Chapter 00 – Number Q1 0 < x < 1 0 < x2 < x Option D

Chapter 00 – Number Q2 Let a = 8p and b = 8q where p and q are relative prime integers. a – b = 8(p – q) which is divisible by 8 ab = 64pq which is divisible by 8 If a = 16 and b = 24, then a + b = 48 which is not divisible by 16. Option C Option D

Chapter 00 – Number Q3 There must at least one even number in three consecutive positive integers. So xyz must be even. Option D

Chapter 00 – Number Q4 x2 + 2x + 7 is even x2 + 2x is odd Both x and x + 2 are odd x can be any positive odd number Option C Option D

Chapter 00 – Number Q5 2k must be even when k is a positive integer So 22n+1 and 3(2n) must be even (2n + 1)2 =4n2 + 4n + 1 = 4(n2 + n) + 1 which is an odd number. Option B

Chapter 00 – Number Q6 2k must be even when k is a positive integer So 22n+1, 2n and 3(2n) must be even 2n + 1 is an odd number. Option B

Chapter 00 – Number Q7 2 and 3 are factors of m and 2 and 3 are relative prime, so we have m = (2)(3)(k) where k is an integer, i.e. m = 6k which is divisible by 6. 15 is a factor of n, then n = 15q where q is an integer, i.e. n = 3(5q) = 5(3q) which is divisible by 3 and 5. Let p = 12 which is the multiple of 4 and 6 but not multiple of 24. Option C

Chapter 00 – Number Q8 x < 0 < y x – x < 0 – x < y – x Option C

Chapter 00 – Number Q9 0.044449 = 0.0444 (correct to 3 significant figures) Option D

Chapter 00 – Number Q10 1.15  15 = 0.076666 … = 0.0767 (correct to 3 significant figures) Option D

Chapter 00 – Number Q11 Let m = 3p and n = 4q where p and q are integers. mn = 12pq which is multiple of 12 m = 3 and n = 4, but H.C.F. is 1 Let the L.C.M. of m and n be p, then p = nk where k is an integer, i.e. p = 4qk which is even. Option C

Chapter 00 – Number Q12 2 = 9.86960 … 2 = 9.87 (correct to 3 significant figures) Option B

Chapter 00 – Number Q13 a < b < 0 a2 > ab > 0 and ab > b2 > 0 (a < 0 and b < 0) a2 > ab > b2 Option B

Chapter 00 – Number Q14 a and b are two consecutive integer Then one must odd and the other is even So a + b must be odd And ab must be even Also either a2 is odd or b2 is odd So a2 + b2 must be odd. Option C

Chapter 00 – Number Q15 If m is an odd number, then m2 is odd. Either m is even or m + 1 is even So m(m + 1) is even. If m is odd, then m + 2 is odd, so m(m + 2) is odd. Option B

Chapter 00 – Number Q16 If a > b, then –a < –b. If a = -3, b = -5, then a + b < b and a2 < b2. Option A

Chapter 00 – Number Q17 If 0.8448 < a < 0.8452, then a = 0.8 (correct to 1 significant figure) If a = 0.844801, then a = 0.84 (cor. to 2 s.f.) If a = 0.845199, then a = 0.85 (cor. to 2 s.f.) So we cannot have approximation cor. to 2 s.f. a = 0.845 (correct to 3 significant figures) If a = 0.844801, then a = 0.8448 (cor. to 4 s.f.) If a = 0.845199, then a = 0.8452 (cor. to 4 s.f.) So we cannot have approximation cor. to 4 s.f. Option C

Chapter 00 – Number Q17 213 + 24 + 3 = 1(213) + 0(212) + …+0(25) +1(24) + 0(23) + 0(22) + 1(21) + 1(20) = 100000000100112 Option C

Chapter 00 – Number Q18

Chapter 00 – Number Q20

Chapter 00 – Number Q21 0.0498765 = 0.05 (cor. to 2 d.p.) 0.0498765 = 0.0499 (cor. to 3 s.f.) 0.0498765 = 0.0499 (cor. to 4 d.p.) 0.0498765 = 0.049877 (cor. to 5 s.f.) Option C

Chapter 00 – Number Q22 110000110001112 = 1(213) + 1(212) + 1(27) + 1(26) + 1(22) + 1(21) + 1(20) = 213 + 212 + 27 + 26 + 4 + 2 + 1 = 213 + 212 + 27 + 26 + 7 Option A

Chapter 00 – Number Q23 a  1, b  -4 and c  3 c(a – b)  3(1 – (–4)) = 15 Option D

Chapter 00 – Number Q24 103 = 1000 < 1024 = 210 < 1234 < 10000 = 104 (103)3235 < 12343235 < (104)3235 109705 < 12343235 < 1012940 Option C

Chapter 00 – Number Q25