Did you know? The diagrams show three scales. On each scale there are different objects

6 I C V D X M etc. L = = = = = = = 1 100 5 500 10 1 000 50 , They used these numerals to make up numbers as follows: I VI II VII III VIII IV IX V X etc. = = = = = = = = = = 1 6 2 7 3 8 4 9 5 10, So, for example, XCIX is 99. What is the value of XLVI? A B C D E 42 44 46 64 66 ANSWER: C L = 50 XL = 40 i.e. 10 from 50 = 40 VI = 6 i.e. 1 added to 5 = 6 XLVI = 46 12. Did you know? • An equilateral triangle is a regular polygon with 3 equal sides, and each interior angle is 60 o . • A square is a regular polygon with 4 equal sides, and each interior angle is 90 o . If a regular polygon has n sides, then the formula to find the size of each interior angle is n n − × 2 180 D . If each interior angle of a regular polygon measures 150°, then the number of sides n is A B C D E 6 9 10 11 12 ANSWER: E 7 interior so if then Hence: ∠ = − × ∠ = = − × = − × = − − = − = = ∴ = , n n n n n n n n n n n n 2 180 150 150 2 180 150 2 180 180 360 180 150 360 180 150 360 30 360 12 D D D D D D D D D D D D D D D D . 13. A solid right prism has a square base. The height is twice the length of the side of the base. The surface area of this prism is 160 cm 2 . If 1 cm³ of the prism has a mass of 250 grams, then the mass of the prism in kilograms is A B C D E 28 32 36 40 44 ANSWER: B Surface Area = 2 bases + 4 sides. Let the side of the base be s cm. The height of the prism will then be 2s cm. So the Surface Area = 2 2 s s s s s s + × = + = 4 2 2 8 10 2 2 2 But the Surface Area = cm 160 10 160 16 4 2 2 ∴ = ∴ = ∴ = s s s 8 The volume = length breadth height cm 1 cm has a mass of 250 g 1 4 kg The prism has a mass of 128 1 4 = 32 kg 3 3 s s s s × × × × = = × = = ∴ × 2 2 2 4 128 3 3 9

14. The diagrams show three scales. On each scale there are different objects

on each side which balance each other, as shown. A B C D E 3 4 5 6 7 ANSWER: D C = T + S + S Diagram 1 eq. 1 but T = P + S Diagram 3 eq. 2 ∴ C = P + S + S + S from eq. 1 2 eq. 3 C + S = T + P Diagram 2 eq. 4 ∴ C + S = P + P + S from eq. 4 2 eq. 5 ∴ C = P + P eq. 6 P = S + S + S from eq. 3 6 eq. 7 ∴ C = S + S + S+ S + S + S from eq. 6 7 15. 1 3 1 2 ; ; ; a b These numbers are arranged from smallest to largest. The difference between any two adjacent next to each other numbers is the same. The value of b is A B C D E 5 12 7 18 4 9 5 6 1 4 ANSWER: C Let the common difference be r. i e a r r b r r r . . − = ∴ − = = − ∴ = = − ∴ = = − = 1 3 1 2 1 3 3 1 2 3 1 6 1 2 1 18 1 18 1 9 1 18 4 9 now from 1: Diagram 1 Diagram 2 Diagram 3 How many -shaped objects will balance a -object? Let = C = T = S = P 10 Or a b a and b a b a b a b a b a b b b b b b b − = − − = − ∴ = + ∴− = − ∴ = + ∴ = − ∴ + = − ∴ + = − × ∴ = ∴ = 1 3 1 2 2 1 3 1 2 2 1 2 1 6 2 1 2 1 2 1 6 2 1 2 1 3 1 12 3 1 6 9 4 4 9 eq. eq. Or Multiply through by 18. Then 6 18 18 9 ; ; ; a b corresponds to 6; 7; 8; 9. ∴ = = 18 8 4 9 b b , hence .

16. Given the set of six numbers below: