1. You are going to make a chocolate cake for some friends. Find out: how much of each ingredient you need The oven temperature and cooking time The size of the cake The number of friends you can invite. 1 Ingredients: 250g plain choc 225g butter 5 eggs 90g caster sugar 15ml cocoa powder 10ml vanilla essence Serves: 15 Oven temp: Gas mark 3 Cake diameter = 24cm Cake height = 15cm Cooking time: 45 mins

2. ? The number of portions you can make is the 5 th triangular number. 13 6 2 =1,3,6,10,15 So 15 portions

3. You need ‘y’ grams of plain chocolate y is proportional to the cube of x y = 432 when x =6 Find y when x = 5 to find out how much chocolate you need ! y  x 3y  x 3  y = kx 3 Find k 3 432 = 216k  k = 2  y =2x 3 y = 2  125 = 250 250g of plain chocolate

4. The amount of butter in grams is the 15 th term in this sequence 1, 4, 9, 16, …….. Square numbers ~ find 15 2 4 15 x 15 = 225 So 225g of butter

5. Simplify to find the number of eggs  875  35  35 =  (5  7) =  5   7 =  5  7 5  875 =  5  7  5  5  875 =  5  7  5  5 =  5  5 = 5  35  5  7 So 5 eggs Or note that 875 = 35  25 so 875/35 = 25 then result follows.

6. A triangle Three sides of a triangle measure: 15cm, 36cm and 39cm. What is the largest angle ? The size of this angle in degrees is the number of grams of caster sugar you need. Pythagoras’ theorem 6 39 36 15 15 2 + 36 2 = 39 2 225 + 1296 = 1521 (This is a 5, 12, 13 triangle!) So largest angle = 90  So you need 90g caster sugar

7. Ingredients List: Plain chocolate ~ 250g Butter ~ 225g Eggs ~ 5 Caster sugar ~ 90g Cocoa powder ~ 15ml Vanilla essence ~ 10ml …but how much of each do you need? 7

8. A B C O Angle ABC (in°) is the amount of cocoa in ml Circle centre = O Angle OAC = 75  Cocoa ! The angle at the centre is twice the angle at the circumference 8 75  30  15  AO = OC = radii AOC = isos triangle OAC = OCA = 75  so AOC= 180-2(75) = 30  ABC is half AOC =15  so 15ml cocoa needed

9. Vanilla essence The base of this pyramid is a regular hexagon with side lengths of 4cm The volume of this pyramid is 80  3 cm 3 The amount of vanilla essence in ml is the same as the perpendicular height of the pyramid in cm. Pyramid volume = 1/3  base area  perpendicular height Regular hexagon = 6 equilateral triangles 9 Area of one triangle = 4  3 (by Pythagoras) Area of six triangles = area of base = 24  3 1/3 area of base = 8  3 Vol = 80  3 = 1/3 base area  height So height = 10cm So 10ml of vanilla essence is needed. 22 44 h h=  (4 2 -2 2 ) h=  (16-4) h=  12 h=  4  3 h= 2  3 Area = 2  2  3

10. x is the gas mark temperature for baking the cake (you are looking for the positive solution…obviously) ((otherwise the cake wouldn’t cook!)) 2 + 1 = 2(x+2) + 1(x+1) = 7 x+1 x+2 (x+2) (x+1) 10 = 3x + 5 = 7 (x+2)(x+1) 10 See card 10a! 10 2 + 1 = 7 x + 1 x + 2 10

11. 10a Algebraic fractions …continued! Now ‘cross multiply’ 10(3x + 5) = 7(x + 1)(x + 2) 30x + 50 = 7(x 2 + 3x + 2) 30x + 50 = 7x 2 +21x + 14) 0 = 7x 2 - 9x – 36 0 = (7x + 12)(x – 3) So x = -12/7 or 3 You need gas mark 3 !!

12. The surface area of the baking tin (sides and base) is 504  cm 2 The height of the tin is 15cm 15cm Find the diameter of the cake h 2r2r +  r 2 11 Solution see 11a

13. 11a Surface area =  r 2 + 2  rh (area of circular base plus area of ‘wrap around’ rectangle) SA = 504  =  r 2 + 2  r  15 504 = r 2 + 30r 0 = r 2 + 30r – 504 0 = (r + 42)(r – 12)  r = -42 or 12 So radius = 12cm and diameter = 24cm (size of cake = 24cm diameter, 15cm height)

14. x is the cooking time in minutes sin x = 1  2 1 1 45  SOH CAH TOA 12 sin 45° = 1  2 22 By Pythagoras So the cooking time is 45 mins