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

2. 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

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

4. 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

5. 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

6. 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

7. 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

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

9. Chapter 00 – Number Q9
= (correct to 3 significant figures)
Option D

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

11. 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

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

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

14. 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

15. 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

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

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

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

19. Chapter 00 – Number Q18

20. Chapter 00 – Number Q20

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

22. Chapter 00 – Number Q22
= 1(213) + 1(212) + 1(27) + 1(26) + 1(22) + 1(21) + 1(20) = =
Option A

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

24. Chapter 00 – Number Q24
103 = 1000 < 1024 = 210 < 1234 < = 104
(103)3235 < < (104)3235
< <
Option C

25. Chapter 00 – Number Q25