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

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