Kerala Syllabus Class 10 Maths Model Question Paper Set 1

Practicing Maths Question Paper Class 10 Kerala Syllabus Set 1 English Medium helps identify strengths and weaknesses in a subject.

Maths Class 10 Kerala Syllabus Model Question Paper Set 1

Time: 2½ Hours
Total Score: 80 Marks

(Answer any 3 questions from 1 to 4. Each question carries 2 scores)

Question 1.
a) Write the sequence got by multiplying consecutive natural numbers by 3 and adding 2.
b) What is the 10^th term of this sequence?
Answer:
a) 5, 8, 11, ……
b) 10^th term = 5 + (9 × 3)
= 5 + 27 = 32

Question 2.

a) From the figure, find the values of x and y.
b) Find the measure of ∠ADC?
Answer:
a) 70 + x + 40 + x = 180
2x + 110 = 180
2x = 180 – 110 = 70
x = 35 (Angles at the opposit vertices of a cyclic quadrilateral are supplementary)
50 + y + 60 + 2y = 180
3y + 110 = 180, 3y = 70
y = \(\frac{70}{3}\)

b) ∠ADC = 50 + \(\frac{70}{3}\)

Question 3.
a) Find the slope of the line joining the points (3, 5) and (5, 6).
b) Write the coordinates of two other points on this line.
Answer:
a. Slope of the line = \(\frac{6 – 5}{5 – 3}\) = \(\frac{1}{2}\)
b. The other two points on the circle (7, 7), (9, 8)

Question 4.
If p(x) = x³ – 5x² + 3x + 6.
a) Find p(2)
b) If (x – 2) is a factor of the polynomial x³ – 5x² + 3x + k, what is the value of k?
Answer:
a. p(2) = 2³ – 5 × 2² + 3 × 2 + 6
= 8 – 20 + 6 + 6 = 0

b. If (x – 2) is a factor, p(2) = 0
2³ – 5 × 2² + 3 × 2 + k = 0
8 – 20 + 6 + k = 0
k – 6 = 0, k = 6

(Answer any 5 questions from 5 to 10. Each question carries 3 scores)

Question 5.
The sum of the first n terms of an arithmetic sequence is 3x² + 5n.
a) Find the first term.
b) Find the common difference of this sequence.
c) Write the sequence.
Answer:
a) If n = 1, the first term a₁
= 3 × 1² + 5 × 1 = 8
If n = 2, a₁ + a₂
= 3 × 2² + 5 × 2 = 22
2^nd term = 22 – 8 = 14

b) Common difference = 14 – 8 = 6

c) 8, 16, 22, 28, ……………..

Question 6.
The sum of the squares of three consecutive natural numbers is 110.
a) Find the numbers.
b) Write the sequence of odd numbers greater than 1.
c) What is the algebraic form of this sequence?
Answer:
a) Consecutive numbers: x, x + 1, x + 2
x² + (x + 1)² + (x + 2)² = 110
x² + x² + 2x + 1 + x² + 4x + 4 = 110
3x² + 6x + 5 = 110
3x² + 6x- 105 = 0,
x² + 2x – 35 = 0
(x + 7) (x – 5) = 0
x = -7, 5
Numbers: 5, 6, 7

b) 3, 5, 7, 9, 11, ……………

c) x_n = dn + (f – d) = 2n + (3 – 2) = 2n + 1

Question 7.
In the figure, PA and PB are tangents. O is the centre of the circle.

a) If ∠PAB = 50°, find the measures of ∠AQB,
b) Find ∠AOB and
c) Find ∠APB.
Answer:

a) ∠AQB = 50°
(The angle which a chord(AB) makes with the tangent is equal to the angle which it makes on the other side.)
b) ∠AOB = 100°
c) ∠APB = 180- 100 = 80°

Question 8.
In the figure, B = 90°, A = 36°

a) What is the measure of ∠C?
b) If AB = 3 cm, BC = 4 cm, find the length of AC.
c) What is the value of sin A × cos C?
Answer:

a. ∠C = 54°
b. AC = 5 cm
c. sin A × cos C = \(\frac{4}{5}\) × \(\frac{4}{5}\) = \(\frac{16}{25}\)

Question 9.
a) Draw a triangle of circum radiu? 3 centimetres and two of the angles 64° and 75°.
b) Measure the length of the sides.
Answer:
Draw a circle of radius 3 centimetres. Draw radius OA. Draw radius OB which make an angle 128° with OA and draw OC which make 150° with OB. Join AB, BC and AC
The required triangle is ready.

Question 10.
One lateral face of a square pyramid is given in the picture below.

a) What is the slant height?
b) What is the total surface area?
Answer:
a. Slant height l² = e² – (a/2)²
= 17² – 8² = 225
I = \(\sqrt{225}\) = 15 cm

b. Total surface arfea = a² + 2al
= 16² + 2 × 16 × 15
= 256 + 480 = 736 cm²

(Answer any 8 questions from 11 to 21. Each question carries 4 scores)

Question 11.
a) Write the coordinates of the vertices of equilateral triangle given in the figure,
b) What is the height from vertex to OX?

Answer:
a)

BC² = 4² – 2² = 16 – 4 = 12
BC = \(\sqrt{12}\) = 2√3
B(2, 2√3)
Vertices: O(0, 0), A (4, 0), B (2, 2√3 )

b) Height = BC = 2√3

Question 12.

In the figure, x-coordinate of one point on the slanted line is 3.
a) What is the y-coordinate of this point?
b) What is the slope of the line?
c) What is the equation of the line?
Answer:

a. y coordinate: 2√3
b. Slope of the line = \(\frac{2 \sqrt{3}-0}{3-1}\) = \(\frac{2 \sqrt{3}}{2}\) = √3

c. Equations the line: y – 0 = √3 (x – 1)
That is, y = √3(x – 1)

Question 13.
a) Draw a triangle of sides 5 centimetres. 6 centimetres and 7 centimetres
b) Draw its incircle
c) Measure its radius
Answer:
a) Draw a triangle of sides 5 centimetres, 6 centimetres and 7 centimetres.
b) Draw bisectors of any two angles. Taking their point of intersection as centre and the perpendicular distance to the side as radius, draw the incircle.

Measure the radius and write it down.

Question 14.
What is
a) the centre of the circle.
b) radius of the circle.
c) the equation of the circle given below.

Answer:

∠AOB = 90°, AB is a diameter.
AB = \(\sqrt{(4-0)^2+(0-2)^2}\)
= \(\sqrt{16+4}\) = \(\sqrt{20}\) = 2√5
a) Radius = √5
b) Centre: (2, 1)
c) Equation: (x – 2)² (y – 1 )² = 5

Question 15.
In class 10 A, there are 15 boys and 18 girls. In 10 B, there are 19 boys and 16 girls. One student from each class to be selected for Math club.
a. What is the probability of both being boys?
b. What is the probability of both being girls?
c. What is the probability of at least one girl?
Answer:
10A
Boys – 15
Girls – 18
Total = 33

10B
Boys – 19
Girls – 16
Total = 35
Total number of pairs if one student is selected from each class = 33 × 35 = 1155

a. Probability of both being boys
= \(\frac{15 \times 19}{1155}\) = \(\frac{285}{1155}\)

b. Probability of both being girls
= \(\frac{18 \times 16}{1155}\) = \(\frac{288}{1155}\)

c. Probability of at least one girl
= \(\frac{(15 \times 16)+(18 \times 19)+(18 \times 16)}{1155}\)
= \(\frac{240+342+288}{1155}\) = = \(\frac{870}{1155}\)

Question 16.
a) Write the roots of the polynom ial 4x² – 16x + 15
b) Write it as the product of two first degree polynomials.
Answer:
a) The solutions of the equation 4x² – 16x + 15 = 0 is found out by using formula
x = \(\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}\)
b² – 4ac = (-16)² – 4 × 4 × 15
= 256 – 240 = 16
x = \(\frac{(-16) \pm \sqrt{16}}{2 \times 4}\) = \(\frac{16 \pm 4}{8}\) = \(\frac{20}{8}\), \(\frac{12}{8}\)
= \(\frac{5}{2}\), 3/2
x = 5/2
(2x – 5) is a factor
x = \(\frac{1}{2}\)
(2x – 3) is a factor

b) 4x² – 16x + 15 = (2x – 5) (2x – 3)

Question 17.
In the figure, P, Q and R are the points where the incircle of the triangle ABC touches the sides of the triangle. Find the measures of
a) ∠P, ∠Q and ∠R of triangle PQR.

Answer:

Join OP, OQ and OR.
∠POQ = 180° – 60° = 120°
∠QOR = 180° – 40° = 140°
∠POR = 180° – 80° = 100°

a) ∠P = \(\frac{140}{2}\) = 70°,
b) ∠Q = \(\frac{1}{2}\) × 100 = 50°
c) ∠R = \(\frac{1}{2}\) × 120 = 60° (Half the central angle)

Question 18.
a) What is the sum of the natural numbers from 1 to 30.
b) Algebraic form of an arithmetic sequence is 8n + 7. What is the sum of the first 30 terms of this sequence?
Answer:
a. 1 + 2 + 3 + ………….. + 30 = \(\frac{30 \times 31}{2}\)
= 15 × 31 = 465

b. Sequence having n^th term 8n + 7 is
15, 23, 31, ……………
Sum of the first 30 terms
= \(\frac{30}{2}\) [2 × 15 + 29 × 8]
= 15 × 262 = 3930

Question 19.
A circle of radius 10 centimetres is divided in tha ratio 2 : 3 to form two sectors. Two cones are made by rolling up the two sectors.
a) Find out the ratio between the base perimeters of the cones.
b) What is the ratio between the curved .surfaceareas?
Answer:
Ratio of the central angle is 2 : 3
Central angles 144°, 216°
Ratio of the arc lengths
\(\frac{144}{360}\) × 2π × 10 : \(\frac{216}{360}\) × 2π × 10 = 2 : 3

a) Ratio of the base perimeters of the cones
2 π r₁ : 2 π r₂ = 2 : 3
r₁ : r₂ = 2 : 3

b) Ratio of the lateral surface areas
π r₁ l : π r₂ l = 2 : 3

Question 20.
ABCD is a parallelogram. A, B, E and F are points on the circle.
If ∠DEF = 80°, find the measures of all angles of the quadrilateral AEFB

Answer:

∠AEF = 180 – 80 = 100
∠B = 80° (Cyclic quadrilateral)
∠A = 100°
(ABCD is a parallelogram.
Therefore ∠A + ∠B = 180)
∠D = 80°(Opposite angles of a parallelogram are equal.)
∠C = 100° (∠D + ∠C = 180)
∠F = 80 (ABFE is cyclic)

Question 21.
The one’s place of a two digit number is 4. The product qf the number and the digit sum is 238.
a) If ten’s place digit is taken as x, write the number.
b) Frame a second degree equation and find the number.
Answer:
a. If one’s place of the number is 4 and the ten’s place is x, then the dumber = 10x + 4

b. Product of the number and the digit sum
(x + 4) (10x + 4) = 238
10x² + 44x + 16 = 238
10x² + 44x – 222 = 0, 5x² + 22n – 111 = 0
x = \(\frac{-22 \pm \sqrt{22^2-4 \times 5 x-111}}{2 \times 5}\) = 3
Number = 10 × 3 + 4 = 34

(Answer any 5 questions from 22 to 28. Each question carries 5 scores)

Question 22.
Draw a rectangle of length 5 centimetres and breadth 4 centimetres and draw a rectangle of same area with length 6 centimetres.
Answer:
Draw a rectanglePBED of length 5 centimetres and breadth 4 centimetres.

Extend the bottom side of the rectangle to the left by 4 cm and mark A and extend left side downwards by 6 cm and mark C. Join AC and BC. Draw circumcircle of triangle ABC. The circle cuts PD at F. Extend PB to right by 1 cm and mark H. Complete the rectangle PFGH.

Question 23.
A girl standing on the top of a light house sees two boats on the left side of the light house at $ ^ngle of depressions 30° and 60°. The distance between the boats is 300 metres
a) Draw a rough sketch
b) Find the height of the light house
Answer:

b) In right triangle BAD,
tan 60° = \(\frac{A D}{A B}\) , AB = \(\frac{A D}{\tan 60}\)
x = \(\frac{h}{\sqrt{3}}\)
In Δ CAD, tan 30° = \(\frac{h}{x+300}\)
x + 300 = \(\frac{h}{\tan 30}\), x + 300 = h√3
\(\frac{h}{\sqrt{3}}\) = h√3 – 300
h = 3h – 300√3
2h = 300√3 , h = \(\frac{300 \sqrt{3}}{2}\) = 259. 8 m
Height of the light house = 259.8 m

Question 24.

a) Write the next two lines of the pattern above.
b) Find,the first and the last numbers of the 20^th line.
Answer:
a. The next two lines
34, 37, 40, 43, 46,
49, 52, 55, 58, 61, 64

b. Total numbers in 19 rows
1 + 2 + 3 + …………… + 19 = \(\frac{19 \times 20}{2}\) = 190
The first number of the 20^th row is the 19^th term of the sequence 4, 7, 10, ……
= a + 190d = 4 + 190 × 3 = 574
20^th row is 574, 577, 580 …. (20 numbers)
20^th term of this sequence is the last number of the 20^th row
= 574 + (19 × 3) = 631

Question 25.
In the figure, radius of the circle with centre O is 6 centimetres. Line AB touches the circle at P. If ∠OAB = 30°,
a) Write the coordinates of A and P.
b) Write the equation of AB.

Answer:

In Δ OAP, ∠OPA = 90°, ∠OAP = 30°, ∠AOP = 60°
∴ OA = 12 unit
In Δ OPD, ∠ODP = 90°, ∠DOP = 60°, ∠OPD = 30°
OD = 3 unit (Using the ratio 1 : √3 : 2)
OD = 3 unit
PD = 3√3 unit

a. A (12, 0)
P(3, 3√3)

b. Slope of AB = \(\frac{y_2-y_1}{x_2-x_1}\) = \(\frac{3 \sqrt{3}-0}{3-12}\) = \(\frac{-1}{\sqrt{3}}\)
Equation of the line is y – y₁ = \(\frac{-1}{\sqrt{3}}\) (x – x₁)
y – 0 = \(\frac{-1}{\sqrt{3}}\)(x – 12)
√3y = – x + 12
x + √3y – 12 = 0

Question 26.
A boiler is in the shape of two hemispheres of equal radius joined to the ends of a cylinder.
Total length of the boiler is 8 centimetres and common radius is 1 centimetre. What is the capacity of the boiler?
Answer:

Length of the cylinder 6m, radius 1m Volume = π r²h = π × 1² 6 = 6π m
Radius of hemisphere = 1 m
Volume of two hemispheres
= 2 × \(\frac{2}{3}\) π r³ = \(\frac{4}{3}\) π × 1³
Total volume = 6π + \(\frac{4}{3}\)π = \(\frac{22 \pi}{3}\) m³
= 23.02 m³ = 23.02 × 1000 = 23020 litres

Question 27.
A diameter of a circle joins the points (0, 1) and (2, 3).
a) Find the coordinates of the centre.
b) What is the radius?
c) If (x, y) is a point on the circle, prove that x² + y² – 2x – 4y + 3 = 0.
Answer:
a) Centre of the circle = midpoint of the line joining the points (0, 1) and (2, 3)
= (\(\frac{0+2}{3}\), \(\frac{1+3}{2}\)) = (1, 2)

b) Radius of the circle = distance between the points (1, 2) and (2, 3)
= \(\sqrt{(2-1)^2+(3-2)^2}\) = \(\sqrt{1+1}\) = √2

c) (x, y) is a point on the circle,
the distance between (x, y) and (1, 2) is equal to √2.
\(\sqrt{(x-1)^2+(y-2)^2}\) = √2
(x – 1)² + (y – 2)² = 2
x² + y² – 2x – 4y + 3 = 0

Question 28.
The table below shows the students of a class sorted according to their heights.

Height (cm)	Number of students
135 – 140	5
140 – 145	8
145 – 150	10
150 – 155	9
155 – 160	6
160 – 165	3

a) If the students are arranged in increasing order of their heights, the height of the student at what position is taken as the median?
b) According to table between which measures comes the height of this student?
c) Find the median height.
Answer:

Height(cm)	Number of student	Height( cm)	Number of students
135 – 140	5	below 140	5
140 – 145	8	below 145	13
145 – 150	10	below 150	23
150 – 155	9	below 155	32
155 – 160	6	below 160	38
160 – 165	3	below 165	41

The median height is the height of \(\frac{41+1}{2}\) = \(\frac{42}{2}\) = 21^st student. According to this, heights of 10 students from 14^th to 23^rd positions in the order of heights are between 145 and 150. We divide 5 units between 145 and 150 into 10 equal parts, and assume that each subdivision contains one student whose height is the midvalue of this class. Then the height of the 14^th student is the midvalue of 145 and 145\(\frac{1}{2}\).That is 145.25. The height of each succeeding student increases by \(\frac{1}{2}\) unit. Therefore height of the 21^st student = 145\(\frac{1}{4}\) + (7 × \(\frac{1}{2}\) )
= 145.25 + 3.5 = 148.75

a. Median height is the height of the 21^st student.
b. Between 145 and 150
c. Median height = 148.75 cm

Question 29.
Read the following, understand the mathematical idea expressed in it and answer the questions that follow:
1, 4, 9, 16 are the squares of the counting numbers. The remainders got by dividing the
squafe numbers with natural numbers have a cyclic property. For example, the remainders on dividing these numbers by 4 are tabulated here.

Number	1	4	9	16	25	—	—	—
Remainder	1	0	1	0	1	—	—	—

On dividing by 4, perfect squares leave only 0 and 1 as remainders. From this, we can conclude that an arithmetic sequence whose terms leaves remainder 2 on dividing by 4 do not have a perfect square.
a) Which are the possible remainders.on dividing any number with 4?
b) Which are the numbers we would not get on dividing a perfect square by 4?
c) What is the remainder that leaves on dividing terms of the arithmetic sequence 2, 5, 8, 11, ………………. by 4?
d) Does the arithmetic sequence 3, 7, 11 ……………….. contain perfect squares?
e) Write a sequence with common difference 4 which contains many perfect squares. (6 × 1 = 6)
Answer:
a. The possible remainders on dividing any number with 4 : 0, 1, 2, 3
b. The numbers we would not get on dividing a perfect square by 4 = 2, 3
c. The remainder that leaves on dividing terms of the arithmetic sequence 2, 5, 8, 11, ……………. by 4 : 0, 1, 2, 3
d. 3, 7, 11 is a sequence with common difference 4 which does not contain any perfect squares.
e. 4, 8, 12, 16 is a sequence with common difference 4 which contains many perfect squares.