Introduction: Humidity and temperature are common parameters to measure environmental conditions. In this Arduino based project we are going to measure ambient temperature and humidity and display it on a 16x2 LCD screen. A combined temperature and humidity sensor DHT11 is used with Arduino uno to develop this Celsius scale thermometer and percentage scale humidity measurement project. This project consists of three sections - one senses the humidity and temperature by using humidity and temperature sensor DHT11. The second section reads the DHT sensor module’s output and extracts temperature and humidity values into a suitable number in percentage and Celsius scale. And the third part of the system displays humidity and temperature on LCD Applications: It can be used for measuring humidity and temperature values in heating, ventilation and air conditioning systems. Weather stations also use these sensors to predict weather conditions. The humidity sensor is used as a preventi...
(DIPLOMA ALL BRANCHES)
SCHEME: I SEMESTER: I
NAME OF SUBJECT: BASIC MATHEMATICS
Subject Code: 22103
Question Bank for Multiple Choice Questions
Program: All Programs in Diploma Engineering Program Code: - CE/CO/ME/ET
Scheme:-I Semester:- 1
Course:- Basic Mathematics Course Code:- 22103
01 – Algebra Marks:-20
Content of Chapter:-
1.1 Logarithms
1.2 Determinants
1.3 Matrices
1.4 Partial Fractions
1.1 Logarithms
1. The value of : loga 1
(A) 1 (B) 2
(C) 0 (D) a
Answer: - Option C
Explanation: - Basic Property of logarithm
2. The value of: log10 10
(A) 1 (B) 2
(C) 0 (D) a
Answer: - Option A
Explanation: - Basic Property of logarithm
3. The value of loga a
(A) 1 (B) 2
(C) 0 (D) a
Answer: - Option A
Explanation: - Basic Property of logarithm
4. The value of: log81 3 = .
(A) 3 (B) 1 4
(C) 81 (D) 2 4
Answer: - Option B
Explanation: - Converting into Exponential form
5. The value of: log5 625=
(A) 4 (B) 5
(C)25 (D) 625
Answer: - Option A
Explanation: - Converting into Exponential form
6. The value of: log3 81=
(A) 81 (B) 3
(C) 1 (D) 4
Answer: - Option D
Explanation: - Converting into Exponential form
7. The value of : log343 7
1
(A) 3 1
(B) 4
1
(C) 6 1
(D) 5
Answer: - Option A
Explanation: - Converting into Exponential form
8. The value of x if log3 27 x
(A) 1 (B) 2
(C) 3 (D) 4
Answer: - Option C
Explanation: - Converting into Exponential form
9. The value of x if log 2 (x 3) 3
(A) 3 (B) 2
(C) 11 (D) 10
Answer: - Option C
Explanation:- Converting into Exponential form
10. The value of x if- log2 (x 6x 40) 5
2
(A) 4 (B) 2
(C) 4, 2 (D) 3
Answer: - Option C
Explanation: - Converting into Exponential form
11. The value of x if log3 x 6 2
(A) 3 (B) 6
(C) 2 (D)1
Answer: - Option A
Explanation: -Converting into Exponential form
12. log m
If m, n, a are positive real numbers and a 1 then a n
(A) loga m loga n (B) loga m loga n
(C) ) loga m loga n log a m
(D) log a n
Answer: - Option A
Explanation: - Basic Property of logarithm
13. If m and a are positive real numbers. a 1 then log a m
n
(A) loga m loga n (B) loga m loga n
(C) n loga m (D) loga m loga n
Answer: - Option C
Explanation: - Basic Property of logarithm
14. If m, n, a are positive real numbers then loga (mn)
(A) loga m loga n (B) loga m loga n
(C) loga m loga n log a m
(D) log a n
Answer: - Option B
Explanation: - Basic Property of logarithm
15. The value of x if log3 (x 5) 4
(A) x = 81 (B) x = 86
(C) x = 76 (D) x = 91
Answer: - Option C
Explanation: - Converting into Exponential form
16. log 2 log 4 log 8 .
The value of 3 5 15
(A)3 (B) 2
(C) 1 (D) 0
Answer: - Option D
Explanation: - Use logarithm of product and quotient
17. log 225 log 25 log 64
The value of 32 81 729
(A) log 5 (B) log 4
(C) log 2 (D) log 3
Answer: - Option C
Explanation: - Use logarithm of product and quotient
18. 1 1 1
The value of : log 3 6 log 8 6 log 9 6
(A) 3 (B)6
(C)8 (D) 9
Answer: - Option A
Explanation: - By Rule of change of base
19. 1 1 1
The value of log ab abc log bc abc log ac abc
(A) 4 (B) 3
(C) 2 (D) 1
Answer: - Option C
Explanation: - By Rule of change of base
20. p 2 q 2 r 2
log log log
The value of : qr rp pq
(A) 1 (B) 2
(C) 3 (D) 0
Answer: - Option D
Explanation: Use logarithm of product
21. The value of : log y x log y log z
2 3 4
z x
(A) 24 (B) 34
(C) 44 (D) 54
Answer: - Option A
Explanation: - Use log of power and Rule of change of base
22. 1 1 1
The value of : log 6 24 log12 24 log 8 24
(A) 1 (B) 2
(C) 3 (D) 4
Answer: - Option B
Explanation: - Use log of power and Rule of change of base
1.2 Determinants
23. 𝟓 𝟑
Value of determinant | | = …
𝟐 𝟒
(A) -14 (B) 14
(C) 12 (D) -12
Answer: - Option B
Explanation: - =(5X4)-(3X2)
24. 𝟐 −𝟒
Value of determinant | | = …
𝟐 −𝟏
(A) -2 (B) 0
(C)- 4 (D) 6
Answer: - Option D
Explanation: - =(2X(-1))-(2X(-4))
25. Value of determinant | 𝒔𝒊𝒏𝜽 𝒄𝒐𝒔𝜽| = …
−𝒄𝒐𝒔𝜽 𝒔𝒊𝒏𝜽
(A) 2 (B) 0
(C) -1 (D) 1
Answer: - Option D
Explanation: - By evaluating determinant
26. 𝟐 𝟑 𝟓
Value of determinant |𝟏 𝟒 𝟐| = …
𝟑 𝟏 𝟔
(A)-10 (B) 12
(C) -11 (D)-12
Answer: - Option C
Explanation:- By evaluating determinant
27. 𝟏 𝟎 𝟔
Value of determinant |𝟕 𝟐 𝟓| = …
𝟑 𝟒 𝟔
(A) 138 (B) 124
(C) 110 (D) 120
Answer: - Option B
Explanation: - By evaluating determinant
28. For the equation |𝒙 𝟒| = 0, the value of ‘x’ is …
𝟑 𝟔
(A) 0 (B) 2
(C) 1 (D) -1
Answer: - Option B
Explanation: - By evaluating determinant
29. For the equation |𝒙 𝟐| = |𝟏 𝟏|, the value of ‘x’ is …
𝟖 𝟒 𝟐 𝟐
(A) 4 (B) 8
(C) 2 (D) 6
Answer: - Option A
Explanation: - By evaluating determinant
30. For the equation |𝒙 −𝟒| = 0, the value of ‘x’ is …
𝟒 𝒙
(A) 4 (B) 0
(C) -4 (D) ±4
Answer: - Option D
Explanation: - By evaluating determinant
31. 𝟏 𝟏 𝟏
For the equation |𝟑 𝒙 𝟑| = 0, the value of ‘x’ is …
𝟏 𝒙 𝟐
(A) 0 (B) 5
(C) 9 (D) 3
Answer: - Option D
Explanation: - By evaluating determinant
32. The solution of the system of equations
x + y + z = 6, 2x + y - 2z = -2, x + y - 3z = -6 is
(A) 1,1,1 (B) 1,2,3
(C) 0,1,1 (D) 1,0,-1
Answer: - Option A
Explanation: - Use Cramer’s rule for finding values of x,y,z
33. The solution of the system of equations x + z = 4, y + z = 2, x + y = 0 is
(A) 3,1,0 (B) 1,-1,3
(C) 0,1,3 (D) ) -3,0,-1
Answer: - Option B
Explanation: Use Cramer’s rule for finding values of x,y,z
34. The voltages in an electric circuit are related by the equations. V1 + V2 + V3 = 9, V1 - V2 + V3 = 3, V1
+ V2 - V3 =1 Then values of V1 , V2 , V3 are
(A) 2,3,4 (B) 1,2,3
(C) 1,1,1 (D) 1,3,5
Answer: - Option A
Explanation: - Use Cramer’s rule for finding values of x,y,z
35. The value of determinant ‘D’ in the system of equations x - y - 2z = 1, 2x + 3y + 4z = 4, 3x - 2y - 6z = 5 is
(A) -8 (B)-16
(C) 8 (D) 16
Answer: - Option A
Explanation: -Writing all equations in determinant form and evaluate determinant D.
36. The value of ‘y’ in the system of equations x + y + z = 3, x - y + z = 1, x + y - 2z = 0 is
(A) 0 (B)1
(C) 2 (D) 3
Answer: - Option B
Explanation: -Use Cramer’s rule and solve for y
1.3 Matrices
37. In a square matrix …
(A) number of rows and columns are equal (B) number of rows and columns are not equal
(C) number of rows is greater than columns (D) number of columns is greater than rows
Answer: - Option A
Explanation: -By using definition of square matrix
38. Order of the matrix [1 2 -4 0] is …
(A) 1×1 (B) 1 × 4
(C) 4×1 (D) 4×4
Answer: - Option B
Explanation: - Order = number of rows into number of columns
39. Which of the following is scalar matrix…?
(A) [0 0
0 ]
0 (B) [3 0
0 ]
7
(C) ) [2 0]
0 0 (D) [5 0]
0 5
Answer: - Option D
Explanation: -Use Defination of Scalar matrix
40. If A = 𝟐 𝟑 𝟑 𝟕
[ ], B = [ ], then 2A+3B = …
𝟒 𝟓 𝟏 𝟗
(A) [12 35]
18 19 (B) [13 27
]
11 37
(C) 22 15] [
13 45 (D) [10 20
30 ]
15
Answer: - Option B
Explanation: - Use Multiplication to matrix by scalar and then addition of matrices
41. If [ 𝟒 𝟓] + 𝐗 = 𝟏𝟎 −𝟏], then matrix X = …
−𝟑 𝟔 [ 𝟎 −𝟔
(A) [3 6 ]
2 −10 (B) [6 −6 ]
3 −12
(C) [ 2 8
10 ]
9 (D) [5 −8
1 5 ]
Answer: - Option B
Explanation: -Use Subtraction of matrices
42. If [−𝟑 𝒙] + [ 𝟒 𝟔] = [ 𝟏 𝟕], then values ‘x’& ‘y’ are …
𝟐𝒚 𝟎 −𝟑 𝟏 −𝟓 𝟏
(A) 1,-2 (B) 0,1
(C) 2,0 (D) 1,-1
Answer: - Option D
Explanation: -. Use Addition and Equality of matrices
43. If order of matrix ‘A’ is 2×3 and order of matrix ‘B’ is 3×4, then order of their multiplication matrix
‘AB’ is …
(A) 2×2 (B) 3×3
(C) 2×4 (D) 4×4
Answer: - Option C
Explanation: - Use Inner product of Matrices
44. 𝟐 −𝟏
If A = 𝟑 𝟒 −𝟐
[ ], B = [𝟑 𝟒 ], then (AB) =
𝟐 𝟏 𝟎 𝟎 𝟐
(A) [18 9]
7 2 (B) [6 9 ]
7 −12
(C) 12 8
[ 7 ]
9 (D) 2 18 ]
[
6 −9
Answer: - Option A
Explanation: -. Use Multiplication of 2 Matrices
45. If A = [𝟓 𝟒], B = [−𝟑 𝟒 ], then AB = …
𝟒 𝟑 𝟒 −𝟓
(A) O (B) I
(C) A-1 (D) B-1
Answer: - Option B
Explanation: - Use Multiplication of 2 Matrices
46. If 𝟏 𝟐 𝒙 𝒚 𝟑 𝟕 𝟎 𝟕
[ ] [ ] = [ ], then values of ‘x’ & ‘y’…
𝟑 𝟐 𝟑 −𝟏 𝟐 𝟗 𝟒 𝟏𝟑
(A) 2,2 (B) 1,2
(C) ) 1,1 (D) 0,0
Answer: - Option B
Explanation: - Use Multiplication of 2 Matrices
47. 𝟏 𝟐
If A = [𝟑 𝟒], then AT = …
𝟓 𝟔
(A) [1 2 3] (B) [6 5 4]
4 5 6 3 2 1
(C) 1 3 5 (D) 1 3 5
2 4 6
Answer: - Option D
Explanation: -. By interchanging rows and coloumns
48. (AB)T = …
(A) AT.BT (B) BT.AT
(C) AT+BT (D) none of these
Answer: - Option B
Explanation: - Use Property of Transposition of Matrices.
49. (𝐀𝐓)𝐓= …
(A) O (B) A
(C) I (D) none of these
Answer: - Option B
Explanation: - Use Defination of Transposition of Matrices
50. A matrix ‘A’ is called orthogonal iff A.AT = …
(A) O (B) A
(C) I (D) none of these
Answer: - Option C
Explanation: - Use Property of Transposition of Matrices
51. The matrix ‘A’ is called singular iff |A| …
(A) = 0 (B) = I
(C) ≠ 0 (D) = A
Answer: - Option C
Explanation: - Use Defination of singular Matrix
52. |AB| = …
(A) |A| (B) |B|
(C) ) I (D) |A||B|
Answer: - Option D
Explanation: - Use Property of Determinant of Matrices
53. 𝟒 𝟔
𝟐 𝟑
(A) singular (B) non-singular
(C) symmetric (D) skew- symmetric
Answer: - Option A
Explanation: -. As determinant is zero hence singular matrix
54. 𝟏 𝟐 𝟑
In the matrix [−𝟒 𝟓 𝟔], then minor of element ‘6’ is …
−𝟕 𝟖 𝟗
(A) 12 (B) 32
(C) 42 (D) 22
Answer: - Option D
Explanation: - Solve determinant by eliminating second row and third coloumn.
55. 𝟐 −𝟑 𝟒
In the matrix [𝟎 𝟏 −𝟓], cofactor of element ‘0’ is …
𝟔 𝟐 −𝟒
(A) -4 (B) 4
(C) 20 (D) -20
Answer: - Option A
Explanation: - Solve determinant by eliminating second row and first column
56. The adjoint of matrix [𝟔 𝟓] is …
𝟐 𝟏
(A) [1 −9 (B) [6 9 ]
7 −12
(C) [ 1 −5 (D) [ 2 1 ]
−6 −9
Answer: - Option C
Explanation: - Find Cofactors and matrix of cofactors
57. Inverse of a square matrix ‘A’ exists, iff A is …
(A) singular (B) non-singular
(C) ) symmetric (D) skew- symmetric
Answer: - Option B
Explanation: - Condition Of Inverse of a matrix
58. A.A-1 = …
(A) AT (B) null matrix
(C) I (D) none of these
Answer: - Option C
Explanation: -. Property Of Inverse of a matrix
59. (𝐀−𝟏)−𝟏 = …
(A) O (B) A
(C) I (D) none of these
Answer: - Option B
Explanation: - Property Of Inverse of a matrix.
60. 𝟑
𝟏 𝟓], then A-1 = …
(A) [ 2 −5 (B) [6 1 ]
−1 5 −1
(C) [ 1 −5 (D) [ 2 1 ]
−2 −6 −9
Answer: - Option A
Explanation: - Find adjoint and determinant of matrix
61. 𝟐
𝟔 𝟒], then A-1 = …
(A) [−2 4 ] (B) [8 4
6 −8 2
(C) [ 2 −4 (D) [−8 4 ]
−6 6 −2
Answer: - Option D
Explanation: - Find adjoint and determinant of matrix
62. Which of the following is proper fraction?
(A) 𝑋−2 3
(B) 𝑋 +1
𝑋−2
4
𝑋3+1 3
(D) 𝑋 +1
𝑋4
Answer: - Option A
Explanation: - Defination of proper fraction
63. Which of the following is improper fraction?
(A) 𝑋−2 3
(B) 𝑋 +1
(𝑋−2)4
3
(C) 𝑋 +1
𝑋4 4
(D) 𝑋
𝑋3+1
Answer: - Option D
Explanation: - Defination of Improper fraction
64. Which of the following has irreducible Quadratic Denominator?
(A) 𝑥−2
𝑥2−4 (B) ) 𝑥−1
𝑥2+4
(C) 𝑥−4
𝑥3−9 (D) 𝑥−4
𝑥2−9
Answer: - Option B
Explanation: - Defination of irreducible quadratic factor
65. Which of the following has reducible Quadratic Denominator?
(A) 𝑥−2
𝑥2−4 (B) 𝑥−4
𝑥2+9
(C) 𝑥−4
𝑥3+9 (D) 𝑥−1
𝑥2+4
Answer: - Option A
Explanation: - Defination of reducible quadratic factor
66. The partial fractions of 𝒙 are
𝒙𝟐+𝒙−𝟐
(A) 1 [ 2 + 1 ]
3 𝑥+2 𝑥−1 (B) 1 [ 3 - 1 ]
2 X+2 X−1
(C) 1 [ 3 + 1 ]
2 𝑥+2 𝑥−1 (D) 1 [ 2 - 1 ]
3 𝑥+2 𝑥−1
Answer: - Option C
Explanation: - Denominator has non repeated linear factors
67. 𝒙
Values of A and B in partial fraction of 𝒆 +𝟏 are
(𝒆𝒙+𝟐)(𝒆𝒙+𝟑)
(A) -1,2 (B) 2,-1
(C) 2,1 (D) 1,2
Answer: - Option A
Explanation: - Denominator has non repeated linear factors, put ex = t
68. Partial fraction of 𝟏 are
𝒙𝟑−𝒙
(A) ) 1 [ 1 + 1 ]+ 𝟏
2 𝑥+1 𝑥−1 𝒙 (B) 1 [ 1 - 1 ]- 𝟏
2 𝑥+1 𝑥−1 𝒙
(C) 1 [ 1 - 1 ]+ 𝟏
2 𝑥+1 𝑥−1 𝒙 (D) 1 [ 1 + 1 ]- 𝟏
2 𝑥+1 𝑥−1 𝒙
Answer: - Option D
Explanation: - Denominator has non repeated linear factors
69. Partial fraction of 𝟏 are
𝟏−𝒙𝟐
(A) 1 [ 1 + 1 ]
2 1+𝑥 1−𝑥 (B) 1 [ 1 - 1 ]
2 1+𝑥 𝑥−1
(C) 1 [ 1 - 1 ]
2 1+𝑥 1−𝑥 (D) 1 [ 1 + 1 ]
2 1+𝑥 𝑥−1
Answer: - Option A
Explanation: - Denominator has non repeated linear factors
70. Partial fraction of 𝟏 are
𝒙𝟐−𝒙
(A) 1 - 1
𝑥−1 𝑥 (B) 1 - 1
𝑥 𝑥−1
(C) 1 + 1
𝑥 𝑥−1 (D) 1 + 1
𝑥−1 𝑥
Answer: - Option A
Explanation: - Denominator has non repeated linear factors
71. Partial fraction of 𝟏 are
𝒙𝟐+𝟑𝒙+𝟐
(A) 1 - 1
𝑥+2 𝑥+1 (B) 1 - 1
𝑥+1 𝑥+2
(C) 1 + 1
𝑥+2 𝑥+1 (D) 1 + 2
𝑥+2 𝑥+1
Answer: - Option B
Explanation: - Denominator has non repeated linear factors
72. 𝟐
Proper fraction after polynomial division of 𝒙 +𝟏 is
𝒙𝟐−𝟏
(A) 2
𝑥2−1 (B) 2
𝑥2+1
(C) 1
𝑥2−1 (D) 1
𝑥2+1
Answer: - Option A
Explanation: - Taking actual polynomial division
73. 𝟒
Proper fraction after polynomial division of 𝒙 is
𝒙𝟑+𝟏
(A) 𝑥
𝑥3+1 (B) −𝑥
𝑥3+1
(C) 𝑥
𝑥3−1 (D) −𝑥
𝑥3−1
Answer: - Option B
Explanation: - Converting it into proper fraction
74. Partial fraction of 𝟏 are
𝒙𝟐−𝟏
(A) 1 [ 1 - 1 ]
2 𝑥−1 𝑥+1 (B) 1 [ 1 - 1 ]
2 𝑥+1 𝑥−1
(C) 1 [ 1 + 1 ]
2 X−1 X+1 (D) 1 [ 1 − 1 ]
3 𝑥+1 𝑥−1
Answer: - Option A
Explanation: - Denominator has non repeated linear factors
75. Partial fraction of 𝒙−𝟐 are
𝒙𝟐−𝒙
(A) 2 + 1
𝑥 𝑥2−1 (B) 2 - 1
𝑥 𝑥−1
(C) 2 + 1
𝑥 𝑥2+1 (D) 2 _ 1
𝑥 𝑥2+1
Answer: - Option B
Explanation: - Denominator has non repeated linear factors
76. 𝟐
Partial fraction of 𝒙 −𝒙+𝟑 are
(𝒙−𝟐)(𝒙𝟐+𝟏)
(A) 1 + 1
𝑥−2 𝑥2+1 (B) 1 - 1
𝑥−2 𝑥2+1
(C) 1 - 1
𝑥−2 𝑥2−1 (D) 1 - 1
𝑥+2 𝑥2−1
Answer: - Option B
Explanation: - Denominator has irreducible quadratic factor
77. Partial fraction of 𝒙+𝟒 are
𝒙𝟐+𝒙
(A) 4 - 3
𝑥 𝑥−1 (B) 4 + 3
𝑥 𝑥−1
(C) 4 + 3
𝑥 𝑥+1 (D) 4 - 3
𝑥 𝑥+1
Answer: - Option D
Explanation: - Denominator has non repeated linear factors
78. Partial fraction of 𝒙−𝟏 are
𝒙(𝒙𝟐+𝟏)
(A) 𝑥+1 - 1
𝑥2+1 𝑥 (B) 𝑥+1 - 1
𝑥2−1 𝑥
(C) 𝑥+1 + 1
𝑥2+1 𝑥 (D) 𝑥−1 - 1
𝑥2+1 𝑥
Answer: - Option A
Explanation: - Denominator has irreducible quadratic factor
79. 𝟑
Partial fraction of 𝒙 are
𝒙𝟐−𝟏
(A) 𝑥 + 1 [ 1 + 1 ]
2 𝑥−1 𝑥+1 (B) 𝑥 − 1 [ 1 - 1 ]
2 𝑥−1 𝑥+1
(C) 𝑥 − 1 [ 1 + 1 ]
2 𝑥−1 𝑥+1 (D) 𝑥 + 1 [ 1 - 1 ]
2 𝑥−1 𝑥+1
Answer: - Option A
Explanation: - Converting it into proper fraction
80. Partial fraction of 𝒙−𝟐 are
𝒙𝟐−𝒙
(A) 2 + 1
𝑥 𝑥−1 (B) 2 - 1
𝑥 𝑥−1
(C) 2 - 1
𝑥 𝑥+1 (D) 2 + 3
𝑥 𝑥+1
Answer: - Option B
Explanation: - Denominator has non repeated linear factors
81. Partial fraction of 𝒙+𝟏 are
𝒙𝟑−𝒙𝟐
(A) 2 - 2 + 1
𝑥−1 𝑥 𝑥2 (B) 2 - 2 + 1
𝑥+1 𝑥 𝑥2
(C) 2 - 2 − 1
𝑥−1 𝑥 𝑥2 (D) 2 + 2 - 1
𝑥−1 𝑥 𝑥2
Answer: - Option C
Explanation: - Denominator has repeated linear factors
Question Bank for Multiple Choice Questions
Program: All Programs in Diploma Engineering Program Code: - CE/CO/ME/ET
Scheme: - I Semester: - 1
Course: - Basic Mathematics Course Code: - 22103
02 – Trigonometry Marks: - 14
Content of Chapter:-
5.1 Range, coefficient of range of discrete and grouped data.
5.2 Mean deviation and standard deviation from mean of grouped and ungrouped data, weighted means.
5.3 Variance and coefficient of variance.
5.4 Comparison of two sets of observation.
1. If A and B are two angles then A+B is called …..
(A) Compound Angle (B) Allied Angle
(C) Multiple Angle (D) Sub-multiple angle
Answer: - Option A
Explanation: -. By definition of compound angle
2. 𝐬𝐢𝐧(𝑨 + 𝑩)=
(A) sin 𝐴 cos 𝐵 − cos 𝐴 sin 𝐵 (B) cos 𝐴 cos 𝐵 + sin 𝐴 sin 𝐵
(C) cos 𝐴 cos 𝐵 − sin 𝐴 sin 𝐵 (D) sin 𝐴 cos 𝐵 + cos 𝐴 sin 𝐵
Answer: - Option D
Explanation: - By Formula sin(𝐴 + 𝐵) = sin 𝐴 cos 𝐵 + cos 𝐴 sin 𝐵
3. 𝐬𝐢𝐧(𝑨 − 𝑩) =
(A) sin 𝐴 cos 𝐵 − cos 𝐴 sin 𝐵 (B) cos 𝐴 cos 𝐵 + sin 𝐴 sin 𝐵
(C) cos 𝐴 cos 𝐵 − sin 𝐴 sin 𝐵 (D) sin 𝐴 cos 𝐵 + cos 𝐴 sin 𝐵
Answer: - Option A
Explanation: -. By Formula sin(𝐴 − 𝐵) = sin 𝐴 cos 𝐵 − cos 𝐴 sin 𝐵
4. 𝐜𝐨𝐬(𝑨 − 𝑩) =
(A) sin 𝐴 cos 𝐵 − cos 𝐴 sin 𝐵 (B) cos 𝐴 cos 𝐵 + sin 𝐴 sin 𝐵
(C) cos 𝐴 cos 𝐵 − sin 𝐴 sin 𝐵 (D) sin 𝐴 cos 𝐵 + cos 𝐴 sin 𝐵
Answer: - Option B
Explanation: - By Formula 𝐜𝐨𝐬(𝑨 − 𝑩) = 𝐜𝐨𝐬 𝑨 𝐜𝐨𝐬 𝑩 + 𝐬𝐢𝐧 𝑨 𝐬𝐢𝐧 𝑩
𝐜𝐨𝐬(𝑨 + 𝑩) =
5.
(A) sin 𝐴 cos 𝐵 − cos 𝐴 sin 𝐵 (B) cos 𝐴 cos 𝐵 + sin 𝐴 sin 𝐵
(C) cos 𝐴 cos 𝐵 − sin 𝐴 sin 𝐵 (D) sin 𝐴 cos 𝐵 + cos 𝐴 sin 𝐵
Answer: - Option C
Explanation: - By Formula cos(𝐴 + 𝐵) = cos 𝐴 cos 𝐵 − sin 𝐴 sin 𝐵
6. (𝑨 − 𝑩) =
(A) tan 𝐴 + tan 𝐵 1+tan 𝐴 tan 𝐵
(C) tan 𝐴 − tan 𝐵 1+tan 𝐴 tan 𝐵
Answer: - Option C
(B) tan 𝐴 + tan 𝐵 1−tan 𝐴 tan 𝐵
(D) tan 𝐴 − tan 𝐵 1−tan 𝐴 tan 𝐵
Explanation: -. By Formula tan(𝐴 − 𝐵) = tan 𝐴 − tan 𝐵
1+tan 𝐴 tan 𝐵
7. (𝑨 + 𝑩) = _
(A) tan 𝐴 + tan 𝐵
1+tan 𝐴 tan 𝐵
(C) tan 𝐴 − tan 𝐵 1+tan 𝐴 tan 𝐵
Answer: - Option A
Explanation: - According to definition of Variance.
8. 𝐬𝐢𝐧 𝑨 𝐜𝐨𝐬 𝑩 + 𝐜𝐨𝐬 𝑨 𝐬𝐢𝐧 𝑩 is an expansion of…
(B) tan 𝐴 + tan 𝐵 1−tan 𝐴 tan 𝐵
(D) tan 𝐴 − tan 𝐵 1−tan 𝐴 tan 𝐵
(A) cos(𝐴 − 𝐵) (B) sin(𝐴 − 𝐵)
(C) sin(𝐴 + 𝐵) (D) cos(𝐴 + 𝐵)
Answer: - Option C
Explanation: - By Formula sin(𝐴 + 𝐵) = sin 𝐴 cos 𝐵 + cos 𝐴 sin 𝐵
9. 𝐬𝐢𝐧 𝑨 𝐜𝐨𝐬 𝑩 − 𝐜𝐨𝐬 𝑨 𝐬𝐢𝐧 𝑩 is an expansion of…
(A) cos(𝐴 − 𝐵) (B) sin(𝐴 − 𝐵)
(C) sin(𝐴 + 𝐵) (D) cos(𝐴 + 𝐵)
Answer: - Option B
Explanation: - By Formula sin(𝐴 − 𝐵) = sin 𝐴 cos 𝐵 − cos 𝐴 sin 𝐵
10. 𝐜𝐨𝐬 𝑨 𝐜𝐨𝐬 𝑩 − 𝐬𝐢𝐧 𝑨 𝐬𝐢𝐧 𝑩 is an expansion of…
(A) cos(𝐴 − 𝐵) (B) sin(𝐴 − 𝐵)
(C) sin(𝐴 + 𝐵) (D) cos(𝐴 + 𝐵)
Answer: - Option D
Explanation: - By Formula cos(𝐴 + 𝐵) = cos 𝐴 cos 𝐵 − sin 𝐴 sin 𝐵
11. 𝐜𝐨𝐬 𝑨 𝐜𝐨𝐬 𝑩 + 𝐬𝐢𝐧 𝑨 𝐬𝐢𝐧 𝑩 is an expansion of…
(A) cos(𝐴 − 𝐵) (B) sin(𝐴 − 𝐵)
(C) sin(A + B) (D) cos(𝐴 + 𝐵)
Answer: - Option A
Explanation: - By Formula cos(𝐴 − 𝐵) = cos 𝐴 cos 𝐵 + sin 𝐴 sin 𝐵.
12. 𝐭𝐚𝐧 𝑨 − 𝐭𝐚𝐧 𝑩 is an expansion of
𝟏+𝐭𝐚𝐧 𝑨 𝐭𝐚𝐧 𝑩
(A) tan(𝐴 + 𝐵) (B) tan 2𝐴
(C) tan(𝐴 − 𝐵) (D) tan 2𝐵
Answer: - Option C
Explanation: - By Formula tan(𝐴 − 𝐵) = tan 𝐴 − tan 𝐵
1+tan 𝐴 tan 𝐵
13. 𝐭𝐚𝐧 𝑨 + 𝐭𝐚𝐧 𝑩 is an expansion of
𝟏−𝐭𝐚𝐧 𝑨 𝐭𝐚𝐧 𝑩
(A) tan(𝐴 + 𝐵) (B) tan 2𝐴
(C) tan(𝐴 − 𝐵) (D) tan 2𝐵
Answer: - Option A
Explanation: - By Formula tan(𝐴 + 𝐵) = tan 𝐴 + tan 𝐵
1−tan 𝐴 tan 𝐵
14. Find the value of 𝐜𝐨𝐬(𝟕𝟓°)
(A) 1−√3
2√2 (B) √𝟑−𝟏
𝟐√𝟐
(C) 1+√3
2√2
(D) √3+1
2√2
Answer: - Option B
Explanation: - cos(750) = cos(450 + 300)
Now apply formula cos(𝐴 + 𝐵) = cos 𝐴 cos 𝐵 − sin 𝐴 sin 𝐵 for finding value of cos(75°)
15. If 𝐭𝐚𝐧 𝑨 = 𝟏 and 𝐭𝐚𝐧 𝑩 = 𝟏 then 𝐭𝐚𝐧(𝑨 + 𝑩) is
𝟐 𝟑
(A) 1 (B) -1
(C) 0 (D) 2
Answer: - Option A
Explanation: - Solve by using tan(𝐴 + 𝐵) = tan 𝐴 + tan 𝐵
1−tan 𝐴 tan 𝐵
16. 𝐬𝐢𝐧 𝑎 𝐜𝐨𝐬(𝖰 − 𝑎) + 𝐜𝐨𝐬 𝑎 𝐬𝐢𝐧(𝖰 − 𝑎) is equal to
(A) cos 𝛼 (B) cos 𝛽 − 𝛼
(C) sin 𝛽 − 𝛼 (D) sin 𝛽
Answer: - Option D
Explanation: - Using formula sin(𝐴 + 𝐵) = sin 𝐴 cos 𝐵 + cos 𝐴 sin 𝐵
17. 𝐜𝐨𝐭 𝑨−𝐜𝐨𝐭 𝟐𝑨 = 𝐬𝐢𝐧 𝑨 is
𝐜𝐨𝐭 𝑨+𝐜𝐨𝐭 𝟐𝑨 𝐬𝐢𝐧 𝟑𝑨
(A) 𝐬𝐢𝐧 𝑨
𝐬𝐢𝐧 𝟑𝑨 (B) 𝐜𝐨𝐬 𝑨
𝐜𝐨𝐬 𝟑𝑨
(C) 𝐭𝐚𝐧 𝑨
𝐭𝐚𝐧 𝟑𝑨 (D) None of These
Answer: - Option A
Explanation: - Use cot 𝐴 = cos 𝐴 and cot 2𝐴 = cos 2𝐴 ,
sin 𝐴 sin 2𝐴
Simplify it.
Then use sin(𝐴 + 𝐵) = sin 𝐴 cos 𝐵 + cos 𝐴 sin 𝐵
and sin(𝐴 − 𝐵) = sin 𝐴 cos 𝐵 − cos 𝐴 sin 𝐵
You will find above result is true.
18. If 𝐭𝐚𝐧 𝑨 = 𝟏 and 𝐭𝐚𝐧 𝑩 = 𝟐 then 𝐭𝐚𝐧 𝑪 = , Where A, B, C are the angles of a triangle.
(A) 2 (B) 1
(C) 4 (D) 4
Answer: - Option C
Explanation: - Since A, B, C are angles of triangle ∴ 𝐴 + 𝐵 + 𝐶 = 1800
=> 𝐴 + 𝐵 = 1800 − 𝐶
Operate tangent ratio on both side and use
tan(𝐴 + 𝐵) = tan 𝐴 + tan 𝐵 and tan(𝜋 − 𝜃) = − tan 𝜃
1−tan 𝐴 tan 𝐵
19. Find the value of 𝐬𝐢𝐧(𝟏𝟓°)
(A) 1−√3
2√2 (B) √𝟑−𝟏
𝟐√𝟐
(C) 1+√3
2√2
(D) √3+1
2√2
Answer: - Option B
Explanation: - Using compound angle formula cos(𝐴 + 𝐵)
Find the value of 𝐜𝐨𝐬𝐞𝐜(𝟏𝟎𝟓°)
20. (A) 1−√3
2√2 (B) 2√2
√3+1
(C) 2√2
1+√3 (D) Both b and c
Answer: - Option D
Explanation: - Using relation cosec 𝜃 = 1 and compound angle formula sin(𝐴 + 𝐵)
sin 𝜃
21. Find the value of 𝐭𝐚𝐧(𝟏𝟓°)
(A) √𝟑−𝟏
√𝟑+𝟏 (B) √3+1
√3−1
(C) 1+√3
2√2
(D) √3−1
2√2
Answer: - Option A
Explanation: - Using compound angle formula tan(𝐴 − 𝐵)
22.
If A, B, and C are angles of a triangle, then we can write 𝐜𝐨𝐭 (𝑩+𝑪) as
𝟐
(A)
𝐭𝐚𝐧(
𝑨 )
𝟐
(B) tan( 𝐵+𝐶 )
2
(C)
tan(𝐵+𝐶) 2
(D)
tan( 𝐵 )
2
Answer: - Option A
Explanation: - Since A, B, C are angles of triangle ∴ 𝐴 + 𝐵 + 𝐶 = 1800 => 𝐵 + 𝐶 = 1800 − 𝐴
Operate cotangent ratio on both side
23.
Let 𝐜𝐨𝐬 (𝑎 + 𝖰) = 𝟒
𝟓
(A) ) 25
16
(C) 19
12
Answer: - Option B
and 𝐥𝐞𝐭 𝐬𝐢𝐧 (𝑎 − 𝖰) = 𝟓 , where
𝟏𝟑
(B) 𝟓𝟔
𝟑𝟑
(D) 19
12
𝟎 ≤ 𝑎, 𝖰 ≤
𝝅.
𝟒
Then 𝐭𝐚𝐧 𝟐𝑎 =?
24.
Explanation: - Using trigonometric formulae 𝑠𝑖𝑛2 𝐴 = 𝑐𝑜𝑠2𝐴 − 1 and 𝑐𝑜𝑠2𝐴 = 𝑠𝑖𝑛2 𝐴 − 1, And compound angle formulae
Find the value of (−𝟕𝟔𝟓𝟎)
(A) − 1
√2
(C) − 1
2
Answer: - Option A
(B) 1
√2
(D) 1
2
Explanation: - sin(−7650) = − sin 7650 = − sin(2 × 3600 + 450) Since sin(2𝜋 + 𝜃) = sin 𝜃
Find the value of (𝟏𝟎𝟓𝟎𝟎)
25.
(A) 1
2
(C) 1
3
Answer: - Option D
Explanation: - Use allied angle concept
(B) 1
√3
(D) − 1
√3
26. The value of cos (90° + 8) sec (-8) tan 180° - 8) / sec (360° - 8) sin 180° + 8) cot (90° - 8) is
(A) cos8 (B) 1
(C) sin8 (D) -1
Answer: - Option D
Explanation: -
27. The value of sin (180° + 8) cot (90° - 8) / sec (-8) + sin 28 is
(A) -1 (B) 1
(C) 1 (D) -2
Answer: - Option C
Explanation: -
28. The value of tan 720° - cos 630° - sin 150° cos 120° is
(A) ¼ (B) 1/3
(C) ½ (D) 1
Answer: - Option A
Explanation: -
29. 𝝅
𝐬𝐢𝐧( − 𝜽) = _
𝟐
(A) sin 𝜃 (B) cos 𝜃
(C) − sin 𝜃 (D) − cos 𝜃
Answer: - Option B
Explanation: - Standard Allied angle ratio
30. 𝝅
𝐜𝐨𝐬( + 𝜽) =
𝟐
(A) sin 𝜃 (B) cos 𝜃
(C) − sin 𝜃 (D) − cos 𝜃
Answer: - Option C
Explanation: - Standard Allied angle ratio
31. 𝐭𝐚𝐧(𝝅 − 𝜽) =
(A) tan 𝜃 (B) − tan 𝜃
(C) cot 𝜃 (D) − cot 𝜃
Answer: - Option B
Explanation: - Standard Allied angle ratio
32. 𝐜𝐨𝐭(𝝅 + 𝜽) =
(A) tan 𝜃 (B) − tan 𝜃
(C) cot 𝜃 (D) − cot 𝜃
Answer: - Option C
Explanation: - Standard Allied angle ratio
33. 𝐜𝐨𝐬𝐞𝐜(𝟐𝝅 − 𝜽) =
(A) cosec 𝜃 (B) − sec 𝜃
(C) cot 𝜃 (D) − cosec 𝜃
Answer: - Option D
Explanation: - Standard Allied angle ratio
34.
Find the value of ( 𝟏𝟗𝝅)
𝟔
(A) √3 (B) − √3
(C) 3 (D) -3
Answer: - Option A
Explanation: - Solved by using allied angle formula cot( 3𝜋 + 𝜃) = 𝑐𝑜𝑡𝜃
Find the value of (𝟐𝟐𝟓𝟎) 𝐜𝐨𝐭(𝟒𝟎𝟓𝟎) + 𝐭𝐚𝐧(𝟕𝟔𝟓𝟎) 𝐜𝐨𝐭(𝟕𝟔𝟓𝟎)
35.
(A) 1 (B) 2
(C) -2 (D) -1
Answer: - Option B
Explanation: - Use allied angle formula for tangent 𝑎𝑛𝑑 cotangent ratio also use tan . cot 𝜃 = 1
36. If 𝜽 be the angle then 2𝜽, 3𝜽, 4𝜽, are called as…….
(A) Compound angles (B) Allied Angle
(C) Multiple angle (D) Sub-multiple angles
Answer: - Option C
Explanation: - By definition of Multiple angles
37.
If 𝜽 be the angle then 𝜽
𝟐
, 𝜽
𝟑
, 𝜽 are called as …….
𝟒
(A) Compound angles (B) Allied Angle
(C) Multiple angle (D) Sub-multiple angles
Answer: - Option D
Explanation: - By definition of Multiple angles
38.
Find 𝐬𝐢𝐧 𝑎 if
(A) ) 𝟏
√𝟑
𝐭𝐚𝐧(
𝑎) =
𝟐
𝟏
√𝟑
(B) √𝟑
𝟐
(C) √3 (D) 1
Answer: - Option B
𝛼
Explanation: - By using Multiple and sub-multiple angle formulae sin 𝛼 = 2 tan(2)
1+𝑡𝑎𝑛 (2)
39. What is 𝐜𝐨𝐭𝐀 + 𝐜𝐨𝐬𝐞𝐜𝐀 𝐢𝐬 𝐞𝐪𝐮𝐚𝐥 𝐭
(A)
tan(𝐴)
2
(B)
cot(𝐴)
2
(C)
2cot(A)
2
(D)
2tan(A)
2
Answer: - Option B
Explanation: - Concept cos 2A = cos2A – 1 and sin2A = 2sinA cosA
40. If 𝐬𝐢𝐧 𝑨 = 𝟎. 𝟒 then find value of 𝐜𝐨𝐬 𝟐𝑨
(A) 0.50 (B) 0.68
(C) 0.60 (D) 1
Answer: - Option B
Explanation: - Use cos 2𝐴 = 1 − 2 𝑠𝑖𝑛2𝐴 to solve above example
41. What is 𝒔𝒊𝒏 𝑪 + 𝒔𝒊𝒏 𝑫 equal to
(A)
𝐶+𝐷
2 𝑠𝑖𝑛 (
2
𝐶−𝐷
) . 𝑐𝑜𝑠 (
2
𝐶+𝐷
−2 𝑠𝑖𝑛 (
2
𝐶−𝐷
) . 𝑠𝑖𝑛 ( )
2
(C)
𝐶+𝐷
2 𝑐𝑜𝑠 (
2
𝐶−𝐷
) . 𝑠𝑖𝑛 (
2
𝐶+𝐷
2 𝑐𝑜𝑠 (
2
𝐶−𝐷
) . 𝑐𝑜𝑠 ( )
2
Answer: - Option A
Explanation: - By definition of Multiple angles
42. What is 𝒔𝒊𝒏 𝑪 − 𝒔𝒊𝒏 𝑫 equal to?
(A)
𝐶+𝐷
2 𝑠𝑖𝑛 (
2
𝐶−𝐷
) . 𝑐𝑜𝑠 (
2
𝐶+𝐷
−2 𝑠𝑖𝑛 (
2
𝐶−𝐷
) . 𝑠𝑖𝑛 ( )
2
(C)
𝐶+𝐷
2 𝑐𝑜𝑠 (
2
𝐶−𝐷
) . 𝑠𝑖𝑛 (
2
𝐶+𝐷
−2 𝑐𝑜𝑠 (
2
𝐶−𝐷
) . 𝑐𝑜𝑠 ( )
2
Answer: - Option C
Explanation: - By factorization formula 𝑠𝑖𝑛 𝐶 − 𝑠𝑖𝑛 𝐷 = 2 𝑐𝑜𝑠 (𝐶+𝐷) . 𝑠𝑖𝑛 (𝐶−𝐷)
2 2
43. What is 𝒄𝒐𝒔 𝑪 − 𝒄𝒐𝒔 𝑫 equal to?
(A)
𝐶+𝐷
2 𝑠𝑖𝑛 (
2
𝐶−𝐷
) . 𝑐𝑜𝑠 (
2
𝐶+𝐷
−2 𝑠𝑖𝑛 (
2
𝐶−𝐷
) . 𝑠𝑖𝑛 ( )
2
(C)
𝐶+𝐷
2 𝑐𝑜𝑠 (
2
𝐶−𝐷
) . 𝑠𝑖𝑛 (
2
𝐶+𝐷
−2 𝑐𝑜𝑠 (
2
𝐶−𝐷
) . 𝑐𝑜𝑠 ( )
2
Answer: - Option B
Explanation: - By factorization formula 𝑐𝑜𝑠 𝐶 − 𝑐𝑜𝑠 𝐷 = −2 𝑠𝑖𝑛 (𝐶+𝐷) . 𝑠𝑖𝑛 (𝐶−𝐷)
2 2
44. What is 𝒄𝒐𝒔 𝑪 + 𝒄𝒐𝒔 𝑫 equal to?
(A)
𝐶+𝐷
2 𝑠𝑖𝑛 (
2
𝐶−𝐷
) . 𝑐𝑜𝑠 (
2
𝐶+𝐷
−2 𝑠𝑖𝑛 (
2
𝐶−𝐷
) . 𝑠𝑖𝑛 ( )
2
(C)
𝐶+𝐷
−2 𝑠𝑖𝑛 (
2
𝐶−𝐷
) . 𝑠𝑖𝑛 (
2
𝐶+𝐷
) 2 𝑐𝑜𝑠 (
2
𝐶−𝐷
) . 𝑐𝑜𝑠 ( )
2
Answer: - Option D
Explanation: - By factorization formula 𝑐𝑜𝑠 𝐶 + 𝑐𝑜𝑠 𝐷 = 2 𝑐𝑜𝑠 (𝐶+𝐷) . 𝑐𝑜𝑠 (𝐶−𝐷)
2 2
45. 𝟐𝒔 . 𝒄𝒐𝒔 𝑩 can be expressed as
(A) (𝐴 + 𝐵) + 𝑠𝑖𝑛(𝐴 − 𝐵) (B) 𝑠𝑖𝑛(𝐴 + 𝐵) − 𝑠𝑖𝑛(𝐴 − 𝐵)
(C) (𝐴 + 𝐵) + 𝑐𝑜𝑠(𝐴 − 𝐵) (D) 𝑐𝑜𝑠(𝐴 + 𝐵) − 𝑐𝑜𝑠(𝐴 − 𝐵)
Answer: - Option A
Explanation: - By defactorisation formula 2𝑠 . 𝑐𝑜𝑠 𝐵 = 𝑠𝑖𝑛(𝐴 + 𝐵) + 𝑠𝑖𝑛(𝐴 − 𝐵)
𝟐𝒄𝒐𝒔 𝑨 . 𝒄𝒐𝒔 𝑩 can be expressed as
46. (A) 𝑠𝑖𝑛(𝐴 + 𝐵) + 𝑠𝑖𝑛(𝐴 − 𝐵) (B) 𝑠𝑖𝑛(𝐴 + 𝐵) − 𝑠𝑖𝑛(𝐴 − 𝐵)
(C) 𝑐𝑜𝑠(𝐴 + 𝐵) + 𝑐𝑜𝑠(𝐴 − 𝐵) (D) 𝑐𝑜𝑠(𝐴 + 𝐵) − 𝑐𝑜𝑠(𝐴 − 𝐵)
Answer: - Option C
Explanation: - By defactorisation formula 2 cos 𝐴 𝑐𝑜𝑠 𝐵 = 𝑐𝑜𝑠(𝐴 + 𝐵) + 𝑐𝑜𝑠(𝐴 − 𝐵)
47. 𝟐 𝐬𝐢𝐧 𝟏𝟓𝟎 𝐜𝐨𝐬 𝟓𝟎 can be expressed as
(A) 2 sin 250 cos 50 (B) sin 200 cos 50
(C) sin 200 sin 100 (D) sin 150 sin 250
Answer: - Option C
Explanation: - By defactorisation formula 2𝑠 . 𝑐𝑜𝑠 𝐵 = 𝑠𝑖𝑛(𝐴 + 𝐵) + 𝑠𝑖𝑛(𝐴 − 𝐵)
48. Express 𝐜𝐨𝐬 𝝅 + 𝐜𝐨𝐬 𝝅 into the product form
𝟒 𝟔
(A) cos 5𝜋 + cos 𝜋
24 24 (B) 2 cos 5𝜋 + cos 𝜋
24 24
(C) sin 5𝜋 + sin 𝜋
24 24 (D) 2 sin 5𝜋 + sin 𝜋 s 𝜋
24 24 6
Answer: - Option B
𝐶+𝐷 𝐶−𝐷
Explanation: - By factorization formula 𝑐𝑜𝑠 𝐶 + 𝑐𝑜𝑠 𝐷 = 2 𝑐𝑜𝑠 ( ) . 𝑐𝑜𝑠 ( )
2 2
49. If 𝟐 𝐬𝐢𝐧 𝟒𝟎 𝐜𝐨𝐬 𝟏𝟎 = 𝐬𝐢𝐧 𝑨 + 𝐬𝐢𝐧 𝑩 find A & B
(A) A = 30, B = 50 (B) A = 10, B = 40
(C) A = 40, B = 10 (D) A = 50, B = 30
Answer: - Option B
Explanation: - By defactorisation formula 2𝑠 . 𝑐𝑜𝑠 𝐵 = 𝑠𝑖𝑛(𝐴 + 𝐵) + 𝑠𝑖𝑛(𝐴 − 𝐵)
50. Value for complementary relation 𝒔𝒊𝒏−𝟏𝒙 + 𝒄𝒐𝒔−𝟏𝒙 = is
(A) ) 𝜋
4 (B) 𝜋
2
(C) 1 (D) -1
Answer: - Option B
Explanation: - By complementary relation 𝑠𝑖𝑛−1𝑥 + 𝑐𝑜𝑠−1𝑥 = 𝜋
2
51. 𝒄𝒐𝒔−(−𝒙) is equal to?
(A) 𝑐𝑜𝑠−1𝑥 (B) 𝜋 − 𝑐𝑜𝑠−1𝑥
(C) −𝑐𝑜𝑠−1𝑥 (D) 𝜋 + 𝑐𝑜𝑠−1𝑥
Answer: - Option B
Explanation: - By negative relation 𝑐𝑜𝑠−1(−𝑥) = 𝜋 − 𝑐𝑜𝑠−1𝑥.
52. Evaluate 𝒕𝒂𝒏−𝟏 + 𝒕𝒂𝒏−𝟏 ( ) =…….
( )
𝟕 𝟏𝟑
(A) 𝑡𝑎𝑛−1 2
( )
9 (B) 𝑐𝑜𝑡−1 2
( )
9
(C) 𝑡𝑎𝑛−1 (9)
2 (D) 𝑡𝑎𝑛−1(1)
Answer: - Option B
𝑥+𝑦
Explanation: - By using 𝑡𝑎𝑛−1(𝑥) + 𝑡𝑎𝑛−1(𝑦) = 𝑡𝑎𝑛−1 ( )
1−𝑥.𝑦
53. Find the principal value of cos 𝝅 −𝒔𝒊𝒏−𝟏 𝟏
( )
𝟐 𝟐
(A) 1
2 (B) 1
(C) 0 (D) −1
2
Answer: - Option A
𝜋
Explanation: - Use cos( − θ) = 𝑠𝑖𝑛𝜃 and 𝑠𝑖𝑛sin−1x = 𝑥
2
54. Find the principal value of 𝒕𝒂𝒏−𝟏∞ − 𝒔𝒊𝒏−𝟏 𝟏
√𝟐
(A) 𝜋
4 (B) 2𝜋
3
(C) 𝜋 (D) 5𝜋
6
Answer: - Option A
Explanation: - By using 𝑠𝑖𝑛−1(sin 𝑥) = 𝑥 and 𝑡𝑎𝑛−1(tan 𝑥) = 𝑥
55. Evaluate 𝒄𝒐𝒔−𝟏 𝟑) +𝒔𝒊𝒏−𝟏 𝟑
( ( )
𝟓 𝟓
(A) 𝜋
3 (B) 𝜋
2
(C) 𝜋 (D) 5𝜋
6
Answer: - Option B
Explanation: - We know 𝑠𝑖𝑛−1𝑥 + 𝑐𝑜𝑠−1𝑥 = 𝜋
2
Question Bank for Multiple Choice Questions
Program: All Programs in Diploma Engineering Program Code: - CE/CO/ME/ET
Scheme: - I Semester: - 1
Course: - Basic Mathematics Course Code: - 22103
03 – Straight Line Marks:-12
Content of Chapter:-
3.1 Straight line and slope of straight line.
a. Angle between two lines
b.Condition of parallel and perpendicular lines .
3.2 Various forms of straight lines.
a. Slope point form,two point form b.Two points intercept form. c.General form.
3.2 Perpendicular distance from a point on the line.
3.3 Perpendicular distance between two parallel lines.
1. If the inclination of the line is 45°, then its slope is …
(A) 1 (B) 0
(C) -1 (D) -2
Answer: - Option A
Explanation:- Slope = Tan(𝜃)
2. The slope of y-axis is …
(A) 1 (B) 0
(C) 1 (D) Not Defined
Answer: - Option D
Explanation:- The angle made by Y-axis with the positive direction of X-axis is 900.
3. The slope of x-axis is
(A) 1 (B) 0
(C) -1 (D) Not Defined
Answer: - Option B
Explanation: - Inclination of X-axis is 00
4. The slope of the line 5x+3y+7=0 is …
(A) 5
3 (B) 3
5
(C) − 5
3 (D) − 3
5
Answer: - Option C
Explanation: - Slope of line ax+by+c=0 is , 𝑚 = − 𝑎
𝑏
5. Two lines are parallel to each other is their slopes are …
(A) equal (B) not equal
(C) opposite (D) imaginary
Answer: - Option A
Explanation:- parallel lines slopes are equals.
6. The slope of line passing through origin and and the point (3, 4) is …
(A) 4 3 (B) − 4
3
(C) 3
4 (D) − 3
4
Answer: - Option A
Explanation: - By using formula ,𝑚 = 𝑦2−𝑦1
𝑥2−𝑥1
7. The y-intercept of line 5x-4y+7=0 is …
(A) 5
4 (B) − 5
4
(C) 7
4 (D) − 7
4
Answer: - Option c
Explanation: - By using formula ,𝑚 = − 𝑐
𝐵
8. If the slope of line passing through the points (-1, -4) and (2, k) is -1 then k=…
(A) 7 (B) 0
(C) -7 (D) -2
Answer: - Option C
Explanation: - By using formula ,𝑚 = 𝑦2−𝑦1
𝑥2−𝑥1
9. The lines 2x-y+1=0 and 8x-4y-5=0 are …
(A) perpendicular (B) parallel
(C) intersecting (D) none of these
Answer: - Option B
Explanation: - check 𝑚1. 𝑚2 = −1 𝑜𝑟 𝑚1 = 𝑚2
The equation of line passing through the point (4, 1) and making an angle of 45° with positive
10. direction of x-axis is …
(A) x-y-3=0 (B) x+y-3=0
(C) x-y+3=0 (D) x+y+3=0
Answer: - Option A
Explanation:- Slope intercept form of line is 𝑦 − 𝑦1 = 𝑚(𝑥 − 𝑥1)
11. The line 2x+3y-1=0 and 3x-2y-5=0 are
(A) perpendicular (B) parallel
(C) intersecting (D) none of these
Answer: - Option A
Explanation: - check 𝑚1. 𝑚2 = −1 𝑜𝑟 𝑚1 = 𝑚2
12. The equation of line having slope 3 and making intercept 4 on y-axis is …
(A) 3x+y-3=0 (B) 3x-y+4=0
(C) 2x+y-3=0 (D) 2x-y+3=0
Answer: - Option B
Explanation:- Slope intercept form of line is y=mx+c
13. The equation of line whose slope is −𝟑 and passing through the point (1, 2) is …
𝟐
(A) 3x+2y-7=0 (B) 3x-2y+7=0
(C) 3x-2y-7=0 (D) 3x-2y-5=0
Answer: - Option A
Explanation: - The equation of line in slope intercept form is 𝑦 − 𝑦1 = 𝑚(𝑥 − 𝑥1)
14. The equation of line passing through the points (3, 4) and (5, 6) is
(A) x+y-1=0 (B) x-y+1=0
(C) x-y-1=0 (D) x-y-2=0
Answer: - Option B
Explanation: - The equation line in two point form is 𝑦−𝑦1 = 𝑥−𝑥1
𝑦1−𝑦2 𝑥1−𝑥2
15. The equation of line whose x-intercept is 10 and y-intercept is 3 is given by
(A) 3x+10y-30=0 (B) 3x-10y-30=0
(C) 3x-10y+30=0 (D) 3x+10y+30=0
Answer: - Option A
Explanation: - The equation line in in two intercept form is 𝑥 + 𝑦 = 1
𝑎 𝑏
16. The acute angle between the line y=5x+6 and y=x is …
(A) tan−1 2
( )
3 (B) tan−1 3
( )
2
(C) tan−1(1) (D) tan−1(−1)
Answer: - Option A
Explanation: - If is the acute angle between lines then 𝑡𝑎𝑛𝜃 = | 𝑚1−𝑚2 |
1+𝑚1𝑚2
17. The distance of point (1, -1) from the straight line 3x-4y+8=0 is
(A) 3 unit (B) 4 unit
(C) 5 unit (D) 0 unit
Answer: - Option A
Explanation: - The distance of a point P(𝑥 𝑦 ) from the line ax+by+c=0 is 𝑎𝑥1+𝑏𝑦1+𝑐|
1, 1 | √𝑎2+𝑏2
18. The distance between the two parallel lines 6x+8y+10=0 and 6x+8y-25=0 is
(A) 5
2 (B) 7
2
(C) 3
2 (D) 1
2
Answer: - Option B
Explanation: - Perpendicular distance between two parallel lines 𝑎𝑥 + 𝑏𝑦 + 𝑐1 = 0 and 𝑎𝑥 + 𝑏𝑦 +
𝑐1−𝑐2
𝑐2=0 is | |
√𝑎2+𝑏2
19. Two lines are perpendicular to each other if product of their slope is equal to
(A) 0 (B) 1
(C) -1 (D) none of these
Answer: - Option C
Explanation: - If two lines are perpendicular then 𝑚1. 𝑚2 = −1
20. If inclination of the line is ‘𝜽’, then its slope is given by …
(A) sin𝜃 (B) cos𝜃
(C) tan𝜃 (D) cot𝜃
Answer: - Option C
Explanation: - Slope = Tan(𝜃)
21. Slope of general line ax+by+c=0 is given by
(A) 𝑎
𝑏 (B) − 𝑎
𝑏
(C) 𝑏
𝑎 (D) − 𝑏
𝑎
Answer: - -Option B
Explanation: - 𝒔𝒍𝒐𝒑𝒆 = − 𝑪𝒐𝒆𝒇𝒇𝒊𝒄𝒊𝒆𝒏𝒕 𝒐𝒇 𝒙
𝒄𝒐𝒆𝒇𝒇𝒊𝒄𝒊𝒆𝒏𝒕 𝒐𝒇 𝒚
22. The slope of line whose inclination is 0° is …
(A) 0 (B) 1
(C) -1 (D) none of these
Answer: - Option A
Explanation: - Slope = Tan(𝜃)
23. The equation of ‘x-axis’ is …
(A) x=0 (B) x=1
(C) y=0 (D) y=1
Answer: - Option C
Explanation: - The y- coordinates on x-axis are zero.
24. The equation of ‘y-axis’ is
(A) x=0 (B) x=1
(C) y=0 (D) y=1
Answer: - Option A
Explanation: - The X- coordinates on Y-axis are zero
25. The point of intersection of the lines 4x+3y=8 and x+y=1 is
(A) (5, 4) (B) (4, 5)
(C) (-5, 4) (D) (5, -4)
Answer: - Option D
Explanation: - Solve these simultaneous equations.
26. Let lines ‘L1’ and ‘L2’ are perpendicular to each other. If slope of line L1 is 𝟒, then slope of line L2 is
𝟓
…
(A) 5
4 (B) −5
4
(C) −4
5 (D) none of these
Answer: - Option B
Explanation: - If two lines are perpendicular then 𝑚1. 𝑚2 = −1
27. Let lines ‘L1’ and ‘L2’ are parallel to each other. If slope of line L1 is 1, then slope of line L2 is …
(A) 0 (B) 1
(C) -1 (D) not defined
Answer: - Option B
Explanation: - If the lines are parallel then slopes are equal.
28. Equation of the line passing through point (-3, -5) and perpendicular to y-axis is …
(A) 0 (B) 1
(C) -1 (D) none of these
Answer: - Option C
Explanation: - The equation of line in slope intercept form is 𝑦 − 𝑦1 = 𝑚(𝑥 − 𝑥1)
29. The area of the triangle whose vertices are (3,1) ,( -1,3) and (-3 ,-2)
(A) 12 sq. Unit (B) 28 sq. Unit
(C) 11 sq. Unit (D) 21.5 sq. Unit
Answer: - Option A
𝑥1 𝑦1 1
Explanation: - we can find the area of the triangle by using formula 1 |𝑥 𝑦 1|
2 2 2
𝑥3 𝑦3 1
30. Which of the following points are collinear
(A) (2,3),(-1,0) and (4,5) (B) (3,1),(-1,3) and (-3,4)
(C) (3,5),(3,-2) and (-3,16) (D) None of the above
Answer: - Option A
𝒙𝟏 𝒚𝟏 𝟏
Explanation: - The condition of collinearity is |𝒙𝟐 𝒚𝟐 𝟏|=0
𝒙𝟑 𝒚𝟑 𝟏
31. The value of x if the points (-5,7) ,( x,5) and (2, -7) are collinear
(A) x=0 (B) x=-3
(C) x=-1 (D) x=2
Answer: - Option B
𝑥1 𝑦1 1
Explanation: - The condition of collinearity is |𝑥2 𝑦2 1|=0
𝑥3 𝑦3 1
32. The length of the perpendicular from the point (1,6) on the line x+y+8=0
(A) 15
√2 (B) 15
√3
(C) 15
√4 (D) 15
√8
Answer: - Option A
Explanation: - The distance of a point P(𝑥 𝑦 ) from the line ax+by+c=0 is 𝑎𝑥1+𝑏𝑦1+𝑐|
1, 1 | √𝑎2+𝑏2
04 – Mensuration Marks:- 08
Content of Chapter:-
4.1 Area of regular closed figures, Area of triangle, square, parallelogram, rhombus, trapezium and circle.
4.2 Volume of cuboids, cone, cylinder and sphere.
1. Area of the triangle when base is b and height is h is .
(A) Area = 1 × 𝑏 × ℎ (B) Area = 𝑏 × ℎ
2
(C) Area = 𝑏2 (D) Area = ℎ2
Answer: - Option A
Explanation: - According to Formula: Area =
1 × base × height.
2
2. Area of an equilateral triangle is .
(A) Area = b × h (B) Area = √3 × (side)4
4
(C) Area = (side)2 (D) Area = √3 × (side)2
4
Answer: - Option D
Explanation: - According to Formula: Area = √3 × (side)2
4
3. Area of rectangle is .
(A) Area = base × height (B) Area = length × breadth
(C) Area = base × length (D) Area = length × heigth
Answer: - Option B
Explanation: - According to Formula: Area of rectangle = length × breadth
4. Area of square is
(A) Area = side4 (B) Area = side × side
(C) Area = side3 (D) Area = side5
Answer: - Option B
Explanation: - According to Formula: Area = length × breadth
5. Area of rhombus is .
(A) Area = 1 × b × h (B) Area = side × side
2
1
Area = × d × d
(D) Area = b × h
2 1 2
Answer: - Option C
Explanation: - According to Formula:
Area =
1 × Product of diagonals,
2
where d1
& d2
are diagonals.
6. Area of parallelogram when base is ‘b’ and height is ‘h’ is .
(A) Area = b × h (B) Area = 1 × b × h
2
(C) Area = √3 × (side)2 (D) Area = (side)2
4
Answer: - Option A
Explanation: - According to Formula: Area = base × height
7. If ‘r’ is the radius of circle, then area of circle is .
(A) Area = 2πr (B)
Area =
d , where ′d′is the diameter
2
(C) Area = 2r (D) Area = πr2
Answer: - Option D
Explanation: - According to Formula: Area = π × (radius)2
8. Area of Trapezium is .
(A) Area = 1 × (sum of parallel sides) × height (B) Area = sum of parallel sides) × height
2
(C) Area = 1 × (sum of parallel sides) (D) Area = 1 × (sum of parallel sides) + height
2 2
Answer: - Option A
Explanation: - According to Formula: Area = 1 × (sum of parallel sides) × height
2
9. If ‘R’ and ‘r’ be radius of outer and inner circles, then area of annulus (ring) is
(A) Area = πr2 − πR2 (B) Area = πR2 − πr2
(C) Area = πr2 = πR2 (D) Area = πr2 + πR2
Answer: - Option B
Explanation: - According to Formula: Area = Area of outer circle − Area of inner circle
10. The area of rectangle with one side 8 cm is 172 𝐜𝐦𝟐. Find length of the other side
(A) 26 cm (B) 30 cm
(C) 21.5 cm (D) 72 cm
Answer: - Option C
Explanation: - According to Formula: Area = length × breadth
11. The area of rhombus whose diagonals are of length 10cm and 8.2 cm.
(A) 26 sq.cm (B) 41 sq. cm
(C) 210 sq. cm (D) 82 sq.cm
Answer: - Option B
Explanation: - According to Formula: Area = 1 × d × d
2 1 2
12. The area of the circle whose radius is 7.7 cm.
(A) 126.5 cm2 (B) 130.4 cm2
(C) 121.5 cm2 (D) 186.34 cm2
Answer: - Option D
Explanation: - According to Formula: Area = πr2
13. If the area of circle is 120 𝐜𝐦𝟐, then radius of a circle is .
(A) r = 6.18 cm (B) r = 8.18 cm
(C) r = 9.18 cm (D) r = 4.18 cm
Answer: - Option A
Explanation: - According to Formula: Area = πr2
14. A circle has a diameter of 14cm. Then its area is .
(A) 164 sq.cm (B) 174 sq.cm
(C) 154 sq.cm (D) 184 sq.cm
Answer: - Option D
Explanation: - According to Formula: Area = πr2
15. The area of a trapezium whose parallel sides are 10 cm and 8cm where the perpendicular distance between the sides is 4cm is .
(A) A = 64 sq.cm (B) A = 74 sq.cm
(C) A = 54 sq.cm (D) A = 36 sq.cm
Answer: - Option D
Explanation: - According to Formula: Area = 1 × (sum of parallel sides) × height
2
16. A wall is of the form of a trapezium with height 4 m and parallel sides being 3m and 5m then the cost of painting the wall if it has rate of painting as Rs. 25 per sq. m
(A) 220 Rs (B) 280 Rs
(C) 240 Rs (D) 260 Rs.
Answer: - Option C
Explanation: - According to Formula: Area = 1 × (sum of parallel sides) × height
2
17. The area of a trapezoid with base of 10cms and 14cms and height of 5cms.
(A) 60 sq.cm (B) 70 sq.cm
(C) 50 sq.cm (D) 30 sq.cm
Answer: - Option A
Explanation: - According to Formula: Area = 1 × (sum of parallel sides) × height
2
18. The area of trapezoid is 24 sq.cm and the bases are 9cms and 7cms then the height is .
(A) h = 4cm (B) h = 3cm
(C) h = 5cm (D) h = 6cm
Answer: - Option B
Explanation: - According to Formula: Area = 1 × (sum of parallel sides) × height
2
19. The area of a rectangular garden is 3000𝒎𝟐 Its sides are in the ratio 6:5. Then the perimeter of the garden is .
(A) 220 m (B) 240 m
(C) 260 m (D) 280 m
Answer: - Option A
Explanation: - According to Formula: Area = length × breadth
20. The circumference of circle whose area is 38.5 𝐜𝐦𝟐.
(A) 22 cm (B) 24 cm
(C) 26 cm (D) 28cm
Answer: - Option A
Explanation: - According to Formula: Area = πr2 , Circumference = 2 πr
21. Find the area of triangular plot whose base is 17.2 cm and height 19.60 cm.
(A) 126.5 cm2 (B) 130.4 cm2
(C) 168.56 cm2 (D) 186.34 cm2
Answer: - Option C
Explanation: - According to Formula: Area = base × height.
22. The of a right-angled triangle is 8m and hypotenuse is 100m. Find its area.
(A) 48 m2 (B) 24 m2
(C) 21 m2 (D) 34 m2
Answer: - Option B
Explanation: - According to Formula: Area = base × height
23. A park is in the form of a right-angled triangle with hypotenuse 13m. If one of the side is 12 m, find the cost of leveling at the rate of Rs. 10 per sq. m.
(A) Rs. 30 (B) Rs.60
(C) Rs. 250 (D) Rs. 300
Answer: - Option D
Explanation: - According to Formula: Area = base × height
24. Find the area of triangle whose sides are 4cm, 6cm and 8cm.
(A) 135 cm2 (B) 130.4 cm2
(C) 11.5 cm2 (D) 11.62 cm2
Answer: - Option D
Explanation: - Using Heron’s to Formula: Area = √s(s − a)(s − b)(s − c)
25. Find the area of triangle if a = 51 cm, b = 70cm and ∠𝑪 = 𝟒𝟏°.
(A) 1117.51 cm2 (B) 1304.4 cm2
(C) 1171.07 cm2 (D) 1816.34 cm2
Answer: - Option C
Explanation: - According to Formula: Area = 1
2
× a. b × sin C
26. The area of an Equilateral triangle is √𝟑 cm2. Find its height.
(A) 18 cm (B) 9√3 cm
(C) 3√3 cm (D) √3 cm
Answer: - Option B
Explanation: - According to Formula: Altitude of equilateral triangle = √3
2
× side
27. The adjacent sides of a parallelogram are 10 cm and 8 cm, one of the diagonal is 6cm. Find the area of the parallelogram.
(A) 12 cm2 (B) 24 cm2
(C) 21 cm2 (D) 48 cm2
Answer: - Option D
Explanation: - According to Formula: Area = √s(s − a)(s − b)(s − c)
28. A square grassy plot is of side 100 meters. It has gravel path 10 meters wide all around it on the inside. Find the area of the path.
(A) 3600 m2 (B) 1304 m2
(C) 1215 m2 (D) 1864 m2
Answer: - Option A
Explanation: - According to Formula: Area of rectangle = length × breadth
29. The side of square shaped field is 170m long. Find the cost of leveling the field at the rate of Rs.
1.20 per m2.
(A) Rs. 28900 (B) Rs. 4680
(C) Rs. 34680 (D) Rs. 18634
Answer: - Option C
Explanation: - According to Formula: Area = (side)2
30. In exchange for a square plot of land, one of whose side is 25 meters, a man want to buy a rectangular plot of 50 meters wide and of the same area as the square plot. Determine the length of the rectangular plot.
(A) L = 12 meters (B) L = 12.5 meters
(C) L = 27 meters (D) L = 11.5 meters
Answer: - Option B
Explanation: - According to Formula: Area of rectangle = length × breadth.
Area of square = (side)2
31. Find the area of rhombus whose diagonals are 6cm and 9cm.
(A) A = 54 cm2 (B) A = 45 cm2
(C) A = 27 cm2 (D) A = 15 cm2
Answer: - Option C
Explanation: - According to Formula: Area = 1 × d × d
2 1 2
32. Area of rhombus is 336cm2 and one diagonal is 14cm. Find the length of side.
(A) side = 25cm (B) side = 48cm
(C) side = 52cm (D) side = 62cm
Answer: - Option A
Explanation: - According to Formula: Side of rhombus = 1 √𝑑 2 + 𝑑 2
2 1 2
33. Find the area of rhombus if its side is 13cm and one of its diagonal is 10cm.
(A) Area = 12 cm2 (B) Area = 240 cm2
(C) Area = 270 cm2 (D) Area = 120 cm2
Answer: - Option D
Explanation: - According to Formula: Area = 1 × d × d and Side of rhombus = 1 √𝑑 2 + 𝑑 2
2 1 2 2 1 2
34. The two parallel sides of a trapezium measures 50m and 20m respectively and altitude is 50m. Find the area.
(A) Area = 70 m2 (B) Area = 1240 m2
(C) Area = 1750 m2 (D) Area = 1120 m2
Answer: - Option D
Explanation: - According to Formula: Area = 1 × (sum of parallel sides) × height
2
35. If sum of two parallel sides of a trapezium is 232cm and its area is 928 cm2. Find its altitude.
(A) h = 12 cm (B) h = 8 cm
(C) h = 2 cm (D) h = 16 cm
Answer: - Option B
Explanation: - According to Formula: Area = 1 × (sum of parallel sides) × height
2
36. The two parallel sides of a trapezium measures 58m and 42m respectively. The other two Sides are equal each being 17m. Find its area.
(A) Area = 750 m2 (B) Area = 240 m2
(C) Area = 270 m2 (D) Area = 120 m2
Answer: - Option A
Explanation: - According to Formula: Area = 1 × (sum of parallel sides) × height and Pythagoras
2
theorem.
37. Volume of Cuboid is .
(A) Volume = 𝑙 × 𝑏 × ℎ cubic units. (B) Volume = 𝑙 × 𝑏 cubic units.
(C) Volume = 𝑙 × ℎ cubic units. (D) Volume = 𝑏 × ℎ cubic units.
Answer: - Option A
Explanation: - According to Formula: Volume = 𝑙 × 𝑏 × ℎ cubic units.
38. Let ‘l’, ‘b’ and ‘h’ are the length, breadth and height respectively, then total surface area is .
(A) Surface Area = 2 ( lb + bh) (B) Surface Area = 2 ( lh + bh)
(C) Surface Area = 2 ( lb + lh) (D) Surface Area = 2 ( lb + bh + lh)
Answer: - Option D
Explanation: - According to Formula: Surface Area of cuboid = 2 ( lb + bh + lh)
39. Let ‘l’, ‘b’ and ‘h’ are the length, breadth and height respectively, then the diagonal of cuboid is .
(A) Diagonal = √𝑙2 + 𝑏2 (B) Diagonal = 𝑙2 + 𝑏2 + ℎ2
(C) Diagonal = √𝑙2 + 𝑏2 + ℎ2 (D) Diagonal = √𝑙 + 𝑏 + ℎ
Answer: - Option C
Explanation: - According to Formula: Diagonal of cuboid = √𝑙2 + 𝑏2 + ℎ2
40. Volume of cube is .
(A) Volume = (side)3. (B) Volume = (side)2.
(C) Volume = side (D) Volume = length × breadth
Answer: - Option A
Explanation: - According to Formula: Volume = (side)3
41. Surface area of cube is
(A) Surface Area = (side)2 (B) Surface Area = 2 (side)2
(C) Surface Area = 6 (side)2 (D) Surface Area = 6 (side)
Answer: - Option C
Explanation: - According to Formula: Surface Area of cube = 6 (side)2
42. Diagonal of cube is .
(A) Diagonal of cube = √3 (side)2 (B) Diagonal of cube = √3 (side)
(C) Diagonal of cube = √3 (side)3 (D) Diagonal of cube = √6 (side)2
Answer: - Option B
Explanation: - According to Formula: Diagonal of cube = √3 (side)
43. If ‘r’ is the radius of cylinder and ‘h’ is the height of cylinder, then volume is .
(A) Volume = π r2h (B) Volume = 2πrh
(C) Volume = π r2 (D) Volume = r2h
Answer: - Option A
Explanation: - According to Formula: Volume of cylinder = π r2h
44. If ‘r’ is the radius of cylinder and ‘h’ is the height of cylinder, then curved surface area is .
(A) Curved Surface Area = π r2h (B) Curved Surface Area = 2πrh
(C) Curved Surface Area = π r2 (D) Curved Surface Area = r2h
Answer: - Option B
Explanation: - According to Formula: Curved Surface Area of cylinder = 2πrh
45. If ‘r’ is the radius of cylinder and ‘h’ is the height of cylinder, then total surface area is .
(A) Total Surface Area = 2πr( r + h) (B) Total Surface Area = π r h
(C) Total Surface Area = 2π( r + h) (C) Total Surface Area = 2r( r + h)
Answer: - Option A
Explanation: - According to Formula: Total Surface Area of cylinder = 2πr( r + h)
46. If ‘r’, ‘h’ and ‘l’ is the radius, height and slant height of right circular cone respectively, then
volume is .
(A) Volume = π r2h (B) Volume = 2πrhl
(C) Volume = 1 π r2h (D) Volume = r2hl 3
Answer: - Option C
Explanation: - According to Formula: Volume of cone = 1 π r2h
3
47. If ‘r’, ‘h’ and ‘l’ is the radius, height and slant height of right circular cone respectively, then
curved surface area is .
(A) Curved Surface Area = π r2l (B) Curved Surface Area = 2πrl
(C) Curved Surface Area = π r2 (D) Curved Surface Area = πrl
Answer: - Option D
Explanation: - According to Formula: Curved Surface Area of cone = πrl
48. If ‘r’, ‘h’ and ‘l’ is the radius, height and slant height of right circular cone respectively, then Total
surface area is .
(A) Total Surface Area = πr( r + h) (B) Total Surface Area = π r l
(C) Total Surface Area = 2π( r + l) (D) Total Surface Area = πr( r + l)
Answer: - Option D
Explanation: - According to Formula: Total Surface Area of cone = πr( r + l)
49. If ‘r’, ‘h’ and ‘l’ is the radius, height and slant height of right circular cone respectively, then slant height 𝒍 = .
(A) Slant Height (l) = √ℎ2 + 𝑙2 (B) Slant Height (l) = √ℎ2 + 𝑟2
(C) Slant Height (l) = √ℎ2 − 𝑙2 (D) Slant Height (l) = √ℎ2 − 𝑟2
Answer: - Option B
Explanation: - According to Formula: Slant Height (l) = √ℎ2 + 𝑟2
50. Volume of Sphere is .
(A) Volume = 4 π r3 (B) Volume = 2πr
3
(C) Volume = 4 π r2 (D) Volume = π r2 3
Answer: - Option A
Explanation: - According to Formula: Volume of sphere = 4 π r3
3
51. Surface Area of Sphere is .
(A) Surface Area = π r2 (B) Surface Area = 2πr
(C) Surface Area = 4 π r2 (D) Surface Area = 4 π r2 3
Answer: - Option C
Explanation: - According to Formula: Surface Area of Sphere = 4 π r2
52. Volume of Hemisphere is .
(A) Volume = 4 π r3 (B) Volume = 2πr
3
(C) Volume = 4 π r2 (D) Volume = 2 π r3
3 3
Answer: - Option D
Explanation: - According to Formula: Volume of Hemisphere = 2 π r3
3
53. Curved Surface Area of Hemisphere is .
(A) Curved Surface Area = π r2 (B) Curved Surface Area = π r2
(C) Curved Surface Area = 2 π r2 (D) Curved Surface Area = 4 π r2
Answer: - Option B
Explanation: - According to Formula: Curved Surface Area of Sphere = 2 π r2
54. Total Surface Area of Hemisphere is .
(A) Total Surface Area = π r2 (A) Total Surface Area = r2
(C) Total Surface Area = 2 π r2 (D) Total Surface Area = 3 π r2
Answer: - Option D
Explanation: - According to Formula: Total Surface Area = 3 π r2
55. Find the Volume of Cuboid if the length, breadth and height are 25cm, 51cm, and 52cm respectively.
(A) 66300 cm3 (B) 6300 cm3
(C) 36300 cm3 (D) 65300 cm3
Answer: - Option A
Explanation: - According to Formula: Volume = 𝑙 × 𝑏 × ℎ
05 – Statistics Marks: - 14
Content of Chapter:-
5.1 Range, coefficient of range of discrete and grouped data.
5.2 Mean deviation and standard deviation from mean of grouped and ungrouped data, weighted means.
5.3 Variance and coefficient of variance.
5.4 Comparison of two sets of observation.
1. The distribution 3, 5, 7, 8, 3, 9, 5, 7, 10 is
(A) Grouped data (B) Ungrouped data
(C) Raw data (D) None of these
Answer: - Option C
Explanation: - According to definition of Raw data.
2. The following data is type.
Marks 3 - 5 5 - 7
No. of students 4 3
7 - 9
10
9 - 11
12
11 - 13
7
(A) Grouped data (B) Ungrouped data
(C) Raw data (D) None of these
Answer: - Option A
Explanation: - According to definition of Grouped data.
3. The following data is type.
Wt. of items in gms No. of items
50
4
100
10
150
15
200
20
250
7
(A) Grouped data (B) Ungrouped data
(C) Raw data (D) None of these
Answer: - Option B
Explanation: - According to definition of Ungrouped data.
4. The 5 is the frequency of observation from the data 1.2, 1.21,1.5 ,1.2, 1.5, 1.4, 1.41 ,1.21,
1.5, 1.2, 1.8, 1.7, 1.8, 1.81, 1.4, 1.5, 1.2, 1.6, 1.7, 1.5, 1.8, 1.31 ,1.2, 1.2.
(A) 1.8 (B) 1.2
(C) 1.5 (D) 1.21
Answer: - Option C
Explanation: - Frequency means number of occurrence (or number of repetitions) of observation in the given data. Here 1.5 is repeated 5 times, so 5 is the frequency of 1.5.
5. The correct formula to find class-mark for grouped frequency distribution is .
(A) Upper limit−Lower limit
2 (B) Lower limit−Upper limit
2
(C) Upper limit + Lower limit
2 (D) None of these
Answer: - Option C
Explanation: - Class marks is the average of upper and lower limit of the class interval.
6. The correct formula for class length of grouped frequency distribution is
(A) U. L + L. L (B) L.L−U.L
2
(C) L.L+U.L
2 (D) U. L − L. L
Answer: - Option D
Explanation: - According to definition of class length.
7. is the relative measure.
(A) Variance (B) Standard deviation
(C) Range (D) Mean Deviation
Answer: - Option A
Explanation: - According to definition of Variance.
8. is the absolute measure.
(A) Variance (B) Standard deviation
(C) Range (D) Mean Deviation
Answer: - Option B
Explanation: - According to definition of Standard deviation.
9. Range of the distribution is given by
(A) L − S (B) L + S
(C) L + S
L − S (D) L − S
L + S
Answer: - Option A
Explanation: - According to definition of range.
10. Coefficient of Range =
(A) Range
L − S (B) L + S
L − S
(C) L + S (D) L − S
L + S
Answer: - Option D
Explanation: - According to formula.
11. 𝐑𝐚𝐧𝐠𝐞 =
𝐋 − 𝐒
(A) 1 (B) Range
(C) −1 (D) 0
Answer: - Option A
Explanation: - According to formula.
12. Coefficient of Range =
(A) L + S
L − S (B) Range
L − S
(C) Range
Range + 2s (D) Range
Range + s
Answer: - Option C
Explanation: - According to formula.
13. The Range of 10, 5, 12, 2, 15, 20, 8, 10 is .
(A) 18 (B) 22
(C) 20 (D) 2
Answer: - Option A
Explanation: - According to formula L − S
14. The Range and coefficient of Range of the data 120, 100, 130, 50, 150 are respectively.
(A) 5.5, 50 (B) 50, 0.5
(C) 2, 100 (D) 100, 0.5
Answer: - Option D
Explanation: - According to formula Range = L − S and Coefficient of Range = L − S
L + S
15. The class marks of a certain frequency distribution are 15, 25, 35, 45, 55, 65 then the range = .
(A) 25 (B) 50
(C) 55 (D) 65
Answer: - Option B
Explanation: - According to formula Range L – S.
16. The Range and coefficient of Range of 5, 7, 9, 13, 11, 5, 3 are .
(A) 5, 11 (B) 10, 0.61
(C) 10, 0.625 (D) 5, 0.5
Answer: - Option C
Explanation: - According to formula Range = L − S and Coefficient of Range = L − S
L + S
17. The coefficient of Range of 50, 90, 120, 40, 180, 200, 80 is .
(A) 0.60 (B) 0.69
(C) 0.65 (D) 0.67
Answer: - Option D
Explanation: - According to formula Coefficient of Range = L − S
L + S
18. The Range of the following distribution is .
𝒙𝒊 3 8 13 18 23 28 33
𝒇𝒊 1 4 5 7 2 3 10
(A) 30 (B) 36
(C) 11 (D) 9
Answer: - Option A
Explanation: - According to formula Range = L – S.
19. The Range and coefficient of Range of the following distribution are
Marks 5 15 25 35 45 55
No. of students 10 20 30 40 50 60
(A) 50, 0.7142 (B) 50, 0.833
(C) 55, 0.833 (D) 55, 0.7142
Answer: - Option B
Explanation: - According to formula Range = L − S and Coefficient of Range = L − S
L + S
20. The Range and coefficient of Range of the following distribution are .
Marks 0 - 10 10 - 20 20 - 30 30 - 40 40 - 50 50 - 60
No. of students 10 20 15 25 30 25
(A) 60, 0 (B) 15, 0.4285
(C) 60, 1 (D) None of these
Answer: - Option C
Explanation: - According to formula Range = L − S and Coefficient of Range = L − S
L + S
21. The Range of the following distribution is .
Max Temp 25 - 26 27 - 28 29 - 30 31 - 32 33 - 34 35 - 36
No. of Days 2 11 12 10 4 1
(A) 12 (B) 11
(C) 13 (D) 10
Answer: - Option A
Explanation: - According to formula Range L – S.
22. The Range and coefficient of Range of the following distribution are .
Marks 10 - 19 20 - 29 30 - 39 40 - 49 50 - 59 60 - 69
No. of students 6 10 16 14 8 4
(A) 59, 0.7468 (B) 60, 0.76
(C) 58, 0.76 (D) 59, 0.716
Answer: - Option B
Explanation: - According to formula Range = L − S and Coefficient of Range = L − S
L + S
23. In two factories A and B engaged in the same industrial area, the average weekly wages and the
S.D. are as follows.
Factory Average wages Standard Deviation A 34.5 5.0
B 28.5 4.5
Which factory A or B is more consistent?
(A) Factory A (B) Factory B
(C) Both A and B (D) None of these
Answer: - Option A
Explanation: - According to formula c.v. = σ × 100, c.v. of A < c. v. of B.
X̅
24. Find standard deviation of the following data 6, 7, 10, 12, 13, 4, 8, 12.
(A) 4.04 (B) 3.04
(C) 5.04 (D) 6.04
Answer: - Option B
Explanation: - According to formula S.D. = √(Xi − X̅)2
n
25. Find standard deviation of the following data 12, 6, 7, 3, 15, 10, 18, 5.
(A) 4.87 (B) 3.87
(C) 5.87 (D) 6.87
Answer: - Option A
Explanation: - According to formula S.D. = √(Xi − X̅)2
n
26. The class marks of a certain frequency distribution are 15, 25, 35, 45, 55, 65 then the range = .
(A) 25 (B) 50
(C) 55 (D) 65
Answer: - Option B
Explanation: - According to formula Range L – S.
27. The class marks of a certain frequency distribution are 15, 25, 35, 45, 55, 65 then the range = .
(A) 25 (B) 50
(C) 55 (D) 65
Answer: - Option B
Explanation: - According to formula Range L – S.
28. The class marks of a certain frequency distribution are 15, 25, 35, 45, 55, 65 then the range = .
(A) 25 (B) 50
(C) 55 (D) 65
Answer: - Option B
Explanation: - According to formula Range L – S.
29. The class marks of a certain frequency distribution are 15, 25, 35, 45, 55, 65 then the range = .
(A) 25 (B) 50
(C) 55 (D) 65
Answer: - Option B
Explanation: - According to formula Range L – S.
30. The class marks of a certain frequency distribution are 15, 25, 35, 45, 55, 65 then the range = .
(A) 25 (B) 50
(C) 55 (D) 65
Answer: - Option B
Explanation: - According to formula Range L – S.
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