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.

Class 10 Maths Kerala Syllabus Model Question Paper Set 1

Time: 2½ Hours
Total Score: 80 Marks

Instructions:

• Use the first 15 minutes to read the questions and think about the answers
• There are 26 questions, split into four parts A, B, C, D
• Answer all questions; but in questions of the type A or B, you need answer only one of those
• You can answer the questions in any order, writing the correct question number
• Answers must be explained, whenever necessary.
• No need to simplify irrationals like √2, √3, etc using approximations unless you are asked to do so.

Section – A

Question 1.
At what points does the graph of the polynomial 4x² – 8x + 3 cross the x-axis? (1 mark)
(a) \(\left(\frac{3}{4}, 0\right),\left(\frac{1}{2}, 0\right)\)
(b) \(\left(\frac{3}{2}, 0\right),\left(\frac{1}{2}, 0\right)\)
(c) \(\left(\frac{3}{4}, 0\right),\left(\frac{1}{6}, 0\right)\)
(d) \(\left(\frac{7}{2}, 0\right),\left(\frac{3}{2}, 0\right)\)
Answer:
(b) \(\left(\frac{3}{2}, 0\right),\left(\frac{1}{2}, 0\right)\)
4x² – 8x + 3 = 0
⇒ a = 4, b = -8, c = 3

So, the points at which the graph of the polynomial
4x² – 8x + 3 cross the x-axis are \(\left(\frac{3}{2}, 0\right),\left(\frac{1}{2}, 0\right)\)

Question 2.
Read the two statements given below.
Statement 1: The scores 10, 18, 14, 20, 12, 16 have mean and median equal.
Statement 2: The numbers form an arithmetic sequence when arranged in order.

Choose the correct answer from those given below. 6.
(a) Statement 1 is true and Statement 2 is false.
(b) Statement 1 is false and Statement 2 is true.
(c) Both the statements are true, and Statement 2 is the correct reason for Statement 1.
(d) Both the statements are true, but Statement 2 is not the correct reason for Statement 1. (1 mark)
Answer:
(c) Both the statements are true, and Statement 2 is the correct reason for Statement 1.

Question 3.
A) What is the probability of getting 5 Sundays in the month December?
Answer:
There are 31 days in December. 28 days decide 4weeks, so four Sundays.
The combinations are (Sunday, Monday, Tuesday), (Monday, Tuesday, Wednesday), (Tuesday, Wednesday, Thursday), (Wednesday, Thusday, Friday), (Thusday, Friday, Saturday), (Friday, Saturday, Sunday), (Saturday, Sunday, Monday).
There are three combinations in which Sundays occur.Probability of occuring five Sundays is \(\frac{3}{7}\)

OR

B) nth term of an arithmetic sequence is 1 – 4n
(a) What is the common difference ?
Answer:
-4

(b) Find the sum of first 25 terms of this sequence.
Answer:
x₂₅ = 1 – 4 × 25 = -99
x₁ = 1 – 4 × 1 = -3
Sum = (x₁ + x₂₅) × \(\frac{25}{2}\)
= (-3 + -99) × \(\frac{25}{2}\)
= -1275

Question 4.
ABC is an equilateral triangle of side 12 cm in which AD is perpendicular to BC.

(a) What are the angles of ∆ADB?
Answer:
∠A = 30°, ∠B = 60°, ∠D = 90°

(b) What are the sides of triangle ADB
Answer:
AB = 12 cm, RD = 6 cm, AD = 6√3 cm

(c) What is the area of the square ADEF? (4 mark)
Answer:
Area = (6√3)² = 36 × 3 = 108 cm²

Question 5.
A) In triangle ABC, a, b, c are the sides opposite to the angles A, B and C and R is the radius of the circumcircle.
(a) Prove that \(\frac{b}{\sin B}\) = 2R ?
Answer:
Draw the diameter AP and join PC.
Triangle APC is a right triangle
∠B = ∠P
sin B = \(\frac{b}{A P}=\frac{b}{2 R}\)
where R is the radius of the circumcircle.
2R = \(\frac{b}{\sin B}\)

(b) Establish the relation the area of the circle A = \(\frac{a b c}{4 R}\) ?
Answer:
We know that Area = \(\frac{1}{2}\) × ac × sin B
= \(\frac{1}{2}\) ac × \(\frac{b}{2 R}=\frac{a b c}{4 R}\)

OR

B) Manju’s age after 15 years will be the square of his age before 15 years.
(a) If the present age is x then what is the age before and after 15 years.
Answer:
x – 15 and x + 15

(b) Write the equation connecting the given conditions.
Answer:
(x – 15)² = (x + 15),
x² – 30x + 225
= x + 15, x² – 31x + 210 = 0

(c) Find the present age (4 mark)
Answer:
Solving x = 21. Present age is 21 years

Question 6.
There are two circles in the picture.One is inside other. Radius of the small circle is half of the radius of the big circle.

(a) If the radius of the small circle is r then what is the area of the small circle and big circle ?
Answer:
Area of small circle πr²
Area of big circle π × (2r)² = 4πr²

(b) If a fine dot is placed into the figure, what is the probability of falling the dot in the small circle?
Answer:
Chance of getting black \(\frac{\pi r^2}{4 \pi r^2}=\frac{1}{4}\)

(c) What is the probability of falling the dot the yellow shaded part in the figure. (5 mark)
Answer:
Probability of falling the dot in the yellow shade is 1 – \(\frac{1}{4}=\frac{3}{4}\)

Section – B

Question 7.
The figure shows one lateral face of a square pyramid. Its sides are 5 centimetres, 5 centimetres and 6 centimetres. What would be the slant height of square pyramid in centimetre? (1 mark)

(a) 3 cm
(b) 4 cm
(c) 5 cm
(d) 6 cm
Answer:
(b) 4 cm

In the figure, height of the triangle is the slant height of the pyramid.
Slant height = \(\sqrt{5^2-3^2}\)
= \(\sqrt{16}\)
= 4cm

Question 8.
The end points of the diameter of a circle are (1, 1) and (7, 9). What is the center of the circle?
(a) (4, 5)
(b) (1, 3)
(c) (3, 1)
(d) (0, 2) (1 mark)
Answer:
(a) (4, 5)

Question 9.
(3, 4) is a point on a circle with center at the origin.
(a) What is the radius of the circle?
(b) What are the points where the circle cut the axes? (3 mark)
Answer:
(a) r = \(\sqrt{3^2+4^2}=\sqrt{25}\) = 5
(b) (5, 0) (0, 5) (-5, 0) (0, -5)

Question 10.
Sum of the first n terms of an arithmetic sequence is 3n² + 2n
(a) What is its common difference?
Answer:
6

(b) Find the sum of first 20 terms (3 mark)
Answer:
3 × 20² + 2 × 20 = 1240

Question 11.
One side of a rectangle is 4 more than the other side. Area of the rectangle is 140 sq.cm
(a) If the small side is x then write the equation.
Answer:
(x + 4) x =140

(b) Find the sides of the rectangle. (4 mark)
Answer:
x(x + 4) = 140
x² + 4x = 140,
x² + 4x + 4 = 144,
(x + 2)² = 144,
x + 2 = 12, x = 10
Sides are 10 cm and 14 cm

Question 12.
A) Angles of triangle OAP are in the ratio 1: 2: 3, O is the center of the circle and PA is the tangent from P, ∠P is the smallest angle and OP =12 cm.

(a) What is the ratio of the sides of this triangle?
Answer:
Angles are x, 2x, 3x
6x = 180, x = 30
Angles are 30°, 60°, 90.
Ratio of the sides is 1: √3 : 2

(b) Find other two sides of the triangle?
Answer:
OT = 6, PT = 6√3

OR

B) If the chords AB and CD are extended, they intersect at P outside the circle. If PA = PC then prove AB CD.
Answer:
PA × PB = PC × PD
If we remove PA and PC on both sides
⇒ PB = PD,
⇒ PA – AB = PC – CD
If we remove PA and PC on both sides
⇒ – AB = -CD
AB = CD

Question 13.
A(1, 1), B(4, 1) and C(1, 5) are the vertices of a triangle
(a) By observing the vertices write the largest angle of this triangle.
Answer:
90°
Since AB is parallel to x axis and AC parallel toy axis, ∠A = 90°

(b) What is the length of BC ?
Answer:
BC = \(\sqrt{(1-4)^2+(5-1)^2}\)
= \(\sqrt{(-3)^2+4^2}\)
= \(\sqrt{25}\)
= 5

(c) What is the radius of the circle passing through the vertices? (5 mark)
Answer:
\(\frac{5}{2}\)

Question 14.
A) A(-4, -3), B(4, -3), C(7, 5), D(-7, 5) are the vertices of a quadrilateral
(a) Observing the coordinates of the vertices suggest a suitable name to ABCD
Answer:
AB and CD are parallel to x axis. So these are prallel sides.Other two sides BC and AD are not parallel.
ABCD is an isosceles trapezium.

(b) Find the length of its parallel sides
Answer:
AB = |4 – (-4)| = 8, CD = |7 – (-7)| = 14

(c) What is the distance between the parallel sides?
Answer:
Distance between the parallel sides is |5 – (-3)| = 8

(d) Calculate the area of ABCD
Answer:
Area = \(\frac{1}{2}\) × h × (a + b)
= \(\frac{1}{2}\) × 8 × (8 + 14)
= 88 sq.units

OR

B) 5th term of an arithmetic sequence is 10 and 10″‘ th term is 5.
(a) What is the common difference of the sequence?
Answer:
Difference between 10th term and 5th term is five times common difference.
5 – 10 = 5 common difference
common difference = \(\frac{-5}{5}\) = -1

(b) Find the first term?
Answer:
First term = 10 – 4 × -1 = 14

(c) What is its 15th term? (5 mark)
Answer:
Fifteenth term = tenth term + 5x common difference = 5 + -1 × 5 = o

Section – C

Question 15.
The length of the rectangular sheet shown in the figure is 13 centimetres. From this sheet, cut off two square sheets of maximum size. The area of remaining sheet isiS square metres.

(a) If breadth of the sheet be x, then what is the length of the remaining sheet?
Answer:
Length of the remaining sheet = 13 – 2x

(b) Form a second degree equation and find the length and breadth of the remaining sheet.
Answer:
Area of the remaining sheet =15 square metres
x(13 – 2x) = 15
⇒ 13x – 2x² =15
⇒ 2x² – 13x + 15 = 0
⇒ a = 2,b = – 13, c = 15

If breadth of the sheet is x = 5,
then length = 13 – 2x = 13 – 2 × 5 = 3
If breadth of the sheet is x = \(\frac{3}{2}\),
then length = 13 – 2x = 13 – 2 × \(\frac{3}{2}\) = 10

Question 16.
The surface area of a solid sphere is 100 p square centimetres.
(a) What is the radius of the sphere?
Answer:
4πr² =100π ⇒ r² = 25 ⇒ r = 5 cm

(b) What is the volume of the sphere? (5 mark)
Answer:
Volume = \(\frac{4}{3}\)πr
= \(\frac{4}{3}\) × π ×5³
= \(\frac{500π}{3}\)cm³

Question 17.
O is the centre of the circle with diametre AB. Another circle is drawn with AO as the diametre

(a) What are the measure of ∠APO, ∠ACB
Answer:
90°. Reason ∠APO, ∠ACB are the angles in the semicircle

(b) Outer circle has radius 5 cm and BC = 8 cm. What is the length OP?
Answer:
Triangle APO, triangle ACB are similar.
\(\frac{A O}{A B}=\frac{O P}{B C}\)
\(\frac{5}{10}=\frac{O P}{8}\), OP = 4 cm

(c) Is AP = PCI Why?
Answer:
AC is the chord of big circle. OP is perpendicular from centre to this chord. OP bisect AC. Therefore
AP = PC

(d) What is the length of AC? (3 mark)
Answer:
AC = \(\sqrt{10-8}\) = 6 cm.

Question 18.
The sum of first 10 terms is 230. The sum of first 15 terms is 495.
(a) If pn² + qn is the sum of first n terms then write a pair of equations
Answer:
100p + 10q = 230
⇒ 10p + q = 23

225 + 15q = 495
⇒15p + q = 33
5p = 10, q = 2 ,q = 3

(b) Find p and q and write the expression of the sum of first n terms
Answer:
2n² + 3n

(c) Find the algebraic form of the sequence (4 mark)
Answer:
x_n = 4n + 1

Question 19.
A) There are two squares in the figure.The perimetre of the outer square is 28 cm , the perimetre of the inner square is 20 cm

(a) What is the area of the outer square?
Answer:
Side of outer square = \(\frac{28}{4}\) = 7 cm
Area = 7² = 49 sq.cm

(b) What is the area of inner square?
Answer:
Side of inner square \(\frac{20}{4}\) = 5 cm
Area = 5² = 25 sq.cm

(c) What is the area of the shaded triangle ?
Answer:
Area of the region in between the squares
= 49 – 25
= 24 sq.cm
Area of shaded part = \(\frac{24}{4}\) = 6 sq.cm

(d) If a fine dot is placed into the figure then what is the probability of falling the dot in the shaded triangle?
Answer:
Probability = \(\frac{6}{49}\)

OR

B) Sum of the first n terms of an arithmetic sequence is n² + n
(a) Find the common difference of the arithmetic sequence.
Answer:
x₁ = 1² + 1 = 2,
x₁, + x₂ = 6 => x₂ = 4
d = 4 – 2 = 2

(b) Write the sequence argebraically
Answer:
x_n = 2n

(c) Find the sum : 3 + 5 + 7 + …. + 51 (5 mark)
Answer:
25th term is 50. Each term of 3, 5, 7…. is one more than 2, 4, 6….
3 + 5 + 7 + …… + 51
= 25² + 25 + 25
= 675

Section – D

Question 20.
Choose the correct answer from the options given below.
In the circumcircle of triangle PQR the tangent at P makes 40° with PQ. What is x ?

(a) 40°
(b) 50°
(c) 80°
(d) 20° (1 mark)
Answer:
(a) 40°

Question 21.
Read the two statements given below.
Statement 1: If angle between the tangents from an outer point is 90 + x then angle between the radii to the point where the tangents touch the circle is 90 – x.
Statement 2: Tangents from the outer point and radii to the points where touch the circle makes a cyclic quadrilateral

Choose the correct answer from those given below.
(a) Statement 1 is true and Statement 2 is false.
(b) Statement 1 is false and Statement 2 is true.
(c) Both the statements are true, and Statement 2 is the correct reason for Statement 1.
(d) Both the statements are true, but Statement 2 is not the correct reason for Statement 1 (1 mark)
Answer:
(c) Both the statements are true, and Statement 2 is the correct reason for Statement 1.

Question 22.
A tree of height 12 m is broken by the wind. The top struck the ground at an angle of 60°.
(a) Draw a rough diagram
Answer:
Diagram

(b) At what height from the bottom of the tree is broken by the wind?
Answer:
sin 60° = \(\frac{A C}{D C}\)
\(\frac{\sqrt{3}}{2}=\frac{x}{12-x}\)
12√3 – √3x = 2x, 12√3 =(2 + √3)x
x = 5.56 metre x

(c) Calculate the distance from the foot of the tree to its tip on the on the ground. (3 mark)
Answer:
tan 60° = \(\frac{x}{DA}\)
3 = \(\frac{x}{D A}\), 1.73 = \(\frac{5.56}{D A}\)
DA = \(\frac{5.56}{1.73}\) = 3.2 m

Question 23.
A) The perpendicular sides of a right triangle differ by 2. Area of the triangle is 24
(a) Form an equation on x connecting the sides and area ?
Answer:
Perpendicular sides are x and x + 2
\(\frac{1}{2}\) × x(x + 2) = 24,
x² + 2x = 48,
x² + 2x + 1 = 49

(b) Find the sides of the triangle
Answer:
(x + 1)² = 49, x + 1 = 7, x = 6
Perpendicular sides are 6 and 8.
Hypotenuse is \(\sqrt{6^2+8^2}\) = 10

OR

B) As you know if a and b are the parallel sides and h is the distance between the parallel sides then the area of the trapezium is \(\frac{1}{2}\) h (a + b)
In the figure you can see three trapeziums and a triangle with vertices A(1, 3), B(8, 5) and C(4, 10)

(a) Write the length of PA, QC and RB which are the parallel sides of the trapeziums
Answer:
PA = 3, QC = 10, RB = 5

(b) Calculate the area of APQC, CQRB.
Answer:
PQ = 4 – 1 = 3
Area = \(\frac{1}{2}\) × 3 × (3 +10)
= \(\frac{39}{2}\)
= 19.5 sq. units

QR = 8 – 4 = 4
Area = \(\frac{1}{2}\) × 4 × (10 + 5)
= \(\frac{60}{2}\)
= 30 sq. units

(c) Calculate the area of PABR
Answer:
PR = 8 – 1 = 7
Area = \(\frac{1}{2}\) × 7 × (3 + 5)
= \(\frac{56}{2}\)
= 28 sq. units

(d) Find the area of triangle ABC (3 mark)
Answer:
Area = 19.5 + 30 – 28 = 21.5 sq. units

Question 24.
A) In the figure ∠ABC, ∠AOC, ∠ADC are in an arithmetic sequence

(a) What is the relation between angle ABC and angle AOC
Answer:
∠AOC = 2 × ∠ABC

(b) What is the relation between angle ABC and ADC
Answer:
∠ABC + ∠ADC = 180°

(c) Find the measure of these angles
Answer:
Let ∠ABC = x, ∠AOC = y, ∠ADC = z
x, y, z are in an arithmetic sequence.
Therefore 2y = x + z
From the relations noted above ,
y = 2x, x + z = 180
2y = x + z
2 × 2x = x + y = 180
4x= 180 x = 45
x = 45°,y = 90°, z = 135°
∠ABC = 45°, ∠AOC = 90°, ∠ADC = 135°

OR

Prove that any cyclic parallelogram is a rectangle. (4 mark)
Answer:

• ABCD is a parallelogram. (Draw rough figure)
• Opposite angles are equal.
  ∠A = ∠C, ∠B = ∠D
• Sum of the opposite angles is 180°
• ∠A + ∠C = 180, ∠A = ∠C
  ∠A = 90°, ∠C = 90°
• ∠B + ∠D = 180°, ∠B = ∠D
  ∠B = 90°, ∠D = 90°
• ABCD is a square

Question 25.
A line passes through the points (a, 0) and (0, b)
(a) Write the equation of the line.
Answer:
Let (x, y) be a point on the line.
\(\frac{y-0}{x-a}=\frac{b-0}{0-a} \Rightarrow \frac{y}{x-a}=\frac{-b}{a}\)
ay + bx = ab

(b) Rewrite the equation as \(\frac{x}{a}+\frac{y}{b}\) = 1
Answer:
Dividing both sides by ab we get
\(\frac{a y}{a b}+\frac{b x}{a b}=\frac{a b}{a b}\)
\(\frac{x}{a}+\frac{y}{b}\) = 1

(c) If (p, q) is the mid point of the line joining (a, 0) and (0, b) then prove that \(\frac{x}{p}+\frac{y}{q}\) = 2 (5 mark)
Answer:
If (p, q) is the mid point of the line joining (a, 0) and (0, b) then
p = \(\frac{a+0}{2}\), q = \(\frac{0+b}{2}\)
a = 2p , b = 2q

Equation becomes \(\frac{x}{2 p}+\frac{y}{2 q}\)
That is \(\frac{x}{p}+\frac{y}{q}\) = 2

Question 26.
Curved surface area of a cone is 135 π sq.cm, radius 9 cm.
(a) What is the slant height?
Answer:
π × 9 × 1 = 135π
⇒ l = 15 cm

(b) What is the height of the cone?
Answer:
h = \(\sqrt{15^2-9^2}\) = 12 cm

(c) What is the volume of the cone? (5 mark)
Answer:
Volume = \(\frac{1}{3}\) × π × 9² × 12
= 324π cm³