TEST BANK FOR Engineering Mathematics 4th Edition By John Bird
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Problem 1. Simplify (a) 223 ÷ 313 (b) 147214×⎛⎝⎜⎞⎠⎟÷ 1315+⎛⎝⎜⎞⎠⎟ + 2724
Marks
(a) 223 ÷ 313 = 83103÷ = 83310× = 810 = 45 4
(b) 147214×⎛⎝⎜⎞⎠⎟÷ 1315+⎛⎝⎜⎞⎠⎟ + 2724 = 1479453152724×⎛⎝⎜⎞⎠⎟÷+⎛⎝⎜⎞⎠⎟+ 2
= 1978152724÷+
= 791582724×+ 1
= 35242724+ = 111242724+
= 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 57629193.. (a) correct to 4 significant figures
(b) correct to 1 decimal place
Marks 57629193.. = 29.859585... by calculator
Hence (a) 57629193.. = 29.86, correct to 4 significant figures 1
(b) 57629193.. = 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 = 571001764×. 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: 5471150100%.× = 4.76%, correct to 3 significant figures 3
total : 3
Problem 6.Evaluate the following: (a) 2222324×× (b) ()()21682323×× (c) 1421⎛⎝⎜⎞⎠⎟−
(d) (27)−13 (e) 32292322⎛⎝⎜⎞⎠⎟−⎛⎝⎜⎞⎠⎟−
2
Marks
(a) 2222324×× = 231 = 2 = 4 2 24++−2
(b) ()()21682323×× = ()()222234233×× = ()()227243 = 221412 = 214 = 2 = 4 3 12−2
(c) 1421⎛⎝⎜⎞⎠⎟−= (4) = 42 = 16 3 2+1
(d) (27)−13 = 12713 = 1273 = 13 3
(e) 32292322⎛⎝⎜⎞⎠⎟−⎛⎝⎜⎞⎠⎟− = 23292322⎛⎝⎜⎞⎠⎟−⎛⎝⎜⎞⎠⎟ = 492949− = 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) 101101.0101
Marks
(a) 1101 = 1 × 2 + 1 × 2 + 0 × 21 + 1 × 20 232
= 8 + 4 + 0 + 1 = 13 2 10
(b) 101101.0101 = 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
= 45.312510 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 = 11011 2 102
(b) 2 44 Remainder
2 22 0
2 11 0
2 5 1
2 2 1
2 1 0
0 1
Hence 44 = 101100 2 102
4
0.1875 × 2 = 0.375
0.375 × 2 = 0.75
[Solved] TEST BANK FOR Engineering Mathematics 4th Edition By John Bird
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