Download 2 - aucksecmaths

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.
Serves: 15
Oven temp: Gas mark 3
Cake diameter = 24cm
Cake height = 15cm
Cooking time: 45 mins
Ingredients:
250g plain choc
225g butter
5 eggs
90g caster sugar
15ml cocoa powder
10ml vanilla essence
2
The number of portions you can make
is the 5th triangular number.
=1,3,6,10,15
So 15 portions
?
1
3
6
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x3
 y = kx3
432 = 216k  k = 2
y = 2  125 = 250
Find k
 y =2x3
250g of plain chocolate
4
The amount of butter in grams is the 15th term in this sequence
Square numbers ~ find 152
15 x 15 = 225
1 , 4 , 9 , 16 , ……..
So 225g of butter
5
Simplify to find the number of eggs
875
35
35 = (5  7) = 5  7 = 5 7
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
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
A triangle
152 + 362 = 392
225 + 1296 = 1521
So largest angle = 90
So you need 90g caster sugar
39
15
(This is a 5, 12, 13 triangle!)
36
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?
8 Angle ABC (in°) is the amount of cocoa in ml
AO = OC = radii
AOC = isos triangle
OAC = OCA = 75
so AOC= 180-2(75)
= 30
ABC is half AOC
=15
so 15ml cocoa needed
Circle centre = O
Angle OAC = 75
B
15
O
30
Cocoa !
A
75
75
C
The angle at the centre is twice the angle at the circumference
9
Pyramid volume = 1/3  base area  perpendicular height
Regular hexagon = 6 equilateral triangles
Vanilla essence
The volume of this pyramid is 803 cm3
The base of this pyramid is a regular hexagon with side lengths of 4cm
h= (42-22)
h= (16-4)
h= 12
h= 4 3
h= 23
Area = 2  23
The amount of vanilla essence in ml
is the same as the perpendicular height
of the pyramid in cm.
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.
4
4
h
2
2
10
x is the gas mark temperature for baking the cake
2 + 1 = 7
x+1 x+2
10
(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!
10a
Algebraic fractions …continued!
Now ‘cross multiply’
10(3x + 5) = 7(x + 1)(x + 2)
30x + 50 = 7(x2 + 3x + 2)
30x + 50 = 7x2 +21x + 14)
0 = 7x2 - 9x – 36
0 = (7x + 12)(x – 3)
So x = -12/7 or 3
You need gas mark 3 !!
11
The height of the tin is 15cm
15cm
The surface area of the baking tin (sides and base) is 504 cm2
Find the diameter of the cake
Solution see 11a
h
+
2r
r2
11a
Surface area = r2 + 2rh
(area of circular base plus area of ‘wrap around’ rectangle)
SA = 504 = r2 + 2r  15
504 = r2 + 30r
0 = r2 + 30r – 504
0 = (r + 42)(r – 12)
 r = -42 or 12
So radius = 12cm and diameter = 24cm
(size of cake = 24cm diameter, 15cm height)
12
x is the cooking time in minutes
By Pythagoras
sin x = 1
2
45
2
1
SOH CAH TOA
45
sin 45° = 1
2
1
So the cooking time is 45 mins