Samacheer Kalvi Class 10 Maths Practice Question Papers with Answers (2025-26)

Brain Grain · braingrain.in

Maths — Practice Paper · Set 1

Class: 10Samacheer KalviMax Marks: 87

Name: ____________________Reg No: ____________

Part I — Multiple Choice Questions 15 × 1 = 15

Choose the correct answer. (Answer all questions.)

1.Choose the correct answer: The slope of the line which is perpendicular to the line joining the points (0, 0) and (−8, 8) is (A) −1 (B) 1 (C) 1/3 (D) −8.[1]

2.Which of the following should be added to make[1]

3.Let f(x) = x + 1/2. Which of the following holds for all x,y? (A) f(x+y) = f(x)·f(y) (B) f(x+y) = f(x) + f(y) (C) f(x+y) ≤ f(x)·f(y) (D) None of these[1]

4.Which of the following sequences are in G.P.?[1]

5.Which of the following can be calculated from the given matrices?[1]

6.Choose the correct answer: The area of the triangle formed by the points (−5, 0), (0, −5) and (5, 0) is (A) 0 sq.units (B) 25 sq.units (C) 5 sq.units (D) none of these.[1]

7.Choose the correct answer: A tower is 60 m high. Its shadow reduces by x metres when the angle of elevation of the sun increases from 30° to 45°. Then x is (A) 41.92 m (B) 43.92 m (C) 43 m (D) 45.6 m.[1]

8.Choose the correct answer: If the ratio of the height of a tower and the length of its shadow is √3 : 1, then the angle of elevation of the sun is (A) 45° (B) 30° (C) 90° (D) 60°.[1]

9.Choose the correct answer: If (5, 7), (3, p) and (6, 6) are collinear, then the value of p is (A) 3 (B) 6 (C) 9 (D) 12.[1]

10.Choose the correct answer: A man walks near a wall such that the distance between him and the wall is 10 units. Considering the wall as the Y-axis, the path travelled by the man is (A) x = 10 (B) y = 10 (C) x = 0 (D) y = 0.[1]

11.Choose the correct answer: The value of sin²θ + 1/(1 + tan²θ) is equal to (A) tan²θ (B) 1 (C) cot²θ (D) 0.[1]

12.Choose the correct answer: If 5x = sec θ and 5/y = tan θ, then 25x² − 25/y² is equal to (A) 25 (B) 1 (C) 5 (D) 1/25.[1]

13.Let A = {1,2,3,7} and B = {3,0,-1,7}. Which of the following are relations from A to B? (i) R1 = {(2,1), (7,1)} (ii) R2 = {(-1,1)} (iii) R3 = {(2,-1), (7,7), (1,3)} (iv) R4 = {(7,-1), (0,3), (3,3), (0,7)}[1]

14.Choose the correct answer: When proving that a quadrilateral is a parallelogram by using slopes, you must find (A) the slopes of two sides (B) the slopes of two pairs of opposite sides (C) the lengths of all sides (D) both the lengths and slopes of two sides.[1]

15.Choose the correct answer: If x = a tan θ and y = b sec θ, then (A) y²/b² − x²/a² = 1 (B) x²/a² − y²/b² = 1 (C) x²/a² + y²/b² = 1 (D) x²/a² − y²/b² = 0.[1]

Part II — Fill in the Blanks 1 × 1 = 1

Fill in the blanks. (Answer all questions.)

16.Which rational expression should be subtracted from ______ ?[1]

Part III — Short Answer Questions 18 × 2 = 36

Answer briefly. (Answer all questions.)

17.The height of a right circular cone whose radius is $5$ cm and slant height is $13$ cm will be[2]

18.Let f(x) = 2x + 5. If x ≠ 0, find (f(x) - 5)/x.[2]

19.The two tangents from external point $P$ touch the circle at $A$ and $B$.[2]

20.The ratio of the volumes of a cylinder, a cone and a sphere, if each has the same diameter and same height is?[2]

21.Rekha has 15 square colour papers of sizes 10 cm, 11 cm, 12 cm, …, 24 cm. How much area can be decorated?[2]

22.Let A = {1, 2}, B = {1, 2, 3, 4}, C = {5, 6} and D = {5, 6, 7, 8}. Verify whether A × C ⊆ B × D.[2]

23.If the standard deviation of a data is 4.5 and if each value of the data is decreased by 5, then find the new standard deviation.[2]

24.Show that the function[2]

25.If $A=\begin{bmatrix} ... \end{bmatrix}$, verify that[2]

26.A man walks 18 m east and 24 m north[2]

27.In a hollow cylinder the sum of the external and internal radii is 14 cm and the wall thickness (difference of radii) is 4 cm. If its height is 20 cm, the volume of the material in it is:[2]

28.Let A = {1, 2}, B = {1, 2, 3, 4}, C = {5, 6} and D = {5, 6, 7, 8}. Verify whether A × C is a subset of B × D.[2]

29.If in triangles ABC and EDF,[2]

30.What is the inclination of a line whose slope is (i) 0 (ii) 1?[2]

31.Kumar writes a letter to four friends...[2]

32.How many tangents can be drawn from an exterior point?[2]

33.Draw triangle ABC[2]

34.If 1^3 + 2^3 + 3^3 + … + k^3 = 44100, then find 1 + 2 + 3 + … + k.[2]

Part IV — Long Answer Questions 7 × 5 = 35

Answer in detail. (Answer all questions.)

35.Find the equations of the lines whose sum and product of intercepts are 1 and −6 respectively.[5]

36.A solid sphere and a solid hemisphere have equal total surface area.[5]

37.Find the equation of a straight line through the point of intersection of the lines 8x + 3y = 18 and 4x + 5y = 9, and bisecting the line segment joining the points (5, −4) and (−7, 6).[5]

38.Find the next three terms of the following sequences.[5]

39.Construct triangle PQR given PQ = 6.8 cm, vertical angle = 50° and the bisector of the vertical angle meets the base at D where PD = 5.2 cm. Draw the triangle.[5]

40.Determine whether the sets of points are collinear: (i) (−1/2, 3), (−5, 6) and (−8, 8) (ii) (a, b+c), (b, c+a) and (c, a+b).[5]

41.Temperature of Ooty from Monday to Friday is in A.P.[5]

🔑 Show Answer Key — Set 1

1. Slope of the join = 8/(−8) = −1, so the perpendicular slope is 1. (B) 1 .
2. $$ x^4+16x^2+64=(x^2+8)^2 $$ Answer $$ \boxed{(2)\ 16x^2} $$ <div
3. (D) None of these. Explanation: f(x+y)=x+y+1/2, f(x)+f(y)=x+y+1, and f(x)·f(y)=(x+1/2)(y+1/2)=xy+½(x+y)+1/4. None of the equalities/inequalities hold for all x,y (for example x=1,y=1 gives f(2)=2.5, f(1)+f(1)=3, f(1)·f(1)=2.25).
4. (i) Common ratio: $$ \frac93=\frac{27}9=\frac{81}{27}=3 $$ Answer G.P. (ii) Ratios are not equal. Answer Not a G.P. (iii) Common ratio: $$ \frac{0.05}{0.5}=0.1 $$ Answer G.P. (iv) Common ratio: $$ \frac12 $$ Answer G.P. (v) Common ratio: $$ -5 $$ Answer G.P. (vi) Ratios are not equal. Answer Not a G.P. (vii) Common ratio: $$ \frac14 $$ Answer G.P.
5. $$ \boxed{(2)\ (ii)\ \text{and}\ (iii)\ \text{only}} $$ <div
6. Area = ½ × base × height = ½ × 10 × 5 = 25. (B) 25 sq.units .
7. x = 60(cot30° − cot45°) = 60(√3 − 1) ≈ 60(0.732) = 43.92 m. (B) 43.92 m .
8. tan(elevation) = height/shadow = √3, so the angle = 60°. (D) 60° .
9. Collinear ⇒ equal slopes: (p − 7)/(3 − 5) = (6 − 7)/(6 − 5) ⇒ (p − 7)/(−2) = −1 ⇒ p = 9. (C) 9 .
10. The distance from the Y-axis stays constant at 10, so the path is the vertical line (A) x = 10 .
11. 1/(1 + tan²θ) = 1/sec²θ = cos²θ, so the value = sin²θ + cos²θ = 1. (B) 1 .
12. 25x² − 25/y² = (5x)² − (5/y)² = sec²θ − tan²θ = 1. (B) 1 .
13. (i) R1 = {(2,1),(7,1)} — Not a relation, since 1 ∉ B. (ii) R2 = {(-1,1)} — Not a relation, since -1 ∉ A (and 1 ∉ B). (iii) R3 = {(2,-1),(7,7),(1,3)} — This is a relation: all first elements are in A and all second elements are in B. (iv) R4 = {(7,-1),(0,3),(3,3),(0,7)} — Not a relation, since 0 ∉ A.
14. You must show both pairs of opposite sides are parallel — i.e. the slopes of two pairs of opposite sides. (B) .
15. tan²θ = x²/a² and sec²θ = y²/b². Since sec²θ − tan²θ = 1, we get y²/b² − x²/a² = 1. (A) y²/b² − x²/a² = 1 .
16. > Note: > The original rational expression was incomplete in the provided OCR/source text. > Full numerator and denominator were not visible. General Method If $$ A-B=C $$ then $$ B=A-C $$ So, the rational expression to be subtracted can be found by: 1. Taking LCM of denominators 2. Simplifying 3. Subtracting appropriately
17. $$ l^2=r^2+h^2 $$ $$ 13^2=5^2+h^2 $$ $$ 169=25+h^2 $$ $$ h^2=144 $$ $$ h=12 $$ Answer $$ \boxed{(1)\ 12\text{ cm}} $$
18. (f(x)-5)/x = (2x+5-5)/x = (2x)/x = 2, for x ≠ 0. Answer: 2 (with x ≠ 0).
19. $$ \angle AOB=180^\circ-\angle APB $$ $$ =180^\circ-70^\circ $$ $$ =110^\circ $$ Answer $$ \boxed{(2)\ 110^\circ} $$ <div
20. Let the common radius be r and height = 2r. Cylinder: V1 = πr²(2r) = 2πr³. Cone: V2 = (1/3)πr²(2r) = (2/3)πr³. Sphere: V3 = (4/3)πr³. Ratio V1:V2:V3 = 2 : 2/3 : 4/3 = 6:2:4 = 3:1:2.
21. Area required: $$ 10^2+11^2+\dots+24^2 $$ $$ = \sum_{1}^{24}n^2 - \sum_{1}^{9}n^2 $$ $$ = \frac{24(25)(49)}{6} - \frac{9(10)(19)}{6} $$ $$ =4900-285 $$ $$ =4615 $$ Answer $$ \boxed{4615\text{ cm}^2} $$
22. A×C = {(1,5), (1,6), (2,5), (2,6)}. Each first component (1 or 2) belongs to B and each second component (5 or 6) belongs to D, so every pair of A×C is also in B×D. Therefore A×C ⊆ B×D.
23. Subtracting a constant from every observation does not change standard deviation. Answer $$ 4.5 $$
24. Suppose $$ f(a)=f(b) $$ Then, $$ 2a-1=2b-1 $$ $$ 2a=2b $$ $$ a=b $$ Hence $f$ is one-one. The range is $$ \{1,3,5,7,\dots\} $$ Even natural numbers are not images of any element. Hence the function is not onto.
25. $$ \boxed{ (A^T)^T=A } $$ Verified.
26. Using Pythagoras theorem: $$ d^2 = 18^2 + 24^2 $$ $$ =324+576 $$ $$ =900 $$ $$ d=\sqrt{900} $$ $$ d=30 $$ Answer $$ \boxed{30\text{ m}} $$
27. Let external radius $R$, internal radius $r$ $$ R+r=14 $$ Width: $$ R-r=4 $$ Using identity: $$ R^2-r^2=(R+r)(R-r) $$ $$ =14\times4 $$ $$ =56 $$ Volume: $$ V=\pi h(R^2-r^2) $$ $$ =\pi(20)(56) $$ $$ =1120\pi $$ Answer $$ \boxed{1120\pi\text{ cm}^3} $$
28. Compute A × C = {(1,5), (1,6), (2,5), (2,6)}. Since every first component (1 or 2) belongs to B and every second component (5 or 6) belongs to D, each ordered pair of A × C is also an element of B × D. Hence A × C ⊆ B × D.
29. $$ \boxed{(3)\ \angle B=\angle D} $$ <div
30. (i) Inclination θ = tan⁻¹(0) = 0° . (ii) Inclination θ = tan⁻¹(1) = 45° .
31. Number of letters in 8th set: $$ 4^8 $$ Cost per letter: $$ ₹2 $$ Total cost: $$ 2\times4^8 $$ $$ =2\times65536 $$ $$ =131072 $$ Answer $$ \boxed{₹131072} $$
32. $$ \boxed{(2)\ \text{Two}} $$ <div
33. Given: $BC=5.6\text{ cm}$ $\angle A=40^\circ$ Angle bisector meets BC at D such that $CD=4\text{ cm}$ Construction Steps 1. Draw BC. 2. Mark D on BC such that $CD=4\text{ cm}$. 3. Using angle bisector theorem locate A. 4. Join AB and AC. Required triangle obtained.
34. Let S = 1 + 2 + … + k. Then 1^3+…+k^3 = S^2 = 44100, so S = √44100 = 210. Answer: 210
35. Let the intercepts be a and b: a + b = 1, ab = −6 ⇒ a, b are roots of t² − t − 6 = 0 ⇒ (t − 3)(t + 2) = 0 ⇒ {3, −2}. Intercept form x/a + y/b = 1 gives: x/3 + y/(−2) = 1 ⇒ 2x − 3y − 6 = 0 , and x/(−2) + y/3 = 1 ⇒ 3x − 2y + 6 = 0 .
36. Prove that the ratio of their volumes is $3\sqrt3:4$. Proof Let radius of sphere be $R$. Let radius of hemisphere be $r$. Surface area of sphere: $$ 4\pi R^2 $$ Total surface area of hemisphere: $$ 3\pi r^2 $$ Given equal: $$ 4\pi R^2=3\pi r^2 $$ $$ R^2=\frac34r^2 $$ $$ R=\frac{\sqrt3}{2}r $$ Volume of sphere: $$ V_1=\frac43\pi R^3 $$ Volume of hemisphere: $$ V_2=\frac23\pi r^3 $$ Ratio: $$ V_1:V_2 = \frac43\pi R^3:\frac23\pi r^3 $$ $$ =2R^3:r^3 $$ Substitute: $$ R=\frac{\sqrt3}{2}r $$ $$ =2\left(\frac{\sqrt3}{2}\right)^3 $$ $$ =\frac{3\sqrt3}{4} $$ Hence, $$ 3\sqrt3:4 $$ Hence proved.
37. Solving 8x + 3y = 18 and 4x + 5y = 9: 2(4x + 5y) − (8x + 3y) = 18 − 18 ⇒ 7y = 0 ⇒ y = 0, then x = 9/4. Intersection (9/4, 0). Midpoint of (5, −4) and (−7, 6) = (−1, 1). The required line passes through (9/4, 0) and (−1, 1). Slope = (1 − 0)/(−1 − 9/4) = 1/(−13/4) = −4/13. Through (9/4, 0): y = (−4/13)(x − 9/4) ⇒ 13y = −4x + 9. Therefore 4x + 13y − 9 = 0 .
38. (i) Each term is multiplied by 3. $$ 72\times3=216 $$ $$ 216\times3=648 $$ $$ 648\times3=1944 $$ Answer $$ 216,\ 648,\ 1944 $$ (ii) Common difference: $$ 1-5=-4 $$ $$ -3-1=-4 $$ Continue subtracting 4. Answer $$ -7,\ -11,\ -15 $$ (iii) Pattern: $$ \frac{1}{2^2},\frac{2}{3^2},\frac{3}{4^2} $$ Thus next terms are: $$ \frac4{5^2},\frac5{6^2},\frac6{7^2} $$ Answer $$ \frac4{25},\ \frac5{36},\ \frac6{49} $$
39. Construction outline: Draw base PQ of required length 6.8 cm. At the midpoint (or the chosen vertex for the vertical angle), construct the vertical angle of 50°. From the vertex, draw the angle bisector and mark point D on the base such that PD = 5.2 cm. Using these constraints, locate the third vertex so that the bisector meets the base at D. Join vertices to complete ΔPQR. Note: steps summarise the standard ruler-and-compass approach; precise drawing requires graphical construction on paper.
40. Points are collinear when the area of the triangle they form is 0. (i) ½|(−1/2)(6 − 8) + (−5)(8 − 3) + (−8)(3 − 6)| = ½|1 − 25 + 24| = 0 ⇒ collinear . (ii) ½|a((c+a) − (a+b)) + b((a+b) − (b+c)) + c((b+c) − (c+a))| = ½|a(c−b) + b(a−c) + c(b−a)| = 0 ⇒ collinear .
41. Let temperatures be: $$ a-2d,\ a-d,\ a,\ a+d,\ a+2d $$ First condition: $$ (a-2d)+(a-d)+a=0 $$ $$ 3a-3d=0 $$ $$ a=d $$ Second condition: $$ a+(a+d)+(a+2d)=18 $$ $$ 3a+3d=18 $$ Since $a=d$, $$ 6a=18 $$ $$ a=3 $$ Thus: $$ -3^\circ C,\ 0^\circ C,\ 3^\circ C,\ 6^\circ C,\ 9^\circ C $$ Answer $$ -3^\circ C,\ 0^\circ C,\ 3^\circ C,\ 6^\circ C,\ 9^\circ C $$

Brain Grain · braingrain.in

Maths — Practice Paper · Set 2

Class: 10Samacheer KalviMax Marks: 87

Name: ____________________Reg No: ____________

Part I — Multiple Choice Questions 15 × 1 = 15

Choose the correct answer. (Answer all questions.)

1.Choose the correct answer: If sin θ + cos θ = a and sec θ + cosec θ = b, then the value of b(a² − 1) is equal to (A) 2a (B) 3a (C) 0 (D) 2ab.[1]

2.Choose the correct answer: If A is a point on the Y-axis whose ordinate is 8 and B is a point on the X-axis whose abscissa is 5, then the equation of the line AB is (A) 8x + 5y = 40 (B) 8x − 5y = 40 (C) x = 8 (D) y = 5.[1]

3.Represent each of the following relations by (a) an arrow diagram, (b) a graph and (c) a set in roster form, wherever possible. (i) {(x,y) | x = 2y, x ∈ {2,3,4,5}, y ∈ {1,2,3,4}} (ii) {(x,y) | y = x+3, x, y are natural numbers < 10}[1]

4.Choose the correct answer: tan θ·cosec²θ − tan θ is equal to (A) sec θ (B) cot²θ (C) sin θ (D) cot θ.[1]

5.Choose the correct answer: A straight line has the equation 8y = 4x + 21. Which of the following is true? (A) slope 0.5, y-intercept 2.6 (B) slope 5, y-intercept 1.6 (C) slope 0.5, y-intercept 1.6 (D) slope 5, y-intercept 2.6.[1]

6.Choose the correct answer: If a cot θ + b cosec θ = p and b cot θ + a cosec θ = q, then p² − q² is equal to (A) a² − b² (B) b² − a² (C) a² + b² (D) b − a.[1]

7.Choose the correct answer: The slope of the line joining (12, 3) and (4, a) is 1/8. The value of a is (A) 1 (B) 4 (C) −5 (D) 2.[1]

8.Choose the correct answer: The angle of elevation of a cloud from a point h metres above a lake is β, and the angle of depression of its reflection in the lake is 45°. The height of the cloud from the lake is (A) h(1 + tan β)/(1 − tan β) (B) h(1 − tan β)/(1 + tan β) (C) h tan(45° − β) (D) none of these.[1]

9.A company has four categories of employees: Assistants (A), Clerks (C), Managers (M) and an Executive Officer (E). The company provides ₹10,000, ₹25,000, ₹50,000 and ₹1,00,000 as salaries to the people who work in the categories A, C, M and E respectively. If A1, A2, A3, A4, A5 are Assistants; C1,C2,C3,C4 are Clerks; M1,M2,M3 are Managers and E1,E2 are Executive officers, and if the relation R is defined by xRy where x is the salary given to person y, express the relation R through an ordered pair notation and by an arrow diagram.[1]

10.Choose the correct answer: Two persons are standing x metres apart and the height of the first is double that of the other. From the midpoint of the line joining their feet, the angular elevations of their tops are complementary. The height of the shorter person (in metres) is (A) 2√2·x (B) x/(2√2) (C) x/2 (D) 2x.[1]

11.Choose the correct answer: Consider four straight lines (i) l₁: 3y = 4x + 5 (ii) l₂: 4y = 3x − 1 (iii) l₃: 4y + 3x = 7 (iv) l₄: 4x + 3y = 2. Which statement is true? (A) l₁ and l₂ are perpendicular (B) l₁ and l₄ are parallel (C) l₂ and l₄ are perpendicular (D) l₂ and l₃ are parallel.[1]

12.Choose the correct answer: The straight line given by the equation x = 11 is (A) parallel to the X-axis (B) parallel to the Y-axis (C) passing through the origin (D) passing through the point (0, 11).[1]

13.Choose the correct answer: (1 + tan θ + sec θ)(1 + cot θ − cosec θ) is equal to (A) 0 (B) 1 (C) 2 (D) −1.[1]

14.Choose the correct answer: If sin θ = cos θ, then 2 tan²θ + sin²θ − 1 is equal to (A) −3/2 (B) 3/2 (C) 2/3 (D) −2/3.[1]

15.Choose the correct answer: The angles of depression of the top and bottom of a 20 m tall building from the top of a multistoried building are 30° and 60° respectively. The height of the multistoried building and the distance between the two buildings (in metres) is (A) 20, 10√3 (B) 30, 5√3 (C) 20, 10 (D) 30, 10√3.[1]

Part II — Fill in the Blanks 1 × 1 = 1

Fill in the blanks. (Answer all questions.)

16.Which rational expression should be subtracted from ______ ?[1]

Part III — Short Answer Questions 18 × 2 = 36

Answer briefly. (Answer all questions.)

17.A solid sphere of radius x cm is melted and cast into the shape of a solid cone having the same base radius x. The height of the cone is:[2]

18.Let g = {(0,2), (1,0), (2,4), (-4,2), (7,0)} and let f be the inverse relation of g (that is, f = g^{-1}). Find the range of f.[2]

19.Find the slope of a line joining the points (i) (5, 5) with the origin (ii) (sin θ, −cos θ) and (−sin θ, cos θ).[2]

20.The sum of squares of first $n$ natural numbers is 285 while the sum of cubes is 2025. Find $n$.[2]

21.The external radius and the length of a hollow wooden log are $16$ cm and $13$ cm respectively.[2]

22.The outer and inner surface areas of a spherical copper shell are $576\pi\text{ cm}^2$ and $324\pi\text{ cm}^2$ respectively.[2]

23.Find the LCM of each pair of the following polynomials: (i) a^2 - 4a + 1, a^2 - 5a + 6 (given GCD = a-2) (ii) x^3 - 4ax + 3a^2 - 7, (x-a)^3 - 3(x-a) (given GCD = (x-a) - 3) [source OCR ambiguous][2]

24.If f(x) = 2x^2 and g(x) = x/3, find (f ∘ g)(x).[2]

25.Find the largest number which divides 1230 and 1926 leaving remainder 12 in each case.[2]

26.Question 1[2]

27.If the ordered pairs (a + 2, 2a + b) and (5, 4) are equal, find a and b.[2]

28.Draw the graph of the quadratic equation x^2 + 2x + 1 = 0 and state its root(s).[2]

29.If $-4$ is a root of[2]

30.If the function f is defined by f(x) = { x + 2, for x ≥ 0; 2x + 1, for x < 0 }, find: (i) f(3) (ii) f(0) (iii) f(−1.5) (iv) f(2) + f(−2).[2]

31.The total surface area of a hemisphere is how much times the square of its radius.[2]

32.Which term of the A.P.[2]

33.The volume (in cm^3) of the greatest sphere that can be cut off from a cylindrical log of wood of base radius 1 cm and height 5 cm is?[2]

34.The solution of the system[2]

Part IV — Long Answer Questions 7 × 5 = 35

Answer in detail. (Answer all questions.)

35.Solve the following quadratic equations by factorization method[5]

36.Use Euclid’s Division Algorithm to find the HCF.[5]

37.A solid sphere of radius 6 cm is melted into a hollow cylinder of uniform thickness.[5]

38.Find the GCD of the given polynomials[5]

39.Solve the following quadratic equations by formula method[5]

40.Brick staircase problem.[5]

41.Let f(x) = 2x + 1 and g(x) = x^2. Find (f ∘ g)(x) and its range for x ∈ ℝ. Then find (g ∘ f)(x) and its range for x ∈ ℝ.[5]

🔑 Show Answer Key — Set 2

1. a² − 1 = 2 sin θ cos θ, and b = (sin θ + cos θ)/(sin θ cos θ) = a/(sin θ cos θ). So b(a² − 1) = 2a. (A) 2a .
2. A = (0, 8), B = (5, 0). Intercept form x/5 + y/8 = 1 ⇒ 8x + 5y = 40. (A) 8x + 5y = 40 .
3. (i) Valid ordered pairs: {(2,1),(4,2)}. Arrow diagram: 2 → 1, 4 → 2. Graph points: (2,1), (4,2). (ii) For natural numbers Ordered pairs: {(1,4),(2,5),(3,6),(4,7),(5,8),(6,9)}. Arrow diagram: 1→4, 2→5, 3→6, 4→7, 5→8, 6→9. Graph points: (1,4),(2,5),(3,6),(4,7),(5,8),(6,9).
4. tan θ(cosec²θ − 1) = tan θ·cot²θ = (sin θ/cos θ)(cos²θ/sin²θ) = cos θ/sin θ = cot θ. (D) cot θ .
5. y = (4/8)x + 21/8 = 0.5x + 2.625, so slope = 0.5 and y-intercept ≈ 2.6. (A) slope 0.5, y-intercept 2.6 .
6. p² − q² = (cot²θ − cosec²θ)(a² − b²) = (−1)(a² − b²) = b² − a². (B) b² − a² .
7. Slope = (a − 3)/(4 − 12) = (a − 3)/(−8) = 1/8 ⇒ a − 3 = −1 ⇒ a = 2. (D) 2 .
8. Taking the cloud height H above the lake with horizontal distance d: tan β = (H − h)/d and tan45° = (H + h)/d. Eliminating d gives H = h(1 + tan β)/(1 − tan β). (A) h(1 + tan β)/(1 − tan β) .
9. The relation R (as ordered pairs salary → person) is: {(10000,A1),(10000,A2),(10000,A3),(10000,A4),(10000,A5), (25000,C1),(25000,C2),(25000,C3),(25000,C4), (50000,M1),(50000,M2),(50000,M3), (100000,E1),(100000,E2)}. Arrow representation: 10000 → A1,A2,A3,A4,A5 25000 → C1,C2,C3,C4 50000 → M1,M2,M3 100000 → E1,E2
10. Let the shorter height be a (taller = 2a). tan α = a/(x/2) and cot α = 2a/(x/2); multiplying gives x/(2a) = 4a/x ⇒ x² = 8a² ⇒ a = x/(2√2). (B) x/(2√2) .
11. Slopes: l₁ = 4/3, l₂ = 3/4, l₃ = −3/4, l₄ = −4/3. Since l₂ × l₄ = (3/4)(−4/3) = −1, they are perpendicular. (C) l₂ and l₄ are perpendicular .
12. x = 11 is a vertical line, so it is (B) parallel to the Y-axis .
13. Writing in terms of sin/cos and simplifying gives 2. (C) 2 .
14. sin θ = cos θ ⇒ θ = 45°. 2 tan²45° + sin²45° − 1 = 2(1) + 1/2 − 1 = 3/2. (B) 3/2 .
15. Let H be the height and d the distance. tan60° = H/d and tan30° = (H − 20)/d give H = 30 m and d = 10√3 m. (D) 30, 10√3 .
16. > Note: > The original rational expression was incomplete in the provided OCR/source text. > Full numerator and denominator were not visible. General Method If $$ A-B=C $$ then $$ B=A-C $$ So, the rational expression to be subtracted can be found by: 1. Taking LCM of denominators 2. Simplifying 3. Subtracting appropriately
17. Volume of sphere: $$ \frac43\pi x^3 $$ Volume of cone: $$ \frac13\pi x^2 h $$ Equating: $$ \frac43\pi x^3=\frac13\pi x^2 h $$ $$ 4x=h $$ Answer $$ \boxed{(3)\ 4x\text{ cm}} $$
18. Since f = g^{-1}, each ordered pair (a,b) in g gives (b,a) in f. From g: g(0)=2 ⇒ f(2)=0 g(1)=0 ⇒ f(0)=1 g(2)=4 ⇒ f(4)=2 g(-4)=2 ⇒ f(2)=0 (repeat) g(7)=0 ⇒ f(0)=1 (repeat) Thus the values (images) of f are {0,1,2}. Hence the range of f is {0, 1, 2} .
19. (i) Slope = (5 − 0)/(5 − 0) = 1 . (ii) Slope = (cos θ − (−cos θ))/(−sin θ − sin θ) = (2 cos θ)/(−2 sin θ) = −cot θ .
20. Using cube formula: $$ \left( \frac{n(n+1)}{2} \right)^2=2025 $$ $$ \frac{n(n+1)}{2}=45 $$ $$ n(n+1)=90 $$ $$ n^2+n-90=0 $$ $$ (n-9)(n+10)=0 $$ $$ n=9 $$ Verification $$ 1^2+2^2+\dots+9^2 = \frac{9(10)(19)}{6} =285 $$ Answer $$ \boxed{9} $$
21. External radius: $$ R=16\text{ cm} $$ Thickness: $$ 4\text{ cm} $$ Internal radius: $$ r=16-4=12\text{ cm} $$ Length: $$ h=13\text{ cm} $$ Total surface area of hollow cylinder: $$ TSA=2\pi h(R+r)+2\pi(R^2-r^2) $$ $$ =2\times\frac{22}{7}\times13(16+12) + 2\times\frac{22}{7}(256-144) $$ $$ =2288+704 $$ $$ =2992 $$ Answer $$ 2992\text{ cm}^2 $$
22. Outer surface area: $$ 4\pi R^2=576\pi $$ $$ R^2=144 $$ $$ R=12 $$ Inner surface area: $$ 4\pi r^2=324\pi $$ $$ r^2=81 $$ $$ r=9 $$ Volume of shell: $$ V=\frac43\pi(R^3-r^3) $$ $$ =\frac43\pi(1728-729) $$ $$ =\frac43\pi(999) $$ $$ =4186.29 $$ Answer $$ 4186.29\text{ cm}^3 $$
23. (i) LCM = (a+6)(a-2)(a-3) (ii) LCM = x(x-3a)^2(x^2+3ax+9a^2)
24. $$ (f\circ g)(x)=2\left(\frac{x}{3}\right)^2 $$ $$ =\frac{2x^2}{9} $$ Correct option: (3)
25. Subtract the remainder from each number: [ 1230-12=1218 ] [ 1926-12=1914 ] Required number: [ \gcd(1218,1914) ] Using Euclid’s Algorithm: [ 1914=1218\times1+696 ] [ 1218=696\times1+522 ] [ 696=522\times1+174 ] [ 522=174\times3+0 ] Answer [ \boxed{174} ]
26. Mathematics : Mensuration : Volume and Surface Area of Combined Solids : Exercise Questions with Answers
27. From equality of ordered pairs: a + 2 = 5 ⇒ a = 3. Also 2a + b = 4 ⇒ 2·3 + b = 4 ⇒ 6 + b = 4 ⇒ b = −2. Hence a = 3, b = −2.
28. Factorize: x^2 + 2x + 1 = (x + 1)^2 = 0 ⇒ x = -1 (double root). Answer: Root x = -1 (real and equal roots; the parabola touches the x-axis at x = -1).
29. Substitute: $$ 16-4p-4=0 $$ $$ 12-4p=0 $$ $$ p=3 $$ Equal roots condition: $$ p^2-4q=0 $$ $$ 9-4q=0 $$ $$ q=\frac94 $$ Answer $$ \boxed{p=3,\ q=\frac94} $$
30. (i) f(3) = 3 + 2 = 5. (ii) f(0) = 0 + 2 = 2. (iii) For x = −1.5 ( (iv) f(2) = 2 + 2 = 4; f(−2) = 2(−2) + 1 = −4 + 1 = −3. So f(2) + f(−2) = 4 + (−3) = 1.
31. TSA of hemisphere: $$ 3\pi r^2 $$ Answer $$ \boxed{(3)\ 3\pi} $$
32. $$ a=16,\quad d=-5 $$ $$ t_n=a+(n-1)d $$ $$ -54=16+(n-1)(-5) $$ $$ -70=-5(n-1) $$ $$ 14=n-1 $$ $$ n=15 $$ Answer $$ 15^{th}\text{ term} $$
33. The sphere that fits must have radius ≤ min(cylinder radius, half the height) = min(1, 2.5) = 1 cm. Volume = (4/3)πr^3 = (4/3)π(1)^3 = 4/3 π cm³.
34. From: $$ 3z=9 $$ $$ z=3 $$ Then: $$ -7y+21=7 $$ $$ -7y=-14 $$ $$ y=2 $$ Now: $$ x+2-9=-6 $$ $$ x=1 $$ Answer $$ \boxed{(1)\ x=1,\ y=2,\ z=3} $$ <div
35. (i) Multiply coefficient of $x^2$ and constant term: $$ 4\times(-2)=-8 $$ We need two numbers whose product is $-8$ and sum is $-7$. Those numbers are: $$ -8,\ 1 $$ Split the middle term: $$ 4x^2-8x+x-2=0 $$ Group terms: $$ 4x(x-2)+1(x-2)=0 $$ $$ (x-2)(4x+1)=0 $$ Therefore, $$ x-2=0 \quad \text{or} \quad 4x+1=0 $$ $$ x=2 \quad \text{or} \quad x=-\frac14 $$ Answer $$ \boxed{x=2,\ -\frac14} $$ (ii) Expand: $$ 3p^2-18=p^2+5p $$ Bring all terms to one side: $$ 2p^2-5p-18=0 $$ Find two numbers whose product is: $$ 2\times(-18)=-36 $$ and sum is $-5$. Those numbers are: $$ -9,\ 4 $$ Split middle term: $$ 2p^2-9p+4p-18=0 $$ Group terms: $$ p(2p-9)+2(2p-9)=0 $$ $$ (2p-9)(p+2)=0 $$ Hence, $$ 2p-9=…
36. (i) 340 and 412 [ 412=340\times1+72 ] [ 340=72\times4+52 ] [ 72=52\times1+20 ] [ 52=20\times2+12 ] [ 20=12\times1+8 ] [ 12=8\times1+4 ] [ 8=4\times2+0 ] Answer [ \boxed{4} ] (ii) 867 and 255 [ 867=255\times3+102 ] [ 255=102\times2+51 ] [ 102=51\times2+0 ] Answer [ \boxed{51} ] (iii) 10224 and 9648 [ 10224=9648\times1+576 ] [ 9648=576\times16+432 ] [ 576=432\times1+144 ] [ 432=144\times3+0 ] Answer [ \boxed{144} ] (iv) 84, 90 and 120 First find HCF of 84 and 90. [ 90=84\times1+6 ] [ 84=6\times14+0 ] So, [ \gcd(84,90)=6 ] Now, [ 120=6\times20+0 ] Thus, [ \gcd(84,90,120)=6 ] Answer [ \boxed{6} ]
37. Sphere radius: $$ R=6\text{ cm} $$ Volume of sphere: $$ V=\frac43\pi R^3 $$ $$ =\frac43\pi(6)^3 $$ $$ =288\pi $$ External radius of cylinder: $$ 5\text{ cm} $$ Let internal radius be $r$. Height: $$ 32\text{ cm} $$ Volume of hollow cylinder: $$ \pi(5^2-r^2)(32) $$ Equating: $$ 288\pi = 32\pi(25-r^2) $$ Cancel $\pi$: $$ 288 = 32(25-r^2) $$ $$ 9 = 25-r^2 $$ $$ r^2=16 $$ $$ r=4 $$ Thickness: $$ 5-4=1 $$ Answer $$ 1\text{ cm} $$
38. (i) Find the GCD of Factor the first polynomial: $$ x^4+3x^3-x-3 $$ Grouping: $$ x^3(x+3)-1(x+3) $$ $$ =(x+3)(x^3-1) $$ $$ =(x+3)(x-1)(x^2+x+1) $$ Factor the second polynomial: $$ x^3+x^2-5x+3 $$ Testing roots: $$ x=1 $$ gives zero. So, $$ =(x-1)(x^2+2x-3) $$ $$ =(x-1)^2(x+3) $$ Common factors: $$ (x-1)(x+3) $$ $$ =x^2+2x-3 $$ Answer $$ \boxed{x^2+2x-3} $$ (ii) Find the GCD of $$ x^4-1=(x^2-1)(x^2+1) $$ $$ =(x-1)(x+1)(x^2+1) $$ Factor second polynomial: $$ x^3-11x^2+x-11 $$ Grouping: $$ x^2(x-11)+1(x-11) $$ $$ =(x^2+1)(x-11) $$ Common factor: $$ x^2+1 $$ Answer $$ \boxed{x^2+1} $$ (iii) Find the GCD of Factor first polynomial: $$ 3x(x^3+2x^2-4x-8) $$ $$ =3x(x+2)(x^2-4) $$ $$ =3x(x+2)^2(x-…
39. (i) Here, $$ a=2,\quad b=-5,\quad c=2 $$ Using formula: $$ x=\frac{-(-5)\pm\sqrt{(-5)^2-4(2)(2)}}{2(2)} $$ $$ x=\frac{5\pm\sqrt{25-16}}{4} $$ $$ x=\frac{5\pm3}{4} $$ Thus, $$ x=2 $$ or $$ x=\frac12 $$ Answer $$ \boxed{x=2,\ \frac12} $$ (ii) Here, $$ a=\sqrt2,\quad b=-6,\quad c=3\sqrt2 $$ Using formula: $$ f=\frac{6\pm\sqrt{36-4(\sqrt2)(3\sqrt2)}}{2\sqrt2} $$ $$ f=\frac{6\pm\sqrt{36-24}}{2\sqrt2} $$ $$ f=\frac{6\pm\sqrt{12}}{2\sqrt2} $$ $$ f=\frac{6\pm2\sqrt3}{2\sqrt2} $$ $$ f=\frac{3\pm\sqrt3}{\sqrt2} $$ Answer $$ \boxed{f=\frac{3+\sqrt3}{\sqrt2},\ \frac{3-\sqrt3}{\sqrt2}} $$ (iii) Here, $$ a=3,\quad b=-20,\quad c=-23 $$ Using formula: $$ y=\frac{20\pm\sqrt{(-20)^2-4(3)(-23)}}{6} $$ $$ y=…
40. Bottom step requires 100 bricks. Each successive step requires 2 bricks less. Total steps: $$ 30 $$ (i) Topmost step $$ a=100,\quad d=-2 $$ $$ t_{30}=100+29(-2) $$ $$ =42 $$ Answer $$ 42 $$ (ii) Total bricks required $$ S_{30}=\frac{30}{2}(100+42) $$ $$ =15(142) $$ $$ =2130 $$ Answer $$ 2130 $$
41. (f ∘ g)(x) = f(g(x)) = 2x^2 + 1. For x ∈ ℝ, x^2 ≥ 0, so 2x^2 ≥ 0 and 2x^2 + 1 ≥ 1. Range: [1, ∞). (g ∘ f)(x) = g(f(x)) = (2x + 1)^2 = 4x^2 + 4x + 1 = 4(x + 1/2)^2. This expression ≥ 0 for all real x and attains 0 at x = −1/2. Range: [0, ∞).

Brain Grain · braingrain.in

Maths — Practice Paper · Set 3

Class: 10Samacheer KalviMax Marks: 87

Name: ____________________Reg No: ____________

Part I — Multiple Choice Questions 15 × 1 = 15

Choose the correct answer. (Answer all questions.)

1.Choose the correct answer: (2, 1) is the point of intersection of which two lines? (A) x − y − 3 = 0; 3x − y − 7 = 0 (B) x + y = 3; 3x + y = 7 (C) 3x + y = 3; x + y = 7 (D) x + 3y − 3 = 0; x − y − 7 = 0.[1]

2.f(x) = (x+1)^3 − (x−1)^3 represents a function which is(a) linear(b) cubic(c) reciprocal(d) quadratic[1]

3.Choose the correct answer: If (sin α + cosec α)² + (cos α + sec α)² = k + tan²α + cot²α, then the value of k is (A) 9 (B) 7 (C) 5 (D) 3.[1]

4.Choose the correct answer: The equation of a line passing through the origin and perpendicular to the line 7x − 3y + 4 = 0 is (A) 7x − 3y + 4 = 0 (B) 3x − 7y + 4 = 0 (C) 3x + 7y = 0 (D) 7x − 3y = 0.[1]

5.Choose the correct answer: When proving that a quadrilateral is a trapezium, it is necessary to show (A) two sides are parallel (B) two parallel and two non-parallel sides (C) opposite sides are parallel (D) all sides are of equal length.[1]

6.If there exists a bijection (a one-to-one and onto function) f : A → B and n(A) = 7, what is n(B)?[1]

7.Choose the correct answer: The point of intersection of 3x − y = 4 and x + y = 8 is (A) (5, 3) (B) (2, 4) (C) (3, 5) (D) (4, 4).[1]

8.Choose the correct answer: An electric pole subtends an angle of 30° at a point on the same level as its foot. At a second point b metres above the first, the depression of the foot of the pole is 60°. The height of the pole (in metres) is (A) √3 b (B) b/3 (C) b/2 (D) b/√3.[1]

9.Choose the correct answer: If the slope of the line PQ is 1/3, then the slope of the perpendicular bisector of PQ is (A) 3 (B) −3 (C) 1/3 (D) 0.[1]

10.Choose the correct answer: The slope of the line which is perpendicular to the line joining the points (0, 0) and (−8, 8) is (A) −1 (B) 1 (C) 1/3 (D) −8.[1]

11.Which of the following should be added to make[1]

12.Let f(x) = x + 1/2. Which of the following holds for all x,y? (A) f(x+y) = f(x)·f(y) (B) f(x+y) = f(x) + f(y) (C) f(x+y) ≤ f(x)·f(y) (D) None of these[1]

13.Which of the following sequences are in G.P.?[1]

14.Which of the following can be calculated from the given matrices?[1]

15.Choose the correct answer: The area of the triangle formed by the points (−5, 0), (0, −5) and (5, 0) is (A) 0 sq.units (B) 25 sq.units (C) 5 sq.units (D) none of these.[1]

Part II — Fill in the Blanks 1 × 1 = 1

Fill in the blanks. (Answer all questions.)

16.Which rational expression should be subtracted from ______ ?[1]

Part III — Short Answer Questions 18 × 2 = 36

Answer briefly. (Answer all questions.)

17.If in $\triangle ABC$,[2]

18.Given the function f(x) = x^2 - 5x + 6, evaluate: (i) f(-1) (ii) f(2a) (iii) f(2) (iv) f(x-1).[2]

19.Matrix problem involving sales[2]

20.If $(x-6)$ is the HCF of[2]

21.A plane is flying at a speed of 500 km per hour.[2]

22.Graph of a linear polynomial is a[2]

23.Given A = {1,2,3}, B = {2,3,5}, C = {3,4} and D = {1,3,5}, check whether (A × B) ∩ (C × D) = (A ∩ C) × (B ∩ D).[2]

24.Students distributed among sections A, B and C[2]

25.A system of three linear equations in three variables is inconsistent if their planes[2]

26.Question 1[2]

27.Let A = {−1, 1} and B = {0, 2}. A function f: A → B is defined by f(x) = ax + b and is onto. Find a and b.[2]

28.Sum of first $n$ odd natural numbers[2]

29.Discuss the nature of solutions of the following system of equations[2]

30.Let f(x) = (x + 6)/8 and g(x) = (x − 2)/3. (i) Calculate g(g(1/2)). (ii) Find (g ∘ f)(x) in simplest form.[2]

31.If $A$ is a $2\times3$ matrix and $B$ is a $3\times4$ matrix, how many columns does $AB$ have?[2]

32.Music Gallery Problem[2]

33.Find the rational form of the number[2]

34.Construction of ΔPQR[2]

Part IV — Long Answer Questions 7 × 5 = 35

Answer in detail. (Answer all questions.)

35.Prove that two consecutive positive integers are always coprime.[5]

36.A cylindrical glass with diameter $20$ cm has water to a height of $9$ cm.[5]

37.Prove that $$\left(\frac{1+\sin\theta-\cos\theta}{1+\sin\theta+\cos\theta}\right)^2=\frac{1-\cos\theta}{1+\cos\theta}.$$[5]

38.Question 1[5]

39.A person standing at the junction of two straight paths 2x − 3y + 4 = 0 and 3x + 4y − 5 = 0 seeks to reach the path 6x − 7y + 8 = 0 in the least time. Find the equation of the path he should follow.[5]

40.Nathan, an engineering student was asked to make a model shaped like a cylinder with two cones attached at its two ends.[5]

41.Find the equation of a straight line passing through the point P(−5, 2) and parallel to the line joining the points Q(3, −2) and R(−5, 4).[5]

🔑 Show Answer Key — Set 3

1. Substituting (2, 1): x + y = 2 + 1 = 3 ✓ and 3x + y = 6 + 1 = 7 ✓. (B) x + y = 3; 3x + y = 7 .
2. (D) Quadratic. Work: (x+1)^3 = x^3+3x^2+3x+1 and (x−1)^3 = x^3−3x^2+3x−1. Subtracting gives f(x)=6x^2+2, which is a quadratic function.
3. Expanding: (sin²α + cos²α) + 4 + (cosec²α + sec²α) = 1 + 4 + (2 + cot²α + tan²α) = 7 + tan²α + cot²α. So k = 7. (B) 7 .
4. Slope of the given line = 7/3, so the perpendicular slope is −3/7. Through the origin: y = (−3/7)x ⇒ 3x + 7y = 0. (C) 3x + 7y = 0 .
5. A trapezium has exactly one pair of parallel sides — i.e. two parallel and two non-parallel sides. (B) .
6. Bijective functions have equal cardinalities. $$ n(A)=n(B)=7 $$ Correct option: (1)
7. Adding the equations: 4x = 12 ⇒ x = 3, then y = 5. (C) (3, 5) .
8. If d is the horizontal distance, height = d·tan30° = d/√3, and tan60° = b/d ⇒ d = b/√3. So height = (b/√3)(1/√3) = b/3. (B) b/3 .
9. The perpendicular bisector is perpendicular to PQ, so its slope = −1 ÷ (1/3) = −3. (B) −3 .
10. Slope of the join = 8/(−8) = −1, so the perpendicular slope is 1. (B) 1 .
11. $$ x^4+16x^2+64=(x^2+8)^2 $$ Answer $$ \boxed{(2)\ 16x^2} $$ <div
12. (D) None of these. Explanation: f(x+y)=x+y+1/2, f(x)+f(y)=x+y+1, and f(x)·f(y)=(x+1/2)(y+1/2)=xy+½(x+y)+1/4. None of the equalities/inequalities hold for all x,y (for example x=1,y=1 gives f(2)=2.5, f(1)+f(1)=3, f(1)·f(1)=2.25).
13. (i) Common ratio: $$ \frac93=\frac{27}9=\frac{81}{27}=3 $$ Answer G.P. (ii) Ratios are not equal. Answer Not a G.P. (iii) Common ratio: $$ \frac{0.05}{0.5}=0.1 $$ Answer G.P. (iv) Common ratio: $$ \frac12 $$ Answer G.P. (v) Common ratio: $$ -5 $$ Answer G.P. (vi) Ratios are not equal. Answer Not a G.P. (vii) Common ratio: $$ \frac14 $$ Answer G.P.
14. $$ \boxed{(2)\ (ii)\ \text{and}\ (iii)\ \text{only}} $$ <div
15. Area = ½ × base × height = ½ × 10 × 5 = 25. (B) 25 sq.units .
16. > Note: > The original rational expression was incomplete in the provided OCR/source text. > Full numerator and denominator were not visible. General Method If $$ A-B=C $$ then $$ B=A-C $$ So, the rational expression to be subtracted can be found by: 1. Taking LCM of denominators 2. Simplifying 3. Subtracting appropriately
17. By Basic Proportionality Theorem: $$ \frac{AD}{AB}=\frac{AE}{AC} $$ $$ \frac{2.1}{3.6}=\frac{AE}{2.4} $$ $$ AE=\frac{2.1\times2.4}{3.6} $$ $$ =1.4 $$ Answer $$ \boxed{(1)\ 1.4\text{ cm}} $$ <div
18. (i) f(-1) = (-1)^2 -5(-1) +6 = 1+5+6 = 12. (ii) f(2a) = (2a)^2 -5(2a) +6 = 4a^2 -10a +6. (iii) f(2) = 4 -10 +6 = 0. (iv) f(x-1) = (x-1)^2 -5(x-1) +6 = x^2 -7x +12.
19. May sales are double April sales. (i) Average sales $$ \boxed{ \frac{3A}{2} } $$ (ii) August sales $$ \boxed{ 16A } $$
20. Since $(x-6)$ is a factor, Substitute $x=6$: $$ 36-6k-6=0 $$ $$ 30-6k=0 $$ $$ k=5 $$ Answer $$ \boxed{(2)\ 5} $$ <div
21. $$ \text{Distance}=\text{Speed}\times\text{Time} $$ $$ d=500t $$ Answer $$ d(t)=500t $$
22. $$ \boxed{(1)\ \text{straight line}} $$ <div
23. A ∩ C = {3} B ∩ D = {3,5} (A ∩ C) × (B ∩ D) = {(3,3), (3,5)} A × B = {(1,2),(1,3),(1,5),(2,2),(2,3),(2,5),(3,2),(3,3),(3,5)} C × D = {(3,1),(3,3),(3,5),(4,1),(4,3),(4,5)} (A × B) ∩ (C × D) = {(3,3),(3,5)} Hence the equality holds.
24. $$ \boxed{ A=56,\ B=50,\ C=44 } $$
25. $$ \boxed{(4)\ \text{do not intersect}} $$ <div
26. Mathematics : Mensuration : Surface Area of Right Circular Cylinder, Hollow Cylinder, Cone, Sphere and Frustum : Exercise Questions with Answers
27. Onto means both elements of B are attained. Use f(−1)=0 and f(1)=2. f(−1)= −a + b = 0 f(1)= a + b = 2 Add: 2b = 2 ⇒ b = 1. Then −a + 1 = 0 ⇒ a = 1. Answer: a = 1, b = 1.
28. $$ 1+3+5+\dots+(2n-1)=n^2 $$
29. (i) $$ $$ \boxed{\text{Infinitely many solutions}} $$ (ii) $$ $$ \boxed{\text{No solution}} $$ (iii) $$ $$ \boxed{\text{Unique solution}} $$
30. (i) g(1/2) = (1/2 − 2)/3 = (−3/2)/3 = −1/2. Then g(g(1/2)) = g(−1/2) = (−1/2 − 2)/3 = (−5/2)/3 = −5/6. (ii) (g ∘ f)(x) = g(f(x)) = (f(x) − 2)/3 = ((x + 6)/8 − 2)/3 = ((x + 6 − 16)/8)/3 = (x − 10)/24.
31. $$ AB \text{ has order } 2\times4 $$ Columns = 4. Answer $$ \boxed{(2)\ 4} $$ <div
32. Let distances from galleries be $x$ and $70-x$. Given: $$ \frac{4}{9}=\frac{x^2}{(70-x)^2} $$ Taking square root: $$ \frac23=\frac{x}{70-x} $$ $$ 3x=140-2x $$ $$ 5x=140 $$ $$ x=28 $$ Other distance: $$ 70-28=42 $$ Answer $$ \boxed{28m,\ 42m} $$
33. (Source expression corrupted in OCR text.) Note The original expression is incomplete in the provided source.
34. Given: $PQ=4.5\text{ cm}$ $\angle R=35^\circ$ Median $RG=6\text{ cm}$ Construction Steps 1. Draw $PQ=4.5\text{ cm}$. 2. Find midpoint $G$ of $PQ$. 3. At $G$, construct angle $35^\circ$. 4. Mark $RG=6\text{ cm}$. 5. Join $RP$ and $RQ$. Required triangle obtained.
35. Proof Let two consecutive positive integers be: [ n \text{ and } n+1 ] Suppose (d) is a common divisor of both. Then, [ d\mid n ] and [ d\mid(n+1) ] Therefore, [ d\mid[(n+1)-n] ] [ d\mid1 ] Thus the only common divisor is 1. Hence the integers are coprime. Proved.
36. Radius of glass: $$ R=10\text{ cm} $$ Radius of metal cylinder: $$ r=5\text{ cm} $$ Height of metal cylinder: $$ h=4\text{ cm} $$ Volume displaced: $$ V=\pi r^2h $$ $$ =\pi(5)^2(4) $$ $$ =100\pi $$ Let rise in water level be $x$. Volume rise in glass: $$ \pi R^2x $$ $$ =\pi(10)^2x $$ $$ =100\pi x $$ Equating, $$ 100\pi x=100\pi $$ $$ x=1 $$ Answer $$ 1\text{ cm} $$
37. Let $$ R=\frac{1+\sin\theta-\cos\theta}{1+\sin\theta+\cos\theta} $$ We first show that $$ R=\frac{1-\cos\theta}{\sin\theta} $$ Cross multiplication gives $$ \sin\theta(1+\sin\theta-\cos\theta) =(1-\cos\theta)(1+\sin\theta+\cos\theta) $$ The right hand side is $$ 1+\sin\theta-\sin\theta\cos\theta-\cos^2\theta =\sin^2\theta+\sin\theta-\sin\theta\cos\theta $$ $$ =\sin\theta(1+\sin\theta-\cos\theta) $$ So $$ R=\frac{1-\cos\theta}{\sin\theta} $$ Now square both sides: $$ R^2=\frac{(1-\cos\theta)^2}{\sin^2\theta} $$ Using $\sin^2\theta=1-\cos^2\theta=(1-\cos\theta)(1+\cos\theta)$, $$ R^2=\frac{(1-\cos\theta)^2}{(1-\cos\theta)(1+\cos\theta)} =\frac{1-\cos\theta}{1+\cos\theta} $$ Hence proved.
38. Mathematics : Numbers and Sequences Important Formulae For a G.P. with first term $a$ and common ratio $r$, Sum of first $n$ terms $$ S_n = \frac{a(r^n-1)}{r-1}, \quad r \ne 1 $$ or $$ S_n = \frac{a(1-r^n)}{1-r}, \quad r \ne 1 $$ Sum to infinity $$ S_\infty = \frac{a}{1-r}, \quad |r|\lt1 $$ If $r=1$, the finite sum is $S_n=na$.
39. The least-time path is the perpendicular from the junction to the line 6x − 7y + 8 = 0. Junction of 2x − 3y + 4 = 0 and 3x + 4y − 5 = 0 is (−1/17, 22/17). The required line has slope −7/6 (perpendicular to slope 6/7). Through (−1/17, 22/17): 119x + 102y − 125 = 0 .
40. Radius: $$ r=\frac32=1.5\text{ cm} $$ Each cone height: $$ 2\text{ cm} $$ Total cone height: $$ 4\text{ cm} $$ Cylinder height: $$ 12-4=8\text{ cm} $$ Volume of cylinder $$ V_1=\pi r^2h $$ $$ =\pi(1.5)^2(8) $$ $$ =18\pi $$ Volume of two cones $$ V_2=2\left(\frac13\pi r^2h\right) $$ $$ =2\left(\frac13\pi(1.5)^2(2)\right) $$ $$ =3\pi $$ Total volume $$ V=18\pi+3\pi $$ $$ =21\pi $$ $$ =66 $$ Answer $$ 66\text{ cm}^3 $$
41. Slope of QR = (4 − (−2))/(−5 − 3) = 6/(−8) = −3/4. A line through P(−5, 2) parallel to QR has the same slope: y − 2 = (−3/4)(x + 5). 4(y − 2) = −3(x + 5) ⇒ 4y − 8 = −3x − 15. Therefore 3x + 4y + 7 = 0 .

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