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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.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 has (x-1) as a factor?[1]
3.Which of the following is not a linear equation in two variables?[1]
4.Which of the following is a solution of (2x-y=6)[1]
5.Find the value of (a) and (b) if[1]
6.Which of the following sets are equivalent, unequal, or equal?[1]
7.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]
8.Which one of the following regarding the sum of two irrational numbers is true?[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.Find the value of 3 sin 70° sec 20° + 2 sin 49° sec 51°.....[1]
11.In quadrilateral ABCD, AB = BC and AD = DC. The diagram gives ∠ABC = 108° and ∠ADC = 42°. Find ∠BCD.....[1]
12.Which one of the following is not a rational number?[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 statement is false?[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 (p(a)=0), then ((x-a)) is a ___________ of (p(x))[1]
18.Cubic polynomial may have maximum of ___________ linear factors[1]
19.GCD of any two prime numbers is __________[1]
20.If a1/a2 ≠ b1/b2, then the pair has _________ solution(s). Options: (1) No solution; (2) Two solutions; (3) Unique; (4) Infinite.[1]
Part III — Short Answer Questions 18 × 2 = 36

Answer briefly. (Answer all questions.)

21.One angle of cyclic quadrilateral is 75°. Find opposite angle.[2]
22.A chord is 15 cm from the centre of a circle of radius 25 cm. Find the length of the chord.[2]
23.Find any two rational numbers between[2]
24.The volume of a container is 1440 m³. If its length is 15 m and breadth is 8 m, find its height.[2]
25.Find the GCD for the following: (i) p^5, p^11, p^9 (ii) 4x^3, y^3, z^3 (iii) 9a^2b^2c^3, 15a^3b^2c^4 (iv) 64x^8, 240x^6 (v) ab^2c^3, a^2b^3c, a^3bc^2 (vi) 35x^5y^3z^4, 49x^2yz^3, 14xy^2z^2 (vii) 25ab^3c, 100a^2bc, 125ab (viii) 3abc, 5xyz, 7pqr[2]
26.PQRS and PTVS are cyclic quadrilaterals.[2]
27.Mean of five positive integers is twice their median. Four integers are 3, 4, 6, 9. Find the fifth integer.[2]
28.Probability of selecting a good Laptop[2]
29.Probability of guessing correctly is x/3 and probability of not guessing correctly is x/5; find x[2]
30.Chocolate Problem: If the cost of one chocolate is \((x+y)\) and the number of chocolates is \((x+y)\), find the total amount in terms of x and y, and evaluate the amount when x = 10, y = 5.[2]
31.Distance between (5,−1) and origin[2]
32.Probability of winning tennis match = 0.72. Find probability of losing.[2]
33.Factorise: (i) \(4x^2+9y^2+25z^2+12xy+30yz+20xz\) (ii) \(25x^2+4y^2+9z^2-20xy+12yz-30xz\).[2]
34.Show that the bisectors of angles of a parallelogram form a rectangle.[2]
35.Height of tree problem[2]
36.The probability of guessing correctly is x/3 and the probability of not guessing correctly is x/5. Find x.[2]
37.Find the zero of the polynomial 2x + 5.[2]
38.Mean tumor volume of mice[2]
Part IV — Long Answer Questions 7 × 5 = 35

Answer in detail. (Answer all questions.)

39.Park in shape of quadrilateral[5]
40.Draw Venn diagrams for the following: (i) A ∪ B (ii) A ∩ B (iii) (A ∩ B)' (iv) (B − A)' (v) A' ∪ B' (vi) A' ∩ B' (vii) State any observation.[5]
41.Is the figure ∠A supplementary to ∠B? Give reasons.[5]
42.Represent the following numbers in scientific notation[5]
43.Find the distance between the following pairs of points[5]
44.Monthly income problem[5]
45.Draw ΔPQR with PQ = 7 cm, QR = 8 cm, PR = 5 cm. Construct its orthocentre.[5]
🔑 Show Answer Key — Set 1
  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. 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 $$\boxed{(2)\ 3x-3}$$
  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. Options: 1. ((2,4)) 2. ((4,2)) 3. ((3,-1)) 4. ((0,6)) Solution For ((4,2)): $$2(4)-2=8-2=6$$ Answer $$\boxed{(2)\ (4,2)}$$
  5. 5. $$\frac{\sqrt7-2}{\sqrt7+2}=a\sqrt7+b$$ 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 $$a=-\frac43,\qquad b=\frac{11}{3}$$
  6. 6. (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
  7. 7. $$n(A\cup B)=n(A)+n(B)-n(A\cap B)$$ 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.
  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 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. 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.
  11. 11. (3) 105°.
  12. 12. 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 $$\boxed{(4)\ \sqrt{13}}$$
  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. 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 $$\boxed{(4)\ \sqrt{25}=\pm5}$$
  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. Options: 1. Divisor 2. Quotient 3. Remainder 4. Factor Answer $$\boxed{(4)\ \text{Factor}}$$
  18. 18. Options: 1. 1 2. 2 3. 3 4. 4 Answer $$\boxed{(3)\ 3}$$
  19. 19. Options: 1. (-1) 2. (0) 3. (1) 4. (2) Answer $$\boxed{(3)\ 1}$$
  20. 20. 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.
  21. 21. Opposite angles are supplementary. \[ 180^\circ-75^\circ=105^\circ \] ✓ Answer: \[ \boxed{(2)\ 105^\circ} \]
  22. 22. Half the chord = √(25^2 − 15^2) = √(625 − 225) = 20 cm. Full chord = 2×20 = 40 cm.
  23. 23. $$2.2360679\ldots$$ and $$2.236505500\ldots$$ Solution Any terminating decimals between the given numbers are rational numbers. Possible answers: $$2.2361$$ $$2.2364$$ Both lie between the given numbers. Answer Two rational numbers are: $$2.2361$$ and $$2.2364$$
  24. 24. V = l × b × h ⇒ 1440 = 15 × 8 × h = 120 h h = 1440 / 120 = 12 m Answer: 12 m
  25. 25. (i) p^5 (ii) 1 (iii) 3a^2b^2c^3 (iv) 16x^6 (v) abc (vi) 7xyz^2 (vii) 25ab (viii) 1
  26. 26. If ∠QRS = 100°, find ∠TVS. Opposite angles in cyclic quadrilateral are supplementary. \[ 180^\circ-100^\circ=80^\circ \] ✓ Answer: \[ \boxed{(1)\ 80^\circ} \]
  27. 27. Mean = 2 × median = 2 × 6 = 12, so total of five integers = 5 × 12 = 60. Sum of the four given integers = 3 + 4 + 6 + 9 = 22. Fifth integer = 60 − 22 = 38 .
  28. 28. Total laptops manufactured: \[ 10000 \] Defective laptops: \[ 25 \] Good laptops: \[ 10000-25=9975 \] Probability of selecting a good laptop: \[ P(\text{good}) = \frac{9975}{10000} \] \[ =0.9975 \] ✓ Answer: \[ \boxed{0.9975} \]
  29. 29. Assuming these two probabilities are complementary: x/3 + x/5 = 1. LCM 15: (5x+3x)/15 =1 → 8x/15 =1 → x = 15/8.
  30. 30. Total = \((x+y)(x+y)=(x+y)^2=x^2+2xy+y^2\). For x=10, y=5: \(10^2+2\cdot10\cdot5+5^2=100+100+25=225\). Amount paid: Rs. 225.
  31. 31. \[ d = \sqrt{5^2+(-1)^2} \] \[ = \sqrt{25+1} \] \[ = \sqrt{26} \] ✓ Answer: \[ \boxed{(3)\ \sqrt{26}} \]
  32. 32. P(lose) = 1 − P(win) = 1 − 0.72 = 0.28.
  33. 33. (i) Recognise perfect square: \((2x+3y+5z)^2\). (ii) Recognise perfect square: \((5x-2y-3z)^2\).
  34. 34. The four angle bisectors form a rectangle (each interior angle 90°).
  35. 35. Distance = 60 m, angle of elevation = 42°. tan 42° ≈ 0.9004 ⇒ h = 60·tan42° ≈ 60·0.9004 = 54.024 → 54.02 m (rounded).
  36. 36. The probabilities of an event and its complement add to 1, so x/3 + x/5 = 1. LCM of 3 and 5 is 15: (5x + 3x)/15 = 1 ⇒ 8x/15 = 1 ⇒ 8x = 15 ⇒ x = 15/8. Answer: 15/8 .
  37. 37. 2x + 5 = 0 ⇒ x = -5/2.
  38. 38. Tumor volumes (mm³): \[ 145,\ 158,\ 142,\ 141,\ 139,\ 140 \] Using arithmetic mean: :contentReference[oaicite:0]{index=0} Sum of observations \[ 145+158+142+141+139+140 \] \[ =865 \] Number of mice: \[ 6 \] Mean: \[ \bar{x} = \frac{865}{6} \] \[ =144.17 \] ✓ Mean tumor volume: \[ \boxed{144.17\text{ mm}^3} \]
  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. Venn shading instructions (i) A ∪ B — Shade both sets A and B (all regions inside A or B). (ii) A ∩ B — Shade only the common overlapping region of A and B. (iii) (A ∩ B)' — Shade every region except the intersection (i.e., all regions not in A ∩ B). (iv) (B − A)' — Shade everything except the part belonging only to B (i.e., the complement of B\A). (v) A' ∪ B' — Shade all regions except the intersection (equivalent to (A ∩ B)'). (vi) A' ∩ B' — Shade the region outside both sets (elements in neither A nor B). (vii) Observation — (A ∩ B)' = A' ∪ B' (De Morgan's law).
  41. 41. Yes. (i) 70° and 110° are supplementary. (ii) 50° and 130° are supplementary. (iii) 40° and 140° are supplementary.
  42. 42. # (i) $$(300000)^2\times(20000)^4$$ Step 1: Write in scientific notation $$300000=3\times10^5$$ $$20000=2\times10^4$$ Step 2: Apply powers $$(3\times10^5)^2(2\times10^4)^4$$ $$=3^2\times10^{10}\times2^4\times10^{16}$$ $$=9\times16\times10^{26}$$ $$=144\times10^{26}$$ $$=1.44\times10^{28}$$ Answer $$1.44\times10^{28}$$ # (ii) $$(0.000001)^{11}\div(0.005)^3$$ Step 1: Convert to scientific notation $$0.000001=10^{-6}$$ $$0.005=5\times10^{-3}$$ Step 2: Simplify $$(10^{-6})^{11}\div(5\times10^{-3})^3$$ $$=10^{-66}\div(125\times10^{-9})$$ $$=\frac{10^{-66}}{125\times10^{-9}}$$ $$=\frac1{125}\times10^{-57}$$ $$=0.008\times10^{-57}$$ $$=8\times10^{-60}$$ Answer $$8\times10^{-60}$$ # (iii) $$\frac…
  43. 43. (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: \[ \boxed{13} \]
  44. 44. 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: $$₹40,000$$
  45. 45. Construction steps (brief): Draw side QR = 8 cm. With centre Q and radius 7 cm and with centre R and radius 5 cm, draw arcs that meet at P. Join P to Q and R to form ΔPQR. Draw the altitude from P to QR (i.e. the perpendicular from P to QR). Draw the altitude from Q to PR (perpendicular from Q to PR). The intersection of these altitudes is the orthocentre H of ΔPQR (the third altitude will pass through H as well). Result: Point H is the orthocentre.
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.Which one of the following has a terminating decimal expansion?[1]
2.Which of the following are sets?[1]
3.Which of the following is not true?[1]
4.Without actual division, find which of the following rational numbers have terminating decimal expansion[1]
5.Which of the following is a linear equation?[1]
6.Which of the following expressions are polynomials? If not, give reason.[1]
7.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]
8.Which of the following points lie in the fourth quadrant? Q(3,−4) and R(1,−1).[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 has (x-1) as a factor?[1]
11.Which of the following is not a linear equation in two variables?[1]
12.Which of the following is a solution of (2x-y=6)[1]
13.Find the value of (a) and (b) if[1]
14.Which of the following sets are equivalent, unequal, or equal?[1]
15.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]
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.Zero of ((2-3x)) is ___________[1]
18.((a+b-c)^2) is equal to __________[1]
19.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]
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.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]
22.The smallest rational number by which 1/3 should be multiplied so that its decimal expansion terminates with one place of decimal is: (1) 1/10; (2) 3/10; (3) 3; (4) 30.[2]
23.Angles of triangle are: 3x−40, x+20, 2x−10. Find x.[2]
24.PQ and RS are equal chords of a circle with centre O. Given ∠POQ = 70°. Find ∠ORS.[2]
25.Find the value of (m) from (2x+3y=m)[2]
26.If \(x^2+\dfrac{1}{x^2}=23\), find (1) \(x+\dfrac{1}{x}\) and (2) \(x^3+\dfrac{1}{x^3}\).[2]
27.Student Record Analysis[2]
28.Value of (k) for parallel lines[2]
29.Simplify: (i) \((2a+3b+4c)(4a^2+9b^2+16c^2-6ab-12bc-8ca)\) (ii) \((x-2y+3z)(x^2+4y^2+9z^2+2xy+6yz-3xz)\).[2]
30.Probability that a youngster does NOT have voter ID[2]
31.Express the following decimal expressions into rational numbers[2]
32.AD is a diameter of length 30 cm and AB is a chord of length 24 cm. Find the distance of chord AB from the centre.[2]
33.Write the following in the form of (5^n)[2]
34.A(2,3), B(2,-4). P lies on x-axis such that[2]
35.Commutative Laws[2]
36.Point whose ordinate is 4 and lies on y-axis[2]
37.Double Complement Law[2]
38.Which statement is correct?[2]
Part IV — Long Answer Questions 7 × 5 = 35

Answer in detail. (Answer all questions.)

39.A circle has radius 6 m. Two chords AB and CD are perpendicular. AB = 8 m, CD = 10 m. If the chords intersect at P, find OP (distance from centre O to P).[5]
40.Let U = {a,b,c,d,e,f,g,h}, A = {b,d,f,h} and B = {a,d,e,h}. Find: (i) A' (ii) B' (iii) A' ∪ B' (iv) A' ∩ B' (v) (A ∪ B)' (vi) (A ∩ B)' (vii) (A')' (viii) (B')'.[5]
41.Construct ΔPQR such that[5]
42.Construct isosceles triangle PQR where[5]
43.Find the centroid of the triangle[5]
44.Draw ΔABC where[5]
45.In quadrilateral ABCD, ∠A = 72° and ∠C is supplementary to ∠A.[5]
🔑 Show Answer Key — Set 2
  1. 1. 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 $$\boxed{(1)\ \frac5{64}}$$
  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. 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. 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 denom…
  5. 5. Options: 1. (x+\frac1x=2) 2. (x(x-1)=2) 3. (3x+5=\frac23) 4. (x^3-x=5) Answer $$\boxed{(3)\ 3x+5=\frac23}$$
  6. 6. # (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. Henc…
  7. 7. 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.
  8. 8. 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.
  9. 9. Given: $$n(A)=25$$ $$n(B)=40$$ $$n(A\cup B)=50$$ $$n(B&#x27;)=25$$ 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: $$n(B&#x27;)=n(U)-n(B)$$ $$25=n(U)-40$$ $$n(U)=65$$ Answer $$n(A\cap B)=15$$ $$n(U)=65$$
  10. 10. 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 $$\boxed{(2)\ 3x-3}$$
  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. Options: 1. ((2,4)) 2. ((4,2)) 3. ((3,-1)) 4. ((0,6)) Solution For ((4,2)): $$2(4)-2=8-2=6$$ Answer $$\boxed{(2)\ (4,2)}$$
  13. 13. $$\frac{\sqrt7-2}{\sqrt7+2}=a\sqrt7+b$$ 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 $$a=-\frac43,\qquad b=\frac{11}{3}$$
  14. 14. (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
  15. 15. $$n(A\cup B)=n(A)+n(B)-n(A\cap B)$$ 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.
  16. 16. Options: 1. 3 2. 2 3. 1 4. 0 Answer $$\boxed{(4)\ 0}$$
  17. 17. Options: 1. (3) 2. (2) 3. (\frac23) 4. (\frac32) Solution $$2-3x=0$$ $$3x=2$$ $$x=\frac23$$ Answer $$\boxed{(3)\ \frac23}$$
  18. 18. 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 $$\boxed{(2)\ (-a-b+c)^2}$$
  19. 19. 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.
  20. 20. Options: 1. Divisor 2. Quotient 3. Remainder 4. Factor Answer $$\boxed{(4)\ \text{Factor}}$$
  21. 21. 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 .
  22. 22. (2) 3/10. Reason: (1/3) × (3/10) = 1/10 = 0.1, which is a terminating decimal with one decimal place.
  23. 23. Sum: (3x−40)+(x+20)+(2x−10)=180 ⇒ 6x−30=180 ⇒ 6x=210 ⇒ x=35°.
  24. 24. Equal chords subtend equal central angles, so ∠ROS = ∠POQ = 70°. In isosceles triangle ORS (OR = OS = radius) the other two angles are equal. Each = (180° − 70°)/2 = 55°.
  25. 25. If one solution is (x=2,\ y=-2) Options: 1. (2) 2. (-2) 3. (10) 4. (0) Solution $$m=2(2)+3(-2)$$ $$=4-6$$ $$=-2$$ Answer $$\boxed{(2)\ -2}$$
  26. 26. 1) \(\left(x+\dfrac1x\right)^2=x^2+\dfrac{1}{x^2}+2=23+2=25\Rightarrow x+\dfrac1x=5.\) 2) \(\left(x+\dfrac1x\right)^3=x^3+\dfrac{1}{x^3}+3\left(x+\dfrac1x\right)\Rightarrow 125=x^3+\dfrac{1}{x^3}+15\). Hence \(x^3+\dfrac{1}{x^3}=125-15=110.\)
  27. 27. Using class records: (i) Mean age of class Group ages into intervals Use grouped mean formula (ii) Mean height of class Group heights into intervals Calculate grouped mean ---# UNIT 9 : Probability
  28. 28. $$4x+6y-1=0$$ $$2x+ky-7=0$$ 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 $$\boxed{(1)\ k=3}$$
  29. 29. (i) This matches identity (x+y+z)(x^2+y^2+z^2-xy-yz-zx)=x^3+y^3+z^3-3xyz with x=2a,y=3b,z=4c. Result: \(8a^3+27b^3+64c^3-72abc\). (ii) With a=x, b=-2y, c=3z the identity gives: \(x^3-8y^3+27z^3+18xyz\).
  30. 30. Total youngsters surveyed: \[ 400 \] Youngsters having voter ID: \[ 191 \] Youngsters without voter ID: \[ 400-191=209 \] Probability: \[ P(\text{No voter ID}) = \frac{209}{400} \] ✓ Answer: \[ \boxed{\frac{209}{400}} \]
  31. 31. The decimal expressions are not visible in the question provided. Please share the decimal numbers clearly to convert them into rational numbers.
  32. 32. Radius = 30/2 = 15 cm. Half chord = 24/2 = 12 cm. If d is distance from centre to chord: d^2 + 12^2 = 15^2 ⇒ d^2 + 144 = 225 ⇒ d^2 = 81 ⇒ d = 9 cm.
  33. 33. # (i) (625) Solution $$625=5\times5\times5\times5$$ $$625=5^4$$ Answer $$625=5^4$$ # (ii) (\frac15) Solution $$\frac15=5^{-1}$$ Answer $$\frac15=5^{-1}$$ # (iii) (\sqrt5) Using: $$\sqrt a=a^{\frac12}$$ Solution $$\sqrt5=5^{\frac12}$$ Answer $$\sqrt5=5^{\frac12}$$ # (iv) (\sqrt[3]{125}) Since: $$125=5^3$$ Solution $$\sqrt[3]{125}=(5^3)^{\frac13}$$ $$=5^1$$ $$=5$$ Answer $$5=5^1$$
  34. 34. \[ AP=\frac37 AB \] Find P. \[ AB=|3-(-4)|=7 \] \[ AP=\frac37\times7=3 \] Since P lies on x-axis: \[ P=(2,0) \] Check: \[ AP=3 \] ✓ Coordinates: \[ \boxed{(2,0)} \]
  35. 35. $$A\cup B = B\cup A$$ $$A\cap B = B\cap A$$ <div
  36. 36. Ordinate = y-coordinate = 4 Point on y-axis has x-coordinate = 0 ✓ Point: \[ \boxed{(2)\ (0,4)} \]
  37. 37. $$(A&#x27;)&#x27; = A$$
  38. 38. Properties of parallelogram: Opposite sides are equal. Opposite angles are equal. Adjacent angles are supplementary. ✓ Correct statement: \[ \boxed{(4)\ \text{Both pairs of opposite sides are equal}} \]
  39. 39. OP = √31 m
  40. 40. A' = {a, c, e, g} B' = {b, c, f, g} A' ∪ B' = {a, b, c, e, f, g} A' ∩ B' = {c, g} (A ∪ B)' = {c, g} (A ∩ B)' = {a, b, c, e, f, g} (A')' = A = {b, d, f, h} (B')' = B = {a, d, e, h}
  41. 41. \(PQ = 5\) cm \(PR = 6\) cm \(\angle QPR = 60^\circ\) Locate the centroid. Construction Steps 1. Draw: \[ PQ = 5 \text{ cm} \] 2. At point \(P\), construct: \[ \angle QPR = 60^\circ \] 3. On the ray mark: \[ PR = 6 \text{ cm} \] 4. Join: \[ QR \] Triangle \(PQR\) is formed. To Locate the Centroid 5. Find midpoint of \(QR\). Let it be \(A\). 6. Join: \[ PA \] 7. Find midpoint of \(PR\). Let it be \(B\). 8. Join: \[ QB \] 9. The medians intersect at \(G\). ✓ \(G\) is the centroid.
  42. 42. \(PQ = PR\) \(\angle Q = 50^\circ\) \(QR = 7\) cm Draw its circumcircle. Construction Steps 1. Draw: \[ QR = 7 \text{ cm} \] 2. At points \(Q\) and \(R\), construct: \[ 50^\circ \] angles. 3. Let the two rays intersect at \(P\). 4. Join: \[ PQ \text{ and } PR \] Triangle formed. Circumcentre 5. Draw perpendicular bisector of \(PQ\). 6. Draw perpendicular bisector of \(QR\). 7. Let them intersect at \(O\). ✓ \(O\) is the circumcentre. Circumcircle 8. With centre \(O\) and radius \(OQ\), draw the circle. ✓ Circumcircle obtained.
  43. 43. (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: \[ \boxed{\left(-\frac83,-\frac{11}{3}\right)} \]
  44. 44. \(AB = 6\) 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.
  45. 45. x = 62; angles: 72°, 108°, 114°, 66°.
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.Which one of the following regarding the sum of two irrational numbers is true?[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.Find the value of 3 sin 70° sec 20° + 2 sin 49° sec 51°.....[1]
4.In quadrilateral ABCD, AB = BC and AD = DC. The diagram gives ∠ABC = 108° and ∠ADC = 42°. Find ∠BCD.....[1]
5.Which one of the following is not a rational number?[1]
6.Which one of the following is an irrational number?[1]
7.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]
8.Which of the following statement is false?[1]
9.Which one of the following has a terminating decimal expansion?[1]
10.Which of the following are sets?[1]
11.Which of the following is not true?[1]
12.Without actual division, find which of the following rational numbers have terminating decimal expansion[1]
13.Which of the following is a linear equation?[1]
14.Which of the following expressions are polynomials? If not, give reason.[1]
15.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]
Part II — Fill in the Blanks 5 × 1 = 5

Fill in the blanks. (Answer all questions.)

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

Answer briefly. (Answer all questions.)

21.Probability of winning tennis match = 0.72[2]
22.Cubical milk tank holds[2]
23.Rewrite in standard form[2]
24.Plotting O(0,0), A(3,−4), B(3,4), C(0,4)[2]
25.Cuboid dimensions: Length = 20 cm, Breadth = 15 cm, Height = 8 cm. Find the Total Surface Area (TSA) and Lateral Surface Area (LSA).[2]
26.Find the following trigonometric table values: (i) sin 49°, (ii) cos 74°39', (iii) tan 54°26', (iv) sin 21°21', (v) cos 33°53', (vi) tan 70°17'.[2]
27.Two dice are rolled[2]
28.If bisectors of ∠A and ∠B of quadrilateral ABCD meet at O, then ∠AOB is[2]
29.Goalkeeper stops goal 32 times out of 40 attempts[2]
30.Mean weight of 4 family members is 60 kg[2]
31.Factorise the following: (i) (p - q)^2 - 6(p - q) - 16 (ii) m^2 + 2mn - 24n^2 (iii) √5 a^2 + 2a - 3√5 (iv) a^4 - 3a^2 + 2 (v) 8m^3 - 2m^2n - 15mn^2 (vi) 1/x^2 + 2/y^2 + 2/(xy)[2]
32.The mid-points of the sides of a triangle are (2,4), (−2,3) and (5,2). Find the vertices of the triangle.[2]
33.In Fig. 4.39, ∠A = 64°, ∠ABC = 58°. BO and CO are angle bisectors. Find x° and y°.[2]
34.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]
35.Find the remainder[2]
36.Represent the following sets in roster form.[2]
37.Ratio of sides of two cubes = 2:3[2]
38.If \(\sqrt{9x}=3\sqrt[3]{9^2}\), find \(x\).[2]
Part IV — Long Answer Questions 7 × 5 = 35

Answer in detail. (Answer all questions.)

39.Consider the following sets[5]
40.Perimeter of triangular plot = 600 m[5]
41.Draw triangle ABC where AB = 8 cm, BC = 6 cm, ∠B = 70°. Locate its circumcentre and draw the circumcircle.[5]
42.For the data[5]
43.In each case determine whether ∠A is supplementary to ∠B. Give reasons. (i) ∠A = 70°, ∠B = 110°. (ii) ∠A = 50°, ∠B = 130°. (iii) ∠A = 40°, ∠B = 140°.[5]
44.Solve by cross multiplication method[5]
45.Plot the points and identify the geometrical shape formed[5]
🔑 Show Answer Key — Set 3
  1. 1. 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 $$\boxed{(2)\ \text{may be a rational or irrational number}}$$
  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. 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.
  4. 4. (3) 105°.
  5. 5. 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 $$\boxed{(4)\ \sqrt{13}}$$
  6. 6. 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 $$\boxed{(4)\ \pi}$$
  7. 7. $$n(A\cup B\cup C)$$ $$=n(A)+n(B)+n(C)$$ $$-n(A\cap B)-n(B\cap C)-n(A\cap C)$$ $$+n(A\cap B\cap C)$$
  8. 8. 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 $$\boxed{(4)\ \sqrt{25}=\pm5}$$
  9. 9. 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 $$\boxed{(1)\ \frac5{64}}$$
  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. 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 $$\boxed{(3)\ \text{Every real number is an irrational number}}$$
  12. 12. 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 denom…
  13. 13. Options: 1. (x+\frac1x=2) 2. (x(x-1)=2) 3. (3x+5=\frac23) 4. (x^3-x=5) Answer $$\boxed{(3)\ 3x+5=\frac23}$$
  14. 14. # (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. Henc…
  15. 15. 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.
  16. 16. Options: 1. 1 2. 2 3. 3 4. 4 Answer $$\boxed{(3)\ 3}$$
  17. 17. Options: 1. (-1) 2. (0) 3. (1) 4. (2) Answer $$\boxed{(3)\ 1}$$
  18. 18. 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.
  19. 19. Options: 1. 3 2. 2 3. 1 4. 0 Answer $$\boxed{(4)\ 0}$$
  20. 20. Options: 1. (3) 2. (2) 3. (\frac23) 4. (\frac32) Solution $$2-3x=0$$ $$3x=2$$ $$x=\frac23$$ Answer $$\boxed{(3)\ \frac23}$$
  21. 21. Find probability of losing. Using: \[ P(\text{lose})=1-P(\text{win}) \] \[ =1-0.72 \] \[ =0.28 \] ✓ Answer: \[ \boxed{0.28} \]
  22. 22. \[ 125000\text{ litres} \] Find side length. Convert litres to cubic metres: \[ 125000\text{ litres}=125\text{ m}^3 \] Let side be \(a\). \[ a^3=125 \] \[ a=5 \] ✓ Side length: \[ \boxed{5\text{ m}} \]
  23. 23. # (i) $$x-9+\sqrt7x^3+6x^2$$ Standard form $$\sqrt7x^3+6x^2+x-9$$ # (ii) $$\sqrt2x^2-\frac72x^4+x-5x^3$$ Standard form $$-\frac72x^4-5x^3+\sqrt2x^2+x$$ # (iii) $$7x^3-\frac65x^2+4x-1$$ Already in standard form. # (iv) $$y^2+\sqrt5y^3-11-\frac73y+9y^4$$ Standard form $$9y^4+\sqrt5y^3+y^2-\frac73y-11$$
  24. 24. Observation: One pair of opposite sides are parallel. ✓ Figure formed: \[ \boxed{(3)\ \text{Trapezium}} \]
  25. 25. TSA = 2(lb + bh + hl) = 2[(20×15) + (15×8) + (8×20)] = 2(300 + 120 + 160) = 2(580) = 1160 cm² LSA = 2h(l + b) = 2×8×(20 + 15) = 16×35 = 560 cm² Answers: TSA = 1160 cm², LSA = 560 cm²
  26. 26. (i) sin49° ≈ 0.7547. (ii) cos74°39' ≈ 0.2647. (iii) tan54°26' ≈ 1.4010. (iv) sin21°21' ≈ 0.3642. (v) cos33°53' ≈ 0.8300. (vi) tan70°17' ≈ 2.7948.
  27. 27. (i) 0 (ii) \tfrac{1}{12} (iii) 1
  28. 28. (2) 1/2(∠C + ∠D)
  29. 29. Goals conceded = 40−32 = 8. Probability that a shot is a goal = 8/40 = 1/5.
  30. 30. Three weights: \[ 56\text{ kg},\ 68\text{ kg},\ 72\text{ kg} \] Find fourth weight. Total weight of family \[ 4\times60 = 240 \] Sum of known weights \[ 56+68+72 = 196 \] Fourth weight: \[ 240-196 = 44 \] ✓ Weight of fourth member: \[ \boxed{44\text{ kg}} \]
  31. 31. (i) Let t = p - q: (t-8)(t+2) ⇒ (p-q-8)(p-q+2) (ii) (m+6n)(m-4n) (iii) (√5 a - 3)(a + √5) (iv) (a^2 - 1)(a^2 - 2) = (a-1)(a+1)(a^2-2) (v) m(2m-3n)(4m+5n) (vi) 1/x^2 + 2/(xy) + 2/y^2 = (1/x + 1/y)^2 + 1/y^2 — not factorable further over rationals
  32. 32. Let D(2,4), E(−2,3), F(5,2). Vertices: A = E+F−D = (−2+5−2, 3+2−4) = (1,1). B = F+D−E = (5+2+2, 2+4−3) = (9,3). C = D+E−F = (2−2−5, 4+3−2) = (−5,5). Answer: (1,1), (9,3), (−5,5).
  33. 33. x = 29°, y = 29°.
  34. 34. 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).
  35. 35. # (i) $$p(x)=x^3-2x^2-4x-1$$ $$g(x)=x+1$$ Solution $$x+1=x-(-1)$$ Find: $$p(-1)$$ $$=(-1)^3-2(-1)^2-4(-1)-1$$ $$=-1-2+4-1$$ $$=0$$ Remainder $$0$$ # (ii) $$p(x)=4x^3-12x^2+14x-3$$ $$g(x)=2x-1$$ Solution $$2x-1=0$$ $$x=\frac12$$ Find: $$p\left(\frac12\right)$$ $$=4\left(\frac18\right)-12\left(\frac14\right)+14\left(\frac12\right)-3$$ $$=\frac12-3+7-3$$ $$=\frac32$$ Remainder $$\frac32$$ # (iii) $$p(x)=x^3-3x^2+4x+50$$ $$g(x)=x-3$$ Solution Find: $$p(3)$$ $$=27-27+12+50$$ $$=62$$ Remainder $$62$$
  36. 36. (i) $$A = {2,4,6,8,10,12,14,16,18}$$ (ii) $$B = \left{\frac12,\frac22,\frac32,\frac42,\frac52\right}$$ or $$B = \left{\frac12,1,\frac32,2,\frac52\right}$$ (iii) Perfect cubes between 27 and 216 are: $$64,\ 125$$ Hence, $$C = {64,125}$$ (iv) $$D = {-4,-3,-2,-1,0,1,2}$$
  37. 37. Surface area ratio: \[ 2^2:3^2 \] \[ 4:9 \] ✓ Answer: \[ \boxed{(2)\ 4:9} \]
  38. 38. Write powers of 3: \(9=3^2\). Then RHS = \(3\cdot9^{2/3}=3\cdot3^{4/3}=3^{7/3}.\) So \((9x)^{1/2}=3^{7/3}\). Square both sides: \(9x=3^{14/3}\). Since \(9=3^2\), divide: \(x=3^{14/3-2}=3^{8/3}.\) Hence \(x=3^{8/3}\) (approximately \(18.72\)).
  39. 39. $$A = {0, 3, 5, 8}$$ $$B = {2, 4, 6, 10}$$ $$C = {12, 14, 18, 20}$$ (a) State whether True or False (i) (18 \in C) → True (ii) (6 \notin A) → True (iii) (14 \notin C) → False (iv) (10 \in B) → True (v) (5 \in B) → False (vi) (0 \in B) → False (b) Fill in the blanks (i) (3 \in \boxed{A}) (ii) (14 \in \boxed{C}) (iii) (18 \boxed{\notin} B) (iv) (4 \boxed{\in} B)
  40. 40. Sides are in ratio: \[ 5:12:13 \] Let sides be: \[ 5x,\ 12x,\ 13x \] Then: \[ 5x+12x+13x=600 \] \[ 30x=600 \] \[ x=20 \] Sides: \[ 100,\ 240,\ 260 \] Semi-perimeter: \[ s=\frac{600}{2}=300 \] Area: \[ A = \sqrt{300(300-100)(300-240)(300-260)} \] \[ = \sqrt{300\times200\times60\times40} \] \[ = \sqrt{144000000} \] \[ =12000 \] ✓ Area: \[ \boxed{12000\text{ m}^2} \]
  41. 41. Circumcentre O is intersection of perpendicular bisectors of sides (e.g., AB and BC). Draw circumcircle with centre O and radius OA.
  42. 42. \[ 11,\ 15,\ 17,\ x+1,\ 19,\ x-2,\ 3 \] Mean is 14. Find: \(x\) Mode Find x Number of observations: \[ 7 \] Total sum: \[ 7\times14=98 \] Sum of observations: \[ 11+15+17+(x+1)+19+(x-2)+3 \] \[ =64+2x \] Thus: \[ 64+2x=98 \] \[ 2x=34 \] \[ x=17 \] Data values \[ 11,\ 15,\ 17,\ 18,\ 19,\ 15,\ 3 \] Arrange: \[ 3,\ 11,\ 15,\ 15,\ 17,\ 18,\ 19 \] Most frequent value: \[ 15 \] ✓ Value of \(x\): \[ \boxed{17} \] ✓ Mode: \[ \boxed{15} \]
  43. 43. (i) Given: ∠A = 70° ∠B = 110° Check: \[ \angle A + \angle B = 70^\circ + 110^\circ = 180^\circ \] ✓ Therefore, ∠A and ∠B are supplementary angles. (ii) Given: ∠A = 50° ∠B = 130° Check: \[ 50^\circ + 130^\circ = 180^\circ \] ✓ Therefore, the angles are supplementary. (iii) Given: ∠A = 40° ∠B = 140° Check: \[ 40^\circ + 140^\circ = 180^\circ \] ✓ Therefore, the angles are supplementary.
  44. 44. # (i) $$8x-3y=12$$ $$5x=2y+7$$ Rewrite second equation: $$5x-2y-7=0$$ First equation: $$8x-3y-12=0$$ Using cross multiplication $$\frac{x}{(-3)(-7)-(-2)(-12)} ============================ # \frac{y}{(-12)(5)-(-7)(8)} \frac{1}{8(-2)-5(-3)}$$ Simplify For (x) $$21-24=-3$$ For (y) $$-60+56=-4$$ Denominator $$-16+15=-1$$ Thus, $$\frac{x}{-3} ============ # \frac{y}{-4} \frac{1}{-1}$$ Therefore $$x=3$$ $$y=4$$ Answer $$x=3,\qquad y=4$$ # (ii) $$6x+7y-11=0$$ $$5x+2y-13=0$$ Cross multiplication $$\frac{x}{7(-13)-2(-11)} ======================= # \frac{y}{(-11)(5)-(-13)(6)} \frac{1}{6(2)-5(7)}$$ Simplify For (x) $$-91+22=-69$$ For (y) $$-55+78=23$$ Denominator $$12-35=-23$$ Thus, $$\frac{x}{-69}…
  45. 45. (i) Points: \[ (0,0),\ (-4,0),\ (-4,-4),\ (0,-4) \] Lengths: Horizontal side = 4 units Vertical side = 4 units All sides equal and all angles are right angles. ✓ Shape formed: \[ \boxed{\text{Square}} \] (ii) Points: \[ (-3,3),\ (2,3),\ (-6,-1),\ (5,-1) \] Observation: First pair has same y-coordinate: \[ y=3 \] Second pair has same y-coordinate: \[ y=-1 \] Only the horizontal pair of opposite sides is parallel. Their lengths are different: \[ 5\text{ units and }11\text{ units} \] So the figure is not a parallelogram. ✓ Shape formed: \[ \boxed{\text{Trapezium}} \] # Activity – 1 Plot: \(A(1,0)\) \(D(4,0)\) Find: \(AD\) \(DA\) Distance between A and D Since both points lie on x-axis: \[ AD…

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