
The sum and product of zeroes of a quadratic polynomial are - 3 and 2 respectively. Then the polynomial is
- (A) x^2 + 2x - 3
- (B) x^2 + 3x + 2 ✓
- (C) x^2 - 3x - 2
- (D) x^2 - 3x + 2
Answer (the board’s key): (B)
A polynomial of degree one is called
- (A) quadratic polynomial
- (B) cubic polynomial
- (C) linear polynomial ✓
- (D) none of these
Answer (the board’s key): (C)
The most general form of the cubic polynomial in the following is
- (A) ax^2 + bx + c, a != 0
- (B) ax^4 + bx^3 + c, a != 0
- (C) ax^3 + bx^2 + cx + d, a != 0 ✓
- (D) none of these
Answer (the board’s key): (C)
0.49 can be expressed in the form p/q (where p, q are integers q != 0)
- (A) 49/100
- (B) 49/99 ✓
- (C) 49/90
- (D) 49/9
Answer (the board’s key): (B)
Which of the following is an irrational number ?
- (A) sqrt(49)
- (B) sqrt(36)
- (C) sqrt(27) ✓
- (D) sqrt(16)
Answer (the board’s key): (C)
The zero of the polynomial 10x - 25 is
- (A) 5/2 ✓
- (B) -5/2
- (C) -1/2
- (D) 5/4
Answer (the board’s key): (A)
The degree of the polynomial 3x^2 + 8x + 10 is
- (A) 3
- (B) 2 ✓
- (C) 8
- (D) 10
Answer (the board’s key): (B)
Which of the following is a quadratic equation ?
- (A) x^2 + 4x + 4 = 0
- (B) (x^2 + sqrt(x))(x + sqrt(x)) = 5
- (C) (x + 1)^2 = x^2 = 0 ✓
- (D) (x + 1)^3 = 11
Answer (the board’s key): (C)
If p, q are prime numbers then the H.C.F of p^3q and p^2q is
- (A) p^3q^3
- (B) p^2q ✓
- (C) p^3q^2
- (D) pq
Answer (the board’s key): (B)
sin(90° - theta) - cos theta =
- (A) sin theta
- (B) tan theta
- (C) 0 ✓
- (D) 1
Answer (the board’s key): (C)
If a = 2^3 * 3, b = 2 * 3, c = 3^n * 5 and LCM of (a, b, c) = 2^3 * 3^2 * 5, then n =
- (A) 1
- (B) 3
- (C) 2 ✓
- (D) 4
Answer (the board’s key): (C)
The product of two irrational numbers 3 - sqrt(5) and 3 + sqrt(5) is a/an
- (A) rational number ✓
- (B) irrational number
- (C) both (A) and (B)
- (D) none of these
Answer (the board’s key): (A)
The zeros of the quadratic polynomial x^2 - 4x - 12 are
- (A) 2, 5
- (B) -2, 5
- (C) 2, -6 ✓
- (D) 2, 6
Answer (the board’s key): (C)
A bag contains 7 red, 8 green and 5 white balls. One ball is taken at random. The probability that it is neither red nor white is
- (A) 1/20
- (B) 1/5
- (C) 2/5 ✓
- (D) 1/4
Answer (the board’s key): (C)
If cos theta = a/b, then sin theta =
- (A) sqrt(b^2 - a^2) / a
- (B) sqrt(b^2 - a^2) / b ✓
- (C) b / sqrt(b^2 - a^2)
- (D) sqrt(b^2 + a^2) / b
Answer (the board’s key): (B)
sqrt((1 - sin A) / (1 + sin A)) =
- (A) sec A + tan A
- (B) sec A - tan A ✓
- (C) sec A . tan A
- (D) sin A
Answer (the board’s key): (B)
tan 30° sin 30° cot 60° cosec 30° =
- (A) 1/sqrt(3)
- (B) sqrt(3)
- (C) 1/(2sqrt(3))
- (D) 1/3 ✓
Answer (the board’s key): (D)
If A = 45°, then the value of sec A + cosec A is
- (A) 1
- (B) sqrt(2)
- (C) 2
- (D) 2sqrt(2) ✓
Answer (the board’s key): (D)
If the angle of elevation of the sun is 60°, then the height of the vertical tower that casts 30 m long shadow is
- (A) 30/sqrt(3) m
- (B) 15sqrt(3) m
- (C) 30sqrt(3) m ✓
- (D) 15 m
Answer (the board’s key): (C)
If alpha = beta = 30°, then the value of cos(alpha + beta) is
- (A) sqrt(3)/2
- (B) 1/sqrt(2)
- (C) 1/2 ✓
- (D) 1
Answer (the board’s key): (C)
(2 tan 30°)/(1 - tan^2 30°) =
- (A) 0
- (B) 1
- (C) sqrt(3) ✓
- (D) 1/sqrt(3)
Answer (the board’s key): (C)
The value of tan 65° / cot 25° is
- (A) 0
- (B) 1 ✓
- (C) -1
- (D) 2
Answer (the board’s key): (B)
tan 5° x tan 30° x 4 tan 85° =
- (A) 1/sqrt(3) ✓
- (B) 4sqrt(3)
- (C) 1
- (D) 4
Answer (the board’s key): (A)
The ratio of the volumes of two cones is 4 : 5 and the ratio of their base radii is 2 : 3. The ratio of their vertical heights is
- (A) 4 : 7
- (B) 3 : 5
- (C) 9 : 5 ✓
- (D) 5 : 9
Answer (the board’s key): (C)
If the n th term of an AP be 5 - 3n then its common difference is
- (A) 3
- (B) 3 ✓
- (C) 4
- (D) 5
Answer (the board’s key): (B)
The coordinates of the vertices of a triangle are (4, 6), (0, 4) and (5, 5). The coordinates of its centroid will be
- (A) (5, 2)
- (B) (1, 3)
- (C) (4, 4)
- (D) (3, 5) ✓
Answer (the board’s key): (D)
The curved surface area of a cylinder is 264 m^2 and its volume is 924 m^3. The ratio of its diameter to its height is
- (A) 6 : 7
- (B) 7 : 6
- (C) 4 : 5
- (D) 7 : 3 ✓
Answer (the board’s key): (D)
If the centroid of the triangle formed by the points (a, b), (b, c) and (c, a) is the origin then a^3 + b^3 + c^3 is equal to
- (A) 0
- (B) abc
- (C) 3abc ✓
- (D) a + b + c
Answer (the board’s key): (C)
If the coordinates of the vertices of a triangle are (0, 4), (0, 0) and (3, 0), then the area of the triangle is
- (A) 3
- (B) 6 ✓
- (C) 9
- (D) 12
Answer (the board’s key): (B)
The ratio of the sides of two similar triangles is 4 : 9. The ratio of their areas will be
- (A) 2 : 3
- (B) 4 : 9
- (C) 81 : 16
- (D) 16 : 81 ✓
Answer (the board’s key): (D)
If ΔABC and ΔDEF are similar and BC = 4 cm, EF = 5 cm and the area of ΔABC is 80 cm^2, then the area of ΔDEF is
- (A) 100 cm^2
- (B) 125 cm^2 ✓
- (C) 150 cm^2
- (D) 200 cm^2
Answer (the board’s key): (B)
If the radius of a circle is 14 cm, then the area of the circle is
- (A) 154 cm^2
- (B) 231 cm^2
- (C) 616 cm^2 ✓
- (D) 516 cm^2
Answer (the board’s key): (C)
If the radius of a circle is 14 cm, then the circumference of the circle is
- (A) 44 cm
- (B) 88 cm ✓
- (C) 42 cm
- (D) 616 cm
Answer (the board’s key): (B)
The area of a circle of diameter 42 cm is
- (A) 1286 cm^2
- (B) 1386 cm^2 ✓
- (C) 1486 cm^2
- (D) None of these
Answer (the board’s key): (B)
The ratio of the radii of two circles is 4:3. Then the ratio of their areas will be
- (A) 3:4
- (B) 4:3
- (C) 16:9 ✓
- (D) 9:16
Answer (the board’s key): (C)
If the area and perimeter of a circle are numerically equal, then the diameter of the circle is
- (A) 2
- (B) 3
- (C) 4 ✓
- (D) 5
Answer (the board’s key): (C)
tan 30° * sin 30° * sec 60° =
- (A) 1/sqrt(3) ✓
- (B) sqrt(3)
- (C) 2/sqrt(3)
- (D) 2/3
Answer (the board’s key): (A)
The length of the arc corresponding to the sector having radius r and angle theta is
- (A) (theta/360) * pi * r^2
- (B) (theta/360) * 2 * pi * r ✓
- (C) (theta/180) * 2 * pi * r
- (D) (theta/180) * pi * r^2
Answer (the board’s key): (B)
In which quadrant is the point (- 2, 5) located ?
- (A) First
- (B) Second ✓
- (C) Third
- (D) Fourth
Answer (the board’s key): (B)
The distance of the point (3, 4) from y-axis is
- (A) 3 ✓
- (B) 4
- (C) 0
- (D) 7
Answer (the board’s key): (A)
The distance of the point (-4, -6) from the origin is
- (A) 2 * sqrt(13) ✓
- (B) 13 * sqrt(2)
- (C) 10
- (D) 24
Answer (the board’s key): (A)
The distance between the points (4, 5) and (- 3, 2) is
- (A) 58
- (B) sqrt(58) ✓
- (C) 29
- (D) sqrt(29)
Answer (the board’s key): (B)
The coordinates of the midpoint of the line segment joining the points (- 3, 4) and (7, 8) are
- (A) (3, 5)
- (B) (2, 6) ✓
- (C) (2, 4)
- (D) (3, 1)
Answer (the board’s key): (B)
Section B · Written questions (38)
Which term of the A.P 7, 13, 19, ... is 205 ?
Medha Ma’am
AskMedha’s AI teacher
Model answer · nth term of an AP
- 1.The first term of the given AP 7, 13, 19, ... is a = 7 and the common difference is d = 13 - 7 = 6.0.5 marks
- 2.The formula to find the nth term of an arithmetic progression is a_n = a + (n - 1)d. Here a = 7 and d = 6, and we have to find n for which a_n = 205.0.5 marks
- 3.Substituting the values in the formula, we get the equation 205 = 7 + (n - 1)6. Solving this gives 205 - 7 = (n - 1)6, i.e. 198 = (n - 1)6.0.5 marks
- 4.Dividing both sides by 6, n - 1 = 33, which gives n = 34. Therefore, the 34th term of the AP 7, 13, 19, ... is 205.0.5 marks
- Write a = 7 and d = 6 clearly before using the formula.
- State the formula a_n = a + (n - 1)d before substituting values.
- Show each step of solving the equation to reach n = 34 and state the 34th term is 205.
Marks split from a structure derived from past papers (checked twice); not Bihar Board’s official scheme
Find the value of k for which 2x^2 - kx + 3 = 0 has two real and equal roots.
Medha Ma’am
AskMedha’s AI teacher
Model answer · nature of roots, discriminant
- 1.Comparing the quadratic equation 2x^2 - kx + 3 = 0 with ax^2 + bx + c = 0, we get a = 2, b = -k, c = 3.0.5 marks
- 2.A quadratic equation ax^2 + bx + c = 0 has two equal real roots if its discriminant b^2 - 4ac = 0. So for the given equation to have two real and equal roots, we must have D = b^2 - 4ac = 0.0.5 marks
- 3.Substituting a = 2, b = -k, c = 3 in b^2 - 4ac = 0, we get (-k)^2 - 4(2)(3) = 0, i.e., k^2 - 24 = 0.0.5 marks
- 4.From k^2 - 24 = 0, we get k^2 = 24, so k = ±√24, i.e., k = ±2√6.0.5 marks
- Write a = 2, b = -k, c = 3 clearly before using the discriminant formula.
- State the condition b^2 - 4ac = 0 for two equal real roots before substituting values.
- Show the final simplification k^2 = 24 leading to k = ±2√6.
Marks split from a structure derived from past papers (checked twice); not Bihar Board’s official scheme
Find the sum of 5 + 9 + 13 + 17 + ... + 81.
Medha Ma’am
AskMedha’s AI teacher
Model answer · sum of n terms of AP
- 1.The given AP is 5 + 9 + 13 + 17 + ... + 81. Here, the first term a = 5, the common difference d = 9 - 5 = 4, and the last term l = 81.0.5 marks
- 2.Using the formula an = a + (n - 1)d to find the number of terms n, we have 81 = 5 + (n - 1) × 4, so 76 = (n - 1) × 4, giving n - 1 = 19, so n = 20.0.5 marks
- 3.Using the formula for the sum of the first n terms of an AP, Sn = (n/2)(a + l), we substitute a = 5, l = 81 and n = 20 to get S20 = (20/2)(5 + 81).0.5 marks
- 4.On calculation, S20 = (20/2)(5 + 81) = 10 × 86 = 860. So, the required sum of 5 + 9 + 13 + 17 + ... + 81 is 860.0.5 marks
- Write a = 5, d = 4 and l = 81 clearly before using any formula.
- Show the step an = a + (n - 1)d substituted to get n = 20.
- Use Sn = (n/2)(a + l) and write the substitution S20 = (20/2)(5 + 81) before the final answer 860.
Marks split from a structure derived from past papers (checked twice); not Bihar Board’s official scheme
The difference of two positive numbers is 3 and their product is 504. Find the numbers.
How the model answer begins
1. Let the two positive numbers be x and x + 3, since their difference is 3. Their product is 504, so x(x + 3) = 504.
The full answer from Medha Ma’am →If the numbers (2n - 1), (3n + 2) and (6n - 1) are in A.P., find the numbers.
How the model answer begins
1. The numbers (2n - 1), (3n + 2) and (6n - 1) are in arithmetic progression, so the middle term is equal to half the sum of the other two terms, which gives the equation 2(3n + 2) = (2n - 1) + (6n - 1). Simplifying the…
The full answer from Medha Ma’am →Prove that one and only one tangent can be drawn to any point on the circumference of a circle
How the model answer begins
1. The tangent drawn at any point on the circle is perpendicular to the radius through the point of contact. Let O be the centre of the circle and P a point on the circle where the tangent XY touches it. OP is the shortest…
The full answer from Medha Ma’am →Find the H.C.F. and L.C.M. of 24 and 36 using prime factorisation.
How the model answer begins
1. The prime factorisation of 24 and 36 is found as follows. 24 = 2 × 2 × 2 × 3 36 = 2 × 2 × 3 × 3 So, 24 = 2^3 × 3^1 and 36 = 2^2 × 3^2.
The full answer from Medha Ma’am →Find the roots of the equation 2x^2 - 3x - 5 = 0.
How the model answer begins
1. To find the roots of 2x^2 - 3x - 5 = 0, we split the middle term. We write -3x as -5x + 2x, because (-5x) × (2x) = -10x^2 = (2x^2) × (-5). So, 2x^2 - 3x - 5 = 2x^2 - 5x + 2x - 5 = x(2x - 5) + 1(2x - 5) = (2x - 5)(x +…
The full answer from Medha Ma’am →Find a quadratic polynomial, the sum and product of whose zeroes are 0 and - 4 respectively.
How the model answer begins
1. A quadratic polynomial with sum of zeroes (α + β) and product of zeroes αβ is given by the formula x² - (α + β)x + αβ. Here the sum of zeroes is given as 0 and the product of zeroes is given as -4.
The full answer from Medha Ma’am →ABC is an isosceles triangle whose angle C is right angle. Prove that AB^2 = 2AC^2.
How the model answer begins
1. ABC is a triangle right angled at C. Using the Pythagoras Theorem in triangle ABC, the square of the hypotenuse is equal to the sum of the squares of the other two sides. Therefore, AB2 = AC2 + BC2.
The full answer from Medha Ma’am →Solve : (x + 4)(x - 4) = 20.
How the model answer begins
1. (x + 4)(x - 4) = 20 Expanding the left hand side using the identity for product of sum and difference, we get x² - 16 = 20.
The full answer from Medha Ma’am →ABC and ADEF are similar and their areas are respectively 64 cm^2 and 121 cm^2. If EF = 15.4 cm, then find the length of BC.
How the model answer begins
1. The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides. Since triangle ABC is similar to triangle DEF, BC and EF are corresponding sides, so ar(ABC)/ar(DEF) =…
The full answer from Medha Ma’am →Find a relation between x and y such that the point P(x, y) is equidistant from the points A(1, 4) and B(-1, 2).
How the model answer begins
1. Let P(x, y) be equidistant from the points A(1, 4) and B(-1, 2). We are given that PA = PB. So, using the distance formula, PA² = PB², which gives (x-1)² + (y-4)² = (x+1)² + (y-2)².
The full answer from Medha Ma’am →Find the common difference of an A.P. in which (a_21 - a_7) = 84.
How the model answer begins
1. Using the formula for the nth term of an arithmetic progression, an = a + (n – 1) d, the 21st term is a21 = a + (21 – 1) d = a + 20d, and the 7th term is a7 = a + (7 – 1) d = a + 6d.
The full answer from Medha Ma’am →Prove that tan^2 theta + cot^2 theta + 2 = sec^2 theta.cosec^2 theta.
How the model answer begins
1. To prove tan²θ + cot²θ + 2 = sec²θ.cosec²θ, we take the LHS = tan²θ + cot²θ + 2. Writing this as tan²θ + 1 + cot²θ + 1, and using the identities 1 + tan²θ = sec²θ and 1 + cot²θ = cosec²θ, we get LHS = sec²θ + cosec²θ.…
The full answer from Medha Ma’am →Find the discriminant of quadratic equation bx^2 + ax + c = 0.
How the model answer begins
1. Comparing the given equation bx^2 + ax + c = 0 with the standard form of a quadratic equation ax^2 + bx + c = 0, we get A = b, B = a and C = c.
The full answer from Medha Ma’am →Show that x = - 3 is a solution of equation 2x^2 + 5x - 3 = 0.
How the model answer begins
1. Substituting the value x = -3 into the equation 2x^2 + 5x - 3 = 0, we get 2(-3)^2 + 5(-3) - 3.
The full answer from Medha Ma’am →Find the nature of the roots of the quadratic equation 3x^2 - 4sqrt(3)x + 4 = 0.
How the model answer begins
1. Comparing the given quadratic equation 3x^2 - 4\sqrt{3}x + 4 = 0 with the standard form ax^2 + bx + c = 0, we get a = 3, b = -4\sqrt{3} and c = 4.
The full answer from Medha Ma’am →Find the volume of a sphere whose radius is 7 cm.
How the model answer begins
1. The formula for the volume of a sphere is V = (4/3)πr³, where r is the radius of the sphere.
The full answer from Medha Ma’am →Find the mean of the following data : | Marks obtained | 10 | 15 | 20 | 25 | 30 | | --- | --- | --- | --- | --- | --- | | Number of students | 4 | 6 | 3 | 2 | 5 |
Find the area of the triangle whose vertices are (3, 4), (- 4, 3) and (8, 6) respectively.
How the model answer begins
1. The area of a triangle formed by the points (x1, y1), (x2, y2) and (x3, y3) is the numerical value of the expression 1/2 [x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2)]. Here the vertices are A(3, 4), B(-4, 3) and C(8, 6), so…
The full answer from Medha Ma’am →By comparing the ratios a_1/a_2, b_1/b_2 and c_1/c_2, find out that the following pair 3x + 12y = 5, 2x + 8y = 7 is inconsistent.
How the model answer begins
1. The given pair of linear equations is 3x + 12y - 5 = 0 and 2x + 8y - 7 = 0. Here a1 = 3, b1 = 12, c1 = -5 and a2 = 2, b2 = 8, c2 = -7. a1/a2 = 3/2 b1/b2 = 12/8 = 3/2 c1/c2 = -5/-7 = 5/7
The full answer from Medha Ma’am →Find the zeroes of 4t^2 - 4t + 1 and verify the relation between zeroes and coefficients.
How the model answer begins
1. The zeroes of 4t^2 - 4t + 1 are found by splitting the middle term. We write 4t^2 - 4t + 1 = 4t^2 - 2t - 2t + 1 = 2t(2t - 1) - 1(2t - 1) = (2t - 1)(2t - 1) = (2t - 1)^2. So, the value of 4t^2 - 4t + 1 is zero when 2t -…
The full answer from Medha Ma’am →Find the coordinates of a point on the y-axis which is equidistant from the points (- 5, - 2) and (3, 2).
How the model answer begins
1. Let the point on the y-axis be P(0, y); PA = PB gives (0 + 5)² + (y + 2)² = (0 − 3)² + (y − 2)².
The full answer from Medha Ma’am →The length of the tangent drawn from an external point P to a circle is 15 cm and the distance of the point P from the centre of the circle is 17 cm. Find the radius of the circle.
How the model answer begins
1. The tangent to a circle is perpendicular to the radius through the point of contact. So the triangle OTP formed by the centre O, the point of contact T and the external point P is a right-angled triangle, right-angled…
The full answer from Medha Ma’am →Find the area of a quadrant of a circle whose circumference is 22 cm.
How the model answer begins
1. The circumference of the circle is given as 22 cm. Circumference of a circle = 2πr So, 2πr = 22 2 × 22/7 × r = 22 r = 22 × 7 / (2 × 22) r = 7/2 cm = 3.5 cm
The full answer from Medha Ma’am →If A = 90° and B = 45°, then prove that cos(A - B) = cos A.cos B + sin A.sin B.
How the model answer begins
1. Substituting A = 90° and B = 45° in the left hand side cos(A - B) gives cos(90° - 45°) = cos 45° = 1/√2.
The full answer from Medha Ma’am →If sec 4A = cosec(A - 20°), where 4A is an acute angle, then find the value of A.
How the model answer begins
1. We are given that sec 4A = cosec (A - 20°). Since sec 4A = cosec (90° - 4A), we can write this as cosec (90° - 4A) = cosec (A - 20°). Since 90° - 4A and A - 20° are both acute angles, therefore, 90° - 4A = A - 20°.
The full answer from Medha Ma’am →If the perimeter of a semi-circular protractor is 36 cm, then find its diameter.
How the model answer begins
1. The perimeter of a semi-circular protractor is given by πr + 2r, where r is the radius of the protractor. Here the perimeter is 36 cm, so πr + 2r = 36, which can be written as r(π + 2) = 36.
The full answer from Medha Ma’am →Prove that (2 tan 30°) / (1 - tan^2 30°) = sqrt(3).
How the model answer begins
1. The value of tan 30° is 1/√3. Substituting tan 30° = 1/√3 in the given expression, LHS = (2 tan 30°)/(1 − tan² 30°) = (2 × 1/√3)/(1 − (1/√3)²).
The full answer from Medha Ma’am →Using graphical method solve the pair of equations 3x - 5y - 1 = 0 and 2x - y - 3 = 0.
How the model answer begins
1. The first equation is 3x - 5y - 1 = 0, which can be written as 3x - 5y = 1. To represent this equation graphically, we need at least two solutions. We write y = (3x-1)/5. When x = 2, y = (3(2)-1)/5 = 5/5 = 1. When x =…
The full answer from Medha Ma’am →Divide 50 into two parts so that the sum of their reciprocals is 1/12.
How the model answer begins
1. Let the first part be x. Since the two parts together make 50, the second part will be 50 - x.
The full answer from Medha Ma’am →Find the mean of the following distribution: | Class-interval | 5-15 | 15-25 | 25-35 | 35-45 | 45-55 | 55-65 | | --- | --- | --- | --- | --- | --- | --- | | Frequency | 6 | 11 | 21 | 23 | 14 | 5 |
How the model answer begins
1. Class marks xᵢ: 10, 20, 30, 40, 50, 60.
The full answer from Medha Ma’am →Two poles of equal heights are standing opposite to each other on either sides of the road which is 80 m wide. From a point P between them on the road the angle of elevation of the top of one pole is 60° and the angle of depression of the point P from the top of another pole is 30°. Find the height of each pole and distances of the point P from the poles.
How the model answer begins
1. Let the two poles AB and CD be of equal height h metres, standing on opposite sides of a road of width 80 m. Let P be a point on the road between the two poles such that the angle of elevation of the top of one pole…
The full answer from Medha Ma’am →Draw a right angled triangle ABC in which AB = 6 cm, BC = 8 cm and angle B = 90°. D is the foot of perpendicular BD from B on AC. Draw circumcircle of ABCD and draw a tangent from A on this circle.
How the model answer begins
1. Draw BC = 8 cm; at B make 90° and cut BA = 6 cm; join AC — right triangle ABC with ∠B = 90°.
The full answer from Medha Ma’am →A well of diameter 3 m is dug 14 m deep. The soil taken out of it has been spread evenly all around it in the shape of a circular ring of width 4 m to form an embankment. Find the height of the embankment.
How the model answer begins
1. The diameter of the well is 3 m, so the radius of the well is r = 3/2 = 1.5 metres. The depth of the well is h = 14 metres.
The full answer from Medha Ma’am →Prove that tan 7°. tan 23°. tan 60°. tan 67°. tan 83° = sqrt(3).
How the model answer begins
1. Using the identity tan (90° – θ) = cot θ, we write tan 83° as tan 83° = tan (90° – 7°) = cot 7°. So the term tan 7° remains as tan 7°, and tan 83° is written as cot 7°.
The full answer from Medha Ma’am →Prove that the angle between the two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line segment joining the points of contact at the centre.
How the model answer begins
1. We are given a circle with centre O and an external point P from which two tangents PA and PB are drawn to the circle, touching it at points A and B. Since the tangent to a circle is perpendicular to the radius through…
The full answer from Medha Ma’am →
The same preparation, with Medha Ma’am — on your phone
Every question of every chapter with its model answer, practice questions in the board’s own shapes, 60 minutes of class a day, written answers marked, a test every Sunday and a WhatsApp report to parents.
Medha Ma’am is an AI teacher, not a live tutor. Every answer is checked twice; a person takes responsibility.