4 Solutions to 2018 Regular BECE Maths Paper 1

1. B
Numbers are in ascending order if they are arranged from smallest to largest. Of the four lists given, only −64, −25, 4, 17 is arranged from smalllest to largest.

2. D
The set of even numbers greater than two and less 2 or equal to 12 are 4, 6, 8, 10, and 12.

3. C
An infinite set of integers does not have a largest element or a smallest element (or both).

Of the four sets given, only {2, 3, 5, 7, 11, …} is infinite. Observe that it does not have a largest element.

4. D
18 = 2 × 3²
36 = 2² × 3²
60 = 2² × 3 × 5

The HCF of the three numbers is the largest integer that divides all of them.

All the prime factorizations have 2 in them. The smallest power of 2 for any of them is 1.

Similarly, all the prime factorizations have 3 in them. The smallest power of 3 for any of them is 1.

2 and 3 are the only integers that divide all of the numbers and the smallest power of 2 and 3 that divides all of them is 2¹ and 3¹ so the HCF is 2 × 3.

5. A

6. C
Dividing 207 by 17 leaves a remainder of 3.

To get a number that is divisible by 17, we have to make the remainder divisible by 17 (so that there is no remainder or that there is a remainder of 0).

Of the four numbers in teh list, 14 is the one that gives a sum divisible by 17 when added to 3.

7. B
P = {1, 2, 3, 4, 6, 9, 12, 18, 36}
Q = {4, 8, 12, 16, 20, 24, 28, 32, 36}
P ∩ Q = {4, 12, 36}

Since P ∩ Q has three elements, it has 2³ = 8 subsets.

8. C

9. A

10. D

11. B

12. A

13. C

14. C

15. B

16. C

17. B

18. C

19. A

20. C

21. C

22. B

23. C

24. D

25. A

26. B

27. D

28. B

29. A

30. B

31. C

32. B

33. A

34. C

35. D

36. C

37. D

38. D

39. D

40. B