Kerala Syllabus Class 10 Maths Model Question Paper Set 4

Practicing Maths Question Paper Class 10 Kerala Syllabus Set 4 English Medium helps identify strengths and weaknesses in a subject.

Maths Class 10 Kerala Syllabus Model Question Paper Set 4

Time: 2½ Hours
Total Score: 80 Marks

Answer any three from questions 1 to 4. Each question carries 2 marks.

Question 1.
a) Write the sequence of numbers which leaves the remainder 3 on dividing by 5 and 10.
b) Is 510 a term of this sequennce?
c) Check it.
Answer:
a) The number which is divided by 5 and 10 must be divided by 10. So take the numbers when divided by 10 leaving a remainder 3.
3, 13, 23, 43, …….

b) 510 – 3 = 507 is not a multiple of 10.
∴ 510 is not a term of this sequence.

Question 2.
In the figure, the shaded triangle is drawn by joining the mid point of the sides of large triangle.

a) Calculate the probability of a dot put on larger triangle to be within the shaded triangle.
b) Calculate the probability of the dot put on the larger triangle outside the shaded triangle
Answer:
4 triangles are equal triangles.
a) Probability of a dot put on large triangle to be within the shaded triangle = \(\frac{1}{4}\)

b) Probability of a dot outside the shaded triangle = I – \(\frac{1}{4}\) = \(\frac{3}{4}\)

Question 3.
In the figure, the sides of the square are parallel to the axes and the origin is the midpoint. Coordinates of one vertex of the square is (3, 3). Write the coordinates of two other vertices of the square.

Answer:
(-3, -3), (3, -3)

Question 4.
The ages of 10 members of a club are 20, 25, 22, 32, 42, 27, 35, 27, 35 and 30.
a) Find the median age.
b) Find the mean.
Answer:
Write in ascending order
a) 20, 22, 25, 27, 27, 30, 32, 35, 35, 42
Median = \(\frac{27+30}{2}\) = 28.5

b) Mean = \(\frac{20+25+22+32+42+27+35+27+35+30}{10}\)
= \(\frac{295}{10}\) = 29.5

Answer any 4 from questions 5 to 10. Each question carries 3 marks.

Question 5.

a) Draw a circle with radius 4 centimetres. Draw a triangle with two of its angles 65° and 78° and all vertices on the circle.
b) What is the measure of the third angle?
Answer:
a) Draw a circle of radius AO, 4 cm, such that ∠AOB = 1300 and ∠BOC = 156° mark points
B and C. Join AB, BC and AC. Complete the triangle.

b) Third angle = 180 – (78 + 65) = 37°

Question 6.
In the figure, the length of the chord AB is 18 centimetre. The chord is extended to P and the tangents drawn from that point, have length 12 centimetres

a) Find the length of BR
b) In the figure what is equal toPA PB?
Answer:
a) PQ = 12 cm
AB = 18 cm
If PB = x, AP = 18 + x
PA × PB = PQ²
(18 + x) x = 12²
18x + x² = 144
x² + 18x = 144
Completing the square
x² + 18x + 81 = 144 + 81
(x + 9)² = 225
x + 9 = ±15
x = ±15 – 9
x = 6 OR x – 24
BP = 6 cm

b) PA × PB = PQ²

Question 7.
In triangle ABC, the length of AP is 10 centimetres.
a) What is the length of BP?

b) What is the length of PC?
c) Calculate the length of BC.
Answer:
a) In the figure ∠ABP = 45°,
∠BAP = 45°,
AP = 10 cm
ie BP = 10 cm

b) ∠ACP = 30°, ∠CAP = 60°,

c) AP = 10 cm
ie PC = \(10^{\sqrt{8^2 \cdot x^2}}\) cm
ie BC = BP + PC = (10 + 10√3) cm

Question 8.
AP is the tangent to the circle with centre at O and radius 4 centimetres. AB = 3 cm
a) Find the length of OA

b) Find the length of the tangent AP
Answer:
a) In the figure OP = 4 cm
OP = OB = 4 cm
BA = 3 cm
OA = OB + BA = 4 + 3 = 7 cm

b) AP² = OA² – OP² = 7² – 4² = 49 – 16 = 33
AP = \(\sqrt{33}\) cm

Question 9.
The radius of two spherical tanks are in the ratio 3 : 4.
a) What is the ratio of the volume?
b) The volume of the first tank is 540 litres. Find the volume of the second tank.
Answer:
Ratio, between the radii = 3 : 4
r₁ : r₂ = 3 : 4
r₁ = 3k, r₂ = 4k
Volume of the first sphere V₁ = 540 litre

\(\frac{S 40}{V_2}\) = \(\frac{27}{64}\)
ie 27 × V₂ = 540 × 64
V₂ = \(\frac{540 \times 64}{27}\) = 1280 litre

Question 10.
a) Write p(x) = x² – 9x + 20 as a product of two first degree polynomials.
b) Write also the solutions of the equation p(x) = 0.
Answer:
a) p(x) = x² – 9x + 20
Sum = -9, product = 20
Numbers = -4 & -5
ie x² – 9x + 20 = (x – 4) (x – 5)

b) p(x) = 0
ie x² – 9x + 20 = 0
ie (x – 4) (x – 5) = 0
x – 4 = 0, x – 5 = 0 x = 4 & x = 5
Solutions = 4, 5

Answer any 8 from questions 11 to 21. Each question carries 4 marks.

Question 11.
a) Find the slope of the line joining (2, 4) and (4, 7).
b) Write the coordinates of another point on the line.
c) Check whether (5, 8) is on this line.
Answer:
a) Slope of the line joining the points (2, 4) and (4, 7) = \(\frac{y_2-y_1}{x_2-x_1}\) = \(\frac{7 – 4}{4 – 2}\) = \(\frac{3}{2}\)

b) Another point = (6, 10)

c) If we consider the points (2, 4) and (5, 8)
Slope
\(\frac{y_2-y_1}{x_2-x_1}\) = \(\frac{5 – 2}{8 – 4}\) = \(\frac{3}{4}\)
Slopes are not equal. So (5, 8) is not a point on the line.

Question 12.
Sum of the first five terms of an arithmetic sequence is 45.
a) What is the third term?
b) The common difference of the sequence is 4. Write the first two terms.
c) Write another arithmetic sequence having the sum of the first five terms 45.
Answer:
Sum of the five terms S₅ = 45
Third term X₃ = \(\frac{\dot{S}_5}{5}\) = \(\frac{45}{5}\) = 9
d = 4
X₂ = 9 – 4 = 5
X₁ = 5 – 4 = 1
First two terms = 1, 5
The third term of the arithmetic sequence having the sum of the first five terms 45 is 9.
Another arithmetic sequetice having the sum of the first five terms 45 = 3, 6, 9, 12, 15

Question 13.
a) Draw a rectangle of area 18 square centimetre.
b) Draw a square of the same area.
Answer:

Question 14.
From all two digit numbers with each digit 1, 2, 3, 4 or 5 one number is chosen.
(a) What is the probability of both digits being the same?
(b) What is the probability of the sum of the digits being 8?
(c) What is the probability that it is a multiple of 5?
Answer:
Total outcomes = 5 × 5 = 25
a) Being two digits are equal
= (1, 1), (2, 2), (3, 3), (4, 4), (5, 5)
∴ Probability = \(\frac{5}{25}\) = \(\frac{1}{5}\)

b) Sum of two digits be 8
= (3, 5), (4, 4), (5, 3)
∴ Probability = \(\frac{3}{25}\)

c) Product of two digits be a multiple of 5
= (1, 5), (2, 5), (3, 5), (4, 5), (5, 5)
∴ ProbabIlity = \(\frac{5}{25}\) = \(\frac{1}{5}\)

Question 15.
In triangle ABC, length of AB = 6 cm, ∠A = 70°, ∠B = 55°.

(a) Find ∠C.
(b) Find AC.
(c) Find the area of triangle ABC. (sin 70° = 0.93)
Answer:
In the figure ∠A = 70°, B = 55°, AB = 6 cm
(a) ∠C = 55°

(b) AC = 6 cm

(c) Area \(\frac{1}{2}\) × AB × AC × sin 70°
\(\frac{1}{2}\) × 6 × 6 × 0.93 = 16.74 cm²

Question 16.
The centre of the circle shown is the origin and the radius is 13.

(a) Check whether each of the points (12, 5), (10, 6) is inside, outside or on the circle.
(b) Write the coordinates of two other points on the circle.
Answer:
a) Distance between the centre (0, 0) and (12, 5)
= \(\sqrt{(12-0)^2+(5-0)^2}\)
= \(\sqrt{144+25}\) = \(\sqrt{169}\) = 13 = Radius
ie (12, 5) is on the circle
Distance the centre (0, 0) and (10, 6)
\(\sqrt{10-0)^2+(6-0)^2}\) = \(\sqrt{100+36}\)
= \(\sqrt{136}\) < 13
Distance between (0, 0) and (10, 6) is less than the radius.
So (10, 6) is inside the circle.

(b) (-13, 0), (0, 13)

Question 17.
a) Calculate the area of the triangle.
b) Calculate the perimeter of a triangle as sides 30 centimetre, 28 centimetre and 26 centimetre.
c) Also calculate the radius of the incircle.
Answer:
a = 30, b = 28, c = 26
a) Perimeter = a + b + c = 30 + 28 + 26 = 84
S = \(\frac{a+b+c}{2}\) = \(\frac{84}{2}\) = 42

b) Area = \(\sqrt{s(s-a)(s-b)(s-c)}\)
= \(\sqrt{42(42-30)(42-28)(42-26)}\)
= \(\sqrt{42 \times 12 \times 14 \times 16}\) = 336 cm²
Radius of the circle
= \(\frac{\text { Area }}{\text { Perimeter }}\) = \(\frac{336}{42}\) = 8 cm

Question 18.
A square pyramid of base edge TO centimeters and height 12 centimetres is to be made of paper.
(a) Calculate the slant height of the pyramid.
(b) What is the area of the paper needed to make square pyramid?
Answer:
a = 10 cm, h = 12 cm
a) R = h² + (\(\frac{a}{2}\))² = 12² + (\(\frac{10}{2}\))²
= 144 + 25 = 169
Slant height l = \(\sqrt{169}\) = 13 cm

b) Total surface area
= a² + 2al = 10² + 2 × 10 × 13
= 100 + 260 = 360 cm²
Area of paper needed to make square pyramid = 360 cm²

Question 19.
p(x) = ax³ + bx² + cx + d ,
(a) Find p(-1) .
(b) If (x +1) is a factor of p(x) then prove that a + c = b + d
(c) Write a third degree polynomial having (x + 1) as a factor.
Answer:
P(x) = ax³ + bx² + cx + d
a) p(-1), = a(-1)³ + b(-1)² + c(-1) + d
= – a + b – c + d

b) If (x + 1) is a factor of p(x),
P(-1) = 0
ie – a + b – c + d = 0
ie b + d = a + c

c) p(x) = 6x³ + 2x² – 3x + 1

Question 20.
In the picture, midpoints of the sides of the quadrilateral ABCD are joined to draw quadrilateral PQRS.

(a) Find coordinates of R
(b) Write coordinates of all vertices of quadrilateral ABCD.
Answer:
a) R (6, 9)
b) B(7, 5), D(3, 7), C(9, 11)

Question 21.
Some households in a locality are sorted according to their electricity usage in the table below.

Usage of electricity (Unit)	No. of households
80 – 100	8
100 – 120	12
120 – 140	10
140 – 160	9
160 – 180	6

(a) If the household using the least unit of electricity is numbered as one and second least as two and so on, what is assumed to be usage of electricity of the 21st household.
(b) Calculate the median usage of electricity
Answer:

Usage of electricity (unit)	Number of houses	Usage (unit)	Number of house holds
80 – 100	8	less than 100	8
100 – 120	12	less than 120	20
120 – 140	10	less than 140	30
140 – 160	9	less than 160	39
160 – 180	6	less than 180	45

a) Total number of houses = 45
Median is the consumption of electricity
(\(\frac{45+1}{2}\)) = 23^rd family.

According to this 10 units from 21 to 30 is divided into 20 equal parts, and imagine that each part contains each house, each house is in the middle of the sub division. Then, consumption of each sub division is = \(\frac{20}{10}\) = 2

According to this electricity consumption of the house in the 21^th place is in between 120 and 122 unit.
That is, 121 unit.

b) Use of electricity in the next onwards increases 2 unit each.
Then, consumption of 22^nd house = 121 + 2 = 123 unit
Consumption of 23^rd house = 123 + 2 = 125 unit

Answer any 5 from questions 22 to 28. Each question carries 5 marks.

Question 22.
a) Find the least and highest three digit number which leaves a remainder 1 on division by 9.
b) How many three digit numbers are there, which leaves a remainder one on division by 9?
c) Find the sum of all such numbers.
Answer:
a) The least three digit number which leaves a remainder 1 on division by 9 = X₁ = 100
The highest three digit number which leaves a remainder 1 on division by 9 = X_n = 991

b) n = \(\frac{X_n-X_1}{d}\) + 1 = \(\frac{991 – 100}{9}\) + 1
= \(\frac{891}{29}\) + 1 = 99 + 1 = 100

c) Sum = \(\frac{n}{2}\)(X₁ + X_n) = 50(100 + 991) = 50 × 1091 = 54550

Question 23.
In the figure AB is the diameter of the circle with centre O
If radius 20 cm. BC = 13 cm. QC = 15 cm and PQ = 9 cm, then

a) What is the length of PC?
b) What is the length of PB?
c) What is the length of OQ.
Answer:
a) In ΔQPC. QC = 15 cm, QP = 9 cm
PC = \(\sqrt{15^2-9^2}\)
= \(\sqrt{144}\) = 12 cm

b) In ΔBPC, PC = 12 cm, BC = 13 cm
PB = \(\sqrt{13^2+12^2}\) – \(\sqrt{169-144}\) – \(\sqrt{25}\)
= 5 cm

c) OB = 20 cm
OQ = 20 – (9 + 5) = 6 cm

Question 24.
From the rectangle ABCD of breadth 1 metre, the largest possible square APQD is cut off. The remaining rectangle is PBCQ.

(a) Taking the length ABCD as x, write the length and breadth of PBCQ.
(b) The ratio of length and breadth of the rectangles. ABCD and PBCQ are same. Find length of AB.
Answer:
a) Length of ABCD = x
b) Length of PBCQ = 1
Breadth = x – 1

b) Since the ratio of length and breadth of the rectangles ABCD and PBCQ are same
x : 1 = 1 : x – 1
ie x(x – 1) = 1
x² – x = 1
x² – x – 1 = 0

Question 25.
A man standing on the top of a light house sees a ship approaching the seashore at an angle of depression of 22°. After the ship has travelled 100 metres more. Towards the sea shore, he sees it at an angle of depression of 31 °. The ship stops there.
(a) Draw a rough-sketch.
(b) How far is the ship from the light house.
(c) Find the height of light house.
tan 22° = 0.4, tan 31° = 0.6 1
Answer:

Question 26.
a) Draw a triangle of-sides 6 centimetre, 7 centimetre and 8 centimetre.
b) Draw a circle which touches all sides of the triangle.
c) Measure its radius.
Answer:

Radius = 1.96 cm

Question 27.
A cone is made from sector of radius 10 centimetre and central angle 216°.
(a) What is slant height and radius of the cone?
(b) Find the volume of the cone.
Answer:
a) Slant height of the cone = 10 cm
\(\frac{x}{l}\) = \(\frac{x}{360}\)
\(\frac{x}{10}\) = \(\frac{216}{360}\)
r = \(\frac{216}{36}\) = 6 cm

b) h² = R – r²
= 10² – 6² = 100 – 36 = 64
Height h = \(\sqrt{64}\) = 8 cm
Volume = \(\frac{1}{3}\)πr²h
= \(\frac{1}{3}\) × π × 6² × 8
= 96 π cm³

Question 28.
Aline is drawn joining (2, 5) and (8, 9).
a) Among the following which one is the midpoint of this line [(10, 14); (6, 4); (5, 7); (4, 4)]
b) Find the radius of the circle having the above line, as diameter.
c) Write the equation of the circle.
d) Is (7, 10) a point on this line?
Answer:
Coordinate of the mid point = (\(\frac{2+8}{2}\), \(\frac{5+9}{2}\)) = (5, 7)
Centre of the circle (5, 7)

b) Radius = \(\sqrt{(7-5)^2+(5-2)^2}\) = \(\sqrt{13}\)

c) Since the centre is (5, 7) and (x, y) be the point on the circle
(x – 5)² + (y – 7)² = 13
x² + y² – 10x – 14y + 61 = 0

d) If (7, 10) be the point on the circle then equation (x – 5)² + (y – 7)² = 13 will satinty
(7 – 5)² + (10 – 7)² = 13
So, (7, 10) is the print on the circle.

Question 29.
Read the mathematical concept given below carefully and understand it. Then answer the following questions.

Diagonal of a polygon is a line joining two non-adjacent vertices. See this table.

Number of sides of polygon	Number of diagonals drawn from one vertex	Total number of diagonals
Triangle 3	0	0 = \(\frac{3 \times 0}{2}\)
Quadrilateral 4	1	2 = \(\frac{4 \times 1}{2}\)
Pentagon 5	2	5 = \(\frac{5 \times 2}{2}\)

From the above table, we see the relationship between the number of sides of a polygon and the number of diagonals.
Answer the questions given below.
a) Which polygon has the same number of sides and diagonals?
b) How many diagonals can be drawn from a single vertex of an 8 – sided polygon?
c) How many diagonals does 20 sided polygon have?
Answer:
a) Pentagon – 5 sides, 5 diagonals

b) Number of diagonals can be drawn from a single vertex of an 8 sided polygon
= 8 – 3 = 5

c) Total number of diagonals
= \(\frac{20 \times 17}{2}\) = 170