Kerala Syllabus Class 10 Maths Model Question Paper Set 5

Practicing Maths Question Paper Class 10 Kerala Syllabus Set 5 English Medium helps identify strengths and weaknesses in a subject.

Maths Class 10 Kerala Syllabus Model Question Paper Set 5

Time: 2½ Hours
Total Score: 80 Marks

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

Question 1.
ABC is an equilateral triangle with A(0, 3), AD is the height. If G is the centroid and D is the origin.

• find the coordinates of B, C, D and G
Answer:
a) D is the origin
∴ Coordinate is (0, 0)
G divides AD in the ratio 2 : 1.
∴ Coordinate of G is (0, 1)
ΔADB is a right triangle with 30°, 60°, 90°
∴ BD = \(\frac{3}{\sqrt{3}}\) = √3
∴ Coordinates of B (-√3, 0)
∴ Coordinates of C (√3, 0)

Question 2.
The letters of the word MATHEMATICS are written in paper slips and put into a box. A child is asked to take one slip from the box without looking.
a) What is the probability of getting the letter M?
b) What is the probability of not getting M?
Answer:
Total outcomes = 11
Favourable outcomes = 2
a) Probability of getting M = \(\frac{2}{11}\)
b) Probability of not getting M = 1 – \(\frac{2}{11}\) – \(\frac{9}{11}\)

Question 3.
The algebraic form of an arithmetic sequence is 8n – 3
a) What is the first term of the sequence?
b) What is the commqn difference?
Answer:
a) First term = 8 × 1 – 3 = 5
b) Common difference = 8

Question 4.
Weekly wages of workers in a factory are given bleow. Find the median
4100, 4500, 4100, 4500, 4800, 4900, 4300, 4200, 5000
Answer:
Write the weekly wages in ascending order.
4100, 4100, 4200, 4300, 4500, 4500, 4800, 4900, 5000
Median = 4500

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

Question 5.
PQ and RS are two mutually perpendicular chords of a circle

If ∠QPR = 50°.
Find ∠PRS
Find ∠PQS.
Answer:
Draw the picture and mark the angles.
∠PRS = 90 – 50 = 40°
∠PQS = 40° (angles in the same segment are equal)

Question 6.
a) How many natural numbers between 300 and 700 gives remainder 1 when divided by 6?
b) Find their sum.
Answer:
a. First term = 301 and last tern = 697 Common difference = 6
Number of terms = \(\frac{697 – 301}{6}\) + 1 = 67
b. Sum = \(\frac{67}{2}\) (301 + 697) = 33433

Question 7.
ABCD is a parallelogram. AB = 6 cm., AD = 4 cm. ∠B = 150°

a) What is ∠A?
b) What is the perpendicular distance from D to AB?
c) What is the area of ABCD
Answer:

a) ∠A = 30°,
b) If the perpendicular distance from D to AB is DE, sin 30 = \(\frac{D B}{A D}\)
DE = AD × sin 30 = 4 × \(\frac{1}{2}\) = 2 cm
Area of ABCD = AB × DE
= 6 × 2 = 12 cm²

Question 8.
a) Draw a circle of radius 3.2 centimetres and mark a point P which is 8 cm away from its centre. Draw the tangents from P to the circle.
b) Measure its length.
Answer:
Draw a circle with centre O and radius

3.2 cm. Mark P, 8 cm away from O. Find the midpoint of OP. Midpoint as centre and OP as diameter draw a circle.

This circle intersects the original circle at A and B. Join PA and PB. These lines are the tangents from P to the circle PA = PB = 7.3 cm.

Question 9.
Using a rod of length 64 cm, the frame of a square pyramid with equal edge is made. If so find.
a) One base edge
b) Slant height
c) Base area
Answer:
If one edge = a cm
8a = 64 cm

a = \(\frac{64}{8}\) = 8 cm
a) Length of one base edge = 8 cm
b) Slant height = \(\frac{\sqrt{3}}{2}\) × 8 = 4 √3 cm
c) Base area = a² = 8 × 8 = 64 sq.cm

Question 10.
A, B, C are the mid points of the sides of the triangle PQR Q (1, 2), B (6, 4), A (3, 7)
a) What type of the quadrilateral PABC.
b) Find the co ordinates of C, R.

Answer:
a) A and B are the midpoints of PQ and RQ. Hence AB is parallel to PC.

BC is parallel to PA
∴ PABC is a parallelogram.

b) Similarly, ABRC and AQBC are parallelogram.
In the parallelogram AQBC has the coordinates C(6 + 3 – 1, 4 + 7 – 2).
Thus the coordinates of C is (8, 9)
In the parallelogram ABRC,
R has the coordinates
(8 + 6 – 3, 9 + 4 – 7) = (11, 6)

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

Question 11.
p(x) is a second degree polynomial with P(2) = 0 and P (-3) = 0.
a) Find two first degree factors of p(x).
b) Find the polynomial p(x).
Answer:
a. (x – 2), (x + 3)
b. p(x) = x² + x – 6

Question 12.
The ratio between the 4^th term and 8th term of an arithmetic sequence is 5 : 11. Its 10th term is 56.
a) What is the common difference of this sequence?
b) What is its 16^th term?
c) Find the sum of first 25 terms.
Answer:
x₄ : x₈ = 5 : 11
∴ x₈ – x₄ = 11k – 5k = 6k
ie 4d = 6k ∴ d = \(\frac{6 k}{4}\) = \(\frac{3 k}{2}\)
x₁₀ = 56
x₈ + 2d = 56
ie, 11k + 2 × \(\frac{3 k}{2}\) = 56
14k = 56 ⇒ k = \(\frac{56}{14}\) = 4
a) ∴ d = common difference
\(\frac{3 k}{2}\) = \(\frac{3 \times 4}{2}\) = 6

b) x₁₆ = x₁₀ + 6d = 56 + 6 × 6 = 92

c) Sum of first 25 terms
= \(\frac{n}{2}\)(x₁ + x₂₅) = \(\frac{n}{2}\)(x₁₀ + x₁₆)
Since it is even number sum of each pair is equal
= \(\frac{25}{2}\)(56 + 92) = 1850

Question 13.
a) Draw a rectangle of length 5 cm, breadth 4cm.
b) Draw a square of the same area.
Answer:

Question 14.
A bag contains 6 blue beads and 4 green beads. Another bag contains 3 blue beads and 7 green beads. If we take one bead from each bag, without looking into it.
a) What is the total number of possible selections?
b) What is the probability of both being same colours?
c) What is the probability of both being different colours?
d) What is the probability of getting atleast one blue bead?
Answer:

a) Total number of pairs = 10 × 10 = 100

b) Number of outcomes, both being same colour = 6 × 3 + 4 × 7 = 18 + 28 = 46
∴ Probability = \(\frac{46}{100}\) = 0.46

c) Outcomes both being different colours
= 6 × 7 + 4 × 3 = 54
∴ Probability = \(\frac{54}{100}\) = 0.54

d) Outcomes atleast one blue bead
= 6 × 3 + 6 × 7 + 4 × 3
= 18 + 42 + 12 = 72
∴ Probability = \(\frac{72}{100}\) = 0.54

Question 15.
In ΔABC, AB = 10 cm, AC = 6 cm, ∠A = 70°
a) Find the area of the triangle.
b) Find BC. [cos 70° = 0.34 r sin 70° = 0.94]
Answer:

a) sin 70° = \(\frac{C D}{6}\)
CD = 6 × sin 70 = 6 × 0.94 = 5.64 cm
Area of ΔABC
= \(\frac{1}{2}\) × AB × CD = \(\frac{1}{2}\) × 10 × 5.64
= 28.2 sq.cm

b) AD = 6 cos 70° = 6 × 0.34 = 2.04 cm
BD = 10 – 2.04 = 7.96 cm
BC = \(\sqrt{C D^2+B D^2}\)
= \(\sqrt{(5.64)^2+(7.96)^2}\)
= \(\sqrt{31.81+63.36}\)
= \(\sqrt{95.17}\) cm

Question 16.
A (6, 0) is a point on a circle with center (0, 0).

a) Find the radius of the circle.
b) Show that B (-3, 3√3), C (-3, -3√3 ) are the points on this circle.
Answer:

B (-3, 3√3 ), C (-3, -3√3 ) are the points on this circle.

Question 17.
In the picture, the tangents at A and Con the circle meet at B and ∠ABC = 54°. The point D is on the circle.
Such that AD = AC.

a) What is ∠ACB ?
b) What is ∠BCD ?
Answer:
a) BA = BC (tangents)
∴ ΔABC is an isosceles triangle
∴ ∠BAC = ∠BCA
= \(\frac{180 – 54}{2}\) = \(\frac{126}{2}\) = 63°
ie. ∠ACB = 63°
∠D = 63° (angle in the segment on the other side of the chord)
Since AD = AC, ∠ACD = 63°
∴ ∠ECD – 180 – (63 + 63)
– 180 – 126 = 54

Question 18.
Five sectors of central angles 120°, 90°, 60°, 50°, 40° and radius 30 cm are taken from a circular plate to make cone. Calculate the radius of each one. What is the ratio of their lateral surface areas.
Answer:
Radius of the cone taken from the sector of central angle 120° = \(\frac{120}{360}\) part of the radius of the sector.
= 30 × \(\frac{120}{360}\) = 10 cm
Radius of the cone taken from the sector of central angle 90° = \(\frac{90}{360}\) part of the radius of the sector.
= 30 × \(\frac{90}{360}\) = 7.5 cm
Radius of the cone taken from the sector of central angle 60° = \(\frac{60}{360}\) part of the radius of the sector.
= 30 × \(\frac{90}{360}\) = 5 cm
Radius of the cone taken from the sector of central angle 50° = \(\frac{50}{360}\) part of the radius of the sector.
\(\frac{30 \times 50}{360}\) = 4.17 cm
Radius of the cone taken from the sector of central angle 40° = \(\frac{40}{360}\) part of the radius of the sector.
\(\frac{30 \times 50}{360}\) = = 3.33 cm

b) Sum of the radii of cone
= 10 + 7.5 + 5 + 4.17 + 3.33 = 30 cm
Area of the cone = area of the sector

c) Sum of the lateral surface area

\(\frac{\pi}{360}\) ((10)² × 120 + (7.5)² × 90 + (5)² × 60 + (4.17)² × 50 + (3.33)² × 40)
\(\frac{\pi}{360}\) (100 × 120 + 56.25 × 60 + 17.3889 × 50 + 11.089 × 40)
= \(\frac{\pi}{360}\) (12000 + 5062.5 + 1500 + 869.445 + 443.56) = 173.358 sq.cm

Question 19.
P (2, 1), Q (-2, 3), R (4, 5) are the vertices of ΔPQR.
a) Find the equation of the median from R to PQ
b) Find the coordinates of the centroid.
Answer:

a) Coordinates of S
(\(\frac{2 + -2}{2}\), \(\frac{1+3}{2}\)) = (0, 2)

b) If (x, y) is a point on the medians RS then
\(\frac{y – 5}{x – 4}\) = \(\frac{2 – 5}{0 – 4}\) = \(\frac{-3}{-4}\) = \(\frac{3}{4}\)
⇒ 4(y – 5) = 3(x – 4)
⇒ 4y – 20 = 3x – 12
⇒ 3x – 4y + 8 = 0
If G(x, y) is the centroid then
RG : GS = 2 : 1
x = 4 + \(\frac{2}{3}\)(0 – 4), y = 5 + \(\frac{2}{3}\)(2 – 5)
⇒ x = 4 – \(\frac{8}{3}\), y = 5 – \(\frac{6}{3}\)
⇒ x = \(\frac{4}{3}\), y = 3
∴ G(x, y) = (\(\frac{4}{3}\), 3)

Question 20.
p(x) = x³ – 5x² + ax + b
a) Find the relation between a and b for (x – 2) to be a factor of p(x).
b) Find the relation between a and b for (x – 3) to be a factor of p(x).
c) Find a and b so that (x – 2) and (x – 3) are factors of p(x).
Answer:
p(x) = x³ – 5x² + ax + b
a) x – 2 is a factor ⇒ p(2) = 0
p(2) = (2)³ – 5(2) + 2a + b = 0
2a + b = 12 .

b) x – 3 is a factor
⇒ P(3) = 0
p(3) = (3)³ – 5(3)² + 3a + b = 0
3a + b = 18 ……………… (1)

c) 2a + b = 12 …………….. (2)
(2) – (1) ⇒ a = 6
3 × 6 + b = 18
b = 18 – 18 = 0

Question 21.
The daily wages of 49 workers of a village are given below.

Daily wages (Rs.)	No. workers
300 – 350	4
350 – 400	6
400 – 450	10
450 – 500	14
500 – 550	8
550 – 600	4
600 – 650	3

a) The daily wage of the worker of what position is taken as the median.
b) Find the median of daily wages.
Answer:

Daily wages (Rs.)	Number of workers
less than 350	4
less than 400	10
less than 450	20
less than 500	34
less than 550	42
less than 600	46
less than 650	49

Total number of workers = 49
Median daily wage is daily wage of \(\frac{49+1}{2}\) = 25^th worker.
If we divide 50 rupees from 450 to 500 to 14 equal parts then the length of each sub division = \(\frac{50}{14}\) = \(\frac{25}{7}\)
According to daily wage 21^th worker
= (450 + 450\(\frac{25}{7}\)) ÷ 2
= 450\(\frac{25}{14}\)
Daily wage of 25^th worker
= 450\(\frac{25}{14}\) + 4 × \(\frac{25}{7}\) = 466\(\frac{1}{14}\)

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

Question 22.
6n + 3 be the algebraic expression of an arithmetic sequence.
a) Find the sum of the first 20 terms of the sequence.
b) Write the algebraic expression of the sum.
Answer:
x_n = 6n + 3
x₁ = 6 + 3 = 9
x₂₀ = 120 + 3 = 123
a) Sum of first 20 terms
= \(\frac{20}{2}\)(9 + 123) = 1320

b) Sum of n terms = \(\frac{n}{2}\)(9 + 6n + 3)
= \(\frac{n}{2}\)(6n + 12) = 3n² + 6n

Question 23.
Write down mathematical concepts of each statement based on the given figure.

a) ∠A = \(\frac{1}{2}\)∠COE
b) ∠A = ∠G
c) ∠B = 90°
d) ∠A + ∠D = 180°
Answer:
a) ∠A = \(\frac{1}{2}\) ∠COE (The angle made by an arc at any point on the alternate arc is equal to the half the angle made at the centre)
b) ∠A = ∠G (Angles in the same segment are equal)
c) ∠B = 90° (Angle in the semicircle is right angle)
d) ∠A + ∠D = 180° (Angles in alternate segments are supplementary)

Question 24.
a) The reciprocal of a number is subtracted from two times of that number will get \(\frac{73}{18}\). Take the number as x and form the equation.
b) Find the number.
Answer:
Number = x
2x – \(\frac{1}{x}\) = \(\frac{73}{18}\)
2x × 18x – \(\frac{73}{18}\) × 18x = 18x
36x² – 18 = 73x
36x² – 73x – 18 = 0
x = \(\frac{73 \sqrt{73^2-4 \times 36 \times-18}}{2 \times 36}\)
= \(\frac{73 \pm \sqrt{7921}}{72}\)
= \(\frac{73 \pm 89}{72}\) = \(\frac{73+89}{72}\) = \(\frac{162}{72}\) = \(\frac{9}{4}\)
Number = \(\frac{9}{4}\)

Question 25.
A tower is built in a river of width 80 metres. One sees the top of the river at an elevation of 55° and 65° from either banks of the river.
a) Draw a figure using the given measurements.
b) Find the distance, from the water level of river to the top of the tower. ‘
c) Find the distances to either banks from the foot of the tower.
[tan 55° = 1.43; tan 65° = 2.14]
Answer:

b) From the figure CD = x tan 65°
CD = (80 – x) tan 55°
x tan 65° = (80 – x) tan 55°
x tan 65° + x tan 55° = 80 tan 55°
x = \(\frac{80 \tan 55^{\circ}}{\tan 65^{\circ}+\tan 55^{\circ}}\) = \(\frac{80 \times 1.43}{2.14+1.43}\)
= \(\frac{114.4}{3.57}\) = 32 metre
CD = x tan 65° = 32 × 2.14 = 68.48 metre
Height of the tower = 68.48 metre

c) The distance from the foot of the tower to one bank of the river
x = 32 metre
Distance from the foot of the tower to other bank of the river
= 80 – 32 = 48 metre

Question 26.
Draw a circle of radius 3 cm. Construct a triangle with angles 60°, 70° and sides touches the circle.
Answer:
Draw a circle of radius 3 cm.

Draw another two radii such that the central angles are 110° and 120°.
∠POQ = 110°, ∠QOR = 120°
Now draw the tangents at P, Q and R.
Draw a triangle ABC by joining these tangents.
∠A = 180 – ∠POQ = 180° – 110° = 70°
(In a circle, the angle between the radii through two points and the angle between the tangents at these points are supplementary)
∠B = 180° – ∠QOR = 180° – 120° = 60°
(In a circle, the angle between the radii through two points and trie angle between the tangents at these points are supplementary)

Question 27.
The total height of a vessel, in the shape of a cone attached at one end of a cylinder is 10 m. The slant height of the conical part is 5 m and common radius 3 m.
a) What is the height of the cone?
b) What is the capacity of the vessel in litres?
Answer:

a) Height of the cone = \(\sqrt{5^2-3^2}\) = 4 m

b) Capacity of the vessel = Volume of the cylinder + Volume of the cone
= \(\frac{1}{3}\)π(3)² × 6 + \(\frac{1}{3}\)π × (3)² × 4
= 54π + 12π
= 66π m³ = 66 × 3.14 m³
= 66 × 3.14 × 1000 litres
= 207240 litres

Question 28.
a) What is the slope of the line joining the points A(2, -3) and B(6, 3)? Find the equation
of this line.
b) Find the co-ordinates of the point C at which the line cuts the x-axis.
c) Show that C is the mid-point of the line AB.
Answer:
a) Slope of the line joining the points A (2, -3) and B(6, 3)
\(\frac{3-(-3)}{6-2}\) = \(\frac{6}{4}\) = \(\frac{3}{2}\)
If (x, y) be any point on this line,
then \(\frac{y-(-3)}{x-2}\) = \(\frac{3}{2}\) ⇒ \(\frac{y+3}{x-2}\) = \(\frac{3}{2}\)
⇒ 2 (y + 3) = 3(x – 2)
⇒ 2y + 6 = 3x – 6
⇒ 3x – 2y – 6 – 6 = 0
⇒ 3x – 2y – 12 = 0, is the equation of the line joining the points A (2, -3) and 6(6,3).

b) If we take the point where this line cuts the x-axis as as C (x, 0)
\(\frac{0-(-3)}{x-2}\) = \(\frac{3}{2}\)
⇒ \(\frac{3}{x – 2}\) = \(\frac{3}{2}\)
⇒ (x – 2) = 2
and so x = 4
Thus the line cuts the x axis at = C(4, 0)

c) (\(\frac{2 + 6}{2}\), \(\frac{- 3 + 3}{2}\)) = (4, 0)
∴ C is the midpoint of the line AB.

Question 29.
Read the following, understand the mathematical idea expressed in it and answer the questions that follow:
The volume of a cone is the third of the product of its base area and height. The curved surface area of a cone is half the product of the base circumference and the slant height.
A cylinder of wax has radius 6 cm and height 8 cm. A cone of largest size is carved from it.
a) Find the radius of the cone.
b) What is the slant height of the cone?
c) Find the volume of the cone.
d) What is its curved surface area?
e) The remaining wax is used to make candles with 1 cm radius and 8 cm height.
(i) Find the volume of a candle.
(ii) How many candles can be made?
Answer:
a) Radius of the cone = 6 cm

b) Slant height = \(\sqrt{8^2+6^2}\) = 10 cm

c) Volume of the cone
= \(\frac{1}{3}\) × π × 6² × 8 = 96π cu.cm

d) Curved surface area
= π × r × l
= π × 6 × 10
= 60π sq.cm

e) Volume of the wax cylinder
= 3 × 96π = 288π cu.cm
Remaining volume of wax
= 288π – 96π
= 192π cu.cm
(i) Volume of the small cylinder (candle)
= π × 1² × 8 = 8π

ii) Number of candles
= \(\frac{192 \pi}{8 \pi}\) = 24