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GENERAL MATHEMATICS
2024
GENERAL MATHEMATICS OLD PAPER
ALL QUESTION ANSWER
| SECTION – A |
Choose the correct answer:
1. The value of y in the blank space of the following table is—
| x | 1 | 2 | 4 | 8 |
| y | 32 | 16 | 8 | —- |
(A) 8
(B) 6
(C) 4
(D) 2
Ans: (C) 4
2. Which of the following is not a perfect square?
(A) 441
(B) 572
(C) 576
(D) 729
Ans: (B) 572
3. If m is the cube root of n, then the value of n is—
(A) √m
(B) 3√m
(C) m³
(D) m²
Ans: (C) m³
4. Given that the LCM of 306 and 657 is 22338. What is LCM of 102, 306 and 657?
(A) 102
(B) 22338
(C) 22338×3
(D) 22338×102
Ans: (B) 22338
5. Given two statements:
Statement (i) : The square of any positive odd integer 2k+1 is always 1 more than a multiple of 8.
Statement (ii) : The square of any positive odd integer 2k+1 is always 1 more than a multiple of 4.
Choose the correct alternative.
(A) Both (i) and (ii) are true.
(B) (i) is true but (ii) is false.
(C) (i) is false but (ii) is true.
(D) Both (i) and (ii) are false.
Ans: (A) Both (i) and (ii) are true.
6. Under what condition will px³ + qx² + rx + s = 0 be a cubic equation?
(A) p, q, r and s are all non-zero.
(B) p ≠ 0 and q ≠ 0
(C) p ≠ 0 or q ≠ 0
(D) p ≠ 0
Ans: (D) p ≠ 0
7. When x is real number, the graph of the cubic polynomial 8x³ – 1
(A) does not intersect the x-axis.
(B) intersects the x-axis at exactly one point.
(C) intersects the x-axis at two distinct points.
(D) intersects the x-axis at three distinct points.
Ans: (D) intersects the x-axis at three distinct points.
8. If the pair of linear equations a₁x + 3y + c₁ = 0 and 4x + b₂y + c₂ = 0 has a unique solution, then—
(A) a₁ = 3, b₂ = 4
(B) a₁ = 12, b₂ = 1
(C) a₁ = 4, b₂ = 3
(D) a₁ = 5, b₂ = 1
Ans: (D) a₁ = 5, b₂ = 1
9. The coordinates of any point on the x-axis are—
(A) (x, 0)
(B) (0, y)
(C) (x, x)
(D) (x, y)
Ans: (A) (x, 0)
10. The sum of the zeroes of the quadratic polynomial p(x) = 4x² – 1 is—
(A) -1
(B) 0
(C) 2
(D) 4
Ans: (B) 0
11. The common difference of the AP √2, √8, √18, √32 is-
(A) greater than the common difference of the AP 2, 4, 6, 8
(B) equal to the common difference of the AP √32, √18, √8
(C) equal to the common difference of the AP √50, √98, √162
(D) greater than the common difference of the AP 1/√2, 0 -1√2
Ans: (D) greater than the common difference of the AP 1/√2, 0 -1√2
12. If the base of a triangle is increased by 10% and the altitude is decreased by 10%, then the new area of the triangle
(A) remains the same.
(B) decreases by 1%
(C) increases by 10%
(D) increases by 11%
Ans: (A) remains the same.
13. The point R divides the line segments AB, such that AR = 3/4 AB is-
(A) 3:1
(B) 3:4
(C) 4:3
(D) 4:7
Ans: (A) 3:1
14. P(-1,0) is the centre of a circle and Q(2, 4) is a point on this circle. Three other points on this circle are-
(i) (-6, 0)
(ii) (-1, 5)
(iii) (3, -3)
(iv) (0, -5)
Choose the correct alternative.
(A) Any three of the above.
(B) (ii), (iii) and (iv) only.
(C) (i), (ii) and (iii) only.
(D) (i), (iii) and (iv) only.
Ans: (C) (i), (ii) and (iii) only.
15. The longest chord of a circle is called—
(A) radius.
(B) arc.
(C) diameter.
(D) major arc.
Ans: (C) diameter.
(A) 5
(B) 1/5
(C) 0
(D) -1
Ans: (B) 1/5
17. If the length of the tangent drawn from a point Q to a circle is 24 cm and the distance of Q from the centre of the circle is 25 cm, then the radius of the circle is—
(A) 7 cm
(B) 12 cm
(C) 15 cm
(D) 24.5 cm
Ans: (A) 7 cm
18. If two cubes each of volume 64 cm³ are joined end to end, then the surface area of the resulting cuboid is—
(A) 160 cm²
(B) 176 cm²
(C) 128 cm²
(D) 192 cm²
Ans: (A) 160 cm²
19. If the median of the data 25, 30, x + 30, 35 + x, 40 + x arranged in increasing order is 35, then the value of x is—
(A) 35
(B) 5
(C) 25
(D) 10
Ans: (A) 35
20. The probability of getting a prime number greater than 2 in a single throw of a die is—
(A) 1
(B) 1/6
(C) 2/3
(D) 1/3
Ans: (D) 1/3
21. If the area of a trapezium is 1350 m² and the sum of the lengths of its parallel sides is three times the highes, then the height of the trapezium is—
(A) 20 m
(B) 10 m
(C) 60 m
(D) 30 m
Ans: (D) 30 m
22. The smallest perfect square number which is divisible by 4, 9 and 10 is—
(A) 144
(B) 900
(C) 3600
(D) 360
Ans: (B) 900
23. If the sum of the zeroes of f(x) = kx² – 8x + 6 is 4, then the value of k is—
(A) 6
(B) 8
(C) 2
(D) 1
Ans: (C) 2
24. If a quadratic polynomial has two different zeroe, then the number of points in which the graph of the polynomial will intersect the x-axis is—
(A) 2
(B) 3
(C) 1
(D) 4
Ans: (A) 2
25. Which of the following equations is a linear equation in one variable?
(A) 2x = 3y
(B) x² – 3x + 5 = 0
(C) 3x + y = 0
(D) 3t + 7 = 8t – 2
Ans: (D) 3t + 7 = 8t – 2
26. The pair of equations kx + 2y = 5 and 3x + y = 1 will have a unique solution if
(A) k = 0
(B) k ≠ 6
(C) k = 2
(D) k = 3
Ans: (B) k ≠ 6
27. The number of roots of the equation (x + 2)³ = x³ – 4 is—
(A) 3
(B) 1
(C) 4
(D) 2
Ans: (D) 2
28. Which of the following equations has two real and equal roots?
(A) 3x²+14x-5=0
(B) 4x²+2x-1=0
(C) 9x²-6x+1=0
(D) x²-5x+4=0
Ans: (C) 9x²-6x+1=0
29. In an AP, the first term and last term are 1 and 11 respectively. If the sum of the terms in the AP is 36, then the number of terms is—
(A) 6
(B) 8
(C) 10
(D) 12
Ans: (A) 6
30. 42 is a term of which of the following arithmetic progressions?
(A) 92, 86, 80, …
(B) 102, 95, 88, …
(C) 2, 6, 10, …
(D) 0, 8, 16, …
Ans: (C) 2, 6, 10, …
31. In a ∆ABC, XY || BC. If AB = 4BX and YC 2 cm, then the value of AY is—
(A) 4 cm
(B) 6 cm
(C) 5 cm
(D) 8 cm
Ans: (B) 6 cm
32. If CM and RN are respectively the medians of ∆ABC and ∆PQR and if ∆ABC ~ ∆PQR, then which of the following is correct?
(А) ∆АМС ~ ∆PNR
(В) ∆АМС ~ ∆PRN
(C) ∆АМС ~ ∆NRP
(D) ∆AMC ~ ∆RNP
Ans: (А) ∆АМС ~ ∆PNR
33. The coordinates of the point A, where AB is a diameter of the circle whose centre is (2, -3) and B is (1, 4), are—
(A)
(B) (2, 8)
(C) (3, -10)
(D) (-2, 3)
Ans: (C) (3, -10)
34. The ratio in which the point (-4, 6) divides the line segment joining the points A (-6, 10) and B(3, -8) internally is—
(A) 3:2
(B) 2:3
(C) 7:2
(D) 2:7
Ans: (D) 2:7
35. If sinA = cos33°, A < 90°, then the value of A is—
(A) 90°
(B) 33°
(C) 27°
(D) 57°
Ans: (D) 57°
36. If atanθ = x, b cotθ = y, then the value of xy is—
(A) a + b
(B) -1
(C) 1
(D) ab
Ans: (D) ab
37. If tangents PA and PB drawn from a point P to a circle with centre 0 are inclined to each other at an angle of 80°, then the value of ∠POA is—
(A) 60°
(B) 50°
(C) 70°
(D) 80°
Ans: (B) 50°
38. The degree measure of the angle at the centre of a circle of radius r is θ. The length of an arc of the sector is—
(A) θπr/90°
(B) θπr/180°
(C) θπr/270°
(D) θπr/360°
Ans: (B) θπr/180°
39. The number of tangents drawn through a point inside a circle is—
(A) 0
(B) 1
(C) 2
(D) 3
Ans: (A) 0
40. If the circumference of a circle is 22 cm, then the area of a quadrant of the circle is—
(A) 77 cm²
(B) 77/2 cm²
(C) 77/8 cm²
(D) 77/4 cm²
Ans: (C) 77/8 cm²
41. The volume and surface area of a sphere are equa. The diameter of the sphere is—
(A) 3 units.
(B) 6 units.
(C) 2 units.
(D) 4 units.
Ans: (B) 6 units.
42. The ratio of the volume of a cone and a cylinder having same radius and height is—
(A) √3:1
(B) 1:3
(C) 1:2
(D) 3:1
Ans: (B) 1:3
43. If the difference of median and mode is 24, then the difference of mean and median is—
(A) 10
(B) 12
(C) 14
(D) 13
Ans: (B) 12
44. A number is selected from the first 100 natural numbers. The probability that the number is divisible by 8 is—
(A) 3/25
(B) 8/25
(C) 1/6
(D) 1/100
Ans: (A) 3/2
45. Which of the following cannot be the probability of an event?
(A) 0.225
(B) 0.6
(C) 1.2
(D) 1/3
Ans: (C) 1.2
| SECTION – B |
46. Factorise: 4x⁴ + 1
Ans: 4x⁴ + 1
= 4x⁴ + 4x² + 1 – 4x²
= (2x²)² + 2 × 2 × x² × 1 + 1² – (2x)²
= (2x² + 1)² – (2x)²
= (2x² + 1 + 2x) (2x² + 1 – 2x)
47. Two ropes are of lengths 64 cm and 80 cm. Both are to be cut into pieces of equal length. What should be the maximum length of the pieces?
Ans: 64 and 80 will be the maximum length of the piece.
80 = 64 × 1 + 16
64 = 16 × 4 + 0
∴ Maximum length of rope that can be cut out = 16 cm.
48. If ∠B and ∠Q are acute angles of right angled triangles ABC and PQR such that sinB = sinQ, then prove that ∠B = ∠Q.
Ans:
∴ SinB = AC/AB
SinQ = PR/PQ
∴ SinB = SinQ
⇒ AC/AB = PR/PQ
AC/PR = AB/PQ = BC/RQ = k
∵ AC/PR = AB/PQ = BC/RQ
∴ ∆ABC ~ ∆PQR
∴ ∠B = ∠Q
49. In ∆PQR, right angled at Q, PQ = 3 cm and PR = 6 cm. Determine ∠QPR and ∠PRQ.
Ans:
∆PQR ৰ, ∠Q = 90°
∴ Sin∠QPR = 3∫3/6 = ∫3/2 = Sin60°
⇒ ∠QPR = 60°
∴ ∠PRQ = 30°
50. Two dice, one blue and one grey, are thrown at the same time. Write down all the possible outcomes. What is the probability that the sum of the two numbers appearing on the top of the dice is 13?
Ans: All the possible outcomes are-
There is no result in favor of 13.
Therefore, P(Sum = 13) = 0/36 = 0
51. Solve the following pair of equations by reducing them to a pair of linear equation:
Ans:
⇒ 7x/xy – 2y/xy = 5, 8x/xy + 2y/xy = 15
⇒ 7/y – 2/x = 5, 8/y + 2/x = 5
Let, 1/x = u, 1/y = v
∴ 7v – 2u = 5 → (1)
8v + 2u = 15 → (2)
(1) + (2) ⇒ 7v – 2u + 8v + 24 = 5 + 15
⇒ 15v = 20
⇒ v = 20/15
⇒ v = 4/3
⇒ 1/y = 4/3
⇒ y = 3/4
∴ (1) ⇒ 7 × 4/3 – 2u = 5
⇒ 28/3 – 2u = 5
⇒ -2u = 5 – 28/3
⇒ -2u = 15 – 28/3
⇒ -2u = -13/3
⇒ u = 13/6
⇒ 1/x = 13/6
⇒ x = 6/13
52. Find two consecutive odd positive integers, the sum of whose squares is 290.
Ans: Let, there are two sequential odd numbers x and x + 2
x² + (x + 2)² = 290
⇒ x² + x² + 2 × x × 2 + 2² = 290
⇒ 2x² + 4x + 4 – 290 = 0
⇒ 2x² + 4x – 286 = 0
⇒ x² + 2x – 143 = 0
⇒ x² + 13x – 11x – 143 = 0
⇒ x(x + 13) – 11(x + 13) = 0
⇒ (x – 11) (x – 13) = 0
∴ x – 11 = 0 or, x + 13 = 0
⇒ x = 11 ⇒ x = -13 (impossible)
∴ The two decisive numbers are 11 and 13.
53. A sum of ₹1,000 is invested at 8% simple interest per year. Calculate the interest at the end of each year. Do these interests form an AP? If so, find the interest at the end of 30 years making use of this fact.
Ans: Capital = ₹1000
Interest in 1st year = 1000 × 8/100 = ₹80
Interest in 2nd year = 2 × 1000 × 8/100 = ₹160
Interest in 3rd year = 3 × 1000 × 8/100 = ₹240
∴ a₃ – a₂ = a₂ – a₁
⇒ 240 – 160 = 160 – 80
⇒ 80 = 80
a₃₀ = a + (30 – 1)d
= 80 + 29 × 80
= 80 + 2320
= ₹2400
54. Prove that if one angle of a triangle is equal to one angle of the other triangle and the sides including these angles are proportional, then the two triangles are similar.
Ans: Given, ∆ABC and ∆DEF ∠A = ∠D
AB/DE = AC/DF
Prove that, ∆ABC ~ ∆DEF
Drawing: Points P and Q are taken on DE and DF respectively such that DP = AB and DQ = AC PQ are connected.
Prove: ∆ABC and ∆DPQ
∠A = ∠D
Thus, ∆ABC ≌ ∆DPQ (S. A. A)
AB/DE = AC/DF
According to Theld’s contrary theorem,
PQ || EF
∆DPQ ~ ∆DEF (A. A)
∠A = ∠D
∠B = ∠E = ∠P
∠C = ∠F = ∠Q
∴ ∆ABC ~ ∆DEF (A A A)
55. Two opposite vertices of a square are (-1, 2) and (3, 2). Find the coordinates of other two vertices.
Ans: The coordinates of the two opposite vertices of the square ABCD are (-1, 2) and (3, 2). Let C be the vertex coordinate (x, y).
∴ Each side of square is equal.
∴ AC = BC
⇒ (AC)² = (BC)²
⇒ (x + 1)² + (y – 2)² = (x – 3)² + (y – 2)²
⇒ (x + 1)² = (x – 3)²
⇒ x² + 2x + 1 = x² – 6x + 9
⇒ 8x = 8
⇒ x = 8/8 = 1 …………… (1)
Now, from the right triangle ACB we get:
AC² + BC² = AB²
⇒ (x + 1)² + (y – 2)² + (x – 3)² + (y – 2)² = (3 + 1)² + (2 – 2)²
⇒ x² + 2x + 1 + y² – 4y + 4 + x² + 6x + 9 + y² – 4y + 4 = 16
⇒ 2x² + 2y² – 4x – 8y + 2 = 0
⇒ x² + y² – 2x – 4y + 1 = 0 ……………. (ii)
Now, let us put x = 1 in equation (ii)
(1)² + y² – 2 × 1 – 4y + 1 = 0
⇒ y² – 4y = 0
⇒ y(y – 4) = 0
∴ y = 0 ,4
∴ Determinable Vertex Data Coordinates (1, 4) and (1, 0)
56. Two tangents TP and TQ are drawn to a circle with centre 0 from an external point T. Prove that ∠PTQ = 2∠OPQ.
Ans:
Given, ‘O’ is the outer point T of the central circle and the two tangents TP and TQ.
Prove that, ∠PTQ = 2∠OPQ
Let, ∠PTQ = θ
TP = TQ [The lengths of the tangents drawn from the outer point to the circle are equal]
Hence ∆TPQ is an isosceles triangle.
∴ ∠TPQ = ∠TQP
∠PTQ + ∠TPQ + ∠TQP = 180°
⇒ θ + ∠TPQ + ∠TPQ = 180°
⇒ 2∠TPQ = 180 – θ
⇒ ∠TPQ = 90 – θ/2
Again, ∠OPT = 90°
⇒ ∠OPQ + ∠TPQ = 90°
⇒ ∠OPQ + 90° – θ/2 = 90°
⇒ ∠OPQ = 90 – 90 + θ/2
⇒ ∠OPQ = θ/2
⇒ 2∠OPQ = θ
⇒ 2∠OPQ = ∠PTQ
∴ ∠PTQ = 2∠OPQ
The tangent drawn at any point on the circle is perpendicular to the radius through the tangent.
57. The cost of fencing a circular field at the rate of ₹24 per metre is ₹5,280. The field is to be ploughed at the rate of ₹0.50 per m². Find the cost of ploughing the field.
Ans: Length of the field = 5280/24 = 220 m
⇒ Circumference = 220 m
⇒ 2πr = 220
⇒ 2 × 22/7 × r = 220
⇒ r = (220 × 7)/(2 × 22)
⇒ r = 35 m
∴ Area = πr²
= 22/7 × 35 × 35
= 22 × 5 × 35 m²
∴ Total cost = 22 × 5 × 35 × 0.50
= ₹1925
58. In a solid cylinder of height 12 cm and radius 5 cm, a conical hole is made. If the height and radius of the cone are same as that of the cylinder, then find the total surface area of the remaining solid.
Ans: Here, r = 5 cm
h = 12 cm
∴ The total surface area of the remaining aggregate = 2πrh + πr² + πrl
= 2π × 5 × 12 + π × 5² + π × 5 × 13
= 120π + 25π + 65π
= 210π cm²
59. Find the median of the following data which give the marks out of 50, obtained by 100 students in a test:
| Marks Obtained | 20 | 29 | 28 | 33 | 42 | 38 | 43 | 25 |
| Number of Students | 6 | 28 | 24 | 15 | 2 | 4 | 1 | 20 |
Ans:
| Marks Obtained | Number of Students | Cumulative frequency |
| 20 | 6 | 6 |
| 25 | 20 | 6 + 20 = 26 |
| 28 | 24 | 26 + 24 = 50 |
| 29 | 28 | 50 + 28 = 78 |
| 33 | 15 | 78 + 15 = 93 |
| 38 | 4 | 93 + 4 = 97 |
| 42 | 2 | 97 + 2 = 99 |
| 43 | 1 | 99 + 1 = 100 |
| n = 100 | ||
∴ Determinable median 28.5
60. Draw a triangle ABC with sides BC = 7 , ∠B = 45°, ∠A = 105°. Then construct a triangle whose sides are 4/3 times the corresponding sides of ∆ABC.
Ans: Drawing System Pressure:
(1) Right triangle ABC is drawn with AB = 7 cm. ∠A = 105° and ∠B = 45°
From the set properties of the angles of a triangle we get-
∠A + ∠B + ∠C = 180°
⇒ 105° + 45° + ∠C = 180°
⇒ ∠C = 180° – 150° = 30°
(2) Under the arm BC, draw the acute angle CBX at the B-point.
(3) Four points B₁, B₂, B₃, (in the ratio 4/3, the larger between 4 and 3) are placed on the arm BX
So that BB₁ = B₂B₃ = B₄B₅
(4) B₃ and C are connected.
(5) The point B₄ is the sum of the lines B₃C and a line BC is drawn. Intersects BC at the point C.
Rationale for the drawing system:
∆A’B’C’ and ∆ABC
In these two triangles, ∠B = ∠B (common angle)
∠A’C’B = ∠ACB (Drawing angle)
∴ ∆B’BC’ ≅ ∆ABC (A- A)
∴ A’B/AB = BC’/BC = C’A’/CA…………….(1)
Again, ∆B₄BC’ and ∆B₃BC
∠B = ∠B
∠C’B₄B = ∠CB₃B
∴ ∆B₄BC’ ≅ ∆B₃BC (A- A)
∴ B₄B/B₃B = BC’/BC = C’B₄/CB₃
BC’/BC = B₄B/B₃B
But, B₄B/B₃B = 4/3
∴ BC’/BC = 4/3 ……………(2)
Now, from (1) and (2) we get:
∴ The drawing system is reasonable. (Certified)
61. If α, β are the zeroes of the polynomial x² + bx + c, then show that α/β is a zero of the polynomial cx² – (b² – 2c)x + c.
Ans: Here, α + β = -b/a
⇒ α + β = -b/1
⇒ α + β = -b
α × β = c/a
⇒ α × β = c/1
⇒ α × β = c
Ca² – (b² – 2c) × x + c
∴ The Sum of two zeros = -{-b² – 2c)}/c
= b² – 2c/c
= (-b)² – 2c/c
= (α + β)² – 2αβ/αβ
= α² + β²/αβ
= (α × α)/αβ + (β × β)/αβ
= α/β + β/α
∴ Ca² – (b² – 2c) × x + c is a zero α/β
