Karnataka SSLC Model Question Paper 2026 Mathematics
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Page 1
SAMPLE PAPER
Karnataka
Board
Model Paper
Karnataka
Board
Model Paper
Page 2
KARNATAKA SCHOOL EXAMINATION AND ASSESSMENT BOARD
Malleshwaram, Bengaluru – 560003
S.S.L.C. MODEL QUESTION PAPER – 01 – 2025-26
Subject: MATHEMATICS
(English Medium)
Subject Code: 81 – E
[ Time: 3 Hours 15 Minutes]
[ Max. Marks: 80]
General Instructions to the candidate:
1. This question paper consists of 38 questions.
2. Follow the instructions given against the questions.
3. Figures in the right hand margin indicate maximum marks for the questions.
4. The maximum time to answer the paper is given at the top of the question paper.
It includes 15 minutes for reading the question paper.
1
Malleshwaram, Bengaluru – 560003
S.S.L.C. MODEL QUESTION PAPER – 01 – 2025-26
Subject: MATHEMATICS
(English Medium)
Subject Code: 81 – E
[ Time: 3 Hours 15 Minutes]
[ Max. Marks: 80]
General Instructions to the candidate:
1. This question paper consists of 38 questions.
2. Follow the instructions given against the questions.
3. Figures in the right hand margin indicate maximum marks for the questions.
4. The maximum time to answer the paper is given at the top of the question paper.
It includes 15 minutes for reading the question paper.
1
Page 3
I. Four alternatives are given for each of the following questions / incomplete
statements. Choose the correct alternative and write the complete answer
along with its letter of alphabet. 8 1=8
1. The H.C.F of 3 and 5 is,
(A) 1 (B) 3
(C) 5 (D) 15
2. In the figure, the number of zeroes of the polynomial is,
(A) 3 (B) 5
(C) 4 (D) 1
3. If the pair of linear equations a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0 have unique
solution, then the correct relation among the following is,
a1 b1 a1 b1 c1
(A) (B)
a2 b2 a2 b2 c2
a1 b1 a1 b1 c1
(C) (D)
a2 b2 a2 b2 c2
2
statements. Choose the correct alternative and write the complete answer
along with its letter of alphabet. 8 1=8
1. The H.C.F of 3 and 5 is,
(A) 1 (B) 3
(C) 5 (D) 15
2. In the figure, the number of zeroes of the polynomial is,
(A) 3 (B) 5
(C) 4 (D) 1
3. If the pair of linear equations a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0 have unique
solution, then the correct relation among the following is,
a1 b1 a1 b1 c1
(A) (B)
a2 b2 a2 b2 c2
a1 b1 a1 b1 c1
(C) (D)
a2 b2 a2 b2 c2
2
Page 4
4. The arithmetic progression in the following is,
(A 1, 2, 4, 8……. B 3, 7, 10, 14…….
C 1, 4, 9, 16……… (D 5, 9, 13, 17……..
5. In the given frequency distribution table the median class is,
Marks Number of Students Cumulative
Frequency
0– 10 3 3
10 – 20 4 7
20 – 30 7 14
30 – 40 6 20
n = 20
(A) 0 – 10 (B) 10 – 20
(C) 20 – 30 (D) 30 – 40
6. If 3 , then the measure of the angle A is,
(A) 90 (B) 60
(C) 45 (D) 30
7. A perpendicular drawn from a point – 4, – 5 intersects X- axis at the point Q.
The coordinates of point Q are,
(A) (0, - 4) (B) ( –4, 0)
(C) (–5, 0) (D) (0, –5)
3
(A 1, 2, 4, 8……. B 3, 7, 10, 14…….
C 1, 4, 9, 16……… (D 5, 9, 13, 17……..
5. In the given frequency distribution table the median class is,
Marks Number of Students Cumulative
Frequency
0– 10 3 3
10 – 20 4 7
20 – 30 7 14
30 – 40 6 20
n = 20
(A) 0 – 10 (B) 10 – 20
(C) 20 – 30 (D) 30 – 40
6. If 3 , then the measure of the angle A is,
(A) 90 (B) 60
(C) 45 (D) 30
7. A perpendicular drawn from a point – 4, – 5 intersects X- axis at the point Q.
The coordinates of point Q are,
(A) (0, - 4) (B) ( –4, 0)
(C) (–5, 0) (D) (0, –5)
3
Page 5
8. The quadratic equation whose roots are 6 and 1 is
(A) x2 + 5x – 6 = 0 (B) x2 – 5x – 6 = 0
(C) x2 – 5x + 6 = 0 (D) x2 + 5x + 6 = 0
II. Answer the following questions: 8
9. How many solutions does the pair of linear equations in two variables have if they are
inconsistent?
10. In the figure, if DE||BC, AD =2cm and BD = 6cm, then find DE:BC.
11. The circumference of the base of a cylinder is 10cm and its height is 25cm. Find its
curved surface area.
12. Write the value of ̅̅̅̅̅ with reference to probability.
13. In an arithmetic progression, the seventh term is 12 more than its fourth term. Find the
common difference.
4
(A) x2 + 5x – 6 = 0 (B) x2 – 5x – 6 = 0
(C) x2 – 5x + 6 = 0 (D) x2 + 5x + 6 = 0
II. Answer the following questions: 8
9. How many solutions does the pair of linear equations in two variables have if they are
inconsistent?
10. In the figure, if DE||BC, AD =2cm and BD = 6cm, then find DE:BC.
11. The circumference of the base of a cylinder is 10cm and its height is 25cm. Find its
curved surface area.
12. Write the value of ̅̅̅̅̅ with reference to probability.
13. In an arithmetic progression, the seventh term is 12 more than its fourth term. Find the
common difference.
4
Page 6
14. In the figure a cone with radius ‘r’ and slant height ‘l’ is mounted on a hemisphere with
radius ‘r’. Write the formula to find the total surface area of the solid formed.
15. . BC and QR are corresponding sides. If 80 and 40 , then
what is the measure of
16. If the zeroes of the polynomial p(x)= x2 + 3x + k are reciprocal to each other, then find
the value of k.
III. Answer the following questions: 8 x 2 = 16
17. In an arithmetic progression, if S20= 820 and a20= 79, then find its first term.
OR
How many numbers between 201 and 401 are divisible by 6?
18. Solve the given pair of linear equations using suitable method.
2x + y = 8
3x – y = 7
5
radius ‘r’. Write the formula to find the total surface area of the solid formed.
15. . BC and QR are corresponding sides. If 80 and 40 , then
what is the measure of
16. If the zeroes of the polynomial p(x)= x2 + 3x + k are reciprocal to each other, then find
the value of k.
III. Answer the following questions: 8 x 2 = 16
17. In an arithmetic progression, if S20= 820 and a20= 79, then find its first term.
OR
How many numbers between 201 and 401 are divisible by 6?
18. Solve the given pair of linear equations using suitable method.
2x + y = 8
3x – y = 7
5
Page 7
19. Find the discriminant of the quadratic equation 2x2 + 3x – 7 = 0 and determine the
nature of roots.
OR
Express the quadratic equation (2x + 3) x = x2 + 1 in the standard form.
20. Prove that √5 is an irrational number.
21. Find the coordinates of the point which divides the line segment joining the points
(1, 6) and (4, 3) in the ratio 1 2 internally.
22. In the figure, find the value of and .
23. In the figure, CP and CQ are tangents to the circle with centre O. ARB is another tangent at R.
Show that the perimeter of triangle CAB is twice the length of tangent CP.
6
nature of roots.
OR
Express the quadratic equation (2x + 3) x = x2 + 1 in the standard form.
20. Prove that √5 is an irrational number.
21. Find the coordinates of the point which divides the line segment joining the points
(1, 6) and (4, 3) in the ratio 1 2 internally.
22. In the figure, find the value of and .
23. In the figure, CP and CQ are tangents to the circle with centre O. ARB is another tangent at R.
Show that the perimeter of triangle CAB is twice the length of tangent CP.
6
Page 8
24. The product of HCF and LCM of 2 numbers is 2016. If one of the numbers is 42, then find
the other number. Also find the HCF of those two numbers by prime factorization
method.
IV. Answer the following questions: 9 x 3 = 27
25. Find the coordinates of the centre of a circle passing through the points A 3, 0),
B(0, 2) and C(0, 3).
26. Prove that “the length of tangents drawn to a circle from an external point are equal”.
27. An express train takes 1 hour less than a passenger train to travel a distance of
132 km. The Average speed of the express train is 11km/h more than that of the
passenger train. Find the average speed of these two trains.
OR
The ages of two students ‘A’ and ‘B’ are 19 years and 15 years respectively. Find how
many years it will take so that the product of their ages become equal to 480.
28. Prove that: 1 √ cosec A
OR
Prove that
29. Write the quadratic polynomial whose sum and product of the zeroes are 2 and 8
respectively and hence find the zeroes of the polynomial.
OR
If and are the zeroes of the quadratic polynomial p(x) = x² + 3x +1 then,
find the value of (i)
(ii)
7
the other number. Also find the HCF of those two numbers by prime factorization
method.
IV. Answer the following questions: 9 x 3 = 27
25. Find the coordinates of the centre of a circle passing through the points A 3, 0),
B(0, 2) and C(0, 3).
26. Prove that “the length of tangents drawn to a circle from an external point are equal”.
27. An express train takes 1 hour less than a passenger train to travel a distance of
132 km. The Average speed of the express train is 11km/h more than that of the
passenger train. Find the average speed of these two trains.
OR
The ages of two students ‘A’ and ‘B’ are 19 years and 15 years respectively. Find how
many years it will take so that the product of their ages become equal to 480.
28. Prove that: 1 √ cosec A
OR
Prove that
29. Write the quadratic polynomial whose sum and product of the zeroes are 2 and 8
respectively and hence find the zeroes of the polynomial.
OR
If and are the zeroes of the quadratic polynomial p(x) = x² + 3x +1 then,
find the value of (i)
(ii)
7
Page 9
30. ABC is a right angled triangle. If 90 and AD BC, then show that AB2 = BC BD
31. The minute hand of a wall clock is 18cm long. Find the area swept by the minute hand
in 35 minutes. Find the length of the arc formed by the sweep of the minute hand.
32. A box contains cards numbered from 6 to 70. If one card is drawn at random from the
box, then find the probability that the card will have
(i) a perfect square number.
(ii) a number divisible by 5.
(iii) an odd number less than 30.
33. Find the Mean for the following frequency distribution table.
Class interval Frequency
0 – 10 4
10 – 20 9
20 – 30 15
30 – 40 14
40 – 50 8
OR
Find the Mode for the following frequency distribution table.
Class interval Frequency
15-20 6
20-25 9
25-30 15
30-35 9
35-40 1
8
31. The minute hand of a wall clock is 18cm long. Find the area swept by the minute hand
in 35 minutes. Find the length of the arc formed by the sweep of the minute hand.
32. A box contains cards numbered from 6 to 70. If one card is drawn at random from the
box, then find the probability that the card will have
(i) a perfect square number.
(ii) a number divisible by 5.
(iii) an odd number less than 30.
33. Find the Mean for the following frequency distribution table.
Class interval Frequency
0 – 10 4
10 – 20 9
20 – 30 15
30 – 40 14
40 – 50 8
OR
Find the Mode for the following frequency distribution table.
Class interval Frequency
15-20 6
20-25 9
25-30 15
30-35 9
35-40 1
8
Page 10
V. Answer the following questions: 4 x 4 = 16
34. Solve the given pair of linear equations by graphical method.
x + 2y = 6
x+y=5
35. Prove that “If in two triangles, sides of one triangle are proportional to i.e., in the
same ratio of) the sides of the other triangle, then their corresponding angles are
equal and hence the two triangles are similar.”
OR
Prove that “If one angle of a triangle is equal to one angle of the other triangle and the
sides including these angles are proportional, then the two triangles are similar”.
36. The sum of first ‘n’ terms of an arithmetic progression is S n = 3n2 + 2n. Find the nth
term of the progression. If this progression consists of 59 terms then find the sum
of last 10 terms of the progression.
37. The lighthouse [AB] of height 10√3 m stands vertically on a sea shore. A tower [CE]
and a ship [F] are standing 30m and 10m away from the foot of the lighthouse
respectively. The angle of elevation of the top of the tower from the top of the
lighthouse is 300 . Find the height of the tower and distance between the top of the
lighthouse to the top of the tower [AE]. Also find the angle of depression formed from
the top of the lighthouse to the ship.
9
34. Solve the given pair of linear equations by graphical method.
x + 2y = 6
x+y=5
35. Prove that “If in two triangles, sides of one triangle are proportional to i.e., in the
same ratio of) the sides of the other triangle, then their corresponding angles are
equal and hence the two triangles are similar.”
OR
Prove that “If one angle of a triangle is equal to one angle of the other triangle and the
sides including these angles are proportional, then the two triangles are similar”.
36. The sum of first ‘n’ terms of an arithmetic progression is S n = 3n2 + 2n. Find the nth
term of the progression. If this progression consists of 59 terms then find the sum
of last 10 terms of the progression.
37. The lighthouse [AB] of height 10√3 m stands vertically on a sea shore. A tower [CE]
and a ship [F] are standing 30m and 10m away from the foot of the lighthouse
respectively. The angle of elevation of the top of the tower from the top of the
lighthouse is 300 . Find the height of the tower and distance between the top of the
lighthouse to the top of the tower [AE]. Also find the angle of depression formed from
the top of the lighthouse to the ship.
9
Page 11
OR
A tower and a pole stand vertically on the same level ground. It is observed that the
angles of depression of the top and foot of the pole from the top of tower of height 60m
is 300 and 600 respectively. Find the height of the pole.
VI. Answer the following question: 1x5=5
38. A hollow solid is made by placing a cylinder on an inverted cone as shown in the
figure. The radius of the cylinder and the cone are 7cm and their heights are 12cm and
9cm respectively. The solid is filled with water up to the height of rd the height of the
cylinder. Find the amount of water in the solid. Calculate the amount of water required
to fill the solid completely.
10
A tower and a pole stand vertically on the same level ground. It is observed that the
angles of depression of the top and foot of the pole from the top of tower of height 60m
is 300 and 600 respectively. Find the height of the pole.
VI. Answer the following question: 1x5=5
38. A hollow solid is made by placing a cylinder on an inverted cone as shown in the
figure. The radius of the cylinder and the cone are 7cm and their heights are 12cm and
9cm respectively. The solid is filled with water up to the height of rd the height of the
cylinder. Find the amount of water in the solid. Calculate the amount of water required
to fill the solid completely.
10
Page 12
OR
A vessel is in the form of an inverted cone. Its height is 8cm and the radius is 5cm. The
vessel is filled with water up to the brim and when 100 lead shots of same size are dropped
in it, th of the water from the vessel flows out. Find the radius and volume of each lead
shot.
11
A vessel is in the form of an inverted cone. Its height is 8cm and the radius is 5cm. The
vessel is filled with water up to the brim and when 100 lead shots of same size are dropped
in it, th of the water from the vessel flows out. Find the radius and volume of each lead
shot.
11
Page 13
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Page 14
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CªÀÅUÀ¼À°è ¸ÀÆPÀÛªÁzÀ GvÀÛgÀªÀ£ÀÄß Dj¹, CzÀgÀ PÀæªÀiÁPÀëgÀzÉÆqÀ£É ¥ÀÆtð GvÀÛgÀªÀ£ÀÄß §gɬÄj :
8 1=8
1. 3 ªÀÄvÀÄÛ 5 gÀ ªÀÄ.¸Á.C
(A) 1 (B) 3
(C) 5 (D) 15
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À À ¸ÀASÉåAiÀÄÄ,
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(C) 4 (D) 1
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À UÀ¼À eÉÆÃrAiÀÄÄ C£À£Àå
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a1 b1 a1 b1 c1
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a2 b2 a2 b2 c2
a1 b1 a1 b1 c1
(C) (D)
a2 b2 a2 b2 c2
2
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8 1=8
1. 3 ªÀÄvÀÄÛ 5 gÀ ªÀÄ.¸Á.C
(A) 1 (B) 3
(C) 5 (D) 15
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2
Page 15
4. EªÀÅUÀ¼À°è ¸ÀªÀiÁAvÀgÀ ±ÉæÃrüAiÀÄÄ,
(A) 1, 2, 4, 8……. (B) 3, 7, 10, 14…….
(C) 1, 4, 9, 16……… (D) 5, 9, 13, 17……..
5. PÉÆnÖgÀĪÀ DªÀÈwÛ «vÀgu
À Á PÉÆõÀÖPÀzÀ°è ªÀÄzsÁåAPÀzÀ ªÀUÁðAvÀgª
À ÀÅ,
CAPÀU¼
À ÀÄ «zÁåyðUÀ¼À ¸ÀASÉå ¸ÀAavÀ DªÀÈwÛ
0– 10 3 3
10 – 20 4 7
20 – 30 7 14
30 – 40 6 20
n = 20
(A) 0 – 10 (B) 10 – 20
(C) 20 – 30 (D) 30 – 40
6. cot A = 3 tan A DzÀg,É PÉÆãÀ A AiÀÄ C¼ÀvÉAiÀÄÄ,
(A) 90 (B) 60
(C) 45 (D) 30
7. P (–4, –5) ©AzÀÄ«¤AzÀ J¼ÉzÀ ®A§ªÀÅ X CPÀª
ë À£ÀÄß Q ©AzÀÄ«£À°è bÉâ¸ÀÄvÀz
Û É. ºÁUÁzÀgÉ Q ©AzÀÄ«£À
¤zÉÃð±ÁAPÀU¼
À ÀÄ,
(A) (0, 4) (B) ( –4, 0)
(C) (–5, 0) (D) (0, –5)
3
(A) 1, 2, 4, 8……. (B) 3, 7, 10, 14…….
(C) 1, 4, 9, 16……… (D) 5, 9, 13, 17……..
5. PÉÆnÖgÀĪÀ DªÀÈwÛ «vÀgu
À Á PÉÆõÀÖPÀzÀ°è ªÀÄzsÁåAPÀzÀ ªÀUÁðAvÀgª
À ÀÅ,
CAPÀU¼
À ÀÄ «zÁåyðUÀ¼À ¸ÀASÉå ¸ÀAavÀ DªÀÈwÛ
0– 10 3 3
10 – 20 4 7
20 – 30 7 14
30 – 40 6 20
n = 20
(A) 0 – 10 (B) 10 – 20
(C) 20 – 30 (D) 30 – 40
6. cot A = 3 tan A DzÀg,É PÉÆãÀ A AiÀÄ C¼ÀvÉAiÀÄÄ,
(A) 90 (B) 60
(C) 45 (D) 30
7. P (–4, –5) ©AzÀÄ«¤AzÀ J¼ÉzÀ ®A§ªÀÅ X CPÀª
ë À£ÀÄß Q ©AzÀÄ«£À°è bÉâ¸ÀÄvÀz
Û É. ºÁUÁzÀgÉ Q ©AzÀÄ«£À
¤zÉÃð±ÁAPÀU¼
À ÀÄ,
(A) (0, 4) (B) ( –4, 0)
(C) (–5, 0) (D) (0, –5)
3
Page 16
8. 6 ªÀÄvÀÄÛ -1 ªÀÄÆ®UÀ¼ÁVgÀĪÀ ªÀUð
À ¸À«ÄÃPÀgt
À ªÀÅ,
(A) x2 + 5x – 6 = 0 (B) x2 – 5x – 6 = 0
(C) x2 – 5x + 6 = 0 (D) x2 + 5x + 6 = 0
II. PɼÀV£À ¥Àæ±ÉßUÀ½UÉ GvÀÛj¹: 8x1=8
9. JgÀqÀÄ ZÀgÁPÀg
ë ÀªÀżÀî gÉÃSÁvÀäPÀ ¸À«ÄÃPÀgt
À UÀ¼À eÉÆÃrAiÀÄÄ C¹ÜgÀ eÉÆÃrAiÀiÁVzÀÝgÉ, CªÀÅ JμÀÄÖ ¥ÀjºÁgÀU¼
À À£ÀÄß
ºÉÆA¢gÀÄvÀª
Û É?
10. avÀæzÀ°è, DE||BC, AD =2 cm ªÀÄvÀÄÛ BD = 6 cm DzÀg,É DE:BC PÀAqÀÄ»r¬Äj.
11. MAzÀÄ ¹°AqÀj£À ¥ÁzÀzÀ ¥Àj¢üAiÀÄÄ 10 cm ªÀÄvÀÄÛ JvÀÛgÀªÀÅ 25 cm DVzÉ. CzÀgÀ ¥Á±Àéð ªÉÄïÉäöÊ
«¹ÛÃtðªÀ£ÀÄß PÀAqÀÄ»r¬Äj.
12. ¸ÀA¨sÀªÀ¤ÃAiÀÄvÉUÉ ¸ÀA§A¢ü¹zÀAvÉ, P(E) + ̅̅̅̅̅
( ) £À ¨É¯ÉAiÀÄ£ÀÄß §gɬÄj.
13. MAzÀÄ ¸ÀªÀiÁAvÀgÀ ±ÉæÃrüAiÀÄ°è K¼À£Éà ¥ÀzÀªÀÅ CzÀgÀ £Á®Ì£Éà ¥ÀzÀQÌAvÀ 12 ºÉZÁÑVzÉ. ºÁUÁzÀgÉ ¸ÁªÀiÁ£Àå ªÀåvÁå¸ÀªÀ£ÀÄß
PÀAqÀÄ»r¬Äj.
4
À ¸À«ÄÃPÀgt
À ªÀÅ,
(A) x2 + 5x – 6 = 0 (B) x2 – 5x – 6 = 0
(C) x2 – 5x + 6 = 0 (D) x2 + 5x + 6 = 0
II. PɼÀV£À ¥Àæ±ÉßUÀ½UÉ GvÀÛj¹: 8x1=8
9. JgÀqÀÄ ZÀgÁPÀg
ë ÀªÀżÀî gÉÃSÁvÀäPÀ ¸À«ÄÃPÀgt
À UÀ¼À eÉÆÃrAiÀÄÄ C¹ÜgÀ eÉÆÃrAiÀiÁVzÀÝgÉ, CªÀÅ JμÀÄÖ ¥ÀjºÁgÀU¼
À À£ÀÄß
ºÉÆA¢gÀÄvÀª
Û É?
10. avÀæzÀ°è, DE||BC, AD =2 cm ªÀÄvÀÄÛ BD = 6 cm DzÀg,É DE:BC PÀAqÀÄ»r¬Äj.
11. MAzÀÄ ¹°AqÀj£À ¥ÁzÀzÀ ¥Àj¢üAiÀÄÄ 10 cm ªÀÄvÀÄÛ JvÀÛgÀªÀÅ 25 cm DVzÉ. CzÀgÀ ¥Á±Àéð ªÉÄïÉäöÊ
«¹ÛÃtðªÀ£ÀÄß PÀAqÀÄ»r¬Äj.
12. ¸ÀA¨sÀªÀ¤ÃAiÀÄvÉUÉ ¸ÀA§A¢ü¹zÀAvÉ, P(E) + ̅̅̅̅̅
( ) £À ¨É¯ÉAiÀÄ£ÀÄß §gɬÄj.
13. MAzÀÄ ¸ÀªÀiÁAvÀgÀ ±ÉæÃrüAiÀÄ°è K¼À£Éà ¥ÀzÀªÀÅ CzÀgÀ £Á®Ì£Éà ¥ÀzÀQÌAvÀ 12 ºÉZÁÑVzÉ. ºÁUÁzÀgÉ ¸ÁªÀiÁ£Àå ªÀåvÁå¸ÀªÀ£ÀÄß
PÀAqÀÄ»r¬Äj.
4
Page 17
14. PÉÆnÖgÀĪÀ avÀæzÀ°è ‘r’ wædå ªÀÄvÀÄÛ ‘l’ NgÉ JvÀg
Û À«gÀĪÀ MAzÀÄ ±ÀAPÀĪÀ£ÀÄß ‘r’ wædå«gÀĪÀ CzsÀðUÉÆüÀzÀ ªÉÄïÉ
Ej¹zÉ. GAmÁzÀ WÀ£ÁPÀÈwAiÀÄ ¥ÀÆtð ªÉÄïÉäöÊ «¹ÛÃtðªÀ£ÀÄß PÀAqÀÄ»rAiÀÄĪÀ ¸ÀÆvÀæªÀ£ÀÄß §gɬÄj.
15. ΔABC . ªÀÄvÀÄÛ C£ÀÄgÀÆ¥À ¨ÁºÀÄUÀ¼ÁVªÉ. 80 ªÀÄvÀÄÛ, 40 DzÀgÉ, AiÀÄ
C¼ÀvÉAiÉĵÀÄÖ?
16. p(x)= x2 + 3x + k F §ºÀÄ¥ÀzÉÆÃQÛAiÀÄ ±ÀÆ£ÀåvÉU¼
À ÀÄ ¥Àg¸ Àà À ªÀÅåvÀÌçªÀÄUÀ¼ÁzÀg,É k AiÀÄ ¨É¯ÉAiÀÄ£ÀÄß
À g
PÀAqÀÄ»r¬Äj.
III. PɼÀV£À ¥Àæ±ÉßUÀ½UÉ GvÀÛj¹: 8 x 2 = 16
17. MAzÀÄ ¸ÀªÀiÁAvÀgÀ ±ÉæÃrüAiÀÄ°è S20= 820 ªÀÄvÀÄÛ a20= 79 DzÀg,É CzÀgÀ ªÉÆzÀ® ¥ÀzÀªÀ£ÀÄß PÀAqÀÄ»r¬Äj.
CxÀªÁ
201 jAzÀ 401 gÀ £ÀqÀÄªÉ 6 jAzÀ ¨sÁUÀªÁUÀĪÀ JµÀÄÖ ¸ÀASÉåUÀ½ªÉ?
18. PÉÆnÖgÀĪÀ gÉÃSÁvÀäPÀ ¸À«ÄÃPÀgt
À UÀ¼À eÉÆÃrAiÀÄ£ÀÄß ¸ÀÆPÀÛ «zsÁ£À¢AzÀ ©r¹.
2x + y = 8
3x – y = 7
5
Û À«gÀĪÀ MAzÀÄ ±ÀAPÀĪÀ£ÀÄß ‘r’ wædå«gÀĪÀ CzsÀðUÉÆüÀzÀ ªÉÄïÉ
Ej¹zÉ. GAmÁzÀ WÀ£ÁPÀÈwAiÀÄ ¥ÀÆtð ªÉÄïÉäöÊ «¹ÛÃtðªÀ£ÀÄß PÀAqÀÄ»rAiÀÄĪÀ ¸ÀÆvÀæªÀ£ÀÄß §gɬÄj.
15. ΔABC . ªÀÄvÀÄÛ C£ÀÄgÀÆ¥À ¨ÁºÀÄUÀ¼ÁVªÉ. 80 ªÀÄvÀÄÛ, 40 DzÀgÉ, AiÀÄ
C¼ÀvÉAiÉĵÀÄÖ?
16. p(x)= x2 + 3x + k F §ºÀÄ¥ÀzÉÆÃQÛAiÀÄ ±ÀÆ£ÀåvÉU¼
À ÀÄ ¥Àg¸ Àà À ªÀÅåvÀÌçªÀÄUÀ¼ÁzÀg,É k AiÀÄ ¨É¯ÉAiÀÄ£ÀÄß
À g
PÀAqÀÄ»r¬Äj.
III. PɼÀV£À ¥Àæ±ÉßUÀ½UÉ GvÀÛj¹: 8 x 2 = 16
17. MAzÀÄ ¸ÀªÀiÁAvÀgÀ ±ÉæÃrüAiÀÄ°è S20= 820 ªÀÄvÀÄÛ a20= 79 DzÀg,É CzÀgÀ ªÉÆzÀ® ¥ÀzÀªÀ£ÀÄß PÀAqÀÄ»r¬Äj.
CxÀªÁ
201 jAzÀ 401 gÀ £ÀqÀÄªÉ 6 jAzÀ ¨sÁUÀªÁUÀĪÀ JµÀÄÖ ¸ÀASÉåUÀ½ªÉ?
18. PÉÆnÖgÀĪÀ gÉÃSÁvÀäPÀ ¸À«ÄÃPÀgt
À UÀ¼À eÉÆÃrAiÀÄ£ÀÄß ¸ÀÆPÀÛ «zsÁ£À¢AzÀ ©r¹.
2x + y = 8
3x – y = 7
5
Page 18
19. 2x2 + 3x – 7 = 0 F ªÀUð
À ¸À«ÄÃPÀgt
À zÀ ±ÉÆÃzsÀPÀªÀ£ÀÄß PÀAqÀÄ»r¬Äj ªÀÄvÀÄÛ ªÀÄÆ®UÀ¼À ¸Àé¨sÁªÀªÀ£ÀÄß ¤zsÀðj¹.
CxÀªÁ
(2x + 3) x = x2 + 1 F ªÀUð
À ¸À«ÄÃPÀgt
À ªÀ£ÀÄß DzÀ±Àð gÀÆ¥ÀzÀ°è ªÀåP¥
ÀÛ Àr¹.
20. √5 MAzÀÄ C¨sÁUÀ®§Þ ¸ÀASÉå JAzÀÄ ¸Á¢ü¹.
21. (1, 6) ªÀÄvÀÄÛ (4, 3) ©AzÀÄUÀ¼À£ÀÄß ¸ÉÃj¸ÀĪÀ gÉÃSÁRAqÀªÀ£ÀÄß 1 2 C£ÀÄ¥ÁvÀzÀ°è DAvÀjPÀªÁV «¨sÁV¸ÀĪÀ
©AzÀÄ«£À ¤zÉÃð±ÁAPÀU¼
À À£ÀÄß PÀAqÀÄ»r¬Äj.
22. avÀæzÀ°è, ªÀÄvÀÄÛ ¨É¯ÉU¼
À À£ÀÄß PÀAqÀÄ»r¬Äj.
23. avÀæzÀ°è, ªÀÄvÀÄÛ UÀ¼ÀÄ PÉÃAzÀæªÀżÀî ªÀÈvÀPÛ ÉÌ ¸À±
à ÀðPÀU¼
À ÁVªÉ. AiÀÄÄ ©AzÀÄ«£À°è ªÀÄvÉÆÛAzÀÄ
¸À±
à ÀðPÀªÁVzÉ. wæ¨sÀÄdzÀ ¸ÀÄvÀ¼
Û ÀvÉAiÀÄÄ ¸À±
à ÀðPÀzÀ GzÀÝzÀ JgÀqÀg¶
À ÖzÉ JAzÀÄ vÉÆÃj¹.
24. JgÀqÀÄ ¸ÀASÉåUÀ¼À ªÀÄ.¸Á.C ªÀÄvÀÄÛ ®.¸Á.C UÀ¼À UÀÄt®§Þ 2016 DVzÉ. CªÀÅUÀ¼À°è MAzÀÄ ¸ÀASÉåAiÀÄÄ 42
DzÀgÉ E£ÉÆßAzÀÄ ¸ÀASÉåAiÀÄ£ÀÄß PÀAqÀÄ»r¬Äj ºÁUÀÆ D JgÀqÀÄ ¸ÀASÉåUÀ¼À ªÀÄ.¸Á.C ªÀ£ÀÄß C«¨sÁdå
C¥ÀªÀvÀð£À «zsÁ£À¢AzÀ PÀAqÀÄ»r¬Äj.
6
À ¸À«ÄÃPÀgt
À zÀ ±ÉÆÃzsÀPÀªÀ£ÀÄß PÀAqÀÄ»r¬Äj ªÀÄvÀÄÛ ªÀÄÆ®UÀ¼À ¸Àé¨sÁªÀªÀ£ÀÄß ¤zsÀðj¹.
CxÀªÁ
(2x + 3) x = x2 + 1 F ªÀUð
À ¸À«ÄÃPÀgt
À ªÀ£ÀÄß DzÀ±Àð gÀÆ¥ÀzÀ°è ªÀåP¥
ÀÛ Àr¹.
20. √5 MAzÀÄ C¨sÁUÀ®§Þ ¸ÀASÉå JAzÀÄ ¸Á¢ü¹.
21. (1, 6) ªÀÄvÀÄÛ (4, 3) ©AzÀÄUÀ¼À£ÀÄß ¸ÉÃj¸ÀĪÀ gÉÃSÁRAqÀªÀ£ÀÄß 1 2 C£ÀÄ¥ÁvÀzÀ°è DAvÀjPÀªÁV «¨sÁV¸ÀĪÀ
©AzÀÄ«£À ¤zÉÃð±ÁAPÀU¼
À À£ÀÄß PÀAqÀÄ»r¬Äj.
22. avÀæzÀ°è, ªÀÄvÀÄÛ ¨É¯ÉU¼
À À£ÀÄß PÀAqÀÄ»r¬Äj.
23. avÀæzÀ°è, ªÀÄvÀÄÛ UÀ¼ÀÄ PÉÃAzÀæªÀżÀî ªÀÈvÀPÛ ÉÌ ¸À±
à ÀðPÀU¼
À ÁVªÉ. AiÀÄÄ ©AzÀÄ«£À°è ªÀÄvÉÆÛAzÀÄ
¸À±
à ÀðPÀªÁVzÉ. wæ¨sÀÄdzÀ ¸ÀÄvÀ¼
Û ÀvÉAiÀÄÄ ¸À±
à ÀðPÀzÀ GzÀÝzÀ JgÀqÀg¶
À ÖzÉ JAzÀÄ vÉÆÃj¹.
24. JgÀqÀÄ ¸ÀASÉåUÀ¼À ªÀÄ.¸Á.C ªÀÄvÀÄÛ ®.¸Á.C UÀ¼À UÀÄt®§Þ 2016 DVzÉ. CªÀÅUÀ¼À°è MAzÀÄ ¸ÀASÉåAiÀÄÄ 42
DzÀgÉ E£ÉÆßAzÀÄ ¸ÀASÉåAiÀÄ£ÀÄß PÀAqÀÄ»r¬Äj ºÁUÀÆ D JgÀqÀÄ ¸ÀASÉåUÀ¼À ªÀÄ.¸Á.C ªÀ£ÀÄß C«¨sÁdå
C¥ÀªÀvÀð£À «zsÁ£À¢AzÀ PÀAqÀÄ»r¬Äj.
6
Page 19
IV. PɼÀV£À ¥Àæ±ÉßUÀ½UÉ GvÀÛj¹: 9 x 3 = 27
25. A( 3, 0), B(0, 2) ªÀÄvÀÄÛ C(0, 3) ©AzÀÄUÀ¼À ªÀÄÆ®PÀ ºÁzÀĺÉÆÃUÀĪÀ ªÀÈvÀz
Û À PÉÃAzÀæ©AzÀÄ«£À ¤zÉÃð±ÁAPÀU¼
À À£ÀÄß
PÀAqÀÄ»r¬Äj.
26. “¨ÁºÀå©AzÀÄ«¤AzÀ ªÀÈvÀPÛ ÉÌ J¼ÉzÀ ¸À±
à ÀðPÀU¼
À À GzÀݪÀÅ ¸ÀªÀĪÁVgÀÄvÀz
Û É” JAzÀÄ ¸Á¢ü¹.
27. MAzÀÄ JPïì¥Éæ¸ï gÉ樀 132 zÀÆgÀªÀ£ÀÄß PÀæ«Ä¸À®Ä ¥Áå¸ÉAdgï gÉÊ°VAvÀ 1 UÀAmÉ PÀrªÉÄ ¸ÀªÀÄAiÀĪÀ£ÀÄß
vÉUz î ÀÛzÉ. JPïì¥Éæ¸ï gÉÊ°£À ¸ÀgÁ¸Àj dªÀªÀÅ ¥Áå¸ÉAdgï gÉÊ°£À ¸ÀgÁ¸Àj dªÀQÌAvÀ 11
É ÀÄPÉƼÀÄv ºÉZÁÑVzÀÝgÉ,
D JgÀqÀÄ gÉÊ®ÄUÀ¼À ¸ÀgÁ¸Àj dªÀªÀ£ÀÄß PÀAqÀÄ»r¬Äj.
CxÀªÁ
‘ ’ ªÀÄvÀÄÛ ‘ ’ JA§ E§âgÀÄ «zÁåyðUÀ¼À ªÀAiÀĸÀÄìUÀ¼ÀÄ PÀæªÀĪÁV 19 ªÀμÀðUÀ¼ÀÄ ªÀÄvÀÄÛ 15 ªÀμÀðUÀ¼ÁVªÉ. CªÀgÀ
ªÀAiÀĸÀÄìUÀ¼À UÀÄt®§ÞªÀÅ 480 PÉÌ ¸ÀªÀĪÁUÀ®Ä JμÀÄÖ ªÀμÀðUÀ¼ÀÄ ¨ÉÃPÁUÀÄvÀª
Û É JA§ÄzÀ£ÀÄß PÀAqÀÄ»r¬Äj.
28. (1 )( ) √(1 )(1 ) cosec A JAzÀÄ ¸Á¢ü¹
CxÀªÁ
JAzÀÄ ¸Á¢ü¹.
29. ±ÀÆ£ÀåvÉU¼
À À ªÉÆvÀÛ ªÀÄvÀÄÛ UÀÄt®§ÞUÀ¼ÀÄ PÀæªÀĪÁV 2 ªÀÄvÀÄÛ 8 DVgÀĪÀ ªÀUð
À §ºÀÄ¥ÀzÉÆÃQÛAiÀÄ£ÀÄß §gɬÄj
ªÀÄvÀÄÛ D §ºÀÄ¥ÀzÉÆÃQÛAiÀÄ ±ÀÆ£ÀåvÉU¼
À À£ÀÄß PÀAqÀÄ»r¬Äj.
CxÀªÁ
7
25. A( 3, 0), B(0, 2) ªÀÄvÀÄÛ C(0, 3) ©AzÀÄUÀ¼À ªÀÄÆ®PÀ ºÁzÀĺÉÆÃUÀĪÀ ªÀÈvÀz
Û À PÉÃAzÀæ©AzÀÄ«£À ¤zÉÃð±ÁAPÀU¼
À À£ÀÄß
PÀAqÀÄ»r¬Äj.
26. “¨ÁºÀå©AzÀÄ«¤AzÀ ªÀÈvÀPÛ ÉÌ J¼ÉzÀ ¸À±
à ÀðPÀU¼
À À GzÀݪÀÅ ¸ÀªÀĪÁVgÀÄvÀz
Û É” JAzÀÄ ¸Á¢ü¹.
27. MAzÀÄ JPïì¥Éæ¸ï gÉ樀 132 zÀÆgÀªÀ£ÀÄß PÀæ«Ä¸À®Ä ¥Áå¸ÉAdgï gÉÊ°VAvÀ 1 UÀAmÉ PÀrªÉÄ ¸ÀªÀÄAiÀĪÀ£ÀÄß
vÉUz î ÀÛzÉ. JPïì¥Éæ¸ï gÉÊ°£À ¸ÀgÁ¸Àj dªÀªÀÅ ¥Áå¸ÉAdgï gÉÊ°£À ¸ÀgÁ¸Àj dªÀQÌAvÀ 11
É ÀÄPÉƼÀÄv ºÉZÁÑVzÀÝgÉ,
D JgÀqÀÄ gÉÊ®ÄUÀ¼À ¸ÀgÁ¸Àj dªÀªÀ£ÀÄß PÀAqÀÄ»r¬Äj.
CxÀªÁ
‘ ’ ªÀÄvÀÄÛ ‘ ’ JA§ E§âgÀÄ «zÁåyðUÀ¼À ªÀAiÀĸÀÄìUÀ¼ÀÄ PÀæªÀĪÁV 19 ªÀμÀðUÀ¼ÀÄ ªÀÄvÀÄÛ 15 ªÀμÀðUÀ¼ÁVªÉ. CªÀgÀ
ªÀAiÀĸÀÄìUÀ¼À UÀÄt®§ÞªÀÅ 480 PÉÌ ¸ÀªÀĪÁUÀ®Ä JμÀÄÖ ªÀμÀðUÀ¼ÀÄ ¨ÉÃPÁUÀÄvÀª
Û É JA§ÄzÀ£ÀÄß PÀAqÀÄ»r¬Äj.
28. (1 )( ) √(1 )(1 ) cosec A JAzÀÄ ¸Á¢ü¹
CxÀªÁ
JAzÀÄ ¸Á¢ü¹.
29. ±ÀÆ£ÀåvÉU¼
À À ªÉÆvÀÛ ªÀÄvÀÄÛ UÀÄt®§ÞUÀ¼ÀÄ PÀæªÀĪÁV 2 ªÀÄvÀÄÛ 8 DVgÀĪÀ ªÀUð
À §ºÀÄ¥ÀzÉÆÃQÛAiÀÄ£ÀÄß §gɬÄj
ªÀÄvÀÄÛ D §ºÀÄ¥ÀzÉÆÃQÛAiÀÄ ±ÀÆ£ÀåvÉU¼
À À£ÀÄß PÀAqÀÄ»r¬Äj.
CxÀªÁ
7
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( ) 3 1 F ªÀUð
À §ºÀÄ¥ÀzÉÆÃQÛAiÀÄ ±ÀÆ£ÀåvÉU¼
À ÀÄ ªÀÄvÀÄÛ DzÀg,É
()
2 2
( ) EªÀÅUÀ¼À ¨É¯É PÀAqÀÄ»r¬Äj.
30. ABC MAzÀÄ ®A§PÉÆãÀ wæ¨sÀÄd. 90 ªÀÄvÀÄÛ AD BC DzÀg,É AB2 = BC BD JAzÀÄ
vÉÆÃj¹.
31. UÉÆÃqÉ UÀrAiÀiÁgÀzÀ ¤«ÄµÀzÀ ªÀÄĽî£À GzÀݪÀÅ 18cm EzÉ. 35 ¤«ÄµÀU¼
À À°è ¤«ÄµÀzÀ ªÀÄļÀÄî PÀæ«Ä¸ÀĪÀ «¹ÛÃtðªÀ£ÀÄß
PÀAqÀÄ»r¬Äj. ¤«ÄµÀzÀ ªÀÄĽî£À ZÀ®£É¬ÄAzÀ GAmÁUÀĪÀ PÀA¸ÀzÀ GzÀݪÀ£ÀÄß PÀAqÀÄ»r¬Äj.
32. MAzÀÄ ¥ÉnÖUÉAiÀÄ°è 6 jAzÀ 70 gÀªÀgU
É É £ÀªÀÄÆzÁVgÀĪÀ PÁqÀÄðUÀ½ªÉ. D ¥ÉnÖUɬÄAzÀ MAzÀÄ PÁqÀð
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10 – 20 9
20 – 30 15
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40 – 50 8
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vÉÆÃj¹.
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PÉÆnÖgÀĪÀ DªÀÈwÛ «vÀgu
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15 20 6
20 25 9
25 30 15
30 35 9
35 40 1
V. PɼÀV£À ¥Àæ±ÉßUÀ½UÉ GvÀÛj¹: 4 x 4 = 16
34. PÉÆnÖgÀĪÀ gÉÃSÁvÀäPÀ ¸À«ÄÃPÀgt
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2 6
5
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15 20 6
20 25 9
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30 35 9
35 40 1
V. PɼÀV£À ¥Àæ±ÉßUÀ½UÉ GvÀÛj¹: 4 x 4 = 16
34. PÉÆnÖgÀĪÀ gÉÃSÁvÀäPÀ ¸À«ÄÃPÀgt
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5
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9
Page 22
37. ¸ÀªÀÄÄzÀæ wÃgÀzÀ ªÉÄÃ¯É £ÉÃgÀªÁV ¤AvÀ MAzÀÄ ¢Ã¥À¸A Û ÀªÀÅ 10√3 m DVzÉ. ¢Ã¥À¸ÀÛA¨sÀzÀ
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10
ÀÛ ¨sÀzÀ [AB] JvÀg
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10
Page 23
VI. PɼÀV£À ¥Àæ±ÉßUÀ½UÉ GvÀÛj¹: 1x5=5
38. avÀæzÀ°è vÉÆÃj¹gÀĪÀAvÉ, vÀ¯ÉPɼÀUÁzÀ MAzÀÄ ±ÀAPÀÄ«£À ªÉÄÃ¯É MAzÀÄ ¹°AqÀg£
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WÀ£ÁPÀÈwAiÀÄ£ÀÄß vÀAiÀiÁj¹zÉ. ¹°AqÀgï ªÀÄvÀÄÛ ±ÀAPÀÄ«£À ¥ÁzÀzÀ wædåUÀ¼ÀÄ 7 DVzÀÄÝ, EªÀÅUÀ¼À JvÀg
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12 ªÀÄvÀÄÛ 9 DVªÉ. F DPÀÈwAiÀÄ°è ¹°AqÀj£À JvÀg
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MAzÀÄ ¥ÁvÉæAiÀÄÄ vÀ¯ÉPɼÀUÁzÀ ±ÀAPÀÄ«£ÁPÁgÀzÀ°èzÉ. CzÀgÀ JvÀg
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12
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12
Page 25
Karnataka State Board
SSLC Examination
Model Answers 2026
SSLC Examination
Model Answers 2026
Page 26
KARNATAKA SCHOOL EXAMINATION AND ASSESSMENT BOARD
Malleshwaram, Bengaluru – 560003
S.S.L.C. MODEL QUESTION PAPER – 01 – 2025-26
MODEL ANSWERS
Subject: MATHEMATICS
(English Medium)
Subject Code: 81 – E
[ Time: 3 Hours 15 Minutes]
[ Max. Marks: 80]
Q.No Ans Value Points Marks
Key allotted
1 1. The H.C.F of 3 and 5 is,
(A) 1 (B) 3
(C) 5 (D) 15
1
(A) Ans: 1
Malleshwaram, Bengaluru – 560003
S.S.L.C. MODEL QUESTION PAPER – 01 – 2025-26
MODEL ANSWERS
Subject: MATHEMATICS
(English Medium)
Subject Code: 81 – E
[ Time: 3 Hours 15 Minutes]
[ Max. Marks: 80]
Q.No Ans Value Points Marks
Key allotted
1 1. The H.C.F of 3 and 5 is,
(A) 1 (B) 3
(C) 5 (D) 15
1
(A) Ans: 1
Page 27
2 In the figure, the number of zeroes of the polynomial is,
(A) 3 (B) 5
(C) 4 (D) 1
1
(C) Ans: 4
3 If the pair of linear equations a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0
have unique solution, then the correct relation among the following is,
a1 b1 a1 b1 c1
(A) (B)
a2 b2 a2 b2 c2
a1 b1 a1 b1 c1
(C) (D)
a2 b2 a2 b2 c2
a1 b1
(A) Ans:
a2 b2 1
4 The arithmetic progression in the following is,
(A 1, 2, 4, 8……. B 3, 7, 10, 14…….
C 1, 4, 9, 16……… D 5, 9, 13, 17……..
(D) Ans: 5, 9, 13, 17…….. 1
(A) 3 (B) 5
(C) 4 (D) 1
1
(C) Ans: 4
3 If the pair of linear equations a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0
have unique solution, then the correct relation among the following is,
a1 b1 a1 b1 c1
(A) (B)
a2 b2 a2 b2 c2
a1 b1 a1 b1 c1
(C) (D)
a2 b2 a2 b2 c2
a1 b1
(A) Ans:
a2 b2 1
4 The arithmetic progression in the following is,
(A 1, 2, 4, 8……. B 3, 7, 10, 14…….
C 1, 4, 9, 16……… D 5, 9, 13, 17……..
(D) Ans: 5, 9, 13, 17…….. 1
Page 28
5 In the given frequency distribution table the median class is,
Marks Number of Students Cumulative Frequency
0– 10 3 3
10 – 20 4 7
20 – 30 7 14
30 – 40 6 20
n = 20
(A) 0 – 10 (B) 10 – 20
(C) 20 – 30 (D) 30 – 40
(C) Ans: 20 – 30 1
6 If 3 , then the measure of the angle A is,
(A) 90 (B) 60
(C) 45 (D) 30
(D) Ans: 30
1
7 A perpendicular drawn from a point 4, 5 intersects X- axis at the
point Q. The coordinates of point Q are,
(A) (0, - 4) (B) ( –4, 0)
(C) (–5, 0) (D) (0, –5)
(B) Ans: ( –4, 0) 1
Marks Number of Students Cumulative Frequency
0– 10 3 3
10 – 20 4 7
20 – 30 7 14
30 – 40 6 20
n = 20
(A) 0 – 10 (B) 10 – 20
(C) 20 – 30 (D) 30 – 40
(C) Ans: 20 – 30 1
6 If 3 , then the measure of the angle A is,
(A) 90 (B) 60
(C) 45 (D) 30
(D) Ans: 30
1
7 A perpendicular drawn from a point 4, 5 intersects X- axis at the
point Q. The coordinates of point Q are,
(A) (0, - 4) (B) ( –4, 0)
(C) (–5, 0) (D) (0, –5)
(B) Ans: ( –4, 0) 1
Page 29
8 The quadratic equation whose roots are 6 and -1 is
(A) x2 + 5x – 6 = 0 (B) x2 – 5x – 6 = 0
(C) x2 – 5x + 6 = 0 (D) x2 + 5x + 6 = 0
(B) Ans: x2 – 5x – 6 = 0 1
9 II. Answer the following questions: 8
How many solutions does the pair of linear equations in two variables have if
they are inconsistent?
Ans: No solution 1
10 In the figure, if DE||BC, AD =2cm and BD = 6cm, then find DE:BC.
Ans:
AD:AB = 2:8 = 1:4
Here, AD:AB = DE: BC = 1:4
1
11 The circumference of the base of a cylinder is 10cm and its height is 25cm. Find
its curved surface area.
Ans:
2 = 10 cm and h = 25 cm
1
Curved surface area of Cylinder = 2 h = 10 25 = 250 cm2
12 Write the value of ̅̅̅̅̅ with reference to probability.
Ans: ̅̅̅̅̅ 1 1
(A) x2 + 5x – 6 = 0 (B) x2 – 5x – 6 = 0
(C) x2 – 5x + 6 = 0 (D) x2 + 5x + 6 = 0
(B) Ans: x2 – 5x – 6 = 0 1
9 II. Answer the following questions: 8
How many solutions does the pair of linear equations in two variables have if
they are inconsistent?
Ans: No solution 1
10 In the figure, if DE||BC, AD =2cm and BD = 6cm, then find DE:BC.
Ans:
AD:AB = 2:8 = 1:4
Here, AD:AB = DE: BC = 1:4
1
11 The circumference of the base of a cylinder is 10cm and its height is 25cm. Find
its curved surface area.
Ans:
2 = 10 cm and h = 25 cm
1
Curved surface area of Cylinder = 2 h = 10 25 = 250 cm2
12 Write the value of ̅̅̅̅̅ with reference to probability.
Ans: ̅̅̅̅̅ 1 1
Page 30
13 In an arithmetic progression, the seventh term is 12 more than its fourth term.
Find the common difference.
Ans: a7 = a4 + 12
a + 6d = a +3d +12
6d – 3d = 12
3d = 12
d=4 1
14 In the figure a cone with radius ‘r’ and slant height ‘l’ is mounted on a hemisphere
with radius ‘r’. Write the formula to find the total surface area of the solid formed.
Ans: TSA of solid = CSA of cone + CSA of Hemisphere
= l+2 2 OR l+2
1
15 . BC and QR are corresponding sides. If 80 and 40 ,
then what is the measure of
Ans: 180 80 40 60 1
16 If the zeroes of the polynomial p(x)= x2 + 3x + k are reciprocal to each other,
then find the value of k.
Ans:
If Zeroes of polynomial are reciprocal to each other then, their product is equal
Find the common difference.
Ans: a7 = a4 + 12
a + 6d = a +3d +12
6d – 3d = 12
3d = 12
d=4 1
14 In the figure a cone with radius ‘r’ and slant height ‘l’ is mounted on a hemisphere
with radius ‘r’. Write the formula to find the total surface area of the solid formed.
Ans: TSA of solid = CSA of cone + CSA of Hemisphere
= l+2 2 OR l+2
1
15 . BC and QR are corresponding sides. If 80 and 40 ,
then what is the measure of
Ans: 180 80 40 60 1
16 If the zeroes of the polynomial p(x)= x2 + 3x + k are reciprocal to each other,
then find the value of k.
Ans:
If Zeroes of polynomial are reciprocal to each other then, their product is equal
Page 31
to 1
1
Therefore k = 1
17 III. Answer the following questions: 8 x 2 = 16
In an arithmetic progression, if S20= 820 and a20= 79, then find its first term.
OR
How many numbers between 201 and 401 are divisible by 6?
Ans: S20= 820, a20= 79, n = 20 1/2
Sn = 1/2
820 = 79
1/2
820 = 10(a + 79)
82 = a + 79 1/2 2
a = 82 – 79
a=3
OR
Numbers divisible by 6, between 201 and 401
204, 210, 216 . . . . . . . . . . . . . . . . . 396
a = 204, d= 210 – 204 = 6, an = 396, n = ? 1/2
an = a + (n – 1)d
1/2
396 = 204 +(n – 1) 6
396 – 204 = (n – 1) 6
192 = (n – 1) 6
1/2
1
32 1
1/2
n = 32 + 1 = 33 1/2
2
18 Solve the given pair of linear equations using suitable method.
2x + y = 8
3x – y = 7
Ans:
1
Therefore k = 1
17 III. Answer the following questions: 8 x 2 = 16
In an arithmetic progression, if S20= 820 and a20= 79, then find its first term.
OR
How many numbers between 201 and 401 are divisible by 6?
Ans: S20= 820, a20= 79, n = 20 1/2
Sn = 1/2
820 = 79
1/2
820 = 10(a + 79)
82 = a + 79 1/2 2
a = 82 – 79
a=3
OR
Numbers divisible by 6, between 201 and 401
204, 210, 216 . . . . . . . . . . . . . . . . . 396
a = 204, d= 210 – 204 = 6, an = 396, n = ? 1/2
an = a + (n – 1)d
1/2
396 = 204 +(n – 1) 6
396 – 204 = (n – 1) 6
192 = (n – 1) 6
1/2
1
32 1
1/2
n = 32 + 1 = 33 1/2
2
18 Solve the given pair of linear equations using suitable method.
2x + y = 8
3x – y = 7
Ans:
Page 32
2x y 8 ………. i)
3x – y 7 ………. ii
------------------------
5x=15
x=3 1
Substitute value of x in equation (i)
2x+y=8
2(3)+y = 8
6+y=8
y=8-6 1
2
y=2
19 Find the discriminant of the quadratic equation 2x2 + 3x – 7 = 0 and determine
the nature of roots.
OR
Express the quadratic equation (2x + 3) x = x2 + 1 in the standard form.
Ans: 2 3 7 0
a= 2 , b = +3 , c = - 7
1/2
Discriminant = b – 4ac
2
1/2
= (3)2 – 4×2×-7
= 9 + 56 = 65 > 0 1/2
Roots are real and distinct 1/2 2
OR
(2x + 3) x = x2 + 1 1
2x2 +3x = x2 + 1
2x2 +3x – x2 – 1 = 0 1
1
x2 +3x – 1 = 0 2
20 Prove that √5 is an irrational number.
Ans: √5 is an irrational num er
1/2
√5 = (a and b are co-primes, ≠ 0
3x – y 7 ………. ii
------------------------
5x=15
x=3 1
Substitute value of x in equation (i)
2x+y=8
2(3)+y = 8
6+y=8
y=8-6 1
2
y=2
19 Find the discriminant of the quadratic equation 2x2 + 3x – 7 = 0 and determine
the nature of roots.
OR
Express the quadratic equation (2x + 3) x = x2 + 1 in the standard form.
Ans: 2 3 7 0
a= 2 , b = +3 , c = - 7
1/2
Discriminant = b – 4ac
2
1/2
= (3)2 – 4×2×-7
= 9 + 56 = 65 > 0 1/2
Roots are real and distinct 1/2 2
OR
(2x + 3) x = x2 + 1 1
2x2 +3x = x2 + 1
2x2 +3x – x2 – 1 = 0 1
1
x2 +3x – 1 = 0 2
20 Prove that √5 is an irrational number.
Ans: √5 is an irrational num er
1/2
√5 = (a and b are co-primes, ≠ 0
Page 33
5= ( squaring on both sides)
5 = a ………… i 1/2
5 divides a a
Let a =5k , k is an integer
1/2
substitute a = 5k In equation (i)
5 = (5
=5
5 divides b 1/2
a and b have at least 5 as a common factor
but this contradicts the fact that a and b are co-prime
Hence our assumption is wrong.
2
√5 is a irrational number
21 Find the coordinates of the point which divides the line segment joining the
points
(1, 6) and (4, 3) in the ratio 1 2 internally.
Ans: (1, 6) = (x1, y1), (4, 3) = (x2, y2), m1 : m2 = 1 : 2, P(x, y) = ?
P(x, y) = ( , ) 1/2
1/2
P(x, y) = ( , )
1/2
=( , )
2
=( , ) = ( 2, 5) 1/2
P(x, y) = ( 2, 5)
5 = a ………… i 1/2
5 divides a a
Let a =5k , k is an integer
1/2
substitute a = 5k In equation (i)
5 = (5
=5
5 divides b 1/2
a and b have at least 5 as a common factor
but this contradicts the fact that a and b are co-prime
Hence our assumption is wrong.
2
√5 is a irrational number
21 Find the coordinates of the point which divides the line segment joining the
points
(1, 6) and (4, 3) in the ratio 1 2 internally.
Ans: (1, 6) = (x1, y1), (4, 3) = (x2, y2), m1 : m2 = 1 : 2, P(x, y) = ?
P(x, y) = ( , ) 1/2
1/2
P(x, y) = ( , )
1/2
=( , )
2
=( , ) = ( 2, 5) 1/2
P(x, y) = ( 2, 5)
Page 34
22 In the figure, find the value of and .
1
2
Ans :
1
23 In the figure, CP and CQ are tangents to the circle with centre O. ARB is another
tangent at R. Show that the perimeter of triangle CAB is twice the length of tangent
CP.
Ans: Perimeter of triangle CAB = CA + AB + BC
= CP – AP + AR + BR + CQ – BQ 1
From Theorem 10.2, CP = CQ, AR = AP and BQ = BR
2
= CP – AP + AP + BQ + CP – BQ 1
= 2CP
1
2
Ans :
1
23 In the figure, CP and CQ are tangents to the circle with centre O. ARB is another
tangent at R. Show that the perimeter of triangle CAB is twice the length of tangent
CP.
Ans: Perimeter of triangle CAB = CA + AB + BC
= CP – AP + AR + BR + CQ – BQ 1
From Theorem 10.2, CP = CQ, AR = AP and BQ = BR
2
= CP – AP + AP + BQ + CP – BQ 1
= 2CP
Page 35
24 The product of HCF and LCM of 2 numbers is 2016. If one of the numbers is 42,
then find the other number. Also find the HCF of those two numbers by prime
factorization method.
Ans: Product of 2 numbers = product of their HCF and LCM
42 = 2016 1
B= = 48
42 = 2 3 7
48 = 24 3 1 2
HCF (42, 48) = 2 3=6
then find the other number. Also find the HCF of those two numbers by prime
factorization method.
Ans: Product of 2 numbers = product of their HCF and LCM
42 = 2016 1
B= = 48
42 = 2 3 7
48 = 24 3 1 2
HCF (42, 48) = 2 3=6
Page 36
25 Find the coordinates of the centre of a circle passing through the points A(-3, 0),
B(0, 2) and C(0, -3).
OA = OB = OC (radii)
Consider OA = OB
(x+3)2 +y2 = x2 + (y – 2)2 (Using Distance formula) 1
x2 + 9 + 6x + y2= x2 + y2 + 4 – 4y
6x + 9 = 4 – 4y
6x + 4y = 4 – 9
6x + 4y = – 5 ………………. i
Consider OB = OC
x2 +(y – 2)2 = x2 + (y + 3)2
x2 + y2 + 4 – 4y = x2 + y2 + 9 + 6y 1
4 – 9 = 6y + 4y
10y = –5
y=
Substitute value of y in equation (i)
6x + 4( = –5
6x – 2 = –5
6x = –5 + 2
6x = –3
3
x= ; hence Coordinates of the centre is ( , 1
B(0, 2) and C(0, -3).
OA = OB = OC (radii)
Consider OA = OB
(x+3)2 +y2 = x2 + (y – 2)2 (Using Distance formula) 1
x2 + 9 + 6x + y2= x2 + y2 + 4 – 4y
6x + 9 = 4 – 4y
6x + 4y = 4 – 9
6x + 4y = – 5 ………………. i
Consider OB = OC
x2 +(y – 2)2 = x2 + (y + 3)2
x2 + y2 + 4 – 4y = x2 + y2 + 9 + 6y 1
4 – 9 = 6y + 4y
10y = –5
y=
Substitute value of y in equation (i)
6x + 4( = –5
6x – 2 = –5
6x = –5 + 2
6x = –3
3
x= ; hence Coordinates of the centre is ( , 1
Page 37
26 Prove that “the length of tangents drawn to a circle from an external point are
equal”.
Given : PQ and PR are the two tangents drawn from the external point P to
a circle of center O.
To Prove : PQ = PR
Construction : Join OP, OQ, OR
Proof : In △ OQP and △ORP
OQ = OR (∵ Radii of same circle)
OP = OP (∵Common side )
OQP = ORP =900 ( Theorem 10.1)
OQP ≅ △ORP ( RHS congruence rule)
PQ = PR (CPCT)
Data , Figure: 1 mark
To prove, construction : ½ marks
Proof : 1 ½ marks
3
equal”.
Given : PQ and PR are the two tangents drawn from the external point P to
a circle of center O.
To Prove : PQ = PR
Construction : Join OP, OQ, OR
Proof : In △ OQP and △ORP
OQ = OR (∵ Radii of same circle)
OP = OP (∵Common side )
OQP = ORP =900 ( Theorem 10.1)
OQP ≅ △ORP ( RHS congruence rule)
PQ = PR (CPCT)
Data , Figure: 1 mark
To prove, construction : ½ marks
Proof : 1 ½ marks
3
Page 38
27 An express train takes 1 hour less than a passenger train to travel a distance of
132 km. The Average speed of the express train is 11km/h more than that of the
passenger train. Find the average speed of these two trains.
OR
The ages of two students ‘A’ and ‘B’ are 19 years and 15 years respectively. Find
how many years it will take so that the product of their ages become equal to 480.
Ans:
Let speed of Passenger train be x km/h
Speed of express train = (x+11) km/h
Time =
Time taken by the passenger train to travel 132 km = t1 = hrs
Time taken by the express train to travel 132 km = t2 = hrs 1
- =1
=1
132 1452 132 11
x2 +11x -1452 = 0
-33x+44x-1452 = 0
x(x-33) +44(x-33) = 0
1
(x-33)(x+44) = 0
x-33= 0 , x+44 = 0
x = 33, x = -44
1/2
Speed of Passenger train x = 33 km/h
Speed of Express Train = (x+11) = 33+11 = 44 km/h 1/2 3
OR
Let the required number of years be x
(19+x)(15+x) = 480
1
34x 195 = 0
132 km. The Average speed of the express train is 11km/h more than that of the
passenger train. Find the average speed of these two trains.
OR
The ages of two students ‘A’ and ‘B’ are 19 years and 15 years respectively. Find
how many years it will take so that the product of their ages become equal to 480.
Ans:
Let speed of Passenger train be x km/h
Speed of express train = (x+11) km/h
Time =
Time taken by the passenger train to travel 132 km = t1 = hrs
Time taken by the express train to travel 132 km = t2 = hrs 1
- =1
=1
132 1452 132 11
x2 +11x -1452 = 0
-33x+44x-1452 = 0
x(x-33) +44(x-33) = 0
1
(x-33)(x+44) = 0
x-33= 0 , x+44 = 0
x = 33, x = -44
1/2
Speed of Passenger train x = 33 km/h
Speed of Express Train = (x+11) = 33+11 = 44 km/h 1/2 3
OR
Let the required number of years be x
(19+x)(15+x) = 480
1
34x 195 = 0
Page 39
+39x 5x 195= 0
1
x(x+39) 5(x+39) = 0
(x+39)(x 5) = 0
x+39= 0 , x 5 = 0 1/2
x = 39 , x = 5
Product of their ages becomes 480 after 5 years 1/2
3
28 Prove that:
1 √ cosec A
OR
Prove that
LHS = secA (1-sinA)(secA +tanA)
= (secA – secA.sinA) (secA + tanA)
1
= (secA – tanA) (secA + tanA)
=sec2A – tan2A 1…………….. i
1/2
RHS = √ 1 1
=√1
1
=√
sinA cosecA 1……………. ii 1/2
3
From (i) and (ii)
LHS = RHS
OR
1
x(x+39) 5(x+39) = 0
(x+39)(x 5) = 0
x+39= 0 , x 5 = 0 1/2
x = 39 , x = 5
Product of their ages becomes 480 after 5 years 1/2
3
28 Prove that:
1 √ cosec A
OR
Prove that
LHS = secA (1-sinA)(secA +tanA)
= (secA – secA.sinA) (secA + tanA)
1
= (secA – tanA) (secA + tanA)
=sec2A – tan2A 1…………….. i
1/2
RHS = √ 1 1
=√1
1
=√
sinA cosecA 1……………. ii 1/2
3
From (i) and (ii)
LHS = RHS
OR
Page 40
LHS
= 1
=
1
3
1
= = . = RHS
29 Write the quadratic polynomial whose sum and product of the zeroes are 2 and
8 respectively and hence find the zeroes of the polynomial.
OR
If and are the zeroes of the quadratic polynomial p(x) = x² + 3x +1 then,
find the value of (i)
(ii)
Ans: Sum of zeroes = α β 2
Product of Zeroes αxβ 8
Required polynomial p(x) = x2 – (α β x αβ
= x2 – (2)x + (-8)
1
= x2 – 2x – 8
To find zeroes of the polynomial
p(x) = x2 – 2x – 8
= x2 – 4x + 2x – 8
1
= x (x – 4) + 2( x – 4)
= 1
=
1
3
1
= = . = RHS
29 Write the quadratic polynomial whose sum and product of the zeroes are 2 and
8 respectively and hence find the zeroes of the polynomial.
OR
If and are the zeroes of the quadratic polynomial p(x) = x² + 3x +1 then,
find the value of (i)
(ii)
Ans: Sum of zeroes = α β 2
Product of Zeroes αxβ 8
Required polynomial p(x) = x2 – (α β x αβ
= x2 – (2)x + (-8)
1
= x2 – 2x – 8
To find zeroes of the polynomial
p(x) = x2 – 2x – 8
= x2 – 4x + 2x – 8
1
= x (x – 4) + 2( x – 4)
Page 41
= (x – 4) (x + 2)
x – 4 =0 and x + 2=0
x = 4 and x = – 2
1 3
Zeroes are 4 and –2
OR
p(x) = x2 + 3x + 1
Sum of zeroes = α β = -3
1
Product of zeroes αxβ =1
+ = = = -3 1
+ =αβ α β 1 - 3 )= - 3 1
3
30 ABC is a right angled triangle. If 90 and AD BC, then show that
AB2 = BC BD
Ans:
1
To prove: BD
Proof: In BDA and BAC
is common
90 (Data) 1
BDA BAC (AA criteria)
1 3
BD
x – 4 =0 and x + 2=0
x = 4 and x = – 2
1 3
Zeroes are 4 and –2
OR
p(x) = x2 + 3x + 1
Sum of zeroes = α β = -3
1
Product of zeroes αxβ =1
+ = = = -3 1
+ =αβ α β 1 - 3 )= - 3 1
3
30 ABC is a right angled triangle. If 90 and AD BC, then show that
AB2 = BC BD
Ans:
1
To prove: BD
Proof: In BDA and BAC
is common
90 (Data) 1
BDA BAC (AA criteria)
1 3
BD
Page 42
31 The minute hand of a wall clock is 18 cm long. Find the area swept by the minute
hand in 35 minutes. Find the length of the arc formed by the sweep of the minute
hand.
Ans:
Sector angle formed = 35minutes = 210 1
Area swept =
1
= 18 = 594cm2
Length of the arc = 2 3
1
= 2 18= 66cm
32 A box contains cards numbered from 6 to 70. If one card is drawn at random
from the box, then find the probability that the card will have
(i) a perfect square number.
(ii) a number divisible by 5.
(iii) an odd number less than 30.
Ans:
S = {6, 7, 8, 9, 10 ……………………………..69, 70} n s 65
(i) A = {9, 16, 25, 36, 49, 64} n(A) = 6 1
P(A) =
(ii) B = { 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60, 65, 70} n(B) = 13
1
P(B) = or
(iii) C = {7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29} n(C) = 12
3
P(C) = 1
hand in 35 minutes. Find the length of the arc formed by the sweep of the minute
hand.
Ans:
Sector angle formed = 35minutes = 210 1
Area swept =
1
= 18 = 594cm2
Length of the arc = 2 3
1
= 2 18= 66cm
32 A box contains cards numbered from 6 to 70. If one card is drawn at random
from the box, then find the probability that the card will have
(i) a perfect square number.
(ii) a number divisible by 5.
(iii) an odd number less than 30.
Ans:
S = {6, 7, 8, 9, 10 ……………………………..69, 70} n s 65
(i) A = {9, 16, 25, 36, 49, 64} n(A) = 6 1
P(A) =
(ii) B = { 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60, 65, 70} n(B) = 13
1
P(B) = or
(iii) C = {7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29} n(C) = 12
3
P(C) = 1
Page 43
33. . Find the Mean for the following frequency distribution table.
Class interval Frequency
0 – 10 4
10 – 20 9
20 – 30 15
30 – 40 14
40 – 50 8
OR
Find the Mode for the following frequency distribution table.
Class interval Frequency
15-20 6
20-25 9
25-30 15
30-35 9
35-40 1
Ans:
IC f
0 - 10 4 5 20
10 - 20 9 15 135
20 - 30 15 25 375
30 - 40 14 35 490
40 - 50 8 45 360
Σ,fi =50 1380
∑ 1380
27.6
50
Completing table : 2 marks
3
Finding mean : 1 mark
OR
Class interval Frequency
0 – 10 4
10 – 20 9
20 – 30 15
30 – 40 14
40 – 50 8
OR
Find the Mode for the following frequency distribution table.
Class interval Frequency
15-20 6
20-25 9
25-30 15
30-35 9
35-40 1
Ans:
IC f
0 - 10 4 5 20
10 - 20 9 15 135
20 - 30 15 25 375
30 - 40 14 35 490
40 - 50 8 45 360
Σ,fi =50 1380
∑ 1380
27.6
50
Completing table : 2 marks
3
Finding mean : 1 mark
OR
Page 44
Here class interval of the mode is 25 – 30
1
25, 9, 15, 9, h = 10
[ ]
2
25 * + 5
1
6
25 [ ] 5
30 18
30
25
12
25 2.5
1 3
Mode = 27.5
34 V. Answer the following questions: 4 x 4 = 16
Solve the given pair of linear equations by graphical method.
x + 2y = 6
x+y=5
Ans:
x+2y = 6
x 0 6
y 3 0
(x,y) (0,3) (6,0)
x+y=5
x 0 5
y 5 0
(x,y) (0,5) (5,0)
1
25, 9, 15, 9, h = 10
[ ]
2
25 * + 5
1
6
25 [ ] 5
30 18
30
25
12
25 2.5
1 3
Mode = 27.5
34 V. Answer the following questions: 4 x 4 = 16
Solve the given pair of linear equations by graphical method.
x + 2y = 6
x+y=5
Ans:
x+2y = 6
x 0 6
y 3 0
(x,y) (0,3) (6,0)
x+y=5
x 0 5
y 5 0
(x,y) (0,5) (5,0)
Page 45
Scale : x axis and Y axis: 1 cm = 1unit.
Truth table each : 1mark
Dawing 2 lines : 1 mark 4
Finding values of x and y : 1 mark
Truth table each : 1mark
Dawing 2 lines : 1 mark 4
Finding values of x and y : 1 mark
Page 46
35. Prove that “If in two triangles, sides of one triangle are proportional to i.e., in the
same ratio of) the sides of the other triangle, then their corresponding angles are
equal and hence the two triangles are similar.”
OR
Prove that “If one angle of a triangle is equal to one angle of the other triangle
and the sides including these angles are proportional, then the two triangles are
similar”.
Ans :
Data : In ABC and DEF
= = ----- (1)
To prove : A = D, B = E and C= F and ΔABC ≅ ΔDEF
Construction : Mark Points P and Q on DE and DF so that AB = DP and AC =
DQ then join PQ
Proof : = ⟹ = ( DP = AB, DQ = AC )
⟹ = ( Inverse ratios)
⟹ 1, 1,
⟹ =
⟹ =
⟹ = ( inverse ratios)
PQ ||EF ( Thales theorem )
P = E and Q =F
same ratio of) the sides of the other triangle, then their corresponding angles are
equal and hence the two triangles are similar.”
OR
Prove that “If one angle of a triangle is equal to one angle of the other triangle
and the sides including these angles are proportional, then the two triangles are
similar”.
Ans :
Data : In ABC and DEF
= = ----- (1)
To prove : A = D, B = E and C= F and ΔABC ≅ ΔDEF
Construction : Mark Points P and Q on DE and DF so that AB = DP and AC =
DQ then join PQ
Proof : = ⟹ = ( DP = AB, DQ = AC )
⟹ = ( Inverse ratios)
⟹ 1, 1,
⟹ =
⟹ =
⟹ = ( inverse ratios)
PQ ||EF ( Thales theorem )
P = E and Q =F
Page 47
DPQ = DEF ( AA similarity criteria)
There fore = ------ (2)
If AB = DP, AC = DQ
= ------- (3)
From (2)and (3)
BC = PQ
In ABC and DPQ
BC = PQ
AB = DP
AC = DQ
ABC ≅ DPQ
A = D, B = P, & C = Q,
⟹ A = D, B = E, & C = F
4
& ABC DE F
OR
Data :In ABC and DEF
A = D, and =
To prove : ABC DEF
Construction: Mark Points P and Q on DE and DF so that AB = DP and AC =
DQ then join PQ
There fore = ------ (2)
If AB = DP, AC = DQ
= ------- (3)
From (2)and (3)
BC = PQ
In ABC and DPQ
BC = PQ
AB = DP
AC = DQ
ABC ≅ DPQ
A = D, B = P, & C = Q,
⟹ A = D, B = E, & C = F
4
& ABC DE F
OR
Data :In ABC and DEF
A = D, and =
To prove : ABC DEF
Construction: Mark Points P and Q on DE and DF so that AB = DP and AC =
DQ then join PQ
Page 48
Proof :
In ABC and DPQ
AB = DP (construction)
AC = DQ (construction)
A = D, (data)
ABC ≅ DPQ ( SAS)
B = P,
C = Q, (corresponding angles of similar triangles) -----(1)
= (data)
= ( AB = DP & AC = DQ)
PQ || EF (converse of Thales theorem)
E = P,
F = Q, --------(2)
From (1)and (2)
B = E , C= F , and A = D ,
4
ABC DEF
Data : ½ mark
Figure : ½ mark
To Prove : ½ mark
Construction: ½ mark
Proof : 2 marks
In ABC and DPQ
AB = DP (construction)
AC = DQ (construction)
A = D, (data)
ABC ≅ DPQ ( SAS)
B = P,
C = Q, (corresponding angles of similar triangles) -----(1)
= (data)
= ( AB = DP & AC = DQ)
PQ || EF (converse of Thales theorem)
E = P,
F = Q, --------(2)
From (1)and (2)
B = E , C= F , and A = D ,
4
ABC DEF
Data : ½ mark
Figure : ½ mark
To Prove : ½ mark
Construction: ½ mark
Proof : 2 marks
Page 49
36 The sum of first ‘n’ terms of an arithmetic progression is S n = 3n2 + 2n. Find the
nth term of the progression. If this progression consists of 59 terms then find the
sum of last 10 terms of the progression.
Ans:
Sn = 3n2 + 2n
Let n = 1 , S1 = 3(1)2 + 2(1) = 3 + 2 = 5 1
n = 2, S2 = 3(2)2 + 2(2) = 12 + 4 = 16
n = 3, S3 = 3(3)2 + 2(3) = 27 + 6 = 33
1/2
a1 = S1 = 5
a2 = S2 – S1 = 16 – 5 = 11
a3 = S3 – S2 = 33 – 16 = 17
Required Arithmetic progression is 5, 11, 17,………… 1
To find nth term:
an = a + (n – 1) d
= 5 + ( n – 1) 6
= 5 + 6n – 6
an= 6n – 1
To find the sum of last 10 terms of the AP, if it contains 59 terms
1/2
Sum of last ten terms = sum of first 59 terms – sum of first 49 terms
= S59 – S49
= {2 5 59 1 6} {2 5 49 1 6}
= {10 58 6} {10 48 6}
nth term of the progression. If this progression consists of 59 terms then find the
sum of last 10 terms of the progression.
Ans:
Sn = 3n2 + 2n
Let n = 1 , S1 = 3(1)2 + 2(1) = 3 + 2 = 5 1
n = 2, S2 = 3(2)2 + 2(2) = 12 + 4 = 16
n = 3, S3 = 3(3)2 + 2(3) = 27 + 6 = 33
1/2
a1 = S1 = 5
a2 = S2 – S1 = 16 – 5 = 11
a3 = S3 – S2 = 33 – 16 = 17
Required Arithmetic progression is 5, 11, 17,………… 1
To find nth term:
an = a + (n – 1) d
= 5 + ( n – 1) 6
= 5 + 6n – 6
an= 6n – 1
To find the sum of last 10 terms of the AP, if it contains 59 terms
1/2
Sum of last ten terms = sum of first 59 terms – sum of first 49 terms
= S59 – S49
= {2 5 59 1 6} {2 5 49 1 6}
= {10 58 6} {10 48 6}
Page 50
{10 348} {10 288}
= {358} {298}
= 59 179 49 149
= 10561 – 7301
1
4
= 3260
37 The lighthouse [AB] of height 10√3 m stands vertically on a sea shore. A
tower [CE] and a ship [F] are standing 30m and 10m away from the foot of the
lighthouse respectively. The angle of elevation of the top of the tower from
the top of the lighthouse is 300 . Find the height of the tower and distance
between the top of the lighthouse to the top of the tower [AE]. Also find the
angle of depression formed from the top of the lighthouse to the ship.
OR
A tower and a pole stand vertically on the same level ground. It is observed
that the angles of depression of the top and foot of the pole from the top of
tower of height 60 m is 300 and 600 respectively. Find the height of the pole.
= {358} {298}
= 59 179 49 149
= 10561 – 7301
1
4
= 3260
37 The lighthouse [AB] of height 10√3 m stands vertically on a sea shore. A
tower [CE] and a ship [F] are standing 30m and 10m away from the foot of the
lighthouse respectively. The angle of elevation of the top of the tower from
the top of the lighthouse is 300 . Find the height of the tower and distance
between the top of the lighthouse to the top of the tower [AE]. Also find the
angle of depression formed from the top of the lighthouse to the ship.
OR
A tower and a pole stand vertically on the same level ground. It is observed
that the angles of depression of the top and foot of the pole from the top of
tower of height 60 m is 300 and 600 respectively. Find the height of the pole.
Page 51
Ans: The height of light house = 10√3 m
The distance between foot of the light house to foot of the tower = BC = 30 m
The distance between foot of the light house to ship = BF = 10 m
Height of the tower = CE = ?
Distance between the top of the light house to the top of the tower = AE = ?
Angle of depression = θ
In ADE ,
tan 300 =
=
√
DE = m
√
DE= 10√3 m 1
Height of the tower = CE = DE + CD
= 10√3+ 10√3
= 20√3 m 1
In ADE ,
The distance between foot of the light house to foot of the tower = BC = 30 m
The distance between foot of the light house to ship = BF = 10 m
Height of the tower = CE = ?
Distance between the top of the light house to the top of the tower = AE = ?
Angle of depression = θ
In ADE ,
tan 300 =
=
√
DE = m
√
DE= 10√3 m 1
Height of the tower = CE = DE + CD
= 10√3+ 10√3
= 20√3 m 1
In ADE ,
Page 52
cos 300 =
√
=
1
AE = = 20√3 m
√
Distance between the top of the lighthouse to the top of the tower = AE = 20√3
m
Angle of depression = DAF = AFB = θ
In ABF , tan θ
√
tan θ
tan θ √3
tan θ tan 600 1 4
Angle of depression = θ 600
OR
Height of the tower = 60m.
Height of the pole = CD = h m.
and BE = CD = h m.
Let BD = EC = x
AE = (60 – h) m.
ACE = 300 1
ADB = 600
In AEC
√
=
1
AE = = 20√3 m
√
Distance between the top of the lighthouse to the top of the tower = AE = 20√3
m
Angle of depression = DAF = AFB = θ
In ABF , tan θ
√
tan θ
tan θ √3
tan θ tan 600 1 4
Angle of depression = θ 600
OR
Height of the tower = 60m.
Height of the pole = CD = h m.
and BE = CD = h m.
Let BD = EC = x
AE = (60 – h) m.
ACE = 300 1
ADB = 600
In AEC
Page 53
tan 300 =
=
√
x = √3 (60 – h) ----------(i) 1
In ABD tan 600 =
√3 =
x=√ ----------(ii)
From equations (i) and (ii)
√3 (60 – h) = √ 1
(60 – h) = √ √
(60 – h) =
(60 – h) = 20
h = 60 – 20
1
h = 40 m 4
Height of the pole = CD = h = 40m
=
√
x = √3 (60 – h) ----------(i) 1
In ABD tan 600 =
√3 =
x=√ ----------(ii)
From equations (i) and (ii)
√3 (60 – h) = √ 1
(60 – h) = √ √
(60 – h) =
(60 – h) = 20
h = 60 – 20
1
h = 40 m 4
Height of the pole = CD = h = 40m
Page 54
38 A hollow solid is made by placing a cylinder on an inverted cone as shown in the
figure. The radius of the cylinder and the cone are 7cm and their heights are
12cm and 9cm respectively. The solid is filled with water up to the height of rd
the height of the cylinder. Find the amount of water in the solid. Calculate the
amount of water required to fill the solid completely.
OR
A vessel is in the form of an inverted cone. Its height is 8cm and the radius is 5cm.
The vessel is filled with water up to the brim and when 100 lead shots of same
size are dropped in it, th of the water from the vessel flows out. Find the radius
and volume of each lead shot.
figure. The radius of the cylinder and the cone are 7cm and their heights are
12cm and 9cm respectively. The solid is filled with water up to the height of rd
the height of the cylinder. Find the amount of water in the solid. Calculate the
amount of water required to fill the solid completely.
OR
A vessel is in the form of an inverted cone. Its height is 8cm and the radius is 5cm.
The vessel is filled with water up to the brim and when 100 lead shots of same
size are dropped in it, th of the water from the vessel flows out. Find the radius
and volume of each lead shot.
Page 55
Ans:
Volume of water inside the solid = Vol. of cone + Vol. of water in the cylinder
= 1
= 7 9 7 12
1
= 49 9 49 12
= 49(9 + 2 12)
1
= 49(9 + 24)
= 22 7(33) 1/2
= 22 7 11
1/2
= 1694 cm3
Amount of water required to fill it completely = of Vol. of cylinder
1/2
= 5
= 7 7 12
1/2
= 22 7 4
= 616cm3
OR
Vol. of water displaced = Vol. of 100 lead shots = Vol. of cone 1/2
100 . of lead shots = Vol. of cone
100 1/2
1/2
=
=
1/2
=
Volume of water inside the solid = Vol. of cone + Vol. of water in the cylinder
= 1
= 7 9 7 12
1
= 49 9 49 12
= 49(9 + 2 12)
1
= 49(9 + 24)
= 22 7(33) 1/2
= 22 7 11
1/2
= 1694 cm3
Amount of water required to fill it completely = of Vol. of cylinder
1/2
= 5
= 7 7 12
1/2
= 22 7 4
= 616cm3
OR
Vol. of water displaced = Vol. of 100 lead shots = Vol. of cone 1/2
100 . of lead shots = Vol. of cone
100 1/2
1/2
=
=
1/2
=
Page 56
=
=
rs3= =( ) 1/2
rs = 0.5
1/2
Radius of lead shot is 0.5cm
Volume of each lead shot is = 1/2
1/2
= ( )
1/2
=
5
1/2
0.523
=
rs3= =( ) 1/2
rs = 0.5
1/2
Radius of lead shot is 0.5cm
Volume of each lead shot is = 1/2
1/2
= ( )
1/2
=
5
1/2
0.523
Page 57
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Page 58
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3
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a b a1 b1 c1
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a2 b2 a2 b2 c2
a1 b1 a1 b1 c1
(C) (D)
a2 b2 a2 b2 c2
1
a1 b1
GvÀÛgÀ:
a2 b2
(A)
EªÀÅUÀ¼°
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4
(A) 1, 2, 4, 8……. (B) 3, 7, 10, 14…….
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Page 59
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0– 10 3 3
10 – 20 4 7
20 – 30 7 14
30 – 40 6 20
n = 20
(A) 0 – 10 (B) 10 – 20
(C) 20 – 30 (D) 30 – 40 1
(C) GvÀÛgÀ: 20 – 30
6
cot A = 3 tan A DzÀgÉ, PÉÆãÀ A AiÀÄ C¼ÀvÉAiÀÄÄ,
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(D) 1
GvÀÛgÀ: 30
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bÉâ¸ÀÄvÀz
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Page 60
8
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(C) x2 – 5x + 6 = 0 (D) x2 + 5x + 6 = 0
(B) GvÀÛgÀ: x2 – 5x – 6 = 0 1
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Document Details
| Board / Org | Karnataka Board |
| Exam | Class 10 |
| Type | Sample Paper |
| Pages | 90 |
| Updated | 24 Sep 2026 |