- Given a:b = 2:3 and b:c = 4:8 find a:b:c. (2 marks)
- Using mathematical tables to find; (2 marks)

0.0912^{3} - Solve for x in the equation. ( 3 marks )

27^{x}x 3^{(2x-2)}= 9^{ (x +2)} - Simplify: (4 marks)
- Use logarithms to 4 decimal places to evaluate: (4 marks)
- Use squares, square roots and reciprocals tables only to evaluate; (4 marks)

3 + 4

√(42.15) (3.152)^{2} - Find the equation of a line through point (5, −1) and perpendicular to line 4x + 2y – 3 = 0. (4 marks)
- Four towns P, Q, R, and S are such that the town Q is 120 Km due to East of town P. Town R is 160km due north of town Q, town S is on a bearing of 330° from Q and on a bearing of 300° from R.
- Show the relative position of towns P, Q, R, and S. (5mks)

Take the scale of 1cm to rep. 50km. - Use the drawing to determine
- The distance SP in Km (2mks)
- The bearing of S from P (1mk)

- Show the relative position of towns P, Q, R, and S. (5mks)

## Marking Scheme

- Given a:b = 2:3 and b:c = 4:8 find a:b:c. (2 marks)
**a:b:c****(2:3)4****(4:8)3 m1****8:12:24****=2:3:6 A1** - Using mathematical tables to find; (2 marks)

0.0912^{3 }**(9.123 x 10**^{−2})^{3}758.6 x 10^{−6}7.586 x 10^{−4} - Solve for x in the equation. ( 3 marks )
**27**^{x}x 3^{(2x-2)}= 9^{ (x +2)}**3**^{3x}x 3^{ (2x – 2)}= 3^{ 2(x + 2 )}M1 expressing in index form**3x + 2x – 2 = 2x + 4 M1 relating index****3x = 6****x = 2**A1 C.A.O - Simplify: (4 marks)

**M1****M1****=¼ ×**^{9}/_{1}= 2¼ - Use logarithms to 4 decimal places to evaluate: (4 marks)
**=0.5310** - Use squares, square roots and reciprocals tables only to evaluate; (4 marks)

3 + 4

√(42.15) (3.152)^{2}**√(42.15)=6.4923****3.152**^{2}= 9.9351**3 + 4****6.4923 9.9351****3 x 0.1540 + 4 x 0.1007****0.462 + 0.4028****= 0.8648** - Find the equation of a line through point (5, -1) and perpendicular to line 4x + 2y – 3 = 0. (4 marks)
**4x + 2y = 3****2y = −4x + 3****y = −2x +**^{3}/_{2}**Gradient (m**_{1}) = −2**m**_{1}x m_{2}= −1**−2 x x = −1****x= −1 = ½****−2****(x, y) (5, −1)****y−−1=½****x−5****y + 1 = ½ (x – 5)****y = ½x−**^{5}/_{2}−^{1}/_{1}y = ½x−3½ - Four towns P, Q, R, and S are such that the town Q is 120 Km due to East of town P. Town R is 160km due north of town Q, town S is on a bearing of 330° from Q and on a bearing of 300°from R.
- Show the relative position of towns P, Q, R, and S.

Take the scale of 1cm to rep. 50km. (5mks) - Use the drawing to determine
- The distance SP in Km (2mks)
**4.9 x 50 = 245 Km****M1A1** - The bearing of S from P (1mk)
**356°****B1**

- The distance SP in Km (2mks)

- Show the relative position of towns P, Q, R, and S.

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