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Samacheer Kalvi Class 9 Maths Practice Question Papers

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Brain Grain · braingrain.in
Maths — Practice Paper · Set 1
Class: 9Samacheer KalviMax Marks: 91
Name: ____________________Reg No: ____________
Part I — Multiple Choice Questions 15 × 1 = 15

Choose the correct answer. (Answer all questions.)

1.Find the value of (a) and (b) if[1]
2.Which of the following are sets?[1]
3.Which of the following is not a linear equation in two variables?[1]
4.Which of the following points lie in the fourth quadrant? Q(3,−4) and R(1,−1).[1]
5.Which one of the following is an irrational number?[1]
6.Verify the formula n(A ∪ B ∪ C) = n(A) + n(B) + n(C) − n(A ∩ B) − n(B ∩ C) − n(A ∩ C) + n(A ∩ B ∩ C) for A = {a,c,e,f,h}, B = {c,d,e,f} and C = {a,b,c,f}.[1]
7.Which of the following is a solution of (2x-y=6)[1]
8.Find the value of 3 sin 70° sec 20° + 2 sin 49° sec 51°.....[1]
9.Given A = {a, c, e, f, h}, B = {c, d, e, f}, C = {a, b, c, f}. Verify that n(A ∪ B ∪ C) = n(A) + n(B) + n(C) − n(A ∩ B) − n(B ∩ C) − n(A ∩ C) + n(A ∩ B ∩ C).[1]
10.Which of the following is not true?[1]
11.Given A = {1, 3, 5}, B = {2, 3, 5, 6}, C = {1, 5, 6, 7}. Verify that n(A ∪ B ∪ C) = n(A) + n(B) + n(C) − n(A ∩ B) − n(B ∩ C) − n(A ∩ C) + n(A ∩ B ∩ C).[1]
12.Verify n(A ∪ B) = n(A) + n(B) − n(A ∩ B) for U = {x : x ∈ N, x ≤ 10}, A = {2,3,4,8,10} and B = {1,2,5,8,10}.[1]
13.Without actual division, find which of the following rational numbers have terminating decimal expansion[1]
14.Which one of the following is not a rational number?[1]
15.Which one of the following regarding the sum of two irrational numbers is true?[1]
Part II — Fill in the Blanks 5 × 1 = 5

Fill in the blanks. (Answer all questions.)

16.Zero of ((2-3x)) is ___________[1]
17.Cubic polynomial may have maximum of ___________ linear factors[1]
18.If a1/a2 = b1/b2 ≠ c1/c2, then the pair has _________ solution(s). Options: (1) No solution; (2) Two solutions; (3) Infinite; (4) Unique.[1]
19.If a1/a2 ≠ b1/b2, then the pair has _________ solution(s). Options: (1) No solution; (2) Two solutions; (3) Unique; (4) Infinite.[1]
20.If (p(a)=0), then ((x-a)) is a ___________ of (p(x))[1]
Part III — Short Answer Questions 18 × 2 = 36

Answer briefly. (Answer all questions.)

21.Represent the following numbers in scientific notation[2]
22.Evaluate using identity[2]
23.Distance between (5,−1) and origin[2]
24.The mid-points of the sides of a triangle are (3/2, 5), (7, −9/2) and (13/2, −13/2). Find the centroid of the triangle.[2]
25.Value of (k) for parallel lines[2]
26.Write the following numbers in decimal form[2]
27.Find the cardinal number of the following sets.[2]
28.Number problem: The middle digit of a three-digit number is zero. If the sum of the hundreds and units digits is 13 and the number obtained by reversing the digits exceeds the original number by 495, find the number.[2]
29.Find the mode of: 3.1, 3.2, 3.3, 2.1, 1.3, 3.3, 3.1[2]
30.In a circle with centre O, ∠ABC = 120°. Find ∠OAC.[2]
31.If ((2,3)) is a solution of (2x+3y=k), then (k=)[2]
32.Write the abscissa and ordinate from Fig. 5.11[2]
33.Area of rectangle[2]
34.Find the values correct to 3 decimal places[2]
35.Show that points form a parallelogram[2]
36.Value of tan1°·tan2°·tan3°·…·tan89° ?[2]
37.PQ and RS are equal chords with centre O and ∠POQ = 70°.[2]
38.Use a fractional index to write[2]
Part IV — Long Answer Questions 7 × 5 = 35

Answer in detail. (Answer all questions.)

39.Factorise the following: (i) x^3 + 8y^3 + 6xy - 1 (ii) l^3 - 8m^3 - 27n^3 - 18lmn[5]
40.Find the centroid of the triangle[5]
41.Simplify using addition and subtraction properties of surds[5]
42.Find the distance between the following pairs of points[5]
43.Monthly income problem[5]
44.In a circle with centre O, ∠CAB = 25°. Find ∠BDC, ∠DBA and ∠COB.[5]
45.Represent the following irrational numbers on the number line[5]
🔑 Show Answer Key — Set 1
  1. 1. $$\frac{\sqrt7-2}{\sqrt7+2}=a\sqrt7+b$$ Solution Rationalise the denominator. $$\frac{\sqrt7-2}{\sqrt7+2} \times \frac{\sqrt7-2}{\sqrt7-2}$$ Using: genui{"math_block_widget_always_prefetch_v2":{"content":"(a+b)(a-b)=a^2-b^2"}} Numerator $$(\sqrt7-2)^2$$ $$=7+4-4\sqrt7$$ $$=11-4\sqrt7$$ Denominator $$(\sqrt7+2)(\sqrt7-2)$$ $$=7-4$$ $$=3$$ Therefore $$\frac{\sqrt7-2}{\sqrt7+2} ========================= \frac{11-4\sqrt7}{3}$$ [ -\frac43\sqrt7+\frac{11}{3} ] Comparing with: $$a\sqrt7+b$$ we get: $$a=-\frac43$$ $$b=\frac{11}{3}$$ Answer $$a=-\frac43,\qquad b=\frac{11}{3}$$
  2. 2. (i) The collection of prime numbers up to 100. ✓ Set (ii) The collection of rich people in India. ✕ Not a set (iii) The collection of all rivers in India. ✓ Set (iv) The collection of good Hockey players. ✕ Not a set
  3. 3. Options: 1. (ax+by+c=0) 2. (0x+0y+c=0) 3. (0x+by+c=0) 4. (ax+0y+c=0) Answer $$\boxed{(2)\ 0x+0y+c=0}$$
  4. 4. Fourth quadrant points have coordinates (+, −). Both Q(3,−4) and R(1,−1) have positive x and negative y, so both lie in the fourth quadrant. Answer: Q and R.
  5. 5. 1. (\sqrt{25}) 2. (\sqrt{\frac94}) 3. (\frac7{11}) 4. (\pi) Solution $$\sqrt{25}=5$$ $$\sqrt{\frac94}=\frac32$$ $$\frac7{11}$$ are rational numbers. $$\pi$$ is irrational. Answer $$\boxed{(4)\ \pi}$$
  6. 6. $$n(A\cup B\cup C)$$ $$=n(A)+n(B)+n(C)$$ $$-n(A\cap B)-n(B\cap C)-n(A\cap C)$$ $$+n(A\cap B\cap C)$$
  7. 7. Options: 1. ((2,4)) 2. ((4,2)) 3. ((3,-1)) 4. ((0,6)) Solution For ((4,2)): $$2(4)-2=8-2=6$$ Answer $$\boxed{(2)\ (4,2)}$$
  8. 8. As printed: approximately 5.39849, so none of the options is correct. If sin 49° is the textbook typo and sin 39° was intended, the answer is (3) 5.
  9. 9. Given sets: n(A) = 5 n(B) = 4 n(C) = 4 Intersections: A ∩ B = {c, e, f} ⇒ n(A ∩ B) = 3 B ∩ C = {c, f} ⇒ n(B ∩ C) = 2 A ∩ C = {a, c, f} ⇒ n(A ∩ C) = 3 A ∩ B ∩ C = {c, f} ⇒ n(A ∩ B ∩ C) = 2 Compute RHS: n(A)+n(B)+n(C) − n(A∩B) − n(B∩C) − n(A∩C) + n(A∩B∩C) = 5 + 4 + 4 − 3 − 2 − 3 + 2 = 7 Union: A ∪ B ∪ C = {a, b, c, d, e, f, h} ⇒ n(A ∪ B ∪ C) = 7 Hence both sides equal 7; the formula is verified.
  10. 10. 1. Every rational number is a real number. 2. Every integer is a rational number. 3. Every real number is an irrational number. 4. Every natural number is a whole number. Solution Real numbers include both rational and irrational numbers. Hence statement (3) is false. Answer $$\boxed{(3)\ \text{Every real number is an irrational number}}$$
  11. 11. Given: n(A) = 3 n(B) = 4 n(C) = 4 Intersections: A ∩ B = {3, 5} ⇒ n(A ∩ B) = 2 B ∩ C = {5, 6} ⇒ n(B ∩ C) = 2 A ∩ C = {1, 5} ⇒ n(A ∩ C) = 2 A ∩ B ∩ C = {5} ⇒ n(A ∩ B ∩ C) = 1 Compute RHS: 3 + 4 + 4 − 2 − 2 − 2 + 1 = 6 Union: A ∪ B ∪ C = {1, 2, 3, 5, 6, 7} ⇒ n(A ∪ B ∪ C) = 6 Therefore the formula is verified.
  12. 12. $$n(A\cup B)=n(A)+n(B)-n(A\cap B)$$ Given: $$U={x:x\in N,\ x\le10}$$ $$A={2,3,4,8,10}$$ $$B={1,2,5,8,10}$$ Step 1: Find (n(A)) $$n(A)=5$$ Step 2: Find (n(B)) $$n(B)=5$$ Step 3: Find (A\cap B) $$A\cap B={2,8,10}$$ $$n(A\cap B)=3$$ Step 4: Find (A\cup B) $$A\cup B={1,2,3,4,5,8,10}$$ $$n(A\cup B)=7$$ Verification RHS: $$n(A)+n(B)-n(A\cap B)$$ $$=5+5-3$$ $$=7$$ LHS: $$n(A\cup B)=7$$ Thus, $$n(A\cup B)=n(A)+n(B)-n(A\cap B)$$ Verified.
  13. 13. A rational number has a terminating decimal expansion if the denominator in simplest form contains only the prime factors: $$2 \text{ and/or } 5$$ (i) (\frac{7}{128}) $$128=2^7$$ Only factor (2) is present. Answer $$\frac{7}{128}$$ has a terminating decimal expansion . (ii) (\frac{21}{15}) Simplify: $$\frac{21}{15}=\frac75$$ Denominator: $$5$$ Only factor (5). Answer $$\frac{21}{15}$$ has a terminating decimal expansion . (iii) (4\frac{9}{35}) Convert fractional part: $$\frac{9}{35}$$ Denominator: $$35=5\times7$$ Factor (7) is present. Answer $$4\frac{9}{35}$$ has a non-terminating recurring decimal expansion . (iv) (\frac{219}{2200}) Simplify: $$\frac{219}{2200}$$ Factorize denominator: $$2200=2^3\times5^2\times11$$ Factor (11) is present. Answer $$\frac{219}{2200}$$ has a non-terminating recurring decimal expansion .
  14. 14. 1. (\sqrt{\frac8{18}}) 2. (\frac73) 3. (\sqrt{0.01}) 4. (\sqrt{13}) Solution $$\sqrt{\frac8{18}} ================= # \sqrt{\frac49} \frac23$$ rational. $$\frac73$$ rational. $$\sqrt{0.01} =========== \frac1{10}$$ rational. $$\sqrt{13}$$ irrational. Answer $$\boxed{(4)\ \sqrt{13}}$$
  15. 15. 1. always an irrational number 2. may be a rational or irrational number 3. always a rational number 4. always an integer Solution Example: $$\sqrt2+\sqrt3$$ is irrational. But, $$\sqrt2+(-\sqrt2)=0$$ is rational. Answer $$\boxed{(2)\ \text{may be a rational or irrational number}}$$
  16. 16. Options: 1. (3) 2. (2) 3. (\frac23) 4. (\frac32) Solution $$2-3x=0$$ $$3x=2$$ $$x=\frac23$$ Answer $$\boxed{(3)\ \frac23}$$
  17. 17. Options: 1. 1 2. 2 3. 3 4. 4 Answer $$\boxed{(3)\ 3}$$
  18. 18. No solution. Reason: If the ratios of coefficients of x and y are equal but differ from the ratio of constants, the lines are parallel and distinct — there is no solution.
  19. 19. Unique (one solution). Reason: For a pair of linear equations, if the ratios of coefficients a1/a2 and b1/b2 are not equal, the lines intersect at a single point — a unique solution.
  20. 20. Options: 1. Divisor 2. Quotient 3. Remainder 4. Factor Answer $$\boxed{(4)\ \text{Factor}}$$
  21. 21. (i) (569430000000) Move decimal point after first digit. $$569430000000=5.6943\times10^{11}$$ Answer $$5.6943\times10^{11}$$ (ii) (2000.57) $$2000.57=2.00057\times10^3$$ Answer $$2.00057\times10^3$$ (iii) (0.0000006000) Move decimal 7 places right. $$0.0000006000=6.000\times10^{-7}$$ Answer $$6.000\times10^{-7}$$ (iv) (0.0009000002) $$0.0009000002=9.000002\times10^{-4}$$ Answer $$9.000002\times10^{-4}$$
  22. 22. (i) $$7^3-10^3+3^3$$ Since: $$7-10+3=0$$ Using identity: $$a^3+b^3+c^3=3abc \quad\text{if }a+b+c=0$$ $$=3(7)(-10)(3)$$ $$=-630$$ Answer $$-630$$ (ii) $$1+\frac18-\frac{27}{8}$$ $$=1^3+\left(\frac12\right)^3-\left(\frac32\right)^3$$ Since: $$1+\frac12-\frac32=0$$ Using identity: $$=3\left(1\right)\left(\frac12\right)\left(-\frac32\right)$$ $$=-\frac94$$ Answer $$-\frac94$$
  23. 23. \[ d = \sqrt{5^2+(-1)^2} \] \[ = \sqrt{25+1} \] \[ = \sqrt{26} \] ✓ Answer: \[ \boxed{(3)\ \sqrt{26}} \]
  24. 24. Centroid = average of the three mid-points: x = (3/2 + 7 + 13/2)/3 = 5, y = (5 − 9/2 − 13/2)/3 = −2. Answer: (5, −2).
  25. 25. $$4x+6y-1=0$$ $$2x+ky-7=0$$ Options: 1. (k=3) 2. (k=2) 3. (k=4) 4. (k=-3) Solution For parallel lines: $$\frac{a_1}{a_2}=\frac{b_1}{b_2}$$ $$\frac42=\frac6k$$ $$2=\frac6k$$ $$k=3$$ Answer $$\boxed{(1)\ k=3}$$
  26. 26. (i) (3.459\times10^6) Move decimal 6 places right. $$=3459000$$ Answer $$3459000$$ (ii) (5.678\times10^4) $$=56780$$ Answer $$56780$$ (iii) (1.00005\times10^{-5}) Move decimal 5 places left. $$=0.0000100005$$ Answer $$0.0000100005$$ (iv) (2.530009\times10^{-7}) $$=0.0000002530009$$ Answer $$0.0000002530009$$
  27. 27. (i) $$M = {p,q,r,s,t,u}$$ $$n(M)=6$$ (ii) $$P = {x : x=3n+2,\ n\in W,\ x<15}$$ Values: $$2,5,8,11,14$$ $$n(P)=5$$ (iii) $$Q = \left\{y : y=\frac{4}{3n},\ n\in\mathbb{N},\ 2<n\le5\right\}$$ Values: $$\left\{\frac49,\frac13,\frac4{15}\right\}$$ $$n(Q)=3$$ (iv) $$R={x:x\in\mathbb{Z}, -5\le x<5}$$ Elements: $${-5,-4,-3,-2,-1,0,1,2,3,4}$$ $$n(R)=10$$ (v) Leap years between 1882 and 1906: $$1884,1888,1892,1896,1904$$ $$n(S)=5$$
  28. 28. Let the number be 100x + y with x + y = 13 and 100y + x - (100x + y) = 495 ⇒ y - x = 5. Solving x+y=13 and y-x=5 gives y=9, x=4. The number is 409 .
  29. 29. Arrange and count frequencies: 1.3 → 1 2.1 → 1 3.1 → 2 3.2 → 1 3.3 → 2 Both 3.1 and 3.3 occur most frequently. Mode: 3.1 and 3.3 (bimodal)
  30. 30. Angle at centre is twice angle at circumference. Since angle subtended by major arc: \[ \angle AOC = 2 \times (360^\circ -120^\circ)/2 \] Minor central angle: \[ \angle AOC = 120^\circ \] Triangle AOC is isosceles. Thus: \[ \angle OAC = \frac{180^\circ -120^\circ}{2} \] \[ = 30^\circ \] ✓ \(\angle OAC = 30^\circ\)
  31. 31. Options: 1. (12) 2. (6) 3. (0) 4. (13) Solution $$k=2(2)+3(3)$$ $$=4+9$$ $$=13$$ Answer $$\boxed{(4)\ 13}$$
  32. 32. Definitions Abscissa = x-coordinate Ordinate = y-coordinate ✓ From the graph: Read horizontal value → Abscissa Read vertical value → Ordinate > Exact values require Fig. 5.11. <div
  33. 33. Area: $$x^2+7x+12$$ Breadth: $$x+3$$ Find length. Solution $$\text{Length} ============= \frac{x^2+7x+12}{x+3}$$ Factorise numerator: $$x^2+7x+12=(x+3)(x+4)$$ Thus, $$\text{Length}=x+4$$ Answer $$x+4$$
  34. 34. Given: $$\sqrt2=1.414$$ $$\sqrt3=1.732$$ $$\sqrt5=2.236$$ $$\sqrt{10}=3.162$$ (i) (\sqrt{40}-\sqrt{20}) Step 1: Simplify $$\sqrt{40}=\sqrt{4\times10}=2\sqrt{10}$$ $$\sqrt{20}=\sqrt{4\times5}=2\sqrt5$$ Step 2: Substitute values $$2(3.162)-2(2.236)$$ $$=6.324-4.472$$ $$=1.852$$ Answer $$1.852$$ (ii) (\sqrt{300}+\sqrt{90}-\sqrt8) Step 1: Simplify $$\sqrt{300}=10\sqrt3$$ $$\sqrt{90}=3\sqrt{10}$$ $$\sqrt8=2\sqrt2$$ Step 2: Substitute values $$10(1.732)+3(3.162)-2(1.414)$$ $$=17.320+9.486-2.828$$ $$=23.978$$ Answer $$23.978$$
  35. 35. (i) A(–3,1), B(–6,–7), C(3,–9), D(6,–1) Using distance formula: \[ AB=CD \] \[ BC=AD \] Opposite sides are equal. ✓ Hence ABCD is a parallelogram. (ii) A(–7,–3), B(5,10), C(15,8), D(3,–5) Similarly: \[ AB=CD \] \[ BC=AD \] ✓ Hence ABCD is a parallelogram.
  36. 36. Pair terms tan A · tan(90°−A) = tan A · cot A = 1. Pairs (1°,89°), (2°,88°), … multiply to 1. Middle term tan45° = 1. Product = 1.
  37. 37. Find ∠ORS. Equal chords subtend equal angles. In isosceles triangle ORS: \[ \angle ROS = 70^\circ \] Remaining angle sum: \[ \angle ORS = \frac{180^\circ-70^\circ}{2} \] \[ =55^\circ \] ✓ Answer: \[ \boxed{(3)\ 55^\circ} \]
  38. 38. (i) (\sqrt5) Solution $$\sqrt5=5^{\frac12}$$ Answer $$5^{\frac12}$$ (ii) (\sqrt[2]{7}) Solution $$\sqrt7=7^{\frac12}$$ Answer $$7^{\frac12}$$ (iii) ((\sqrt[3]{49})^5) Solution $$(\sqrt[3]{49})^5 ================ \left(49^{\frac13}\right)^5$$ Using: $$(a^m)^n=a^{mn}$$ $$=49^{\frac53}$$ Answer $$49^{\frac53}$$ (iv) The expression is missing in the question provided. Please share the complete expression.
  39. 39. (i) Using a^3 + b^3 + c^3 - 3abc = (a+b+c)(a^2+b^2+c^2-ab-bc-ca) with a = x, b = 2y, c = -1, (x + 2y - 1)(x^2 + 4y^2 + 1 - 2xy + 2y + x) (ii) Using the same identity with a = l, b = -2m, c = -3n, (l - 2m - 3n)(l^2 + 4m^2 + 9n^2 + 2lm + 3ln - 6mn)
  40. 40. (i) Vertices: \[ (2,-4),\ (-3,-7),\ (7,2) \] \[ G = \left( \frac{2+(-3)+7}{3}, \frac{-4+(-7)+2}{3} \right) \] \[ = \left( \frac{6}{3}, \frac{-9}{3} \right) \] \[ =(2,-3) \] ✓ Centroid: \[ \boxed{(2,-3)} \] (ii) Vertices: \[ (-5,-5),\ (1,-4),\ (-4,-2) \] \[ G = \left( \frac{-5+1+(-4)}{3}, \frac{-5+(-4)+(-2)}{3} \right) \] \[ = \left( \frac{-8}{3}, \frac{-11}{3} \right) \] ✓ Centroid: \[ \boxed{\left(-\frac83,-\frac{11}{3}\right)} \]
  41. 41. (i) (5\sqrt3+18\sqrt3-2\sqrt3) Since all are like surds, combine coefficients. Solution $$= (5+18-2)\sqrt3$$ $$=21\sqrt3$$ Answer $$21\sqrt3$$ (ii) (4\sqrt[3]{5}+2\sqrt[3]{5}-3\sqrt[3]{5}) Solution $$=(4+2-3)\sqrt[3]{5}$$ $$=3\sqrt[3]{5}$$ Answer $$3\sqrt[3]{5}$$ (iii) (3\sqrt{75}+5\sqrt{48}-\sqrt{243}) Step 1: Simplify each surd $$\sqrt{75}=\sqrt{25\times3}=5\sqrt3$$ $$\sqrt{48}=\sqrt{16\times3}=4\sqrt3$$ $$\sqrt{243}=\sqrt{81\times3}=9\sqrt3$$ Step 2: Substitute $$3(5\sqrt3)+5(4\sqrt3)-9\sqrt3$$ $$=15\sqrt3+20\sqrt3-9\sqrt3$$ $$=26\sqrt3$$ Answer $$26\sqrt3$$ (iv) (5\sqrt[3]{40}+2\sqrt[3]{625}-3\sqrt[3]{320}) Step 1: Simplify cube roots $$\sqrt[3]{40}=\sqrt[3]{8\times5}=2\sqrt[3]{5}$$ $$\sqrt[3]{625}=\sqrt[3]{125\times5}=5\sqrt[3]{5}$$ $$\sqrt[3]{320}=\sqrt[3]{64\times5}=4\sqrt[3]{5}$$ Step 2: Substitute $$5(2\sqrt[3]{5})+2(5\sqrt[3]{5})-3(4\sqrt[3]{5})$$ $$=10\sqrt[3]{5}+10\sqrt[3]{5}-12\sqrt[3]{5}$$ $$=8\sqrt[3]{5}$$ Answer $$8\sqrt[3]{5}$$
  42. 42. (i) (1,2) and (4,3) \[ d = \sqrt{(4-1)^2+(3-2)^2} \] \[ = \sqrt{3^2+1^2} \] \[ = \sqrt{9+1} \] \[ = \sqrt{10} \] ✓ Distance: \[ \boxed{\sqrt{10}} \] (ii) (3,4) and (–7,2) \[ d = \sqrt{(-7-3)^2+(2-4)^2} \] \[ = \sqrt{(-10)^2+(-2)^2} \] \[ = \sqrt{100+4} \] \[ = \sqrt{104} \] \[ = 2\sqrt{26} \] ✓ Distance: \[ \boxed{2\sqrt{26}} \] (iii) (a,b) and (c,b) Since y-coordinates are equal: \[ d = \sqrt{(c-a)^2+(b-b)^2} \] \[ = \sqrt{(c-a)^2} \] \[ = |c-a| \] ✓ Distance: \[ \boxed{|c-a|} \] (iv) (3,–9) and (–2,3) \[ d = \sqrt{(-2-3)^2+(3+9)^2} \] \[ = \sqrt{(-5)^2+12^2} \] \[ = \sqrt{25+144} \] \[ = \sqrt{169} \] \[ =13 \] ✓ Distance: \[ \boxed{13} \]
  43. 43. Let monthly incomes be $$3x,\ 4x$$ Monthly expenditures: $$5y,\ 7y$$ Given savings Each saves ₹5000. Thus, $$3x-5y=5000$$ $$4x-7y=5000$$ Step 1: Eliminate (x) Multiply first equation by 4: $$12x-20y=20000$$ Multiply second equation by 3: $$12x-21y=15000$$ Subtract: $$y=5000$$ Step 2: Find (x) $$3x-5(5000)=5000$$ $$3x=30000$$ $$x=10000$$ Monthly incomes $$3x=30000$$ $$4x=40000$$ Answer Monthly income of A: $$₹30,000$$ Monthly income of B: $$₹40,000$$
  44. 44. Angles in the same segment are equal, so ∠BDC = ∠CAB = 25°. Similarly, ∠DBA = 25°. The angle at the centre is twice the angle at the circumference, so ∠COB = 2×25° = 50°. Answers: ∠BDC = 25°, ∠DBA = 25°, ∠COB = 50°.
  45. 45. (i) (\sqrt3) $$\sqrt3 \approx 1.732$$ So, on the number line: Mark (0), (1), and (2) Locate the point approximately at (1.732) Answer $$\sqrt3$$ lies between (1) and (2). (ii) (\sqrt{4.7}) $$\sqrt{4.7}\approx2.168$$ So, on the number line: Mark (2) and (3) Locate the point approximately at (2.168) Answer $$\sqrt{4.7}$$ lies between (2) and (3). (iii) (\sqrt{6.5}) $$\sqrt{6.5}\approx2.549$$ So, on the number line: Mark (2) and (3) Locate the point approximately at (2.549) Answer $$\sqrt{6.5}$$ lies between (2) and (3).
Brain Grain · braingrain.in
Maths — Practice Paper · Set 2
Class: 9Samacheer KalviMax Marks: 91
Name: ____________________Reg No: ____________
Part I — Multiple Choice Questions 15 × 1 = 15

Choose the correct answer. (Answer all questions.)

1.Given: n(A) = 25, n(B) = 40, n(A ∪ B) = 50 and n(B') = 25. Find (1) n(A ∩ B) and (2) n(U).[1]
2.Which of the following sets are equivalent, unequal, or equal?[1]
3.Which one of the following has a terminating decimal expansion?[1]
4.Which of the following is a linear equation?[1]
5.Which of the following statement is false?[1]
6.Which of the following has (x-1) as a factor?[1]
7.Which of the following expressions are polynomials? If not, give reason.[1]
8.In quadrilateral ABCD, AB = BC and AD = DC. The diagram gives ∠ABC = 108° and ∠ADC = 42°. Find ∠BCD.....[1]
9.Find the value of (a) and (b) if[1]
10.Which of the following are sets?[1]
11.Which of the following is not a linear equation in two variables?[1]
12.Which of the following points lie in the fourth quadrant? Q(3,−4) and R(1,−1).[1]
13.Which one of the following is an irrational number?[1]
14.Verify the formula n(A ∪ B ∪ C) = n(A) + n(B) + n(C) − n(A ∩ B) − n(B ∩ C) − n(A ∩ C) + n(A ∩ B ∩ C) for A = {a,c,e,f,h}, B = {c,d,e,f} and C = {a,b,c,f}.[1]
15.Which of the following is a solution of (2x-y=6)[1]
Part II — Fill in the Blanks 5 × 1 = 5

Fill in the blanks. (Answer all questions.)

16.Degree of the constant polynomial is __________[1]
17.((a+b-c)^2) is equal to __________[1]
18.GCD of any two prime numbers is __________[1]
19.Zero of ((2-3x)) is ___________[1]
20.Cubic polynomial may have maximum of ___________ linear factors[1]
Part III — Short Answer Questions 18 × 2 = 36

Answer briefly. (Answer all questions.)

21.Draw ΔABC where AB = 6 cm, ∠B = 110°, BC = 5 cm. Construct its orthocentre.[2]
22.Given: n(A) = 300, n(A ∪ B) = 500, n(A ∩ B) = 50 and n(B') = 350. Find (1) n(B) and (2) n(U).[2]
23.If the y-coordinate of a point is zero, then the point always lies[2]
24.Find any three rational numbers between[2]
25.The lengths of diagonals of a rhombus are 12 cm and 16 cm. Find the side of the rhombus.[2]
26.In what ratio does P(2,−5) divide the line joining[2]
27.Which condition does not satisfy the linear equation (ax+by+c=0)[2]
28.Factorise: (i) \(8x^3+125y^3\) (ii) \(27x^3-8y^3\) (iii) \(a^6-64\).[2]
29.Expand the following: (i) \((x+2y+3z)^2\) (ii) \((-p+2q+3r)^2\) (iii) \((2p+3)(2p-4)(2p-5)\) (iv) \((3a+1)(3a-2)(3a+4)\).[2]
30.Tea and Coffee Problem[2]
31.Find (p,\ q) and remainder[2]
32.Evaluate using identities[2]
33.A parallelogram has adjacent sides 34 m and 20 m and a diagonal of length 42 m. Find the area of the parallelogram.[2]
34.If (x^3+6x^2+kx+6) is exactly divisible by ((x+2)), then (k=\ ?)[2]
35.Associative Laws[2]
36.Goalkeeper stops goal 32 times out of 40 attempts[2]
37.In cyclic quadrilateral ABCD, ∠A = 4x and ∠C = 2x. Find x.[2]
38.The angles of a triangle are 3x − 40, x + 20 and 2x − 10. Find x.[2]
Part IV — Long Answer Questions 7 × 5 = 35

Answer in detail. (Answer all questions.)

39.Find the three angles of ΔABC, given that an exterior angle equals 4x + 10° and the two interior opposite angles are 2x and x + 20°.[5]
40.Find the value of the following: (i) sin 65°39' + cos 24°57' + tan 10°10' (ii) tan 70°58' + cos 15°26' − sin 84°59'[5]
41.Find the mean using the Step Deviation Method for the following distribution: Age class (years): 15–19, 20–24, 25–29, 30–34, 35–39, 40–44 with frequencies 6, 8, 12, 4, 10, 6 respectively.[5]
42.Mean using Assumed Mean Method[5]
43.Height of tree problem[5]
44.Find the mode of the following distribution[5]
45.Verify the associative property of intersection of sets[5]
🔑 Show Answer Key — Set 2
  1. 1. Given: $$n(A)=25$$ $$n(B)=40$$ $$n(A\cup B)=50$$ $$n(B')=25$$ Find: 1. (n(A\cap B)) 2. (n(U)) Step 1: Find (n(A\cap B)) Using: $$n(A\cup B)=n(A)+n(B)-n(A\cap B)$$ $$50=25+40-n(A\cap B)$$ $$50=65-n(A\cap B)$$ $$n(A\cap B)=15$$ Step 2: Find (n(U)) Since: $$n(B')=n(U)-n(B)$$ $$25=n(U)-40$$ $$n(U)=65$$ Answer $$n(A\cap B)=15$$ $$n(U)=65$$
  2. 2. (i) A = vowels in English alphabet $$A={a,e,i,o,u}$$ B = letters in “VOWEL” $$B={V,O,W,E,L}$$ Both have 5 elements. ✓ Equivalent sets (ii) $$C={2,3,4,5}$$ $$D={x:x\in W,\ 1<x<5}={2,3,4}$$ ✕ Unequal sets (iii) $$X={L,I,F,E}$$ $$Y={F,I,L,E}$$ ✓ Equal sets (iv) $$G={5,7,11,13,17,19}$$ $$H={1,2,3,6,9,18}$$ Both contain 6 elements. ✓ Equivalent sets
  3. 3. 1. (\frac5{64}) 2. (\frac89) 3. (\frac{14}{15}) 4. (\frac1{12}) Solution A rational number has a terminating decimal if the denominator contains only factors (2) and/or (5). $$64=2^6$$ Hence, $$\frac5{64}$$ has a terminating decimal. Answer $$\boxed{(1)\ \frac5{64}}$$
  4. 4. Options: 1. (x+\frac1x=2) 2. (x(x-1)=2) 3. (3x+5=\frac23) 4. (x^3-x=5) Answer $$\boxed{(3)\ 3x+5=\frac23}$$
  5. 5. 1. The square root of (25) is (5) or (-5) 2. (-\sqrt{25}=-5) 3. (\sqrt{25}=5) 4. (\sqrt{25}=\pm5) Solution The symbol: $$\sqrt{25}$$ represents only the principal positive square root. Hence, $$\sqrt{25}=5$$ not (\pm5). Answer $$\boxed{(4)\ \sqrt{25}=\pm5}$$
  6. 6. Options: 1. (2x-1) 2. (3x-3) 3. (4x-3) 4. (3x-4) Solution For factor ((x-1)), $$p(1)=0$$ $$3(1)-3=0$$ Hence (3x-3). Answer $$\boxed{(2)\ 3x-3}$$
  7. 7. (i) $$\frac1{x^2}+3x-4$$ Solution $$\frac1{x^2}=x^{-2}$$ The exponent is negative. Hence, it is not a polynomial . (ii) $$x^2(x-1)$$ Solution $$=x^3-x^2$$ All exponents are non-negative integers. Hence, it is a polynomial. (iii) $$\frac1x(x+5)$$ Solution $$=\frac{x+5}{x} =1+\frac5x$$ Contains negative exponent. Hence, it is not a polynomial . (iv) $$\frac1{x-2}+\frac1{x-1}+7$$ Solution Variable occurs in denominator. Hence, it is not a polynomial . (v) $$\sqrt5x^2+\sqrt3x+\sqrt2$$ Solution All exponents are non-negative integers. Irrational coefficients are allowed. Hence, it is a polynomial. (vi) $$m^2-3\sqrt m+7m-10$$ Solution $$\sqrt m=m^{1/2}$$ Exponent is fractional. Hence, it is not a polynomial .
  8. 8. (3) 105°.
  9. 9. $$\frac{\sqrt7-2}{\sqrt7+2}=a\sqrt7+b$$ Solution Rationalise the denominator. $$\frac{\sqrt7-2}{\sqrt7+2} \times \frac{\sqrt7-2}{\sqrt7-2}$$ Using: genui{"math_block_widget_always_prefetch_v2":{"content":"(a+b)(a-b)=a^2-b^2"}} Numerator $$(\sqrt7-2)^2$$ $$=7+4-4\sqrt7$$ $$=11-4\sqrt7$$ Denominator $$(\sqrt7+2)(\sqrt7-2)$$ $$=7-4$$ $$=3$$ Therefore $$\frac{\sqrt7-2}{\sqrt7+2} ========================= \frac{11-4\sqrt7}{3}$$ [ -\frac43\sqrt7+\frac{11}{3} ] Comparing with: $$a\sqrt7+b$$ we get: $$a=-\frac43$$ $$b=\frac{11}{3}$$ Answer $$a=-\frac43,\qquad b=\frac{11}{3}$$
  10. 10. (i) The collection of prime numbers up to 100. ✓ Set (ii) The collection of rich people in India. ✕ Not a set (iii) The collection of all rivers in India. ✓ Set (iv) The collection of good Hockey players. ✕ Not a set
  11. 11. Options: 1. (ax+by+c=0) 2. (0x+0y+c=0) 3. (0x+by+c=0) 4. (ax+0y+c=0) Answer $$\boxed{(2)\ 0x+0y+c=0}$$
  12. 12. Fourth quadrant points have coordinates (+, −). Both Q(3,−4) and R(1,−1) have positive x and negative y, so both lie in the fourth quadrant. Answer: Q and R.
  13. 13. 1. (\sqrt{25}) 2. (\sqrt{\frac94}) 3. (\frac7{11}) 4. (\pi) Solution $$\sqrt{25}=5$$ $$\sqrt{\frac94}=\frac32$$ $$\frac7{11}$$ are rational numbers. $$\pi$$ is irrational. Answer $$\boxed{(4)\ \pi}$$
  14. 14. $$n(A\cup B\cup C)$$ $$=n(A)+n(B)+n(C)$$ $$-n(A\cap B)-n(B\cap C)-n(A\cap C)$$ $$+n(A\cap B\cap C)$$
  15. 15. Options: 1. ((2,4)) 2. ((4,2)) 3. ((3,-1)) 4. ((0,6)) Solution For ((4,2)): $$2(4)-2=8-2=6$$ Answer $$\boxed{(2)\ (4,2)}$$
  16. 16. Options: 1. 3 2. 2 3. 1 4. 0 Answer $$\boxed{(4)\ 0}$$
  17. 17. Options: 1. ((a-b+c)^2) 2. ((-a-b+c)^2) 3. ((a+b+c)^2) 4. ((a-b-c)^2) Solution $$(a+b-c)^2=[-( -a-b+c)]^2$$ $$=(-a-b+c)^2$$ Answer $$\boxed{(2)\ (-a-b+c)^2}$$
  18. 18. Options: 1. (-1) 2. (0) 3. (1) 4. (2) Answer $$\boxed{(3)\ 1}$$
  19. 19. Options: 1. (3) 2. (2) 3. (\frac23) 4. (\frac32) Solution $$2-3x=0$$ $$3x=2$$ $$x=\frac23$$ Answer $$\boxed{(3)\ \frac23}$$
  20. 20. Options: 1. 1 2. 2 3. 3 4. 4 Answer $$\boxed{(3)\ 3}$$
  21. 21. Orthocentre H is intersection of altitudes.
  22. 22. Given: $$n(A)=300$$ $$n(A\cup B)=500$$ $$n(A\cap B)=50$$ $$n(B')=350$$ Find: 1. (n(B)) 2. (n(U)) Step 1: Find (n(B)) $$n(A\cup B)=n(A)+n(B)-n(A\cap B)$$ $$500=300+n(B)-50$$ $$500=250+n(B)$$ $$n(B)=250$$ Step 2: Find (n(U)) $$n(B')=n(U)-n(B)$$ $$350=n(U)-250$$ $$n(U)=600$$ Answer $$n(B)=250$$ $$n(U)=600$$
  23. 23. Points on x-axis have: \[ y=0 \] ✓ Answer: \[ \boxed{(3)\ \text{on x-axis}} \]
  24. 24. $$-\frac{7}{11} \quad \text{and} \quad \frac{2}{11}$$ Solution We need rational numbers greater than $$-\frac{7}{11}$$ and less than $$\frac{2}{11}$$ Possible rational numbers are: $$-\frac{6}{11},; -\frac{5}{11},; -\frac{4}{11}$$ Answer Any three rational numbers are: $$-\frac{6}{11},; -\frac{5}{11},; -\frac{4}{11}$$
  25. 25. Diagonals of a rhombus bisect each other at right angles. Half-diagonals: 6 cm and 8 cm. Side = √(6^2 + 8^2) = √(36 + 64) = √100 = 10 cm.
  26. 26. A(−3,5) and B(4,−9)? Let ratio be: \[ m:n \] Using x-coordinate: \[ 2 = \frac{4m+(-3)n}{m+n} \] \[ 2m+2n=4m-3n \] \[ 5n=2m \] \[ m:n=5:2 \] Check with y-coordinate: \[ -5 = \frac{-9m+5n}{m+n} \] Substituting: \[ m:n=5:2 \] satisfies the equation. ✓ Ratio: \[ \boxed{5:2} \]
  27. 27. Options: 1. (a\ne0,\ b=0) 2. (a=0,\ b\ne0) 3. (a=0,\ b=0,\ c\ne0) 4. (a\ne0,\ b\ne0) Answer $$\boxed{(3)\ a=0,\ b=0,\ c\ne0}$$
  28. 28. (i) \(8x^3+125y^3=(2x+5y)(4x^2-10xy+25y^2)\). (ii) \(27x^3-8y^3=(3x-2y)(9x^2+6xy+4y^2)\). (iii) \(a^6-64=(a^3-8)(a^3+8)=(a-2)(a^2+2a+4)(a+2)(a^2-2a+4)\).
  29. 29. (i) \((x+2y+3z)^2 = x^2+4y^2+9z^2+4xy+12yz+6xz\). (ii) \((-p+2q+3r)^2 = p^2+4q^2+9r^2-4pq+12qr-6pr\). (iii) \((2p+3)(2p-4)(2p-5)=8p^3-24p^2-14p+60\). (iv) \((3a+1)(3a-2)(3a+4)=27a^3+27a^2-18a-8\).
  30. 30. Given: Total people = 45 Tea = 35 Coffee = 20 Everyone likes tea or coffee or both. (i) Like both tea and coffee $$45=35+20-n(T\cap C)$$ $$45=55-n(T\cap C)$$ $$n(T\cap C)=10$$ Answer $$10$$ (ii) Do not like tea $$45-35=10$$ Answer $$10$$ (iii) Do not like coffee $$45-20=25$$ Answer $$25$$
  31. 31. Given: Dividend: $$8x^4-2x^2+6x-7$$ Divisor: $$2x+1$$ Quotient: $$4x^3+px^2-qx+3$$ Use division algorithm $$\text{Dividend} =============== (\text{Divisor})(\text{Quotient})+\text{Remainder}$$ Multiply: $$(2x+1)(4x^3+px^2-qx+3)$$ $$=8x^4+(2p+4)x^3+(p-2q)x^2+(6-q)x+3$$ Compare with: $$8x^4+0x^3-2x^2+6x-7$$ Compare coefficients (x^3) $$2p+4=0$$ $$p=-2$$ (x^2) $$p-2q=-2$$ $$-2-2q=-2$$ $$q=0$$ Constant term $$3+\text{remainder}=-7$$ $$\text{remainder}=-10$$ Answer $$p=-2,\quad q=0$$ Remainder: $$-10$$
  32. 32. (i) (98^3) Using: $$98=100-2$$ $$(100-2)^3$$ $$=1000000-60000+1200-8$$ $$=941192$$ Answer $$941192$$ (ii) (1001^3) Using: $$1001=1000+1$$ $$(1000+1)^3$$ $$=1000000000+3000000+3000+1$$ $$=1003003001$$ Answer $$1003003001$$
  33. 33. The diagonal divides the parallelogram into two congruent triangles with sides 34 m, 20 m and 42 m. Semi-perimeter s = (34 + 20 + 42)/2 = 48 Area of one triangle = √[48(48−34)(48−20)(48−42)] = √[48·14·28·6] = 336 m² Parallelogram area = 2 × 336 = 672 m² Answer: 672 m²
  34. 34. Options: 1. (-6) 2. (-7) 3. (-8) 4. (11) Solution Since divisible by ((x+2)), $$p(-2)=0$$ $$(-2)^3+6(-2)^2+k(-2)+6=0$$ $$-8+24-2k+6=0$$ $$22-2k=0$$ $$2k=22$$ $$k=11$$ Answer $$\boxed{(4)\ 11}$$
  35. 35. $$(A\cup B)\cup C = A\cup(B\cup C)$$ $$(A\cap B)\cap C = A\cap(B\cap C)$$ <div
  36. 36. Goals conceded = 40−32 = 8. Probability that a shot is a goal = 8/40 = 1/5.
  37. 37. Opposite angles of a cyclic quadrilateral are supplementary: 4x + 2x = 180° ⇒ 6x = 180° ⇒ x = 30°.
  38. 38. Sum: (3x−40)+(x+20)+(2x−10)=180 ⇒ 6x−30=180 ⇒ 6x=210 ⇒ x=35°. Answer: 35°.
  39. 39. Using the exterior-angle theorem: exterior angle = sum of the two interior opposite angles. So, 4x + 10 = 2x + (x + 20) => 4x + 10 = 3x + 20 => x = 10. Thus the two opposite interior angles are 2x = 20° and x + 20 = 30°. The third angle is 180° − (20° + 30°) = 130°. Answer: 20°, 30°, 130°.
  40. 40. Using trigonometric tables (values rounded to 4 decimal places): (i) sin 65°39' ≈ 0.9115, cos 24°57' ≈ 0.9067, tan 10°10' ≈ 0.1794 Sum = 0.9115 + 0.9067 + 0.1794 = 1.9976 Answer: 1.9976 (ii) tan 70°58' ≈ 2.9042, cos 15°26' ≈ 0.9639, sin 84°59' ≈ 0.9962 Value = 2.9042 + 0.9639 − 0.9962 = 2.8719 Answer: 2.8719
  41. 41. Class width h = 5, assumed mean A = 27. Mid‑values x_i = 17, 22, 27, 32, 37, 42. Compute u_i = (x_i − A)/h: −2, −1, 0, 1, 2, 3 and f u_i: −12, −8, 0, 4, 20, 18. Σf = 46, Σ(fu) = 22. Mean x̄ = A + h(Σ(fu)/Σf) = 27 + 5(22/46) = 27 + 2.3913 ≈ 29.39 Answer: 29.39
  42. 42. | Class Interval | Frequency | |---|---| | 0–10 | 5 | | 10–20 | 7 | | 20–30 | 15 | | 30–40 | 25 | | 40–50 | 8 | Assumed mean: \[ A=25 \] | Class | \(f\) | Mid value \(x_i\) | \(d_i=x_i-A\) | \(fd_i\) | |---|---|---|---|---| | 0–10 | 5 | 5 | -20 | -100 | | 10–20 | 7 | 15 | -10 | -70 | | 20–30 | 15 | 25 | 0 | 0 | | 30–40 | 25 | 35 | 10 | 250 | | 40–50 | 8 | 45 | 20 | 160 | Totals \[ \sum f=60 \] \[ \sum fd=240 \] Using assumed mean formula: \[ \bar{x} = A+\frac{\sum fd}{\sum f} \] \[ = 25+\frac{240}{60} \] \[ =25+4 \] \[ =29 \] ✓ Mean: \[ \boxed{29} \]
  43. 43. Given: Distance from tree: \[ 60\text{ m} \] Angle of elevation: \[ 42^\circ \] Let height of tree be \(h\). Using: \[ \tan42^\circ=\frac{h}{60} \] \[ h=60\tan42^\circ \] Using tables: \[ \tan42^\circ\approx0.9004 \] \[ h\approx60(0.9004) \] \[ h\approx54.02 \] ✓ Height of tree: \[ \boxed{54.02\text{ m}} \] ---# Exercise 6.5 – Multiple Choice Questions Trigonometry – Validated & Corrected Answers
  44. 44. | Weight (kg) | 25–34 | 35–44 | 45–54 | 55–64 | 65–74 | 75–84 | |---|---|---|---|---|---|---| | No. of students | 4 | 8 | 10 | 14 | 8 | 6 | Modal class Highest frequency: \[ 14 \] Thus modal class: \[ 55-64 \] Formula \[ \text{Mode} = l+ \left( \frac{f_1-f_0}{2f_1-f_0-f_2} \right)\times h \] Where: \(l=55\) \(f_1=14\) \(f_0=10\) \(f_2=8\) \(h=10\) Substitution \[ \text{Mode} = 55+ \left( \frac{14-10}{2(14)-10-8} \right)\times10 \] \[ = 55+ \left( \frac4{28-18} \right)\times10 \] \[ = 55+ \left( \frac4{10} \right)\times10 \] \[ = 55+4 \] \[ =59 \] ✓ Mode: \[ \boxed{59\text{ kg}} \]
  45. 45. Given: $$A={x:x=2n,\ n\in W,\ n<4}$$ $$B={x:x=2n,\ n\in N,\ n\le4}$$ $$C={0,1,2,5,6}$$ Step 1: Write the sets explicitly Set (A) Since (n\in W) and (n<4), $$n=0,1,2,3$$ Therefore, $$A={0,2,4,6}$$ Set (B) Since (n\in N) and (n\le4), $$n=1,2,3,4$$ Therefore, $$B={2,4,6,8}$$ Set (C) $$C={0,1,2,5,6}$$ We verify: $$(A\cap B)\cap C=A\cap(B\cap C)$$ Left Side Step 1: Find (A\cap B) $$A\cap B={2,4,6}$$ Step 2: Find ((A\cap B)\cap C) $${2,4,6}\cap{0,1,2,5,6}$$ $$={2,6}$$ Right Side Step 1: Find (B\cap C) $$B\cap C={2,6}$$ Step 2: Find (A\cap(B\cap C)) $${0,2,4,6}\cap{2,6}$$ $$={2,6}$$ Conclusion $$(A\cap B)\cap C=A\cap(B\cap C)$$ Hence, the associative property of intersection is verified .
Brain Grain · braingrain.in
Maths — Practice Paper · Set 3
Class: 9Samacheer KalviMax Marks: 91
Name: ____________________Reg No: ____________
Part I — Multiple Choice Questions 15 × 1 = 15

Choose the correct answer. (Answer all questions.)

1.Find the value of 3 sin 70° sec 20° + 2 sin 49° sec 51°.....[1]
2.Given A = {a, c, e, f, h}, B = {c, d, e, f}, C = {a, b, c, f}. Verify that n(A ∪ B ∪ C) = n(A) + n(B) + n(C) − n(A ∩ B) − n(B ∩ C) − n(A ∩ C) + n(A ∩ B ∩ C).[1]
3.Which of the following is not true?[1]
4.Given A = {1, 3, 5}, B = {2, 3, 5, 6}, C = {1, 5, 6, 7}. Verify that n(A ∪ B ∪ C) = n(A) + n(B) + n(C) − n(A ∩ B) − n(B ∩ C) − n(A ∩ C) + n(A ∩ B ∩ C).[1]
5.Verify n(A ∪ B) = n(A) + n(B) − n(A ∩ B) for U = {x : x ∈ N, x ≤ 10}, A = {2,3,4,8,10} and B = {1,2,5,8,10}.[1]
6.Without actual division, find which of the following rational numbers have terminating decimal expansion[1]
7.Which one of the following is not a rational number?[1]
8.Which one of the following regarding the sum of two irrational numbers is true?[1]
9.Given: n(A) = 25, n(B) = 40, n(A ∪ B) = 50 and n(B') = 25. Find (1) n(A ∩ B) and (2) n(U).[1]
10.Which of the following sets are equivalent, unequal, or equal?[1]
11.Which one of the following has a terminating decimal expansion?[1]
12.Which of the following is a linear equation?[1]
13.Which of the following statement is false?[1]
14.Which of the following has (x-1) as a factor?[1]
15.Which of the following expressions are polynomials? If not, give reason.[1]
Part II — Fill in the Blanks 5 × 1 = 5

Fill in the blanks. (Answer all questions.)

16.If a1/a2 = b1/b2 ≠ c1/c2, then the pair has _________ solution(s). Options: (1) No solution; (2) Two solutions; (3) Infinite; (4) Unique.[1]
17.If a1/a2 ≠ b1/b2, then the pair has _________ solution(s). Options: (1) No solution; (2) Two solutions; (3) Unique; (4) Infinite.[1]
18.If (p(a)=0), then ((x-a)) is a ___________ of (p(x))[1]
19.Degree of the constant polynomial is __________[1]
20.((a+b-c)^2) is equal to __________[1]
Part III — Short Answer Questions 18 × 2 = 36

Answer briefly. (Answer all questions.)

21.Find (m) if[2]
22.Commutative Laws[2]
23.The following are the marks scored by students in the Summative Assessment exam. Class intervals: 0-10, 10-20, 20-30, 30-40, 40-50, 50-60. Number of students: 2, 7, 15, 10, 11, 5. Calculate the median.[2]
24.Volume of cuboid = 660 cm³[2]
25.State which pairs of sets are disjoint or overlapping.[2]
26.Find the value of the polynomial[2]
27.Probability of getting two heads when two coins are tossed[2]
28.Factorise the following: (i) 2a^2 + 9a + 10 (ii) 5x^2 - 29xy - 42y^2 (iii) 9 - 18x + 8x^2 (iv) 6x^2 + 16xy + 8y^2 (v) 12x^2 + 36x^2y + 27x^2y^2 (vi) (a+b)^2 + 9(a+b) + 18[2]
29.Identify the following sets as null set or singleton set.[2]
30.Find the GCD of the following: (i) (2x+5), (5x+2) (ii) a^{m+1}, a^{m+2}, a^{m+3} (iii) 2a^2+a, 4a^2-1 (iv) 3a^2, 5b^3, 7c^4 (v) x^4-1, x^2-1 (vi) a^3 - 9ax^2, (a-3x)^2[2]
31.If both ((x-2)) and ((x-\frac12)) are factors of[2]
32.Find the area of unshaded region[2]
33.If a pair of linear equations has no solution, what is the graphical representation of their graphs?[2]
34.Diameter AB bisects chord CD at E.[2]
35.An irrational number between 2 and 2.5 is: (1) √11; (2) √5; (3) √2.5; (4) √8.[2]
36.Ratio problem: Two numbers are in the ratio 5 : 6. If 8 is subtracted from each, the ratio becomes 4 : 5. Find the two numbers.[2]
37.The correct congruence statement is[2]
38.Subtract the second polynomial from the first[2]
Part IV — Long Answer Questions 7 × 5 = 35

Answer in detail. (Answer all questions.)

39.Park in shape of quadrilateral[5]
40.Draw ΔABC where[5]
41.In cyclic quadrilateral ABCD, ∠A = 2y + 4°, ∠B = 6x - 4°, ∠C = 4y - 4°, and ∠D = 7x + 2°. Find all four angles.[5]
42.AB and CD are parallel chords.[5]
43.In cyclic quadrilateral ABCD,[5]
44.Diameter of a circle = 52 cm and chord length = 20 cm. Find the distance of the chord from the centre.[5]
45.Find the marked angle in each circle: (i) ∠OBC = 30° and ∠OCB = 60°; find inscribed ∠BAC. (ii) ∠QOR = 80° and ∠ORP = 30°; find ∠OQP. (iii) MN is a diameter and ∠PON = 70°; find ∠OPN. (iv) inscribed ∠YXZ = 120°; find the smaller central ∠YOZ. (v) ∠BOA = 140° and ∠AOC = 100°; find ∠OAC.[5]
🔑 Show Answer Key — Set 3
  1. 1. As printed: approximately 5.39849, so none of the options is correct. If sin 49° is the textbook typo and sin 39° was intended, the answer is (3) 5.
  2. 2. Given sets: n(A) = 5 n(B) = 4 n(C) = 4 Intersections: A ∩ B = {c, e, f} ⇒ n(A ∩ B) = 3 B ∩ C = {c, f} ⇒ n(B ∩ C) = 2 A ∩ C = {a, c, f} ⇒ n(A ∩ C) = 3 A ∩ B ∩ C = {c, f} ⇒ n(A ∩ B ∩ C) = 2 Compute RHS: n(A)+n(B)+n(C) − n(A∩B) − n(B∩C) − n(A∩C) + n(A∩B∩C) = 5 + 4 + 4 − 3 − 2 − 3 + 2 = 7 Union: A ∪ B ∪ C = {a, b, c, d, e, f, h} ⇒ n(A ∪ B ∪ C) = 7 Hence both sides equal 7; the formula is verified.
  3. 3. 1. Every rational number is a real number. 2. Every integer is a rational number. 3. Every real number is an irrational number. 4. Every natural number is a whole number. Solution Real numbers include both rational and irrational numbers. Hence statement (3) is false. Answer $$\boxed{(3)\ \text{Every real number is an irrational number}}$$
  4. 4. Given: n(A) = 3 n(B) = 4 n(C) = 4 Intersections: A ∩ B = {3, 5} ⇒ n(A ∩ B) = 2 B ∩ C = {5, 6} ⇒ n(B ∩ C) = 2 A ∩ C = {1, 5} ⇒ n(A ∩ C) = 2 A ∩ B ∩ C = {5} ⇒ n(A ∩ B ∩ C) = 1 Compute RHS: 3 + 4 + 4 − 2 − 2 − 2 + 1 = 6 Union: A ∪ B ∪ C = {1, 2, 3, 5, 6, 7} ⇒ n(A ∪ B ∪ C) = 6 Therefore the formula is verified.
  5. 5. $$n(A\cup B)=n(A)+n(B)-n(A\cap B)$$ Given: $$U={x:x\in N,\ x\le10}$$ $$A={2,3,4,8,10}$$ $$B={1,2,5,8,10}$$ Step 1: Find (n(A)) $$n(A)=5$$ Step 2: Find (n(B)) $$n(B)=5$$ Step 3: Find (A\cap B) $$A\cap B={2,8,10}$$ $$n(A\cap B)=3$$ Step 4: Find (A\cup B) $$A\cup B={1,2,3,4,5,8,10}$$ $$n(A\cup B)=7$$ Verification RHS: $$n(A)+n(B)-n(A\cap B)$$ $$=5+5-3$$ $$=7$$ LHS: $$n(A\cup B)=7$$ Thus, $$n(A\cup B)=n(A)+n(B)-n(A\cap B)$$ Verified.
  6. 6. A rational number has a terminating decimal expansion if the denominator in simplest form contains only the prime factors: $$2 \text{ and/or } 5$$ (i) (\frac{7}{128}) $$128=2^7$$ Only factor (2) is present. Answer $$\frac{7}{128}$$ has a terminating decimal expansion . (ii) (\frac{21}{15}) Simplify: $$\frac{21}{15}=\frac75$$ Denominator: $$5$$ Only factor (5). Answer $$\frac{21}{15}$$ has a terminating decimal expansion . (iii) (4\frac{9}{35}) Convert fractional part: $$\frac{9}{35}$$ Denominator: $$35=5\times7$$ Factor (7) is present. Answer $$4\frac{9}{35}$$ has a non-terminating recurring decimal expansion . (iv) (\frac{219}{2200}) Simplify: $$\frac{219}{2200}$$ Factorize denominator: $$2200=2^3\times5^2\times11$$ Factor (11) is present. Answer $$\frac{219}{2200}$$ has a non-terminating recurring decimal expansion .
  7. 7. 1. (\sqrt{\frac8{18}}) 2. (\frac73) 3. (\sqrt{0.01}) 4. (\sqrt{13}) Solution $$\sqrt{\frac8{18}} ================= # \sqrt{\frac49} \frac23$$ rational. $$\frac73$$ rational. $$\sqrt{0.01} =========== \frac1{10}$$ rational. $$\sqrt{13}$$ irrational. Answer $$\boxed{(4)\ \sqrt{13}}$$
  8. 8. 1. always an irrational number 2. may be a rational or irrational number 3. always a rational number 4. always an integer Solution Example: $$\sqrt2+\sqrt3$$ is irrational. But, $$\sqrt2+(-\sqrt2)=0$$ is rational. Answer $$\boxed{(2)\ \text{may be a rational or irrational number}}$$
  9. 9. Given: $$n(A)=25$$ $$n(B)=40$$ $$n(A\cup B)=50$$ $$n(B')=25$$ Find: 1. (n(A\cap B)) 2. (n(U)) Step 1: Find (n(A\cap B)) Using: $$n(A\cup B)=n(A)+n(B)-n(A\cap B)$$ $$50=25+40-n(A\cap B)$$ $$50=65-n(A\cap B)$$ $$n(A\cap B)=15$$ Step 2: Find (n(U)) Since: $$n(B')=n(U)-n(B)$$ $$25=n(U)-40$$ $$n(U)=65$$ Answer $$n(A\cap B)=15$$ $$n(U)=65$$
  10. 10. (i) A = vowels in English alphabet $$A={a,e,i,o,u}$$ B = letters in “VOWEL” $$B={V,O,W,E,L}$$ Both have 5 elements. ✓ Equivalent sets (ii) $$C={2,3,4,5}$$ $$D={x:x\in W,\ 1<x<5}={2,3,4}$$ ✕ Unequal sets (iii) $$X={L,I,F,E}$$ $$Y={F,I,L,E}$$ ✓ Equal sets (iv) $$G={5,7,11,13,17,19}$$ $$H={1,2,3,6,9,18}$$ Both contain 6 elements. ✓ Equivalent sets
  11. 11. 1. (\frac5{64}) 2. (\frac89) 3. (\frac{14}{15}) 4. (\frac1{12}) Solution A rational number has a terminating decimal if the denominator contains only factors (2) and/or (5). $$64=2^6$$ Hence, $$\frac5{64}$$ has a terminating decimal. Answer $$\boxed{(1)\ \frac5{64}}$$
  12. 12. Options: 1. (x+\frac1x=2) 2. (x(x-1)=2) 3. (3x+5=\frac23) 4. (x^3-x=5) Answer $$\boxed{(3)\ 3x+5=\frac23}$$
  13. 13. 1. The square root of (25) is (5) or (-5) 2. (-\sqrt{25}=-5) 3. (\sqrt{25}=5) 4. (\sqrt{25}=\pm5) Solution The symbol: $$\sqrt{25}$$ represents only the principal positive square root. Hence, $$\sqrt{25}=5$$ not (\pm5). Answer $$\boxed{(4)\ \sqrt{25}=\pm5}$$
  14. 14. Options: 1. (2x-1) 2. (3x-3) 3. (4x-3) 4. (3x-4) Solution For factor ((x-1)), $$p(1)=0$$ $$3(1)-3=0$$ Hence (3x-3). Answer $$\boxed{(2)\ 3x-3}$$
  15. 15. (i) $$\frac1{x^2}+3x-4$$ Solution $$\frac1{x^2}=x^{-2}$$ The exponent is negative. Hence, it is not a polynomial . (ii) $$x^2(x-1)$$ Solution $$=x^3-x^2$$ All exponents are non-negative integers. Hence, it is a polynomial. (iii) $$\frac1x(x+5)$$ Solution $$=\frac{x+5}{x} =1+\frac5x$$ Contains negative exponent. Hence, it is not a polynomial . (iv) $$\frac1{x-2}+\frac1{x-1}+7$$ Solution Variable occurs in denominator. Hence, it is not a polynomial . (v) $$\sqrt5x^2+\sqrt3x+\sqrt2$$ Solution All exponents are non-negative integers. Irrational coefficients are allowed. Hence, it is a polynomial. (vi) $$m^2-3\sqrt m+7m-10$$ Solution $$\sqrt m=m^{1/2}$$ Exponent is fractional. Hence, it is not a polynomial .
  16. 16. No solution. Reason: If the ratios of coefficients of x and y are equal but differ from the ratio of constants, the lines are parallel and distinct — there is no solution.
  17. 17. Unique (one solution). Reason: For a pair of linear equations, if the ratios of coefficients a1/a2 and b1/b2 are not equal, the lines intersect at a single point — a unique solution.
  18. 18. Options: 1. Divisor 2. Quotient 3. Remainder 4. Factor Answer $$\boxed{(4)\ \text{Factor}}$$
  19. 19. Options: 1. 3 2. 2 3. 1 4. 0 Answer $$\boxed{(4)\ 0}$$
  20. 20. Options: 1. ((a-b+c)^2) 2. ((-a-b+c)^2) 3. ((a+b+c)^2) 4. ((a-b-c)^2) Solution $$(a+b-c)^2=[-( -a-b+c)]^2$$ $$=(-a-b+c)^2$$ Answer $$\boxed{(2)\ (-a-b+c)^2}$$
  21. 21. $$(x+3)$$ is a factor of $$x^3-3x^2-mx+24$$ Solution $$x+3=x-(-3)$$ So, $$p(-3)=0$$ $$(-3)^3-3(-3)^2-m(-3)+24=0$$ $$-27-27+3m+24=0$$ $$3m-30=0$$ $$3m=30$$ $$m=10$$ Answer $$m=10$$
  22. 22. $$A\cup B = B\cup A$$ $$A\cap B = B\cap A$$ <div
  23. 23. Class interval Frequency (f) Cumulative frequency 0-10 2 2 10-20 7 9 20-30 15 24 30-40 10 34 40-50 11 45 50-60 5 50 Total frequency, N = 2 + 7 + 15 + 10 + 11 + 5 = 50, so N/2 = 25. The first cumulative frequency greater than or equal to 25 is 34, so the median class is 30-40. Here l = 30, m = 24 (cumulative frequency before the median class), f = 10 and c = 10. Median = l + ((N/2 - m)/f) x c = 30 + ((25 - 24)/10) x 10 = 31. ✓ Median marks: 31
  24. 24. Area of base = 33 cm² Find height. Using: \[ V=\text{Base area}\times h \] \[ 660=33h \] \[ h=20 \] ✓ Answer: \[ \boxed{(3)\ 20\text{ cm}} \]
  25. 25. (i) $$A={f,i,a,s}$$ $$B={a,n,f,h,s}$$ Common elements: $${a,f,s}$$ ✓ Overlapping sets (ii) C = odd prime numbers greater than 2 D = even prime number No common element. ✓ Disjoint sets (iii) Factors of 24: $$E={1,2,3,4,6,8,12,24}$$ Multiples of 3 less than 30: $$F={3,6,9,12,15,18,21,24,27}$$ Common elements: $${3,6,12,24}$$ ✓ Overlapping sets
  26. 26. $$f(y)=6y-3y^2+3$$ (i) At (y=1) Solution $$f(1)=6(1)-3(1)^2+3$$ $$=6-3+3$$ $$=6$$ Answer $$f(1)=6$$ (ii) At (y=-1) Solution $$f(-1)=6(-1)-3(-1)^2+3$$ $$=-6-3+3$$ $$=-6$$ Answer $$f(-1)=-6$$ (iii) At (y=0) Solution $$f(0)=6(0)-3(0)^2+3$$ $$=3$$ Answer $$f(0)=3$$
  27. 27. Sample space: \[ HH,\ HT,\ TH,\ TT \] Total outcomes: \[ 4 \] Favorable outcome: \[ HH \] Thus: \[ P = \frac14 \] ✓ Answer: \[ \boxed{\frac14} \]
  28. 28. (i) (a+2)(2a+5) (ii) (5x+6y)(x-7y) (iii) (4x-3)(2x-3) (iv) 2(3x+2y)(x+2y) (v) 3x^2(2+3y)^2 (vi) (a+b+6)(a+b+3)
  29. 29. (i) $$A={x:x\in\mathbb{N},1<x<2}$$ No natural number exists. ✓ Null set (ii) Set of even natural numbers not divisible by 2. Impossible. ✓ Null set (iii) $$C={0}$$ ✓ Singleton set (iv) Set of triangles having four sides. Impossible. ✓ Null set
  30. 30. (i) 1 (ii) a^{m+1} (iii) 2a+1 (iv) 1 (v) x^2-1 (vi) a-3x
  31. 31. $$ax^2+5x+b$$ show that (a=b) Solution Since (x=2) is a zero, $$4a+10+b=0$$ $$4a+b=-10$$ Since (x=\frac12) is a zero, $$a\left(\frac14\right)+\frac52+b=0$$ Multiply by 4: $$a+10+4b=0$$ $$a+4b=-10$$ Subtract: $$(4a+b)-(a+4b)=0$$ $$3a-3b=0$$ $$a=b$$ Answer $$a=b$$
  32. 32. Use: Area of larger figure Subtract shaded area ✓ Exact numerical answer requires the figure dimensions.
  33. 33. They are parallel lines (no intersection).
  34. 34. Given: \(CE=ED=8\) cm \(EB=4\) cm Find radius. Let radius be \(r\). Distance from centre to chord: \[ OE=r-4 \] Using Pythagoras: \[ (r-4)^2+8^2=r^2 \] \[ r^2-8r+16+64=r^2 \] \[ 80=8r \] \[ r=10\text{ cm} \] ✓ Answer: \[ \boxed{(4)\ 10\text{ cm}} \]
  35. 35. (2) √5. Reason: 2^2 = 4 and (2.5)^2 = 6.25; 5 lies between 4 and 6.25, so √5 lies between 2 and 2.5 and is irrational.
  36. 36. Let numbers be 5x and 6x. (5x-8)/(6x-8) = 4/5 ⇒ x = 8. Numbers are 40 and 48.
  37. 37. (4) ΔABC ≅ ΔFED
  38. 38. (i) $$p(x)=7x^2+6x-1$$ $$q(x)=6x-9$$ Subtraction $$(7x^2+6x-1)-(6x-9)$$ $$=7x^2+8$$ Degree $$2$$ (ii) $$f(y)=6y^2-7y+2$$ $$g(y)=7y+y^3$$ Subtraction $$6y^2-7y+2-(7y+y^3)$$ $$=-y^3+6y^2-14y+2$$ Degree $$3$$ (iii) $$h(z)=z^5-6z^4+z$$ $$f(z)=6z^2+10z-7$$ Subtraction $$z^5-6z^4-6z^2-9z+7$$ Degree $$5$$
  39. 39. Sides: \[ 15,\ 20,\ 26,\ 17 \] Angle between first two sides is \(90^\circ\). First triangle \[ A_1 = \frac12\times15\times20 \] \[ =150 \] Diagonal: \[ \sqrt{15^2+20^2} = 25 \] Second triangle Sides: \[ 25,\ 26,\ 17 \] Semi-perimeter: \[ s=\frac{25+26+17}{2}=34 \] Area: \[ A_2 = \sqrt{34(9)(8)(17)} \] \[ = \sqrt{41616} \] \[ =204 \] Total area \[ 150+204=354 \] ✓ Area: \[ \boxed{354\text{ m}^2} \]
  40. 40. \(AB = 6\) cm \(\angle B = 110^\circ\) \(BC = 5\) cm Construct its Orthocentre. Construction Steps 1. Draw: \[ AB = 6 \text{ cm} \] 2. At point \(B\), construct: \[ \angle ABC = 110^\circ \] 3. On the ray mark: \[ BC = 5 \text{ cm} \] 4. Join: \[ AC \] Triangle \(ABC\) is formed. To Construct Orthocentre 5. Draw perpendicular from \(A\) to \(BC\). 6. Draw perpendicular from \(C\) to \(AB\). 7. The altitudes intersect at \(H\). ✓ \(H\) is the orthocentre.
  41. 41. ∠A = 64°, ∠B = 80°, ∠C = 116°, ∠D = 100°.
  42. 42. Given: \(AB = 8\) cm \(CD = 6\) cm Distance between perpendiculars \(LM = 7\) cm Find radius. Half chords: \[ \frac{AB}{2} = 4 \text{ cm} \] \[ \frac{CD}{2} = 3 \text{ cm} \] Let distances from centre be: \[ OM = x \] \[ OL = 7 - x \] Using Pythagoras theorem: For AB: \[ r^2 = x^2 + 4^2 \] \[ r^2 = x^2 + 16 \] For CD: \[ r^2 = (7-x)^2 + 3^2 \] \[ r^2 = (7-x)^2 + 9 \] Equating: \[ x^2 + 16 = (7-x)^2 + 9 \] \[ x^2 + 16 = 49 -14x + x^2 + 9 \] \[ 16 = 58 -14x \] \[ 14x = 42 \] \[ x = 3 \] Now: \[ r^2 = 3^2 + 16 \] \[ = 25 \] \[ r = 5 \text{ cm} \] ✓ Radius = \(5\) cm
  43. 43. \(\angle ADC = 80^\circ\) \(DC\) produced to \(E\) \(CF \parallel AB\) \(\angle ECF = 20^\circ\) Find ∠BAD. The text data alone does not determine a unique value for \(\angle BAD\) without the figure orientation. Using the cyclic quadrilateral property: \[ \angle ABC=180^\circ-\angle ADC=100^\circ \] The parallel-line condition with \(\angle ECF=20^\circ\) can lead to different valid configurations, so \(\angle BAD=120^\circ\) is not supported from the visible text alone. ✓ Manual review with the textbook figure is required.
  44. 44. Diameter: \[ 52 \text{ cm} \] Radius: \[ r = \frac{52}{2} = 26 \text{ cm} \] Chord length: \[ 20 \text{ cm} \] Half chord: \[ 10 \text{ cm} \] Let distance from centre to chord be \(d\). Using Pythagoras theorem: ::contentReference[oaicite:0]{index=0} \[ d^2 + 10^2 = 26^2 \] \[ d^2 + 100 = 676 \] \[ d^2 = 576 \] \[ d = 24 \text{ cm} \] ✓ Distance of chord from centre = \(24\) cm
  45. 45. (i) 45°, (ii) 10°, (iii) 55°, (iv) 120°, (v) 40°.

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