Pre-Board Examination 2026
Subject - Mathematics
Subject - Mathematics
Q.1 Choose the correct option to answer the following multiple choice question (i to xviii) and write in the answer book. (1 × 18 = 18)
(i) The sum of the power of the prime factors of 400 is –
✔ Correct Answer : 6
(ii) The HCF of number 72 and 120 is –
✔ Correct Answer : 12
(iii) If α and β are the zeros of the quadratic polynomial 4x² − 6x + 7, then (α + β) is –
✔ Correct Answer : 3/2
(iv) For what value of k will the pair of equations 2x + 3y = 4, 6x + ky = 3 have no solution?
✔ Correct Answer : 6
(v) The common difference for given A.P. is : −11, −6, −1, 4, ………
✔ Correct Answer : 5
(vi) The formula for the sum of the first n even natural number is –
✔ Correct Answer :n(n+1)
(vii) If ΔABC ∼ ΔDEF and AB = 12 cm, DE = 8 cm, BC = 4 cm then EF is –
✔ Correct Answer : 8/3 cm
(viii) The coordinates of the midpoint of point A(2, −4) and point B(6, −2) is –
✔ Correct Answer : (4, −3)
(ix) If cos A = 7/25, then the value of tan A is –
✔ Correct Answer : 24/7
(x) The angle of depression of a point on the earth from the top of a 10 m high tower is 60°. The distance of the point from the base of the tower is –
✔ Correct Answer : 10/√3 m
(xi) Tangent drawn at the ends of the diameter of a circle are mutually –
✔ Correct Answer : Parallel
(xii) A tangent PQ at a point P of a circle of radius 24 cm meets a line through the centre O at a point Q such that OQ = 25 cm. Then the length of PQ is –
✔ Correct Answer : 7 cm
(xiii) If the length of arc is 44 cm and angle is 90°, then the radius is –
✔ Correct Answer : 7 cm
(xiv) The formula for finding the total surface area of hemisphere is –
✔ Correct Answer : 3πr²
(xv) If the volume of cube is 64 cm³ then the side of cube is –
✔ Correct Answer : 4 cm
(xvi) The mean value of the following data is – 4, 6, 5, 9, 8, 6, 7, 9, 4, 2
✔ Correct Answer : 6
(xvii) The relation between mode (x), mean (z) and median (y) is –
✔ Correct Answer : x − y = 3(z − y)
(xviii) Which of the following numbers cannot be the probability of an event?
✔ Correct Answer : 5/2
Score: 0 / 18
Q.2. Fill in the blanks in the following (i to vi) (½ × 10 = 5)
(i) The roots of a quadratic equation are equal if k x² − 4√3 x + 4 = 0, then k = __________ .
(ii) If the first term and the common difference are −7 and −2, then the 10th term of the A.P. is __________ .
(iii) The value of 2 tan² 45° is __________ .
(iv) If in a triangle, the square of one side is equal to the sum of the squares of the other two sides, then the angle opposite to the first side is __________ .
(v) A line that touches a circle at a single point is called a __________ to the circle.
(vi) A card is drawn from a deck of 52 cards. The probability that it is a face card is __________ .
(i) The roots of a quadratic equation are equal if k x² − 4√3 x + 4 = 0, then k = __________ .
(ii) If the first term and the common difference are −7 and −2, then the 10th term of the A.P. is __________ .
(iii) The value of 2 tan² 45° is __________ .
(iv) If in a triangle, the square of one side is equal to the sum of the squares of the other two sides, then the angle opposite to the first side is __________ .
(v) A line that touches a circle at a single point is called a __________ to the circle.
(vi) A card is drawn from a deck of 52 cards. The probability that it is a face card is __________ .
Q.3. Very short Answer Type Question –(i to xii))
(i) Find the LCM of two numbers 300 and 360.
(ii) Express 336 as a product of its prime factors.
(iii) Find a quadratic polynomial whose sum and product of its zeroes are 5 and 2.
(iv) 5 pencils and 7 pens together cost Rupees 50. Express this problem by the linear equations.
(v) Write the first term and common difference of an A.P. 7, 11, 15 ……
(vi) Find the distance between the points (5, 3) and (8, 1).
(vii) In a ∆ABC, right angled at B, AB = 3 cm, BC = 4 cm, determine Sin A and Cos C.
(viii) If tangents PA and PB from a point P to a circle with center O are inclined to each other at angle of 80°, then find ∠POA.
(ix) Find the value of the angle subtended by the minute hand of a clock on the centre in 20 minutes.
(x) Find the length of arc of a circle with radius 3.5 cm, whose angle is 60°.
(xi) Find the median and mode of the distribution 1, 6, 3, 5, 7, 9, 11, 4, 9.
(xii) Find the mean of x, x + 2 and x + 4.
(i) Find the LCM of two numbers 300 and 360.
(ii) Express 336 as a product of its prime factors.
(iii) Find a quadratic polynomial whose sum and product of its zeroes are 5 and 2.
(iv) 5 pencils and 7 pens together cost Rupees 50. Express this problem by the linear equations.
(v) Write the first term and common difference of an A.P. 7, 11, 15 ……
(vi) Find the distance between the points (5, 3) and (8, 1).
(vii) In a ∆ABC, right angled at B, AB = 3 cm, BC = 4 cm, determine Sin A and Cos C.
(viii) If tangents PA and PB from a point P to a circle with center O are inclined to each other at angle of 80°, then find ∠POA.
(ix) Find the value of the angle subtended by the minute hand of a clock on the centre in 20 minutes.
(x) Find the length of arc of a circle with radius 3.5 cm, whose angle is 60°.
(xi) Find the median and mode of the distribution 1, 6, 3, 5, 7, 9, 11, 4, 9.
(xii) Find the mean of x, x + 2 and x + 4.
Q.4. Find the zeroes of the quadratic polynomial x² + 7x + 10 and verify the relationship between the zeroes and the coefficients.
Solution:
Given polynomial: x² + 7x + 10
x² + 7x + 10 = 0
⇒ (x + 5)(x + 2) = 0
∴ x = −5, −2
Zeroes are α = −5, β = −2
Sum of zeroes = α + β = −5 + (−2) = −7
= −b/a = −7/1 = −7 ✔
Product of zeroes = αβ = (−5)(−2) = 10
= c/a = 10/1 = 10 ✔
Hence, the relationship between zeroes and coefficients is verified.
Given polynomial: x² + 7x + 10
x² + 7x + 10 = 0
⇒ (x + 5)(x + 2) = 0
∴ x = −5, −2
Zeroes are α = −5, β = −2
Sum of zeroes = α + β = −5 + (−2) = −7
= −b/a = −7/1 = −7 ✔
Product of zeroes = αβ = (−5)(−2) = 10
= c/a = 10/1 = 10 ✔
Hence, the relationship between zeroes and coefficients is verified.
Q.5. Solve the system of equations:
x − y = −1 and 3x + 2y = 12.
Solution:
x − y = −1 …(1)
3x + 2y = 12 …(2)
From (1): x = y − 1
Substitute in (2):
3(y − 1) + 2y = 12
3y − 3 + 2y = 12
5y = 15
y = 3
x = 3 − 1 = 2
∴ Solution is x = 2, y = 3.
x − y = −1 …(1)
3x + 2y = 12 …(2)
From (1): x = y − 1
Substitute in (2):
3(y − 1) + 2y = 12
3y − 3 + 2y = 12
5y = 15
y = 3
x = 3 − 1 = 2
∴ Solution is x = 2, y = 3.
Q.6. Find the 31st term of an A.P. whose 11th term is 38 and 16th term is 73.
Solution:
a₁₁ = a + 10d = 38 …(1)
a₁₆ = a + 15d = 73 …(2)
Subtract (1) from (2):
5d = 35 ⇒ d = 7
From (1):
a + 10(7) = 38
a = −32
a₃₁ = a + 30d = −32 + 30×7 = 178
∴ 31st term = 178.
a₁₁ = a + 10d = 38 …(1)
a₁₆ = a + 15d = 73 …(2)
Subtract (1) from (2):
5d = 35 ⇒ d = 7
From (1):
a + 10(7) = 38
a = −32
a₃₁ = a + 30d = −32 + 30×7 = 178
∴ 31st term = 178.
Q.7. In the given figure, DE ∥ AC and DF ∥ AE. Prove that BF/FE = BE/EC.
Solution:
Since DE ∥ AC, by Basic Proportionality Theorem:
BF / FE = BE / EC
Hence proved.
Since DE ∥ AC, by Basic Proportionality Theorem:
BF / FE = BE / EC
Hence proved.
Q.8. Find the ratio in which the y-axis divides the line segment joining the points (5, −6) and (−1, −4). Also find the point of intersection.
Solution:
Let the point P divide the line segment in ratio m : n.
Using section formula for x-coordinate:
( m(−1) + n(5) ) / (m + n) = 0
−m + 5n = 0 ⇒ m : n = 5 : 1
y-coordinate of P:
= (5(−4) + 1(−6)) / 6 = −26/6 = −13/3
∴ Ratio = 5 : 1 and point is (0, −13/3).
Let the point P divide the line segment in ratio m : n.
Using section formula for x-coordinate:
( m(−1) + n(5) ) / (m + n) = 0
−m + 5n = 0 ⇒ m : n = 5 : 1
y-coordinate of P:
= (5(−4) + 1(−6)) / 6 = −26/6 = −13/3
∴ Ratio = 5 : 1 and point is (0, −13/3).
Q.9. If tan(A + B) = √3 and tan(A − B) = 1/√3,
0° < A + B ≤ 90°, A > B, find A and B.
Solution:
tan(A + B) = √3 ⇒ A + B = 60°
tan(A − B) = 1/√3 ⇒ A − B = 30°
Add both equations:
2A = 90° ⇒ A = 45°
B = 60° − 45° = 15°
∴ A = 45°, B = 15°.
tan(A + B) = √3 ⇒ A + B = 60°
tan(A − B) = 1/√3 ⇒ A − B = 30°
Add both equations:
2A = 90° ⇒ A = 45°
B = 60° − 45° = 15°
∴ A = 45°, B = 15°.
Q.10. A quadrilateral ABCD is drawn to circumscribe a circle. Prove that AB + CD = AD + BC.
Solution:
Tangents drawn from an external point to a circle are equal.
AB = AD, BC = CD
∴ AB + CD = AD + BC.
Hence proved.
Tangents drawn from an external point to a circle are equal.
AB = AD, BC = CD
∴ AB + CD = AD + BC.
Hence proved.
Q.11. A chord of a circle of radius 21 cm subtends an angle of 120° at the centre. Find the area of the corresponding segment of the circle.
Solution:
Radius r = 21 cm, angle θ = 120°
Area of sector = (120/360) × π × 21² = 462 cm²
Area of triangle = (1/2) r² sin120°
= (1/2) × 441 × (√3/2) = 441√3/4
Area of segment = Area of sector − Area of triangle.
Radius r = 21 cm, angle θ = 120°
Area of sector = (120/360) × π × 21² = 462 cm²
Area of triangle = (1/2) r² sin120°
= (1/2) × 441 × (√3/2) = 441√3/4
Area of segment = Area of sector − Area of triangle.
Q.12. Find the mean of the following frequency distribution:
x : 5, 6, 7, 8, 9, 10
f : 2, 8, 10, 7, 5, 3
x : 5, 6, 7, 8, 9, 10
f : 2, 8, 10, 7, 5, 3
Solution:
Σf = 35
Σfx = 260
Mean = Σfx / Σf = 260 / 35 = 7.43
Σf = 35
Σfx = 260
Mean = Σfx / Σf = 260 / 35 = 7.43
Q.13. A piggy bank contains 100 fifty paise coins, 50 one-rupee coins,
20 two-rupee coins and 10 five-rupee coins.
Find the probability that the coin is
(i) a 50 paise coin
(ii) not a ₹5 coin.
Solution:
Total coins = 100 + 50 + 20 + 10 = 180
(i) Probability of 50 paise coin = 100 / 180 = 5 / 9
(ii) Coins other than ₹5 = 170
Probability = 170 / 180 = 17 / 18
Total coins = 100 + 50 + 20 + 10 = 180
(i) Probability of 50 paise coin = 100 / 180 = 5 / 9
(ii) Coins other than ₹5 = 170
Probability = 170 / 180 = 17 / 18
Q.14. Find two numbers whose sum is 27 and product is 182.
Solution (First Part):
Let the two numbers be x and y.
x + y = 27 …(1)
xy = 182 …(2)
The required quadratic equation is:
t² − 27t + 182 = 0
(t − 13)(t − 14) = 0
∴ The two numbers are 13 and 14.
OR
Solution (Second Part):
Given equation: 3x² − 2x + 1 = 0
Discriminant, D = b² − 4ac
= (−2)² − 4×3×1
= 4 − 12 = −8
Since D < 0, the roots are not real.
Let the two numbers be x and y.
x + y = 27 …(1)
xy = 182 …(2)
The required quadratic equation is:
t² − 27t + 182 = 0
(t − 13)(t − 14) = 0
∴ The two numbers are 13 and 14.
OR
Solution (Second Part):
Given equation: 3x² − 2x + 1 = 0
Discriminant, D = b² − 4ac
= (−2)² − 4×3×1
= 4 − 12 = −8
Since D < 0, the roots are not real.
Q.15. If the points A(6, 1), B(8, 2), C(9, 4) and D(p, 3)
are the vertices of a parallelogram in this order, find the value of p.
Solution (First Part):
In a parallelogram, diagonals bisect each other.
Mid-point of AC = Mid-point of BD
Mid-point of AC = ((6+9)/2 , (1+4)/2) = (15/2 , 5/2)
Mid-point of BD = ((8+p)/2 , (2+3)/2) = ((8+p)/2 , 5/2)
Equating x-coordinates:
(8 + p)/2 = 15/2
8 + p = 15
p = 7
In a parallelogram, diagonals bisect each other.
Mid-point of AC = Mid-point of BD
Mid-point of AC = ((6+9)/2 , (1+4)/2) = (15/2 , 5/2)
Mid-point of BD = ((8+p)/2 , (2+3)/2) = ((8+p)/2 , 5/2)
Equating x-coordinates:
(8 + p)/2 = 15/2
8 + p = 15
p = 7
Q.16. Prove that:
(1 + sin A) / (1 − sin A) = sec A + tan A
(1 + sin A) / (1 − sin A) = sec A + tan A
Solution (First Identity):
LHS = (1 + sin A)/(1 − sin A)
Multiply numerator and denominator by (1 + sin A):
= (1 + sin A)² / (1 − sin² A)
= (1 + sin A)² / cos² A
= (1/cos A + sin A/cos A)²
= (sec A + tan A)²
Taking square root on both sides:
LHS = RHS
Hence proved.
LHS = (1 + sin A)/(1 − sin A)
Multiply numerator and denominator by (1 + sin A):
= (1 + sin A)² / (1 − sin² A)
= (1 + sin A)² / cos² A
= (1/cos A + sin A/cos A)²
= (sec A + tan A)²
Taking square root on both sides:
LHS = RHS
Hence proved.
Q.17. Find the median of the following frequency distribution:
| Class Interval | Frequency |
|---|---|
| 1 – 4 | 6 |
| 4 – 7 | 30 |
| 7 – 10 | 40 |
| 10 – 13 | 16 |
| 13 – 16 | 4 |
| 16 – 19 | 4 |
Solution (First Part):
Total frequency, N = 100
N/2 = 50
Median class = 7–10
Median = l + [(N/2 − cf)/f] × h
= 7 + [(50 − 36)/40] × 3
= 7 + (14/40) × 3
= 8.05
Total frequency, N = 100
N/2 = 50
Median class = 7–10
Median = l + [(N/2 − cf)/f] × h
= 7 + [(50 − 36)/40] × 3
= 7 + (14/40) × 3
= 8.05
Q.18. From a point on the ground, the angles of elevation of the bottom
and the top of a transmission tower fixed at the top of a 20 metre high building
are 45° and 60° respectively. Find the height of the tower.
Solution (First Part):
Let the height of the tower be h metres.
tan 45° = 20/x ⇒ x = 20
tan 60° = (20 + h)/x = √3
20 + h = 20√3
h = 20(√3 − 1) metres.
Let the height of the tower be h metres.
tan 45° = 20/x ⇒ x = 20
tan 60° = (20 + h)/x = √3
20 + h = 20√3
h = 20(√3 − 1) metres.
Q.19. A solid consisting of a right circular cone of height 120 cm
and radius 60 cm standing on a hemisphere of radius 60 cm is placed upright
in a right circular cylinder full of water such that it touches the bottom.
Find the volume of water left in the cylinder, if the radius of the cylinder
is 60 cm and its height is 180 cm.
Solution (First Part):
Volume of cylinder = πr²h = π×60²×180
Volume of cone = (1/3)π×60²×120
Volume of hemisphere = (2/3)π×60³
Volume of water left = Volume of cylinder − (Volume of cone + Volume of hemisphere)
Volume of cylinder = πr²h = π×60²×180
Volume of cone = (1/3)π×60²×120
Volume of hemisphere = (2/3)π×60³
Volume of water left = Volume of cylinder − (Volume of cone + Volume of hemisphere)
Q.20. Find the mode of the following distribution:
| Class Interval | Frequency |
|---|---|
| 2 – 11 | 15 |
| 11 – 20 | 16 |
| 20 – 29 | 17 |
| 29 – 38 | 12 |
| 38 – 47 | 11 |
Modal class = 20–29
Mode = l + [(f₁ − f₀)/(2f₁ − f₀ − f₂)] × h
= 20 + [(17 − 16)/(34 − 16 − 12)] × 9
= 21.5
Mode = l + [(f₁ − f₀)/(2f₁ − f₀ − f₂)] × h
= 20 + [(17 − 16)/(34 − 16 − 12)] × 9
= 21.5
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