The claim is agreed 1AOXFORD UNIVERSITY PRESS 2014 4A Chapter 1 P79

The claim is agreed 1aoxford university press 2014 4a

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The claim is agreed. 1A OXFORD UNIVERSITY PRESS 2014 4A Chapter 1 P.79
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<bk=4A><ch=1><type=L2><mark=3> [10152468] Convert 5 4 0 . 5 into a fraction. (3 marks) Solution: Let x = 5 4 0 . 5 . x = 5.045 454 5 ..................................... (i) 100 x = 504.545 454 5 ................................ (ii) 1M (ii) – (i): 99 x = 499.5 1M x = 990 995 4 = 22 111 5 4 0 . 5 = 22 1 11 22 1 5 or 1A <bk=4A><ch=1><type=L2><mark=3> [10153027] Convert 7 0 2 . 6 into a fraction. (3 marks) Solution: Let x = 7 0 2 . 6 . x = 6.207 207 207 ............................. (i) 1 000 x = 6 207.207 207 207 ...................... (ii) 1M (ii) – (i): 999 x = 6 201 1M x = 999 201 6 = 111 689 7 0 2 . 6 = 111 23 6 or 111 689 1A OXFORD UNIVERSITY PRESS 2014 4A Chapter 1 P.80
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<bk=4A><ch=1><type=L2><mark=7> [10154002] (a) Convert 2 1 . 3 and 5 . 4 into fractions. (b) Using the results of (a), find the least positive integers a and b such that 2 1 . 3 5 . 4 = b a . (7 marks) Solution: (a) Let x = 2 1 . 3 . x = 3.121 212 ................................ (i) 100 x = 312.121 212 ........................... (ii) 1M (ii) – (i): 99 x = 309 1M x = 99 309 = 33 103 2 1 . 3 = 33 103 33 4 3 or 1A Let y = 5 . 4 . y = 4.555 5 .................................... (iii) 10 y = 45.555 5 .................................. (iv) (iv) – (iii): 9 y = 41 y = 9 41 5 . 4 = 9 41 9 5 4 or 1A (b) 2 1 . 3 5 . 4 = 9 41 33 103 1M = 41 9 33 103 = 451 309 a = 309 1A b = 451 1A OXFORD UNIVERSITY PRESS 2014 4A Chapter 1 P.81
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<bk=4A><ch=1><type=L2><mark=7> [10154141] (a) Convert 4 2 . 7 , 5 0 . 2 and 1 3 . 9 into fractions. (b) Using the results of (a), evaluate 4 2 . 7 5 0 . 2 + 1 3 . 9 and express the answer as a decimal. (7 marks) Solution: (a) Let x = 4 2 . 7 . x = 7.242 424 ................................ (i) 100 x = 724.242 424 ........................... (ii) 1M (ii) – (i): 99 x = 717 1M x = 99 717 = 33 239 4 2 . 7 = 33 239 33 8 7 or 1A Let y = 5 0 . 2 . y = 2.055 5 ................................... (iii) 10 y = 20.555 5 ................................. (iv) (iv) – (iii): 9 y = 18.5 y = 90 185 = 18 37 5 0 . 2 = 18 37 18 1 2 or 1A Let z = 1 3 . 9 . z = 9.313 131 ................................ (v) 100 z = 931.313 131 ........................... (vi) (vi) – (v): 99 z = 922 z = 99 922 1 3 . 9 = 99 922 99 31 9 or 1A (b) 4 2 . 7 5 0 . 2 + 1 3 . 9 = 99 922 18 37 33 239 1M = 14.5 1A OXFORD UNIVERSITY PRESS 2014 4A Chapter 1 P.82
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<bk=4A><ch=1><type=L2><mark=5> [10154315] 1 + cos 30 , 16 81 4 1 , ) 3 5 )( 3 5 ( , π 2 + π Consider the result of each of the above expressions. Which of the results is/are (a) rational number(s)? (b) irrational number(s)? (5 marks) Solution: 1 + cos 30 = 1 + 2 3 16 81 4 1 = 16 9 4 1 = 16 5 ) 3 5 )( 3 5 ( = ( 5 ) 2 – 3 2 = 5 – 9 = –4 1M (a) numbers. rational are ) 3 5 )( 3 5 ( and 16 81 4 1 of results The 1A+1A (b) numbers. irrational are π π and 30 cos 1 of results The 2 1A+1A <bk=4A><ch=1><type=L2><mark=7> [10154390] 36 10 , ) 5 1 ( 4 , 7 49 , sin 30 tan 45 , ) 10 8 )( 10 8 ( , 20 5 Consider the result of each of the above expressions. Which of the results is/are (a) integer(s)? (b) rational number(s) but not integer(s)? (c) irrational number(s)? (d) non-real number(s)? (7 marks) Solution: 36 10 = 6 10 = 16 = 4 7 49 = 7 – 7 sin 30 tan 45 = 2 1 1 = 2 1 ) 10 8 )( 10 8 ( = 8 2 – ( 10 ) 2 = 64 – 10 = 54 20 5 = 15 1M OXFORD UNIVERSITY PRESS 2014 4A Chapter 1 P.83
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(a) integers. are ) 10 8 )( 10 8 ( and 36 10 of results The 1A+1A (b) The result of s in 30  tan 45  is a rational number but not an integer . 1A (c) numbers. irrational are 7 49 and ) 5 1 ( 4 of results The 1A+1A (d) number. real non a is 20 5 of result The - 1A <bk=4A><ch=1><type=L2><mark=7> [10154605] Equation 1: 8 – 7 x = 0 Equation 2: 3 x 2 – 2 = 0 Equation 3: x 2 + 4 = 0 Equation 4: 5 x 2 – 45 = 0 Consider the above equations. Which of the equations has/have solution(s) in each of the following number systems?
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