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 powers of prime factors of 400 is
✔ Correct Answer : 6
(ii) If 3 is a zero of the polynomial 2x² + x + k, then the value of k will be
✔ Correct Answer : -21
(iii) In a two digit number, the unit digit is x and the tens digit is y, then that number is
✔ Correct Answer : 10y + x
(iv) If △ABC ~ △DEF and AB = 10 cm, DE = 8 cm, then BC : EF is
✔ Correct Answer : 5 : 4
(v) Distance of point P(−3, 4) from origin O(0, 0) is
✔ Correct Answer : 5
(vi) cosec²45° − cot²45° equals
✔ Correct Answer : 1
(vii) The shadow of a vertical pillar is same as the height of pillar. Then the angle of elevation of sun is
✔ Correct Answer : 45°
(viii) From a point P, the length of the tangent to a circle is 24 cm and the distance of P from the centre is 25 cm. The radius of the circle is
✔ Correct Answer : 7 cm
(ix) The area of a quadrant of a circle whose radius is 7 cm is
✔ Correct Answer : 38.5 cm²
(x) If the radius of a cone is 14 cm and slant height is 10 cm, then the curved surface area of the cone is
✔ Correct Answer : 440 cm²
(xi) Which of the following number cannot be the probability of any event?
✔ Correct Answer : 2
(xii) If HCF = LCM for two rational numbers, then numbers always should be
✔ Correct Answer : Equal
(xiii) If the sum and product of the zeros of a quadratic polynomial are 5 and 6 respectively, then the quadratic polynomial is
✔ Correct Answer : x² − 5x + 6
(xiv) For which value of k, the linear pair x + y − 4 = 0, 2x + ky − 3 = 0 has no solution?
✔ Correct Answer : 2
(xv) In the given figure, if BC = 3 cm, then the measurement of EF is
✔ Correct Answer : 4.5 cm
(xvi) sin 2A = 2 sin A is true when A equals
✔ Correct Answer : 0°
(xvii) If cos A = 12/13, then the value of sin A is
✔ Correct Answer : 5/13
(xviii) The angle subtended at the centre by the minute hand of a clock in 5 minutes is
✔ Correct Answer : 30°
Score: 0 / 18
Q.2. Fill in the blanks in the following (i to vi) (½ × 10 = 5)
(i) If 18, a, 10 are in arithmetic progression, then a = ______ .
(ii) The co-ordinates of the mid point of point P(7, −3) and point Q(3, 9) is ______ .
(iii) The value of sin60° cosec60° + cos30° sec30° is ______ .
(iv) If the diameter of a solid hemisphere is 14 cm, then its total surface area is ______ .
(v) The mode of the distribution 1, 4, 5, 6, 4, 7, 9, 2, 4, 3, 5 is ______ .
(vi) The class mark for any class interval is 17. If the upper class limit is 24, then the lower class limit is ______ .
(i) If 18, a, 10 are in arithmetic progression, then a = ______ .
(ii) The co-ordinates of the mid point of point P(7, −3) and point Q(3, 9) is ______ .
(iii) The value of sin60° cosec60° + cos30° sec30° is ______ .
(iv) If the diameter of a solid hemisphere is 14 cm, then its total surface area is ______ .
(v) The mode of the distribution 1, 4, 5, 6, 4, 7, 9, 2, 4, 3, 5 is ______ .
(vi) The class mark for any class interval is 17. If the upper class limit is 24, then the lower class limit is ______ .
Q.3. Very short Answer Type Question –(i to xii))
(i) If the radius of a circle is 14 cm and the length of the arc is 12 cm, then find the angle subtended by the arc at the centre.
(ii) If the total surface area of a cube is 864 square cm, then find the surface area of one of its faces.
(iii) Find the median of the following frequency distribution.
(iv) In a single throw of a die, determine the probability of getting a number more than 5.
(v) A right circular cone and a cylinder are of equal radius and equal height. If the volume of the cone is 66 cubic cm, then find the volume of the cylinder.
(vi) Find median of the following data: 19, 17, 25, 27, 18, 20, 29
(vii) Two players Ram and Shyam play a chess match. It is given that probability of winning the match by Ram is 4/5. Find the probability of winning the match by Shyam.
(viii) A solid cuboid is formed by joining the adjacent faces of two cubes, each of side 2 cm. Find the volume of the resulting cuboid.
(ix) Find the arithmetic mean of the first ten positive odd natural numbers.
(x) A bag contains 6 red and 7 white balls. One ball is drawn randomly. Find the probability that the ball drawn is white.
(xi) If the curved surface area of a solid hemisphere is 50π square cm, then find the radius of that hemisphere.
(xii) If the arithmetic mean of 5, 7, 9, 4, 3, (x + 2) is 6, then find the value of x.
(i) If the radius of a circle is 14 cm and the length of the arc is 12 cm, then find the angle subtended by the arc at the centre.
(ii) If the total surface area of a cube is 864 square cm, then find the surface area of one of its faces.
(iii) Find the median of the following frequency distribution.
| Class | Frequency |
|---|---|
| 0 – 10 | 3 |
| 10 – 20 | 7 |
| 20 – 30 | 5 |
| 30 – 40 | 6 |
(iv) In a single throw of a die, determine the probability of getting a number more than 5.
(v) A right circular cone and a cylinder are of equal radius and equal height. If the volume of the cone is 66 cubic cm, then find the volume of the cylinder.
(vi) Find median of the following data: 19, 17, 25, 27, 18, 20, 29
(vii) Two players Ram and Shyam play a chess match. It is given that probability of winning the match by Ram is 4/5. Find the probability of winning the match by Shyam.
(viii) A solid cuboid is formed by joining the adjacent faces of two cubes, each of side 2 cm. Find the volume of the resulting cuboid.
(ix) Find the arithmetic mean of the first ten positive odd natural numbers.
(x) A bag contains 6 red and 7 white balls. One ball is drawn randomly. Find the probability that the ball drawn is white.
(xi) If the curved surface area of a solid hemisphere is 50π square cm, then find the radius of that hemisphere.
(xii) If the arithmetic mean of 5, 7, 9, 4, 3, (x + 2) is 6, then find the value of x.
Q.4. Find the HCF and LCM of 12, 15 and 21 using the prime factorisation method.
Solution:
12 = 2 × 2 × 3 = 2² × 3
15 = 3 × 5
21 = 3 × 7
HCF is the product of the common prime factors with the smallest powers.
HCF = 3
LCM is the product of the highest powers of all prime factors.
LCM = 2² × 3 × 5 × 7 = 420
∴ HCF = 3 and LCM = 420.
12 = 2 × 2 × 3 = 2² × 3
15 = 3 × 5
21 = 3 × 7
HCF is the product of the common prime factors with the smallest powers.
HCF = 3
LCM is the product of the highest powers of all prime factors.
LCM = 2² × 3 × 5 × 7 = 420
∴ HCF = 3 and LCM = 420.
Q.5. If α and β are the zeroes of the quadratic polynomial
3x² − 5x + 9, then find (α + β) and αβ.
Solution:
Given quadratic polynomial is:
3x² − 5x + 9
Here, a = 3, b = −5, c = 9
Sum of zeroes = α + β = −b/a
= −(−5)/3
= 5/3
Product of zeroes = αβ = c/a
= 9/3
= 3
∴ (α + β) = 5/3 and αβ = 3.
Given quadratic polynomial is:
3x² − 5x + 9
Here, a = 3, b = −5, c = 9
Sum of zeroes = α + β = −b/a
= −(−5)/3
= 5/3
Product of zeroes = αβ = c/a
= 9/3
= 3
∴ (α + β) = 5/3 and αβ = 3.
Q.6. Use elimination method to find all possible solutions of the following pair of linear equations:
3x + 5y = 7
6x + y = −4
Solution:
3x + 5y = 7 …(1)
6x + y = −4 …(2)
Multiply equation (2) by 5:
30x + 5y = −20 …(3)
Subtract equation (1) from equation (3):
(30x + 5y) − (3x + 5y) = −20 − 7
27x = −27
x = −1
Substitute x = −1 in equation (1):
3(−1) + 5y = 7
−3 + 5y = 7
5y = 10
y = 2
∴ The solution is x = −1, y = 2.
3x + 5y = 7 …(1)
6x + y = −4 …(2)
Multiply equation (2) by 5:
30x + 5y = −20 …(3)
Subtract equation (1) from equation (3):
(30x + 5y) − (3x + 5y) = −20 − 7
27x = −27
x = −1
Substitute x = −1 in equation (1):
3(−1) + 5y = 7
−3 + 5y = 7
5y = 10
y = 2
∴ The solution is x = −1, y = 2.
Q.7. Find the number of terms in arithmetic progression
7, 13, 19, . . . , 205.
Solution:
Given A.P. is: 7, 13, 19, . . . , 205
First term, a = 7
Common difference, d = 13 − 7 = 6
Last term, l = 205
Using the formula:
l = a + (n − 1)d
205 = 7 + (n − 1) × 6
205 − 7 = 6(n − 1)
198 = 6(n − 1)
n − 1 = 33
n = 34
∴ The number of terms is 34.
Given A.P. is: 7, 13, 19, . . . , 205
First term, a = 7
Common difference, d = 13 − 7 = 6
Last term, l = 205
Using the formula:
l = a + (n − 1)d
205 = 7 + (n − 1) × 6
205 − 7 = 6(n − 1)
198 = 6(n − 1)
n − 1 = 33
n = 34
∴ The number of terms is 34.
Q.8. In the given figure, DE ∥ BC, then find the value of x.
Solution:
Since DE ∥ BC, therefore by Basic Proportionality Theorem:
AD / DB = AE / EC
From the figure:
AD = (x + 2), DB = 6
AE = 4, EC = 8
(x + 2) / 6 = 4 / 8
(x + 2) / 6 = 1 / 2
Cross multiplying:
2(x + 2) = 6
x + 2 = 3
x = 1
∴ The value of x is 1.
Since DE ∥ BC, therefore by Basic Proportionality Theorem:
AD / DB = AE / EC
From the figure:
AD = (x + 2), DB = 6
AE = 4, EC = 8
(x + 2) / 6 = 4 / 8
(x + 2) / 6 = 1 / 2
Cross multiplying:
2(x + 2) = 6
x + 2 = 3
x = 1
∴ The value of x is 1.
Q.9. In which ratio, x-axis divides the line segment which joins points
(5, 3) and (−3, −2) ?
Solution:
Let the point P divide the line joining A(5, 3) and B(−3, −2) in the ratio m : n.
Since P lies on the x-axis, its y-coordinate is 0.
Using section formula:
y-coordinate of P = (m y₂ + n y₁) / (m + n)
0 = (m × (−2) + n × 3) / (m + n)
−2m + 3n = 0
2m = 3n
m : n = 3 : 2
∴ The x-axis divides the line segment in the ratio 3 : 2.
Let the point P divide the line joining A(5, 3) and B(−3, −2) in the ratio m : n.
Since P lies on the x-axis, its y-coordinate is 0.
Using section formula:
y-coordinate of P = (m y₂ + n y₁) / (m + n)
0 = (m × (−2) + n × 3) / (m + n)
−2m + 3n = 0
2m = 3n
m : n = 3 : 2
∴ The x-axis divides the line segment in the ratio 3 : 2.
Q.10. Find the value of
4cot²45° − sec²60° + sin²60°.
Solution:
We know that:
cot45° = 1
sec60° = 2
sin60° = √3/2
Substituting the values:
4cot²45° − sec²60° + sin²60°
= 4(1)² − (2)² + (√3/2)²
= 4 − 4 + 3/4
= 3/4
∴ The required value is 3/4.
We know that:
cot45° = 1
sec60° = 2
sin60° = √3/2
Substituting the values:
4cot²45° − sec²60° + sin²60°
= 4(1)² − (2)² + (√3/2)²
= 4 − 4 + 3/4
= 3/4
∴ The required value is 3/4.
Q.11. A 12 meter long ladder touches the top of a vertical wall.
If this ladder makes an angle of 60° with the wall, then find height of the wall.
Solution:
Let the height of the wall be h meters.
Given:
Length of ladder = 12 m
Angle between ladder and wall = 60°
In right angled triangle,
cos 60° = height of wall / length of ladder
cos 60° = h / 12
1/2 = h / 12
h = 12 × 1/2
h = 6 m
∴ The height of the wall is 6 meters.
Let the height of the wall be h meters.
Given:
Length of ladder = 12 m
Angle between ladder and wall = 60°
In right angled triangle,
cos 60° = height of wall / length of ladder
cos 60° = h / 12
1/2 = h / 12
h = 12 × 1/2
h = 6 m
∴ The height of the wall is 6 meters.
Q.12. A quadrilateral ABCD is drawn to circumscribe a circle.
Prove that AB + CD = AD + BC.
Solution:
Given: A quadrilateral ABCD circumscribes a circle touching
AB, BC, CD and DA at points P, Q, R and S respectively.
According to the property of tangents from an external point:
AP = AS
BP = BQ
CQ = CR
DR = DS
Now,
AB = AP + PB
BC = BQ + QC
CD = CR + RD
AD = AS + SD
Adding AB and CD:
AB + CD = (AP + PB) + (CR + RD)
= (AS + BQ) + (CQ + DS)
= (AS + SD) + (BQ + QC)
= AD + BC
∴ AB + CD = AD + BC.
Hence proved.
Given: A quadrilateral ABCD circumscribes a circle touching
AB, BC, CD and DA at points P, Q, R and S respectively.
According to the property of tangents from an external point:
AP = AS
BP = BQ
CQ = CR
DR = DS
Now,
AB = AP + PB
BC = BQ + QC
CD = CR + RD
AD = AS + SD
Adding AB and CD:
AB + CD = (AP + PB) + (CR + RD)
= (AS + BQ) + (CQ + DS)
= (AS + SD) + (BQ + QC)
= AD + BC
∴ AB + CD = AD + BC.
Hence proved.
Q.13. The angle subtended at the centre by an arc of a circle is 50°.
If the length of the arc is 5π cm, then find the radius of that circle.
Solution:
Given:
Angle subtended at the centre, θ = 50°
Length of arc, l = 5π cm
We know that:
l = (θ / 360°) × 2πr
Substitute the given values:
5π = (50 / 360) × 2πr
5π = (5 / 18) × πr
Cancelling π on both sides:
5 = 5r / 18
r = 18 cm
∴ The radius of the circle is 18 cm.
Given:
Angle subtended at the centre, θ = 50°
Length of arc, l = 5π cm
We know that:
l = (θ / 360°) × 2πr
Substitute the given values:
5π = (50 / 360) × 2πr
5π = (5 / 18) × πr
Cancelling π on both sides:
5 = 5r / 18
r = 18 cm
∴ The radius of the circle is 18 cm.
Q.14. Find the sum of first 15 terms of arithmetic progression,
whose nth term is an = 3 + 2n.
OR
There are 60 terms in an arithmetic progression. If its first and last terms are 7 and 125 respectively, then find its 32nd term.
OR
There are 60 terms in an arithmetic progression. If its first and last terms are 7 and 125 respectively, then find its 32nd term.
Solution (First part):
Given:
an = 3 + 2n
For n = 1,
a = 3 + 2(1) = 5
Common difference,
d = a2 − a1 = (3 + 4) − (3 + 2) = 2
Number of terms, n = 15
Sum of n terms of an A.P. is given by:
Sn = n/2 [2a + (n − 1)d]
S15 = 15/2 [2×5 + (15 − 1)×2]
= 15/2 [10 + 28]
= 15/2 × 38
= 285
∴ The sum of first 15 terms is 285.
Given:
an = 3 + 2n
For n = 1,
a = 3 + 2(1) = 5
Common difference,
d = a2 − a1 = (3 + 4) − (3 + 2) = 2
Number of terms, n = 15
Sum of n terms of an A.P. is given by:
Sn = n/2 [2a + (n − 1)d]
S15 = 15/2 [2×5 + (15 − 1)×2]
= 15/2 [10 + 28]
= 15/2 × 38
= 285
∴ The sum of first 15 terms is 285.
Q.15. Find the co-ordinates of points which divide the line segment
joining points (4, 0) and (0, −8) into 4 equal parts.
OR
Find the co-ordinates of a point A, where AB is a diameter of a circle whose centre is (2, −3) and co-ordinate of B is (1, 4).
OR
Find the co-ordinates of a point A, where AB is a diameter of a circle whose centre is (2, −3) and co-ordinate of B is (1, 4).
Solution (First part):
Let A(4, 0) and B(0, −8).
The points divide AB internally in the ratios 1:3, 2:2 and 3:1.
Using section formula:
First point P divides AB in ratio 1 : 3
P = ((1×0 + 3×4)/4 , (1×(−8) + 3×0)/4)
P = (3, −2)
Second point Q divides AB in ratio 2 : 2
Q = ((2×0 + 2×4)/4 , (2×(−8) + 2×0)/4)
Q = (2, −4)
Third point R divides AB in ratio 3 : 1
R = ((3×0 + 1×4)/4 , (3×(−8) + 1×0)/4)
R = (1, −6)
∴ The required points are (3, −2), (2, −4) and (1, −6).
Let A(4, 0) and B(0, −8).
The points divide AB internally in the ratios 1:3, 2:2 and 3:1.
Using section formula:
First point P divides AB in ratio 1 : 3
P = ((1×0 + 3×4)/4 , (1×(−8) + 3×0)/4)
P = (3, −2)
Second point Q divides AB in ratio 2 : 2
Q = ((2×0 + 2×4)/4 , (2×(−8) + 2×0)/4)
Q = (2, −4)
Third point R divides AB in ratio 3 : 1
R = ((3×0 + 1×4)/4 , (3×(−8) + 1×0)/4)
R = (1, −6)
∴ The required points are (3, −2), (2, −4) and (1, −6).
Q.16. Prove that in two concentric circles, the chord of the larger circle,
which touches the smaller circle, is bisected at the point of contact.
OR
Prove that the lengths of tangents drawn from an external point to a circle are equal.
OR
Prove that the lengths of tangents drawn from an external point to a circle are equal.
Solution (First part):
Let two concentric circles have centre O.
Let AB be a chord of the larger circle which touches the smaller circle at point P.
Join OP.
Since OP is a radius of the smaller circle and AB touches it at P,
therefore OP ⟂ AB.
The perpendicular drawn from the centre of a circle to a chord bisects the chord.
Hence, OP bisects AB at point P.
∴ The chord of the larger circle is bisected at the point of contact.
Hence proved.
Let two concentric circles have centre O.
Let AB be a chord of the larger circle which touches the smaller circle at point P.
Join OP.
Since OP is a radius of the smaller circle and AB touches it at P,
therefore OP ⟂ AB.
The perpendicular drawn from the centre of a circle to a chord bisects the chord.
Hence, OP bisects AB at point P.
∴ The chord of the larger circle is bisected at the point of contact.
Hence proved.
Q.17. Find the mean of the following frequency distribution.
OR
If mean of the following distribution is 7, then find the value of P.
| x | 5 | 6 | 7 | 8 | 9 | 10 | 11 |
|---|---|---|---|---|---|---|---|
| f | 5 | 8 | 9 | 12 | 6 | 6 | 4 |
OR
If mean of the following distribution is 7, then find the value of P.
| x | 2 | 5 | P | 9 | 10 |
|---|---|---|---|---|---|
| f | 1 | 5 | 4 | 7 | 3 |
Solution (First part):
Mean = Σfx / Σf
Mean = 390 / 50 = 7.8
OR
Mean = Σfx / Σf
| x | f | fx |
|---|---|---|
| 5 | 5 | 25 |
| 6 | 8 | 48 |
| 7 | 9 | 63 |
| 8 | 12 | 96 |
| 9 | 6 | 54 |
| 10 | 6 | 60 |
| 11 | 4 | 44 |
| Total | 50 | 390 |
Mean = 390 / 50 = 7.8
Solution (Second part):
Mean = 7
Σf = 1 + 5 + 4 + 7 + 3 = 20
Σfx = (2×1) + (5×5) + (P×4) + (9×7) + (10×3)
= 2 + 25 + 4P + 63 + 30
= 120 + 4P
Mean = Σfx / Σf
7 = (120 + 4P) / 20
140 = 120 + 4P
4P = 20
P = 5
∴ The value of P is 5.
Mean = 7
Σf = 1 + 5 + 4 + 7 + 3 = 20
Σfx = (2×1) + (5×5) + (P×4) + (9×7) + (10×3)
= 2 + 25 + 4P + 63 + 30
= 120 + 4P
Mean = Σfx / Σf
7 = (120 + 4P) / 20
140 = 120 + 4P
4P = 20
P = 5
∴ The value of P is 5.
Q.18. The difference of squares of two numbers is 180.
The square of the smaller number is eight times the larger number.
Find both the numbers.
OR
Find two consecutive positive integers, whose sum of squares is 365.
OR
Find two consecutive positive integers, whose sum of squares is 365.
Solution (First part):
Let the larger number be x and the smaller number be y.
According to the question:
x² − y² = 180 …(1)
y² = 8x …(2)
Substitute (2) in (1):
x² − 8x = 180
x² − 8x − 180 = 0
x² − 18x + 10x − 180 = 0
x(x − 18) + 10(x − 18) = 0
(x − 18)(x + 10) = 0
x = 18 (reject −10)
From (2):
y² = 8×18 = 144
y = 12
∴ The two numbers are 18 and 12.
OR
Let the larger number be x and the smaller number be y.
According to the question:
x² − y² = 180 …(1)
y² = 8x …(2)
Substitute (2) in (1):
x² − 8x = 180
x² − 8x − 180 = 0
x² − 18x + 10x − 180 = 0
x(x − 18) + 10(x − 18) = 0
(x − 18)(x + 10) = 0
x = 18 (reject −10)
From (2):
y² = 8×18 = 144
y = 12
∴ The two numbers are 18 and 12.
Solution (Second part):
Let the two consecutive positive integers be x and x + 1.
According to the question:
x² + (x + 1)² = 365
x² + x² + 2x + 1 = 365
2x² + 2x − 364 = 0
x² + x − 182 = 0
x² + 14x − 13x − 182 = 0
x(x + 14) − 13(x + 14) = 0
(x − 13)(x + 14) = 0
x = 13 (reject −14)
∴ The two consecutive positive integers are 13 and 14.
Let the two consecutive positive integers be x and x + 1.
According to the question:
x² + (x + 1)² = 365
x² + x² + 2x + 1 = 365
2x² + 2x − 364 = 0
x² + x − 182 = 0
x² + 14x − 13x − 182 = 0
x(x + 14) − 13(x + 14) = 0
(x − 13)(x + 14) = 0
x = 13 (reject −14)
∴ The two consecutive positive integers are 13 and 14.
Q.19. Prove that
1/(1 + sinθ) + 1/(1 − sinθ) = 2sec²θ.
OR
Prove that
sin²θcosθ + cos³θ + tanθsinθ = secθ.
1/(1 + sinθ) + 1/(1 − sinθ) = 2sec²θ.
OR
Prove that
sin²θcosθ + cos³θ + tanθsinθ = secθ.
Solution (First part):
LHS = 1/(1 + sinθ) + 1/(1 − sinθ)
Taking LCM:
= (1 − sinθ + 1 + sinθ) / (1 − sin²θ)
= 2 / cos²θ
= 2sec²θ
∴ LHS = RHS.
OR
LHS = 1/(1 + sinθ) + 1/(1 − sinθ)
Taking LCM:
= (1 − sinθ + 1 + sinθ) / (1 − sin²θ)
= 2 / cos²θ
= 2sec²θ
∴ LHS = RHS.
Solution (Second part):
LHS = sin²θcosθ + cos³θ + tanθsinθ
= cosθ(sin²θ + cos²θ) + (sinθ/cosθ)sinθ
= cosθ(1) + sin²θ/cosθ
= cosθ + (1 − cos²θ)/cosθ
= cosθ + 1/cosθ − cosθ
= secθ
∴ LHS = RHS.
LHS = sin²θcosθ + cos³θ + tanθsinθ
= cosθ(sin²θ + cos²θ) + (sinθ/cosθ)sinθ
= cosθ(1) + sin²θ/cosθ
= cosθ + (1 − cos²θ)/cosθ
= cosθ + 1/cosθ − cosθ
= secθ
∴ LHS = RHS.
Q.20. Find the median of the following frequency distribution.
OR
Find the mode of the following frequency distribution.
OR
| Class | 7–17 | 17–27 | 27–37 | 37–47 | 47–57 | 57–67 |
|---|---|---|---|---|---|---|
| Frequency | 22 | 18 | 20 | 12 | 15 | 13 |
OR
Find the mode of the following frequency distribution.
| Class | 2–11 | 11–20 | 20–29 | 29–38 | 38–47 |
|---|---|---|---|---|---|
| Frequency | 15 | 16 | 17 | 12 | 11 |
Solution (Median):
Total frequency (N) = 22 + 18 + 20 + 12 + 15 + 13 = 100
N/2 = 50
Cumulative frequencies:
Median class = 27–37
l = 27, h = 10, f = 20, c.f. = 40
Median = l + [(N/2 − c.f.)/f] × h
= 27 + [(50 − 40)/20] × 10
= 27 + 5
= 32
Total frequency (N) = 22 + 18 + 20 + 12 + 15 + 13 = 100
N/2 = 50
Cumulative frequencies:
| Class | Frequency | c.f. |
|---|---|---|
| 7–17 | 22 | 22 |
| 17–27 | 18 | 40 |
| 27–37 | 20 | 60 |
| 37–47 | 12 | 72 |
| 47–57 | 15 | 87 |
| 57–67 | 13 | 100 |
Median class = 27–37
l = 27, h = 10, f = 20, c.f. = 40
Median = l + [(N/2 − c.f.)/f] × h
= 27 + [(50 − 40)/20] × 10
= 27 + 5
= 32
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