Exam (elaborations) TEST BANK FOR Engineering Mathematics 4th Edition By John Bird (Solutions Manual)

Exam (elaborations) TEST BANK FOR Engineering Mathematics 4th Edition By John Bird (Solutions Manual) ENGINEERING MATHEMATICS 4TH EDITION INSTRUCTOR’S MANUAL WORKED SOLUTIONS TO THE ASSIGNMENTS JOHN BIRD CONTENTS Page ASSIGNMENT 1 (chapters 1 to 4) 1 ASSIGNMENT 2 (chapters 5 to 8) 8 ASSIGNMENT 3 (chapters 9 to 12) 14 ASSIGNMENT 4 (chapters 13 to 16) 20 ASSIGNMENT 5 (chapters 17 to 20) 26 ASSIGNMENT 6 (chapters 21 to 23) 35 ASSIGNMENT 7 (chapters 24 to 26) 42 ASSIGNMENT 8 (chapters 27 to 31) 47 ASSIGNMENT 9 (chapters 32 to 35) 56 ASSIGNMENT 10 (chapters 36 to 39) 60 ASSIGNMENT 11 (chapters 40 to 43) 68 ASSIGNMENT 12 (chapters 44 to 46) 75 ASSIGNMENT 13 (chapters 47 to 49) 80 ASSIGNMENT 14 (chapters 50 to 53) 84 ASSIGNMENT 15 (chapters 54 to 58) 90 ASSIGNMENT 16 (chapters 59 to 61) 97 LIST OF FORMULAE 103 ASSIGNMENT 1 (Page 33) This assignment covers the material contained in Chapters 1 to 4. Problem 1. Simplify (a) 223 ÷ 313 (b) ×⎛⎝⎜⎞⎠⎟÷ 1315+⎛⎝⎜⎞⎠⎟ + 2724 Marks (a) 223 ÷ 313 = 83103÷ = 83310× = 810 = 45 4 (b) ×⎛⎝⎜⎞⎠⎟÷ 1315+⎛⎝⎜⎞⎠⎟ + 2724 = 2724×⎛⎝⎜⎞⎠⎟÷+⎛⎝⎜⎞⎠⎟+ 2 = 4÷+ = ×+ 1 = + = + = 31824 = 334 2 total : 9 Problem 2. A piece of steel, 1.69 m long, is cut into three pieces in the ratio 2 to 5 to 6. Determine, in centimetres, the lengths of the three pieces. Marks Number of parts = 2 + 5 + 6 = 13 Length of one part = 16913.m = 16913cm = 13 cm 1 Hence 2 parts ≡ 2 × 13 = 26 5 parts ≡ 5 × 13 = 65 6 parts ≡ 6 × 13 = 78 i.e. 2 : 5 : 6 :: 26 cm : 65 cm : 78 cm 3 total : 4 1 Problem 3. Evaluate .. (a) correct to 4 significant figures (b) correct to 1 decimal place Marks .. = 29.... by calculator Hence (a) .. = 29.86, correct to 4 significant figures 1 (b) .. = 29.9, correct to 1 decimal place 1 total : 2 Problem 4. Determine, correct to 1 decimal place, 57% of 17.64 g. Marks 57% of 17.64 g = ×. g = 10.1 g, correct to 1 decimal place 2 total : 2 Problem 5. Express 54.7 mm as a percentage of 1.15 m, correct to 3 significant figures. Marks 54.7 mm as a percentage of 1.15 m is: 0%.× = 4.76%, correct to 3 significant figures 3 total : 3 Problem 6.Evaluate the following: (a) ×× (b) ()()×× (c) 1421⎛⎝⎜⎞⎠⎟− (d) (27)−13 (e) ⎛⎝⎜⎞⎠⎟−⎛⎝⎜⎞⎠⎟− 2 Marks (a) ×× = 231 = 2 = 4 2 24++−2 (b) ()()×× = ()()×× = ()() = = 214 = 2 = 4 3 12−2 (c) 1421⎛⎝⎜⎞⎠⎟−= (4) = 42 = 16 3 2+1 (d) (27)−13 = 12713 = 1273 = 13 3 (e) ⎛⎝⎜⎞⎠⎟−⎛⎝⎜⎞⎠⎟− = ⎛⎝⎜⎞⎠⎟−⎛⎝⎜⎞⎠⎟ = − = 2949 = 2994× = 12 3 total : 14 Problem 7. Express the following in standard form: (a) 1623 (b) 0.076 (c) 14525 Marks (a) 1623 = 1.623 × 103 1 (b) 0.076 = 7.6 × 10 1 2− (c) 14525 = 145.4 = 1.454 × 102 1 total : 3 Problem 8.Determine the value of the following, giving the answer in standard form: (a) 5.9 × 102 + 7.31 × 102 (b) 2.75 × 10 - 2.65 x 10 −2−3 Marks (a) 5.9 × 102 + 7.31 × 102 = 590 + 731 = 1321 = 1.321 × 10 2 3 (b) 2.75 × 10 - 2.65 x 10 = 0.0275 - 0.00265 −2−3 = 0.02485 = 2.485 × 10 2 2− total : 4 3 Problem 9. Convert the following binary numbers to decimal form: (a) 1101 (b) .0101 Marks (a) 1101 = 1 × 2 + 1 × 2 + 0 × 21 + 1 × 20 232 = 8 + 4 + 0 + 1 = 13 2 10 (b) .0101 = 1 × 2 + 0 × 2 + 1 × 23 + 1 × 22 + 0 × 21 254 + 1 × 20 + 0 × 2 + 1 × 2 + 0 × 2 + 1 × 2 −1−2−3−4 = 32 + 0 + 8 + 4 + 0 + 1 + 0 + 14 + 0 + 116 = 45. 3 total : 5 Problem 10. Convert the following decimal numbers to binary form: (a) 27 (b) 44.1875 Marks (a) 2 27 Remainder 2 13 1 2 6 1 2 3 0 2 1 1 0 1 Hence 27 = (b) 2 44 Remainder 2 22 0 2 11 0 2 5 1 2 2 1 2 1 0 0 1 Hence 44 = 2 102 4 0.1875 × 2 = 0.375 0.375 × 2 = 0.75 0.75 × 2 = 1.50 0.50 × 2 = 1.00 Hence 44.1875 = . total : 6 Problem 11. Convert the following decimal numbers to binary, via octal: (a) 479 (b) 185. Marks (a) 8 479 Remainder 8 59 7 8 7 3 0 7 7 3 7 From Table 3.1, page 19, 737 = (b) 8 185 Remainder 8 23 1 8 2 7 0 2 2 7 1 From Table 3.1, page 19, 271 = 0. × 8 = 2.3125 0.3125 × 8 = 2.5 0.5 × 8 = 4.0 i.e. 0. = .2 2 4 = .from Table 3.1, page 19 8 Hence 185. = . 2 2 total : 6 5 Problem 12. Convert (a) 5F16 into its decimal equivalent (b) 13210 into its hexadecimal equivalent (c) into its hexadecimal equivalent 2 Marks (a) 5F16 = 5 × 161 + F × 160 = 5 × 161 + 15 × 160 = 80 + 15 = 95 2 10 (b) 16 132 Remainder 16 8 4 0 8 i.e. 13210 = 8416 2 (c)Grouping bits in 4’s from the right gives: = and assigning hexadecimal symbols to each group gives: 1 A B Hence = 1AB 2 216 total : 6 Problem 13. Evaluate the following, each correct to 4 significant figure: (a) 61.22 (b) . (c) 00527. Marks (a) 61.22 = 3748, correct to 4 significant figures 1 2 (b) . = 23.87, correct to 4 significant figures 1 (c) 00527. = 0.2296, correct to 4 significant figures 1 total : 3 Problem 14. Evaluate the following, each correct to 2 decimal places: (a) ....××⎛⎝⎜⎞⎠⎟ (b) 7982...×⎛⎝⎜⎞⎠⎟ 6 Marks (a) ....××⎛⎝⎜⎞⎠⎟ = 8.76, correct to 2 decimal places 3 (b) 7982...×⎛⎝⎜⎞⎠⎟ = 1.17, correct to 2 decimal places 4 total : 7 Problem 15. If 1.6 km = 1 mile, determine the speed of 45 miles/hour in kilometres per hour. Marks 45 miles/hour = 45 × 1.6 km/h = 72 km/h 3 total : 3 Problem 16. Evaluate B, correct to 3 significant figures, when W = 7.20, v = 10.0 and g = 9.81, given that B = Wvg22 Marks B = Wvg22 = (.)(.)(.)12 = 36.7, correct to 3 significant figures 3 total : 3 TOTAL ASSIGNMENT MARKS: 80 7 ASSIGNMENT 2 (Page 64) This assignment covers the material contained in Chapters 5 to 8. Problem 1. Evaluate 3xy2z3 - 2yz when x = 43, y = 2 and z = 12 Marks 3xy2z3 - 2yz = 343⎛⎝⎜⎞⎠⎟(2)2123⎛⎝⎜ ⎞⎠ ⎟ - 2(2)12⎛⎝⎜⎞⎠⎟ = 2 - 2 = 0 3 total : 3 Problem 2. Simplify the following: (a) 82232abcabc() (b) 3x + 4 ÷ 2x + 5 × 2 - 4x Marks (a) 82232abcabc() = 2abcabc/// = 2b1c or 22/bc 3 (b) 3x + 4 ÷ 2x + 5 × 2 - 4x = 3x + 42x + 5 × 2 - 4x = 3x + 2x + 10 - 4x = -x + 2x + 10 or 2x - x + 10 3 total : 6 Problem 3. Remove the brackets in the following expressions and simplify: (a) (2x - y) (b) 4ab - [3{2(4a - b) + b(2 - a)}] 2 Marks (a) (2x - y) = (2x - y)(2x - y) = 4x2- 2xy - 2xy + y2 1 2 = 4x- 4xy + y2 1 2 (b) 4ab - [3{2(4a - b) + b(2 - a)}] = 4ab - [3{8a - 2b + 2b - ab}] 1 = 4ab - [3{8a - ab}] = 4ab - [24a - 3ab] 1 = 4ab - 24a + 3ab 8 = 7ab - 24a or a(7b - 24) 1 total : 5 Problem 4. Factorise 3x2y + 9xy2 + 6xy3 Marks 3x2y + 9xy2 + 6xy3 = 3xy(x + 3y + 2y2) 3 total : 3 Problem 5. If x is inversely proportional to y and x = 12 when y = 0.4, determine (a) the value of x when y is 3, and (b) the value of y when x = 2 Marks x α 1y i.e. x = ky x = 12 when y = 0.4, hence 12 = k04. from which, k = (12)(0.4) = 4.8 (a) When y = 3, x = ky = 483. = 1.6 2 (b) When x = 2, 2 = 48.y and y = 482. = 2.4 2 total : 4 Problem 6. Factorise x+ 4x2+ x - 6 using the factor theorem . Hence solve the 3equation x+ 4x2+ x - 6 = 0 3 Marks Let f(x) = x+ 4x2+ x - 6 3 then f(1) = 1 + 4 + 1 - 6 = 0, hence (x - 1) is a factor f(2) = 8 + 16 + 2 - 6 = 20 f(-1) = - 1 + 4 - 1 - 6 = - 4 f(-2) = - 8 + 16 - 2 - 6 = 0, hence (x + 2) is a factor f(-3) = - 27 + 36 - 3 - 6 = 0, hence (x + 3) is a factor Thus x+ 4x2+ x - 6 = (x - 1)(x + 2)(x + 3)