Kerala SSLC or Class 10 Maths Question Paper 2020 with Answers – Free Download
Kerala SSLC (Class 10) maths 2020 question paper with solutions are provided here in a downloadable pdf format and also in the text so that the students can easily access them. Along with the solutions, they can also get the maths question paper 2020 class 10 SSLC for reference. Students are able to access all the Kerala board previous year maths question papers here. SSLC question paper 2020 Kerala for English medium with answers can be downloaded easily and students can practice and verify the answers provided by BYJUâ€™S. Solving 2020 Maths question papers for class 10 will help the students to predict what type of questions will appear in the exam.
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QUESTION PAPER CODE S 1531
Answer any three questions from 1 to 4. Each question carries 2 scores.
[3 * 2 = 6]
Question 1: [a] Write the 6^{th} term of the arithmetic sequence 1, 25, 49, 73, 97 â€¦..
[b] How many perfect square terms are there in the arithmetic sequence 97, 73, 49 â€¦..?
Solution:
[a] Given an arithmetic sequence is 1, 25, 49, 73, 97 , ………….First term (f) = 1
d = 25 â€“ 1 = 24.
T_{6} = a + (n – 1)d
6^{th} term = f + 5d
â‡’ 1 + 5 Ã— 24
â‡’ 1 + 120 = 121
[b] Given an arithmetic sequence is 97, 73, 49, ……….The perfect square numbers are 1, 4, 9, 16, ……….
Hence the given sequence is 97, 73, 49, 25, 1.
From the above sequence, the perfect square numbers are 49, 25, and 1.
âˆ´ The number of perfect square terms are 3.
Question 2: Chords AB and CD are intersecting at P. AB = 10cm, PB = 4cm, PD = 3cm.
[a] What is the length of PA?
[b] Find the length of the PC.
Solution:
Given:
AB = 10cm
PB = 4cm
PD = 3cm
[a] PA = AB â€“ PB= 10 â€“ 4
= 6cm
[b] PC Ã— PD = PA Ã— PBPC = [PA Ã— PB] / [PD]
= [6 Ã— 4] / 3
= 8cm
Question 3: Write the polynomial p(x) = x^{2} – 4 as the product of two first degree polynomials.
Solution:
Given polynomial = p(x) = x^{2 }âˆ’ 4
First-degree polynomial x^{2 }âˆ’ 4
= (x + 2) ( x â€“ 2)
Question 4: In the figure, O is the centre of the circle and x^{2} + y^{2} = 25 is the equation of the circle.
[a] What is the radius of the circle?
[b] Write the equation of the circle whose centre is at origin and the radius is 3.
Solution:
Given circle is x^{2 }+ y^{2} = 25
[a] x^{2 }+ y^{2} = r^{2}r^{2} = 25
r = âˆš25 = 5
âˆ´ r = 5
The radius of the circle is 5 units.
[b] Given radius = 3Hence the equation of the circle = x^{2} + y^{2} = r^{2}.
â‡’ x^{2} + y^{2} = 3^{2}
â‡’ x^{2} + y^{2} = 9
Answer any five questions from 5 to 11. Each question carries 3 scores. [5 * 3 = 15]
Question 5: [a] Write the first term and the common difference of the arithmetic sequence whose algebraic expression is 3n + 5.
[b] The first term of an arithmetic sequence is 8 and the common difference is 5. Write its algebraic form.
Solution:
Given x_{n }= 3n + 5
[a] If n = 1, then the first term can be obtained.First term = 3 Ã— 1 + 5 = 3 + 8 = 8
Common difference = 3 [âˆµ coefficient of n be the d ] [b] Given f = 8
d = 5
x_{n} = d_{n} + f â€“ d)
= 5n + (8 â€“ 5)
= 5n + 3
Question 6: In the figure, âˆ ABC = 90^{o}, âˆ C = âˆ D = 45^{o}, AB = 10cm.
[a] What is the length of AC?
[b] What is the radius of the circumcircle of triangle ABC?
[c] What is the radius of the circumcircle of triangle ABD?
Solution:
In right â–³ ABC, the angles are 450, 450, 900
â‡’ 1 : 1 : âˆš2
â‡’ AB : BC : AC
â‡’ x : x : x âˆš2
â‡’ 10 : 10 : 10âˆš2
[a] The length of AC = 10âˆš2 cm. [b] The radius of the circumcircle of â–³ABC= Half of the hypotenuses AC
= 10 âˆš2 / 2
= 5âˆš2 cm
[c] The radius of the circumcircle of â–³ABD= Half of the hypotenuses AC
= 10 âˆš2 / 2
= 5âˆš2 cm
Question 7: Draw a circle of radius 3cm. Mark a point P at a distance of 6cm from the centre of the circle. Draw tangents from P to the circle.
Solution:
Question 8: [a] What is the common difference of the arithmetic sequence x – 1, x, x + 1, â€¦..
[b] If x – 1 is an even number, what is the next even number?
[c] Prove that the product of 2 consecutive even numbers added to 1 gives a perfect square.
Solution:
[a] Given sequence x â€“ 1 , x, x + 1 , ……d = x â€“ (x â€“ 1)
= x â€“ x + 1
d = 1
[b] Given even number = x â€“ 1Next even number
= x â€“ 1 + 2
= x + 1
[c] Let consecutive two even number be (x â€“ 1 ) and (x+ 1)By question, the product of two consecutive numbers is + 1,
(x â€“ 1) (x+ 1) 1 = x^{1 }â€“ 1 + 1 = x^{2}
Here x^{2 }being a perfect square.
Question 9: In the figure, ABCD is a square. Its diagonals are parallel to the coordinate axes. AC = 6 and the coordinates of A is (3, 2). Write the coordinates of the vertices of C, B and D.
Solution:
Given, the coordinates of A = (3, 2)
AC be parallel to the x-axis.
âˆ´ The coordinates of C
= (3 + 6, 2)
= (9, 2)
The coordinates of the midpoint of AC = (6,2)
It is known that the diagonals are equal in a square.
âˆ´ The coordinates of B
= (6, 2 â€“ 3)
= (6, -1)
The coordinates of D
= (6, 2 + 3)
= (6, 5)
Hence the coordinates of A = (3, 2)
The coordinates of B = (6, -1)
The coordinates of C = (9, 2)
The coordinates of D = (6, 5)
Question 10: In the figure, ABCD is a cyclic quadrilateral. Also, âˆ A + âˆ D = 210^{o}, âˆ D + âˆ C = 250^{o}.
[a] What is âˆ A + âˆ C?
[b] Find the measures of âˆ A and âˆ C.
Solution:
Given that âˆ A + âˆ D = 210^{o}, âˆ D + âˆ C = 250^{o},
[a] A + C = 180^{0 }(cyclic quadrilateral)A + D = 210 â†’ (1)
D + C = 250 â†’ (2)
Adding (1) + (2)
â‡’ A + C + 2D = 460
â‡’ 180 + 2D = 460
â‡’ 2D = 28^{0}
D = 140^{0}
[b] A = 210 â€“ 140A = 70^{0}
C = 250 â€“ 140
C = 110^{0}
Question 11: The figure of a square sheet paper is shown below. Length of one side of the paper sheet is 36cm and AB = 10cm. The shaded portion is cut out and folded into a square pyramid.
[a] What is the length of the base of the pyramid?
[b] What is the slant height of the pyramid?
[c] Find the lateral surface area of the pyramid.
Solution:
Given, Side of the paper sheet = 36cm
AB = 10cm
[a] Base edge of the pyramidAB = 10cm
[b] Slant height of the pyramid= [36 âˆ’ 10] / 2
= 26 / 2
= 13cm [âˆµ a + 2l = 36, side of the larger square] [c] Lateral surface area = 2al
= 2 Ã— 10 Ã— 13
= 260 cm^{2}
Answer any seven questions form 12 to 21. Each question carries 4 scores. [7 * 4 = 28]
Question 12:
[a] What is the sum of the first 5 terms of the arithmetic sequence 1, 3, 5, 7 â€¦..?
[b] What is the sum of the first n terms of the arithmetic sequence 1, 3, 5, 7 â€¦..?
[c] Find the sum of the first n terms of the arithmetic sequence (1 / n), (3 / n), (5 / n), (7 / n) â€¦â€¦
[d] What is the sum of the first 2020 terms of the arithmetic sequence (1 / 2020), (3 / 2020), (5 / 2020), (7 / 2020)……
Solution:
[a] Given sequence = 1, 3, 5, 7, ……Sum = n^{2}
Here n = 5
âˆ´ The sum of the first 5 terms
= 5^{2}
= 25
[b] The sum of the first n terms = n^{2} [c] Given sequence = (1 / n), (3 / n), (5 / n), (7 / n) â€¦â€¦The sum of the first n terms = n^{2} / n = n
[d] Given sequence (1 / 2020), (3 / 2020), (5 / 2020), (7 / 2020)……The sum of the first n terms = n^{2} / n = n
The sum of the first 2020 terms
= n
= 2020
Question 13: Draw a rectangle of length 4cm and breadth 2cm. Draw a square having the same area of the rectangle.
Solution:
- Construct a rectangle ABCD with length 4cm and breadth 2cm.
- Extend the line AB to BE, such that BE = BC.
- Draw a semi-circle with AE as diameter.
- Draw a parallel line BF through B and then BF as length, draw the square BFGH.
Question 14: In a school, the total number of students in 10 A division is equal to the number of students in 10 B. One student is to be selected from each division. The number of boys in 10 A is 20. The probability of selecting a boy from 10 A is (2 / 5) and that of class B is (3 / 5).
[a] How many students are there in 10 A?
[b] What is the probability of selecting a girl from 10 A?
[c] How many boys are there in 10 B?
[d] What is the probability of both the selected students being boys?
Solution:
Class |
XA |
XB |
Boys |
20 |
30 |
Girls |
30 |
20 |
Total |
50 |
50 |
Given the probability of boys in XA = 2 / 5
Given the probability of boys in XB = 3 / 5
[a] Number of boys in XA= 20 Ã— [5 / 2]
= 50
[b] Probability of girl from XA= 1 âˆ’ [2 / 5]
= [5 âˆ’ 2] / 5
= 3 / 5
[c] Number of boys in X B= 50 Ã— [3 / 5]
= 10 Ã— 3
= 30
[d] Both being boys= [2 / 5] Ã— [3 / 5]
= 6 / 25
Question 15: Perimeter of the rectangle in the figure is 36cm. AC = âˆš164 cm.
[a] What is AB + BC?
[b] Find the length of AB.
Solution:
Given the perimeter = 36cm
AC = âˆš164 cm
[a] 2 (l + b) = 36âˆ´ AB + BC = 36 / 2
= 18cm
[b] Let AB = x , BC = 18 â€“ xSince âˆ†ABCis right-angled, according to Pythagoras theorem,
AC^{2} = AB^{2} + BC^{2}
164 = x^{2} + (18 â€“ x)^{2}
x^{2}+ 324 â€“ 36x + x^{2} = 164
2x^{2 }â€“ 36x = 164 â€“ 324 = – 160
Dividing by 2
x^{2} â€“ 18x = â€“ 80 [ square completion method]
x^{2} â€“ 18x + 81 = â€“ 80 + 81
(x â€“ 9)^{2} = 1
x â€“ 9 = Â± 1
x â€“ 9 = 1 or x â€“ 9 = -1
x = 10 or = 8
So, AB = 10 cm.
Question 16: In triangle ABC, âˆ A = âˆ B = 30^{o}, AC = 4cm.
[a] What is the length of BC?
[b] Find the length of AB.
[c] In triangle PQR, PQ = 4âˆš3 cm, âˆ P = âˆ Q = 30^{o}.
Draw the triangle.
Solution:
Given ,
âˆ A = âˆ B = 30^{0}
AC = 4cm
Draw CD âŠ¥ AB.
In right â–³ ADC,
30^{0} : 60^{0} : 90^{0}
1 : âˆš3 : 2
DC : AD : AC
x : xâˆš3 : 2x
2 : 2âˆš3 : 4
DC = 2
AD = 2âˆš3
AC = 4
[a] Length of BC = AC = 4cm [b] Length of AB = AD + DB= 2âˆš3 + 2âˆš3
= 4âˆš3cm
[c] Draw PR = 4cm and make R be 120^{0} and join PQ. â–³PRQ is the required triangle.
Question 17:
[a] If p(x) = x^{2} – 7x + 13, what is p(3)?
[b] Write the polynomial p(x) – p(3) as the product of two first degree polynomials.
[c] Find the solutions of the equation p(x) – p(3) = 0.
Solution:
[a] Given polynomialp(x) = x^{2} – 7x + 13
p(3) = 3^{2 }âˆ’ 7 Ã— 3 + 13
= 9 â€“ 21 + 13
= 1
[b] p(x) â€“ p(3) = x^{2} – 7x + 13 – 1= x^{2 }âˆ’ 7x + 12
= (x â€“ 3) (x â€“ 4)
Hence the product two first degree polynomial = (x â€“ 3) (x â€“ 4)
[c] p(x) â€“ p(3) = 0x^{2} – 7x + 12 = 0
â‡’ (x â€“ 3) (x â€“ 4) = 0
â‡’ (x â€“ 3) = 0 or (x â€“ 4 ) = 0
ie., x = 3 or x = 4.
Hence the solution is x = 3 and 4.
Question 18: In the figure, O is the centre of both circles. AB and AC touch the small circle at P and Q. A, B and C are points on the larger circle.
[a] If AP = 5cm, then what is the length of AQ?
[b] Prove that AB = AC.
[c] If AP = 5cm and âˆ A = 90^{o}, then what is the radius of the small circle?
Solution:
[a] Given AP = 5 cmHence the length of AQ = 5cm. [âˆµ Same tangents from A] [b] AB and AC are tangents
OP âŠ¥ AB and OQ âŠ¥ AC [âˆµ Chord bisector theorem]
AP = BP and AQ = QC
AB = AC
[c] Given âˆ A = 90^{0}So it can be seen that APOQ is a square.
[ âˆµ OP and OQ be radii âˆ´ âˆ APO = âˆ AQO = 90^{0}. ie., âˆ POQ = 90^{0}]Hence the radius of the small circle = 5cm.
Question 19: Draw the coordinate axes and mark the points A (-3, 0), B (3, 0), C (0, 3âˆš3).
Solution:
Question 20: A sector of radius 12cm and the central angle 120^{o}is rolled up to a cone.
[a] What is the slant height of the cone?
[b] Find the radius and height of the cone.
[c] What is the central angle of the sector to be used to make a cone of base radius âˆš2 cm and height 4cm?
Solution:
[a] The radius of the sector = 12cm [The radius of the sector to be the slant height of the cone ] [b] r / l = x^{0} / 360^{o}â‡’ r / 12 = 120 / 360
â‡’ 360r = 12 Ã— 120
â‡’ r = [12 Ã— 120] / 360
âˆ´ r = 4cm
Radius = 4cm
h = âˆšl^{2} âˆ’ r^{2}
= âˆš12^{2 }âˆ’4^{2}
= âˆš144 âˆ’ 16
= âˆš128
= 8âˆš2 cm
[c] r / l = x^{0} / 360^{o}Central angle (x^{0}) = [360 Ã— r] / l
To find â€˜lâ€™
l = âˆšh^{2} + r^{2}
r = âˆš2 , h = 4cm
âˆ´ l = âˆš4^{2 }+ âˆš2^{2}
= âˆš16 + 2
= âˆš18
= 3âˆš2
âˆ´ Center angle (x^{0}) = 360Ã—rl
= [360 Ã— âˆš2] / [3âˆš2]
= 120^{0}
Question 21:
[a] What is the slope of the line passing through the points (5, 0) and (3, 2)? Write the equation of the line.
[b] The x coordinate of a point on the line x – y = 5 is 5. What is the y coordinate of that point?
[c] Write the coordinates of the point of intersection of the lines x + y = 5 and x – y = 5.
Solution:
[a] Given points are (5, 0) and (3, 2).Slope = [y_{2 }âˆ’ y_{1}] / [x_{2} âˆ’ x_{1}]
= [2 âˆ’ 0] / [3 âˆ’ 5]
= 2 / âˆ’2
= – 1
Equation of the line is
y â€“ y_{1} = m( x â€“ x_{1})
= y â€“ 0
= -1 ( x â€“ 5)
y = -x + 5
x + y â€“ 5 = 0 is the equation.
[b] If x = 5, then5 â€“ y = 5
-y = 5 â€“ 5
= 0
The y coordinate of that point is 0.
[c] Given x + y = 5 and x â€“ y = 0.x + y = 5 â†’ (1)
x â€“ y = 0 â†’ (2)
Solve (1) and (2),
x and y = 5, and 0
So the coordinates = (5, 0).
Answer any five questions from 22 to 28. Each question carries 5 scores. [5 * 5 = 25]
Question 22: Sum of the first four terms of an arithmetic sequence is 72. Sum of first n terms is also 72.
[a] What is the 5^{th} term of the sequence?
[b] Find the sum of the first five terms.
[c] Write the sequence.
Solution:
Given sum of the first 4 terms = 72
Sum of the first 9 terms = 72
[a] 5^{th} term (x_{5}) = (S_{9}) / 9= 72 / 9
= 8
[b] Sum of the first 5 terms (S_{5})= S_{4} + x_{5}
= 72 + 8
= 80
[c] X_{3} = (S_{5}) / 5= 80 / 5
= 16
X_{3} + 2d = X_{5}
16 + 2d = 8
2d = â€“ 8
d = â€“ 4
X_{1} = X_{3} â€“ 2d
= 16 â€“ [2 Ã— â€“ 4]
= 24
The sequence = 24, 20, 16, 12, 8,…..
Question 23: A boy standing at the edge of a canal sees the top of a tree on the other edge at an elevation of 60^{o}. Stepping 12m back, he sees it at an elevation of 30^{o}. Find the height of the tree.
Solution:
Let AB be the height of the tree.
B be the first position of the boy.
C be the second position of the boy.
BC = 12
âˆ C = 30^{0}
âˆ BPA = 60^{0}
âˆ PBA = 30^{0}
âˆ A = 90^{0}
â–³CBD is an isosceles triangle.
âˆ´ BC = BD = 12
From right â–³BAP,
30^{0} ; 60^{0 }; 90^{0}
1 : âˆš3 : 2
â‡’ AD : AB : BD
â‡’ x : xâˆš3 : 2x
â‡’ 6 : 6âˆš3 : 12
âˆ´ AB = 6âˆš3
Hence the height of the tree = 6âˆš3m.
Question 24: In â–³ABC, AB = 5cm, âˆ A = 65^{0}, âˆ B = 55^{0}. Draw the triangle ABC and the incircle. Measure the radius of the incircle.
Solution:
Question 25: A circle is drawn with (5, 3) as a centre. (5, 6) is a point on the circle.
[a] What is the radius of the circle?
[b] Write the equation of the circle.
[c] What is the distance from the centre of the circle to the x-axis?
[d] What is the length of the tangents from the origin to the circle?
Solution:
[a] Radius of the circle = 6 â€“ 3 = 3 units [b] Equation of the circle= (x â€“ a)^{2 }+ (y â€“ b)^{2} = r^{2}
â‡’ (x â€“ 5)^{2 }+ (y â€“ 3)^{2 }= 3^{2}
â‡’ x^{2 }â€“ 10x + 25 + y^{2} â€“ 6y + 9 = 9
â‡’ x^{2} + y^{2 }â€“ 10x â€“ 6y + 25 = 0
[c] The distance from the centre of the circle to the x-axis is the radius of the circle = 3 units. [d] The length of the tangents from the origin to the circle = 5 units. [The x-axis being itself a tangent.]Question 26:
[a] The radius of a solid sphere is 6cm. Find its volume and surface area.
[b] It is cut into two equal halves. What is the total surface area of each hemisphere? What is the volume of a hemisphere?
Solution:
[a] Given radius = 6cm;Volume = [4 / 3] Ï€r^{3}
= [4 / 3] Ï€6^{3}
= 288Ï€ cm^{3}
Total Surface Area = 4Ï€r^{2}
= 4Ï€ Ã— 6^{2}
= 144Ï€cm^{2}
[b] Total Surface Area of hemisphere = 3Ï€r^{2}= 3Ï€ Ã— 6^{2}
= 108Ï€cm^{2}
The volume of hemisphere = [2 / 3] Ï€r^{3}
= [2 / 3] Ï€6^{3}
= 144Ï€cm^{3}
Question 27: The table below shows, children of a class sorted according to their marks in the examination.
[a] If we arrange the children from the one with the least mark to the one with the greatest, then what will be the assumed mark of the 12^{th} student?
[b] Compute the median mark.
Solution:
Class |
Frequency |
Marks |
Cumulative Frequency |
0 – 10 |
4 |
<10 |
4 |
10 – 20 |
7 |
<20 |
11 |
20 – 30 |
10 |
<30 |
21 |
30 – 40 |
12 |
<40 |
33 |
40 – 50 |
8 |
<50 |
41 |
Total |
41 |
= 20 + [30 âˆ’ 20] / [10 Ã— 2]
= 20 + 1 / 2
= 20.5
[b] N / 2 = 41 / 2 = 20.5Median class = 20 – 30
l = 20
F = 11
f = 10
Median = l + [(N / 2 âˆ’ F] / f) / c
= 20 + ([20.5 âˆ’ 11] / 10) Ã— 10
= 20 + [9.5 / 10] Ã— 10
= 20 + 9.5
= 29.5
âˆ´ Median mark = 29.5
Question 28: In the figure, O is the centre of the large circle. The centre of the small circle is C. OP is a tangent to the small circle. âˆ BOQ = 50^{o}.
[a] âˆ OAQ = ?
[b] âˆ OCP = ?
[c] âˆ APO = ?
[d] âˆ POQ = ?
Solution:
Given BOQ = 50^{0}
[a] Since â–³AOQ is isosceles, their base angles are equal.AOQ = 180 â€“ 50 = 130
âˆ A = âˆ Q = [180 âˆ’ 130] / 2
= 50 / 2
= 25^{0}
âˆ´ âˆ OAQ = 25^{0}
[b] âˆ OCP = 25^{0} Ã— 2 = 50^{0} [c] âˆ APO = 25^{0} + 90^{o} = 115^{0} [d] âˆ POQ = 180^{0} – ( 50 + âˆ AOP)= 180^{o} â€“ 50^{o} â€“ 40^{o}
= 90^{0}
Question 29: Read the following passage. Understand the mathematical concept in it and answer the questions that follow. Each question carries 1 score. [6 * 1 = 6]
The common difference of the arithmetic sequence 15, 14, 13, 12, 11, 10 â€¦.. is 14 – 15 = -1. The first term of the sequence is 15 and the 15^{th} term is 15 + 14 * -1 = 15 – 14 = 1. Similarly, the 4^{th} term is 12 and the 12^{th} term is 4. Its 16^{th} term is x_{16} = 15 + 15 * -1 = 15 – 15 = 0. So, the sum of the first 31 terms is also 0. That is if the n^{th} term of an arithmetic sequence with common difference -1 is m, then the m^{th} term is n and the (m + n)^{th} term is 0.
[a] 7^{th} term of an arithmetic sequence is 10 and the 10^{th} term is 7. What is the common difference?
[b] What is the 21^{st} term of the arithmetic sequence 21, 20, 19â€¦.
[c] 5^{th} term of an arithmetic sequence is 17 and the 17^{th} term is 5. Which term of the sequence is 0?
[d] 5^{th} term of an arithmetic sequence is 17 and the 17^{th} term is 5. What is the 44^{th} term?
[e] 1^{st} term of an arithmetic sequence is n and the n^{th} term is 1. What is the (n + 1)^{th} term?
[f] The 1^{st} term of an arithmetic sequence is n and the n^{th} term is 1. Sum of how many terms, starting from the first term, of this sequence is 0?
Solution:
[a] The common difference d = – 1 [b] 21^{st} term becomes 1. [c] 22^{nd} term becomes 0. [d] 44^{th} term = 0 â€“ 22 = -22 [e] The (n + 1)^{th} term is 0. [f] 0 is the sum of (2n + 1)^{th} term.
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