Samacheer Kalvi Class 10 Maths Practice Question Papers with Answers (2026-27)

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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: 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]

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

5.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]

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: (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]

8.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]

9.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]

10.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]

11.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]

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 should be added to make[1]

14.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]

15.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]

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 horizontal distance between two buildings is $70$ m.[2]

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

19.The solution of[2]

20.Area of similar triangles[2]

21.Three digit number problem[2]

22.The volumes of two cones of same base radius are $3600\text{ cm}^3$ and $5040\text{ cm}^3$.[2]

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

24.A hemi-spherical hollow bowl has material of volume $\frac{436\pi}{3}$ cubic cm. Its external diameter is 14 cm. Find its thickness.[2]

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

26.Difference between a number and its reciprocal[2]

27.If the radius of the base of a cone is tripled and its height is doubled, then the volume becomes:[2]

28.A man repays a loan of ₹65000 by paying ₹400 in the first month and increasing payment by ₹300 every month.[2]

29.Find the total surface area of a cylinder whose radius is one-third of its height.[2]

30.Sum of squares[2]

31.The solution of the system[2]

32.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]

33.If 1 + 2 + 3 + … + k = 325, then find 1^3 + 2^3 + … + k^3.[2]

34.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]

Part IV — Long Answer Questions 7 × 5 = 35

Answer in detail. (Answer all questions.)

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

36.Let A(3, −4), B(9, −4), C(5, −7) and D(7, −7). Show that ABCD is a trapezium.[5]

37.Construction: Triangle similar to triangle PQR with scale factor $7/3$[5]

38.Find the value of a for which the given points are collinear: (i) (2, 3), (4, a) and (6, −3) (ii) (a, 2−2a), (−a+1, 2a) and (−4−a, 6−2a).[5]

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

40.A $14$ m deep well with inner diameter $10$ m is dug and the earth taken out is evenly spread all around the well to form an embankment of width $5$ m.[5]

41.The volume of a cone is $1005\frac57$ cu.cm. The area of its base is $201\frac17$ sq.cm. Find the slant height of the cone.[5]

🔑 Show Answer Key — Set 1

1. 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 .
2. tan θ(cosec²θ − 1) = tan θ·cot²θ = (sin θ/cos θ)(cos²θ/sin²θ) = cos θ/sin θ = cot θ. (D) cot θ .
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. Expanding: (sin²α + cos²α) + 4 + (cosec²α + sec²α) = 1 + 4 + (2 + cot²α + tan²α) = 7 + tan²α + cot²α. So k = 7. (B) 7 .
5. (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.
6. Bijective functions have equal cardinalities. $$ n(A)=n(B)=7 $$ Correct option: (1)
7. Substituting (2, 1): x + y = 2 + 1 = 3 ✓ and 3x + y = 6 + 1 = 7 ✓. (B) x + y = 3; 3x + y = 7 .
8. Area = ½ × base × height = ½ × 10 × 5 = 25. (B) 25 sq.units .
9. 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) .
10. 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
11. 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 .
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. $$ x^4+16x^2+64=(x^2+8)^2 $$ Answer $$ \boxed{(2)\ 16x^2} $$ <div
14. 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 .
15. tan(elevation) = height/shadow = √3, so the angle = 60°. (D) 60° .
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. Let the height of the first building be $h$ m. Difference in heights: $$ 120-h $$ Using tangent ratio, $$ \tan45^\circ=\frac{120-h}{70} $$ $$ 1=\frac{120-h}{70} $$ $$ 120-h=70 $$ $$ h=50 $$ Answer $$ 50\text{ m} $$
18. 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).
19. $$ 2x-1=\pm3 $$ Case 1: $$ 2x=4 $$ $$ x=2 $$ Case 2: $$ 2x=-2 $$ $$ x=-1 $$ Answer $$ \boxed{(3)\ -1,\ 2} $$ <div
20. $$ \boxed{2.8\text{ cm}} $$
21. $$ \boxed{246} $$
22. For cones with same radius: $$ V\propto h $$ Therefore, $$ h_1:h_2=3600:5040 $$ $$ =5:7 $$ Answer $$ 5:7 $$
23. Using Pythagoras theorem: $$ d^2 = 18^2 + 24^2 $$ $$ =324+576 $$ $$ =900 $$ $$ d=\sqrt{900} $$ $$ d=30 $$ Answer $$ \boxed{30\text{ m}} $$
24. External radius: $$ R=7\text{ cm} $$ Let internal radius be $r$. Volume of material: $$ \frac23\pi(R^3-r^3) $$ $$ =\frac{436\pi}{3} $$ Cancel $\frac{\pi}{3}$: $$ 2(343-r^3)=436 $$ $$ 686-2r^3=436 $$ $$ 2r^3=250 $$ $$ r^3=125 $$ $$ r=5 $$ Thickness: $$ 7-5=2 $$ Answer $$ 2\text{ cm} $$
25. $$ \boxed{ (A^T)^T=A } $$ Verified.
26. Let the number be $x$. Given: $$ x-\frac1x=\frac{24}{5} $$ Multiply throughout by $5x$: $$ 5x^2-5=24x $$ $$ 5x^2-24x-5=0 $$ Factorize: $$ (5x+1)(x-5)=0 $$ Therefore, $$ x=5 $$ or $$ x=-\frac15 $$ Answer $$ \boxed{5,\ -\frac15} $$
27. Volume: $$ V=\frac13\pi r^2 h $$ New volume: $$ =\frac13\pi(3r)^2(2h) $$ $$ =\frac13\pi(9r^2)(2h) $$ $$ =18V $$ Answer $$ \boxed{(2)\ \text{made 18 times}} $$
28. $$ a=400,\quad d=300 $$ $$ 65000=\frac{n}{2}[800+(n-1)300] $$ $$ 130000=n(300n+500) $$ $$ 3n^2+5n-1300=0 $$ $$ (3n+65)(n-20)=0 $$ $$ n=20 $$ Answer $$ 20\text{ months} $$
29. Total surface area (TSA) of a cylinder = 2πr(h + r). Given r = h/3 so h = 3r. Thus TSA = 2πr(3r + r) = 2πr·4r = 8πr². In terms of h, r = h/3 so TSA = 8π(h²/9) = (8/9)π h². Answer: 8π r² = (8/9)π h².
30. $$ 1^2+2^2+3^2+\dots+n^2 = \frac{n(n+1)(2n+1)}{6} $$
31. 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
32. 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.
33. Sum of first k natural numbers = k(k+1)/2 = 325. Sum of cubes formula: (1^3+2^3+…+k^3) = [k(k+1)/2]^2 = 325^2 = 105625. Answer: 105625
34. (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.
35. (i) Divide throughout by 9: $$ x^2-\frac43x+\frac49=0 $$ Move constant term: $$ x^2-\frac43x=-\frac49 $$ Add square of half coefficient of $x$: $$ \left(\frac{-4/3}{2}\right)^2=\left(-\frac23\right)^2=\frac49 $$ Add $\frac49$ on both sides: $$ x^2-\frac43x+\frac49=0 $$ $$ \left(x-\frac23\right)^2=0 $$ Therefore, $$ x-\frac23=0 $$ $$ x=\frac23 $$ Repeated root. Answer $$ \boxed{x=\frac23,\ \frac23} $$ (ii) Cross multiply: $$ 5x+7=(3x+2)(x-1) $$ Expand RHS: $$ 5x+7=3x^2-3x+2x-2 $$ $$ 5x+7=3x^2-x-2 $$ Bring all terms to one side: $$ 3x^2-6x-9=0 $$ Divide by 3: $$ x^2-2x-3=0 $$ Move constant: $$ x^2-2x=3 $$ Add square of half coefficient of $x$: $$ \left(\frac{-2}{2}\right)^2=1 $$ $$ x^2-2x+1=4 $$ $$ (x-1)^2=4 $$ Take square root: $$ x-1=\pm2 $$ Hence, $$ x=3 $$ or $$ x=-1 $$ Answer $$ \boxed{x=3,\ -1} $$
36. Slope AB = (−4 − (−4))/(9 − 3) = 0; slope CD = (−7 − (−7))/(7 − 5) = 0 ⇒ AB ∥ CD. Slope BC = (−7 − (−4))/(5 − 9) = 3/4; slope AD = (−7 − (−4))/(7 − 3) = −3/4 ⇒ BC and AD are not parallel. Exactly one pair of opposite sides is parallel, so ABCD is a trapezium .
37. Construction Steps 1. Draw triangle $PQR$. 2. Draw a ray from $P$. 3. Mark 7 equal segments on the ray. 4. Join the 3rd point to $R$. 5. Through the 7th point draw a line parallel to it. 6. Extend sides to complete the construction. Required triangle obtained with scale factor: $$ \boxed{\frac73} $$ Answers Summary | Question | Answer | |---|---| | 1(i) | Not similar | | 1(ii) | Similar, $x=2.5$ | | 2 | $330\text{ m}$ | | 3 | $42\text{ m}$ | | 5 | $AE=\frac{15}{13},\ DE=\frac{36}{13}$ | | 6 | $CA=5.6\text{ cm},\ AQ=3.25\text{ cm}$ | | 8 | $EF=2.8\text{ cm}$ | | 9 | $2\text{ m}$ |
38. (i) Collinear ⇒ area 0: 2(a − (−3)) + 4((−3) − 3) + 6(3 − a) = 0 ⇒ −4a = 0 ⇒ a = 0 . (ii) Setting the area to 0 and simplifying gives 8a² + 4a − 4 = 0 ⇒ 2a² + a − 1 = 0 ⇒ (2a − 1)(a + 1) = 0 ⇒ a = 1/2 or a = −1 .
39. 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 $$ 1\text{ cm} $$
40. Radius of well: $$ r=5\text{ m} $$ Depth: $$ h=14\text{ m} $$ Volume of earth dug out: $$ V=\pi r^2h $$ $$ =\pi(5)^2(14) $$ $$ =350\pi $$ Outer radius of embankment: $$ R=5+5=10\text{ m} $$ Let height of embankment be $x$. Volume of embankment: $$ \pi(R^2-r^2)x $$ $$ =\pi(100-25)x $$ $$ =75\pi x $$ Equating volumes: $$ 75\pi x=350\pi $$ $$ x=\frac{350}{75} $$ $$ x=4.67 $$ Answer $$ 4.67\text{ m} $$
41. Volume: $$ 1005\frac57=\frac{7040}{7} $$ Base area: $$ 201\frac17=\frac{1408}{7} $$ Using: $$ V=\frac13(\text{base area})\times h $$ $$ \frac{7040}{7}=\frac13\times\frac{1408}{7}\times h $$ $$ 7040=\frac{1408h}{3} $$ $$ h=15 $$ Base area: $$ \pi r^2=\frac{1408}{7} $$ Using $\pi=\frac{22}{7}$: $$ \frac{22}{7}r^2=\frac{1408}{7} $$ $$ 22r^2=1408 $$ $$ r^2=64 $$ $$ r=8 $$ Slant height: $$ l=\sqrt{r^2+h^2} $$ $$ =\sqrt{64+225} $$ $$ =\sqrt{289} $$ $$ =17 $$ Answer $$ 17\text{ cm} $$

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

2.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]

3.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]

4.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]

5.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]

6.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]

7.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]

8.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]

9.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]

10.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]

11.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]

12.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]

13.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]

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

15.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]

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.Find $x,y,z$ if[2]

18.If $\alpha,\beta$ are roots of[2]

19.Check whether AD is bisector[2]

20.Reduce the rational expressions to lowest form[2]

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

22.In right triangle $ABC$, prove that[2]

23.Today is Tuesday. My uncle will come after 45 days.[2]

24.Find the greatest 6-digit number exactly divisible by 24, 15 and 36.[2]

25.Calculate the range of the given data.[2]

26.Find the slope of the following straight lines: (i) 5y − 3 = 0 (ii) 7x − 3 = 0[2]

27.LCM and GCD[2]

28.LCM[2]

29.If the three points (3, −1), (a, 3) and (1, −3) are collinear, find the value of a.[2]

30.Show that the function[2]

31.If matrices are given, find $AB$, $BA$ and check whether $AB=BA$[2]

32.Given the LCM and GCD of the two polynomials $p(x)$ and $q(x)$, find the unknown polynomial in the following table[2]

33.Find the standard deviation of first 21 natural numbers.[2]

34.Write the first three terms of the G.P.[2]

Part IV — Long Answer Questions 7 × 5 = 35

Answer in detail. (Answer all questions.)

35.A quadrilateral swimming pool with vertices E(−3, −5), F(6, −2), G(3, 7) and H(−6, 4) is surrounded by a concrete patio bounded by A(−4, −8), B(8, −4), C(6, 10) and D(−10, 6). Find the area of the patio.[5]

36.A function is defined by f(x) = 2x − 3. (i) Find (f(0) + f(1))/2. (ii) Find x such that f(x) = 0. (iii) Find x such that f(x) = x. (iv) Find x such that f(x) = 1/2.[5]

37.Find the image of the point (3, 8) with respect to the line x + 3y = 7, assuming the line to be a plane mirror.[5]

38.Find the equation of a straight line through the intersection of the lines 7x + 3y = 10 and 5x − 4y = 1 and parallel to the line 13x + 5y + 12 = 0.[5]

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

40.The perpendicular PS on the base QR of ΔPQR intersects QR at S such that QS = 3·SR. Prove that 2·PQ^2 = 2·PR^2 + QR^2 (or equivalently 2PQ^2 = 2PR^2 + QR^2).[5]

41.Prove that $b,a,c$ are in Arithmetic Progression[5]

🔑 Show Answer Key — Set 2

1. $$ \boxed{(2)\ (ii)\ \text{and}\ (iii)\ \text{only}} $$ <div
2. You must show both pairs of opposite sides are parallel — i.e. the slopes of two pairs of opposite sides. (B) .
3. p² − q² = (cot²θ − cosec²θ)(a² − b²) = (−1)(a² − b²) = b² − a². (B) b² − a² .
4. Slope = (a − 3)/(4 − 12) = (a − 3)/(−8) = 1/8 ⇒ a − 3 = −1 ⇒ a = 2. (D) 2 .
5. sin θ = cos θ ⇒ θ = 45°. 2 tan²45° + sin²45° − 1 = 2(1) + 1/2 − 1 = 3/2. (B) 3/2 .
6. Collinear ⇒ equal slopes: (p − 7)/(3 − 5) = (6 − 7)/(6 − 5) ⇒ (p − 7)/(−2) = −1 ⇒ p = 9. (C) 9 .
7. Writing in terms of sin/cos and simplifying gives 2. (C) 2 .
8. The distance from the Y-axis stays constant at 10, so the path is the vertical line (A) x = 10 .
9. A = (0, 8), B = (5, 0). Intercept form x/5 + y/8 = 1 ⇒ 8x + 5y = 40. (A) 8x + 5y = 40 .
10. 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 .
11. The perpendicular bisector is perpendicular to PQ, so its slope = −1 ÷ (1/3) = −3. (B) −3 .
12. 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) .
13. 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 .
14. 1/(1 + tan²θ) = 1/sec²θ = cos²θ, so the value = sin²θ + cos²θ = 1. (B) 1 .
15. 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 .
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. Let common difference be $d$. $$ y=10+d $$ $$ 24=10+3d $$ $$ 14=3d $$ $$ d=\frac{14}{3} $$ Using progression: $$ x=10-d=\frac{16}{3} $$ $$ y=10+d=\frac{44}{3} $$ $$ z=24+d=\frac{86}{3} $$ The source answer $3,17,31$ corresponds to another intended AP. Corrected Answer $$ x=\frac{16}{3},\quad y=\frac{44}{3},\quad z=\frac{86}{3} $$
18. $$ x^2-2x+3 $$ find polynomial whose roots are (i) $\alpha+2,\beta+2$ $$ \boxed{ x^2-6x+11 } $$ (ii) (Expression incomplete in source) Polynomial obtained using transformed roots.
19. (i) $$ $$ \boxed{\text{Not a bisector}} $$ (ii) $$ $$ \boxed{\text{Bisector}} $$
20. Reduced forms obtained after factorization and cancellation.
21. 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.
22. $$ \boxed{ AE=\frac{15}{13} } $$ $$ \boxed{ DE=\frac{36}{13} } $$
23. $$ 45\equiv3\pmod7 $$ 3 days after Tuesday: Wednesday → Thursday → Friday Answer $$ \text{Friday} $$
24. LCM: $$ 24=2^3\times3 $$ $$ 15=3\times5 $$ $$ 36=2^2\times3^2 $$ $$ LCM=2^3\times3^2\times5=360 $$ Largest 6-digit number: $$ 999999 $$ Largest multiple: $$ 999720 $$
25. Range: $$ =L-S $$ After identifying the largest and smallest observations from the table, $$ \text{Range}=250 $$ Answer $$ 250 $$
26. (i) 5y − 3 = 0 ⇒ y = 3/5, a horizontal line, so its slope is 0 . (ii) 7x − 3 = 0 ⇒ x = 3/7, a vertical line, so its slope is undefined .
27. | Question | LCM | GCD | |---|---|---| | 1(i) | $105x^2y^2$ | $7xy$ | | 1(ii) | $(x-1)(x+1)(x^2+x+1)(x^2-x+1)$ | $(x+1)$ | | 1(iii) | $xy(x+y)$ | $x(x+y)$ |
28. | Question | Answer | |---|---| | 2(i) | $8x^3y^2$ | | 2(ii) | $-36a^3b^2c$ | | 2(iii) | $-48m^2n^2$ | | 2(iv) | $(p-1)(p-2)(p+2)$ | | 2(v) | $4(x+3)(2x+1)(x-3)$ | | 2(vi) | $2^3x^2(2x-3y)^3(4x^2+6xy+9y^2)$ | Ex 3.3 Relationship between LCM and GCD 8 questions <div
29. Slope of (3, −1) and (1, −3) = (−3 − (−1))/(1 − 3) = −2/−2 = 1. For collinearity, slope of (3, −1) and (a, 3) must also be 1: (3 − (−1))/(a − 3) = 4/(a − 3) = 1 ⇒ a − 3 = 4. Therefore a = 7 .
30. 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.
31. $$ \boxed{ AB \neq BA } $$ (in general)
32. (i) Answer $$ \boxed{(a+2)(a-7)} $$ (ii) Answer $$ \boxed{x^2+xy+y^2} $$ Answers Summary
33. Natural numbers: $$ 1,2,3,\ldots,21 $$ Mean: $$ \bar{x}=\frac{21+1}{2}=11 $$ Using standard deviation formula, $$ \sigma=\sqrt{\frac{\sum(x-\bar{x})^2}{n}} $$ $$ \sigma\approx6.05 $$ Answer $$ 6.05 $$
34. (i) $$ $$ 6,18,54 $$ (ii) $$ $$ \sqrt2,2,2\sqrt2 $$ (iii) $$ $$ 1000,400,160 $$
35. Patio area = area of the outer quadrilateral ABCD − area of the pool EFGH. By the shoelace formula, area(ABCD) = 212 sq.units and area(EFGH) = 90 sq.units. Therefore the patio area = 212 − 90 = 122 sq.units .
36. Given f(x) = 2x − 3. (i) f(0) = 2·0 − 3 = −3, f(1) = 2·1 − 3 = −1. So (f(0)+f(1))/2 = (−3 + (−1))/2 = −4/2 = −2. (ii) Solve 2x − 3 = 0 ⇒ x = 3/2. (iii) Solve 2x − 3 = x ⇒ x = 3. (iv) Solve 2x − 3 = 1/2 ⇒ 2x = 3 + 1/2 = 7/2 ⇒ x = 7/4. Answers: (i) −2, (ii) 3/2, (iii) 3, (iv) 7/4.
37. For line x + 3y − 7 = 0, compute d = (3 + 3·8 − 7)/(1² + 3²) = 20/10 = 2. Image = (3 − 2·1·2, 8 − 2·3·2) = (−1, −4) .
38. Solving 7x + 3y = 10 and 5x − 4y = 1: 4(7x + 3y) + 3(5x − 4y) = 40 + 3 ⇒ 43x = 43 ⇒ x = 1, then y = 1. Intersection (1, 1). A line parallel to 13x + 5y + 12 = 0 is 13x + 5y + c = 0. Through (1, 1): 13 + 5 + c = 0 ⇒ c = −18. Therefore 13x + 5y − 18 = 0 .
39. (i) Find the GCD of Factor the first polynomial: $$ 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-2) $$ Factor second polynomial: $$ 2x(2x^3+7x^2+4x-4) $$ $$ =2x(x+2)^2(2x-1) $$ Common factors: $$ x(x+2)^2 $$ Answer $$ \boxed{x(x^2+4x+4)} $$ (iv) Find the GCD of Factor first polynomial: $$ 3(x^3+x^2+x+1) $$ $$ =3(x+1)(x^2+1) $$ Factor second polynomial: $$ 6(x^3+2x^2+x+2) $$ $$ =6(x+2)(x^2+1) $$ Common factor: $$ 3(x^2+1) $$ Answer $$ \boxed{3(x^2+1)} $$
40. Let SR = x, so QS = 3x and QR = 4x. Using Pythagoras in the right triangles: PQ^2 = PS^2 + (3x)^2 = PS^2 + 9x^2 PR^2 = PS^2 + x^2 Then 2·PR^2 = 2PS^2 + 2x^2 and 2·PQ^2 = 2PS^2 + 18x^2. Subtracting gives 2·PQ^2 − 2·PR^2 = 16x^2 = (4x)^2 = QR^2, so 2·PQ^2 = 2·PR^2 + QR^2 .
41. For equal roots: $$ D=0 $$ $$ (b-c)^2-4(a-b)(c-a)=0 $$ Expand: $$ b^2-2bc+c^2-4(ac-a^2-bc+ab)=0 $$ $$ b^2-2bc+c^2-4ac+4a^2+4bc-4ab=0 $$ $$ 4a^2+b^2+c^2+2bc-4ab-4ac=0 $$ This simplifies to: $$ (2a-b-c)^2=0 $$ Therefore, $$ 2a=b+c $$ Hence, $$ a=\frac{b+c}{2} $$ So $a$ is the arithmetic mean of $b$ and $c$. Thus $b,a,c$ are in A.P. Result $$ \boxed{b,a,c\text{ are in arithmetic progression}} $$

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

2.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]

3.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]

4.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]

5.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]

6.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]

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

8.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]

9.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]

10.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]

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

12.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]

13.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]

14.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]

15.If there exists a bijection (a one-to-one and onto function) f : A → B and n(A) = 7, what is n(B)?[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 girl wishes to prepare birthday caps in the form of right circular cones for her birthday party, using a sheet of paper whose area is $5720\text{ cm}^2$.[2]

18.A has ‘a’ rows and ‘a+3’ columns.[2]

19.State whether the following functions are bijective.[2]

20.In the G.P.[2]

21.What is the smallest number that when divided by 35, 56 and 91 leaves remainder 7 in each case?[2]

22.Let A = {1,2,3,4,...,45} and let R be the relation “square of a number” on A. Write R as a subset of A × A. Also find the domain and range of R.[2]

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

24.Find the square root of[2]

25.The ratio of the 6th and 8th terms of an A.P. is $7:9$.[2]

26.Show that the matrices satisfy commutative property[2]

27.If S1, S2, ..., Sm are the sums of n terms of m A.P.'s whose first terms are 1, 2, 3, ..., m and whose common differences are 1, 3, 5, ..., (2m−1) respectively, find S_m, the sum of n terms of the m-th A.P.[2]

28.Boat and stream problem[2]

29.Solve 5x ≡ 4 (mod 6).[2]

30.A relation R is given by R = {(x,y) | y = x + 3, x ∈ {0,1,2,3,4,5}}. Determine its domain and range.[2]

31.In the matrix $A$, write[2]

32.The square root of (expression incomplete in source) is equal to[2]

33.In $\triangle LMN$,[2]

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

Part IV — Long Answer Questions 7 × 5 = 35

Answer in detail. (Answer all questions.)

35.Determine the quadratic equations whose sum and product of roots are given[5]

36.Find the sum of first $n$ terms of the G.P.[5]

37.An aeroplane at an altitude of $1800$ m finds that two boats are sailing towards it in the same direction.[5]

38.Find $A \times B$, $A \times A$ and $B \times A$[5]

39.Find the sum to $n$ terms of the series[5]

40.Brick staircase problem.[5]

41.Find the sums.[5]

🔑 Show Answer Key — Set 3

1. (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.
2. Slope of the join = 8/(−8) = −1, so the perpendicular slope is 1. (B) 1 .
3. x = 60(cot30° − cot45°) = 60(√3 − 1) ≈ 60(0.732) = 43.92 m. (B) 43.92 m .
4. x = 11 is a vertical line, so it is (B) parallel to the Y-axis .
5. 25x² − 25/y² = (5x)² − (5/y)² = sec²θ − tan²θ = 1. (B) 1 .
6. Adding the equations: 4x = 12 ⇒ x = 3, then y = 5. (C) (3, 5) .
7. (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.
8. tan²θ = x²/a² and sec²θ = y²/b². Since sec²θ − tan²θ = 1, we get y²/b² − x²/a² = 1. (A) y²/b² − x²/a² = 1 .
9. A trapezium has exactly one pair of parallel sides — i.e. two parallel and two non-parallel sides. (B) .
10. 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 .
11. tan θ(cosec²θ − 1) = tan θ·cot²θ = (sin θ/cos θ)(cos²θ/sin²θ) = cos θ/sin θ = cot θ. (D) cot θ .
12. (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).
13. Expanding: (sin²α + cos²α) + 4 + (cosec²α + sec²α) = 1 + 4 + (2 + cot²α + tan²α) = 7 + tan²α + cot²α. So k = 7. (B) 7 .
14. (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.
15. Bijective functions have equal cardinalities. $$ n(A)=n(B)=7 $$ Correct option: (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. Slant height: $$ l=\sqrt{5^2+12^2} $$ $$ =\sqrt{169}=13 $$ Curved surface area of one cap: $$ \pi rl $$ $$ =\frac{22}{7}\times5\times13 $$ $$ =\frac{1430}{7} $$ Number of caps: $$ \frac{5720}{1430/7} $$ $$ =28 $$ Answer $$ 28\text{ caps} $$
18. Order of $A$: $$ a\times(a+3) $$ Order of $B$: $$ b\times(17-b) $$ For $AB$ to exist: $$ a+3=b $$ For $BA$ to exist: $$ 17-b=a $$ Substitute: $$ 17-(a+3)=a $$ $$ 14-a=a $$ $$ 2a=14 $$ $$ a=7 $$ Then: $$ b=a+3=10 $$ Answer $$ \boxed{a=7,\ b=10} $$
19. (i) The function is linear and every real number has a unique pre-image. Hence it is both one-one and onto. Answer Bijective function. (ii) $$ $$ f(1)=f(-1) $$ Hence it is not one-one. Also range does not cover all real numbers. Answer Not bijective.
20. $$ a=729,\quad r=\frac13 $$ $$ t_n=ar^{n-1} $$ $$ t_7=729\left(\frac13\right)^6 $$ $$ =\frac{729}{729} $$ $$ =1 $$ Answer $$ 1 $$
21. Required number: $$ LCM(35,56,91)+7 $$ Prime factors: $$ 35=5\times7 $$ $$ 56=2^3\times7 $$ $$ 91=7\times13 $$ $$ LCM=2^3\times5\times7\times13=3640 $$ Hence, $$ 3640+7=3647 $$
22. Squares ≤ 45 are: 1, 4, 9, 16, 25, 36. Hence R = {(1,1),(2,4),(3,9),(4,16),(5,25),(6,36)}. Domain: {1,2,3,4,5,6}. Range: {1,4,9,16,25,36}.
23. $$ x^2+1=0 $$ $$ x^2=-1 $$ No real number satisfies this equation. Graph does not intersect x-axis. Answer $$ \boxed{\text{No real roots}} $$ <div
24. Recognize perfect square: $$ (17x^2-18x+19)^2 $$ Answer $$ \boxed{ 17x^2-18x+19 } $$
25. $$ \frac{a+5d}{a+7d}=\frac79 $$ $$ 9a+45d=7a+49d $$ $$ 2a=4d $$ $$ a=2d $$ Now, $$ t_9=a+8d=10d $$ $$ t_{13}=a+12d=14d $$ Ratio: $$ 10:14=5:7 $$ Answer $$ 5:7 $$
26. $$ \boxed{ AB=BA } $$ Hence proved. Answers Summary | Question | Answer | |---|---| | 1 | $P\times R,\ \text{not defined}$ | | 2 | $AB=p\times r$, $BA=q\times q$ (if defined) | | 3 | $a=7,\ b=10$ | | 4 | $AB\neq BA$ generally | | 5 | $A(B+C)=AB+AC$ | | 6 | $AB=BA$ | <div
27. For the m-th A.P.: first term a = m and common difference d = 2m − 1. The sum of n terms is S_m = (n/2)[2a + (n−1)d] = (n/2)[2m + (n−1)(2m−1)]. Simplifying: S_m = (n/2)(2mn − n + 1).
28. $$ \boxed{16\text{ km/hr}} $$
29. We have 5 ≡ −1 (mod 6). So 5x ≡ 4 (mod 6) is equivalent to −x ≡ 4 (mod 6), i.e. x ≡ −4 ≡ 2 (mod 6). General solution: x = 2 + 6n , where n ∈ ℤ. (Particular values: 2, 8, 14, ...)
30. Compute y = x + 3 for each x: (0,3), (1,4), (2,5), (3,6), (4,7), (5,8) Domain: {0, 1, 2, 3, 4, 5} Range: {3, 4, 5, 6, 7, 8}
31. (i) The number of elements $$ \boxed{16} $$ (ii) The order of the matrix $$ \boxed{4 \times 4} $$ (iii) Write the elements $$ \boxed{ \sqrt7,\ \frac{\sqrt3}{2},\ 5,\ 0,\ -11,\ 1 } $$
32. $$ \boxed{(4)} $$ <div
33. $$ \angle N=180^\circ-(60^\circ+50^\circ) $$ $$ =70^\circ $$ Since corresponding angles are equal, $$ \angle R=70^\circ $$ Answer $$ \boxed{(2)\ 70^\circ} $$ <div
34. $$ (f\circ g)(x)=2\left(\frac{x}{3}\right)^2 $$ $$ =\frac{2x^2}{9} $$ Correct option: (3)
35. (i) Sum of roots $= -9$ Using: $$ x^2-Sx+P=0 $$ $$ x^2-(-9)x+20=0 $$ $$ x^2+9x+20=0 $$ Answer $$ \boxed{x^2+9x+20=0} $$ (ii) Sum of roots $= \frac{5}{3}$ $$ x^2-\frac{5}{3}x+4=0 $$ Multiply throughout by 3: $$ 3x^2-5x+12=0 $$ Answer $$ \boxed{3x^2-5x+12=0} $$ (iii) Sum of roots $= -\frac{3}{2}$ $$ x^2+\frac{3}{2}x-1=0 $$ Multiply by 2: $$ 2x^2+3x-2=0 $$ Answer $$ \boxed{2x^2+3x-2=0} $$ (iv) Sum of roots Quadratic equation: $$ x^2-(\text{sum})x+(\text{product})=0 $$ $$ x^2+(2-a)^2x+(a+5)^2=0 $$ Expanding: $$ (2-a)^2=a^2-4a+4 $$ Hence, $$ x^2+(a^2-4a+4)x+(a+5)^2=0 $$ Answer $$ \boxed{x^2+(a^2-4a+4)x+(a+5)^2=0} $$
36. (i) $$ $$ a=5,\quad r=-\frac35 $$ $$ S_n=\frac{a(1-r^n)}{1-r} $$ $$ S_n= \frac{ 5\left(1-\left(-\frac35\right)^n\right) }{ 1+\frac35 } $$ $$ S_n= \frac{25}{8} \left( 1-\left(-\frac35\right)^n \right) $$ Answer $$ \boxed{ S_n= \frac{25}{8} \left( 1-\left(-\frac35\right)^n \right) } $$ (ii) $$ $$ a=256,\quad r=\frac14 $$ $$ S_n= \frac{ 256\left(1-\left(\frac14\right)^n\right) }{ 1-\frac14 } $$ $$ S_n= \frac{1024}{3} \left( 1-\left(\frac14\right)^n \right) $$ Answer $$ \boxed{ S_n= \frac{1024}{3} \left( 1-\left(\frac14\right)^n \right) } $$
37. Let distances of boats from the point vertically below the aeroplane be $x$ and $y$. For first boat: $$ \tan60^\circ=\frac{1800}{x} $$ $$ 1.732=\frac{1800}{x} $$ $$ x\approx1039.2 $$ For second boat: $$ \tan30^\circ=\frac{1800}{y} $$ $$ \frac1{1.732}=\frac{1800}{y} $$ $$ y\approx3117.6 $$ Distance between boats: $$ 3117.6-1039.2 $$ $$ =2078.4 $$ Answer $$ 2078.4\text{ m} $$
38. (i) $A=\{2,-2,3\}$, $B=\{1,-4\}$ $$ A \times B = \{(2,1),(2,-4),(-2,1),(-2,-4),(3,1),(3,-4)\} $$ $$ A \times A = \{(2,2),(2,-2),(2,3),(-2,2),(-2,-2),(-2,3),(3,2),(3,-2),(3,3)\} $$ $$ B \times A = \{(1,2),(1,-2),(1,3),(-4,2),(-4,-2),(-4,3)\} $$ (ii) $A=B=\{p,q\}$ $$ A \times B = \{(p,p),(p,q),(q,p),(q,q)\} $$ $$ A \times A = \{(p,p),(p,q),(q,p),(q,q)\} $$ $$ B \times A = \{(p,p),(p,q),(q,p),(q,q)\} $$ (iii) $A=\{m,n\}, B=\phi$ $$ A \times B = \phi $$ $$ A \times A = \{(m,m),(m,n),(n,m),(n,n)\} $$ $$ B \times A = \phi $$
39. (i) $$ $$ 0.4=\frac4{10} $$ $$ 0.44=\frac{44}{100} $$ General sum: $$ S_n= \frac49 \left( n-\frac{1}{9}(1-10^{-n}) \right) $$ Answer $$ \boxed{ S_n= \frac49 \left( n-\frac{1}{9}(1-10^{-n}) \right) } $$ (ii) $$ $$ 3+33+333+\dots = 3(1+11+111+\dots) $$ Using GP decomposition: $$ S_n= \frac1{27} \left( 10^{n+1}-9n-10 \right) $$ Answer $$ \boxed{ S_n= \frac1{27} \left( 10^{n+1}-9n-10 \right) } $$
40. Bottom step requires 100 bricks. Each successive step requires 2 bricks less. Total steps: $$ 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 $$
41. (i) $$ $$ a=3,\quad d=4,\quad n=40 $$ $$ S_n=\frac{n}{2}[2a+(n-1)d] $$ $$ S_{40}=\frac{40}{2}[6+39(4)] $$ $$ =20(162) $$ $$ =3240 $$ Answer $$ 3240 $$ (ii) $$ $$ a=102,\quad d=-5,\quad n=27 $$ $$ S_{27}=\frac{27}{2}[204+26(-5)] $$ $$ =\frac{27}{2}(74) $$ $$ =999 $$ Answer $$ 999 $$ (iii) $$ $$ a=6,\quad d=7,\quad l=97 $$ Find number of terms: $$ 97=6+(n-1)7 $$ $$ 91=7(n-1) $$ $$ n=14 $$ Now, $$ S_{14}=\frac{14}{2}(6+97) $$ $$ =7(103) $$ $$ =721 $$ Answer $$ 721 $$

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