1. a) 3/4 ÷ 5/6 = 3/4 x 6/4 = 18/16= 1 2/16 = 1 1/8
b) 2/3 ÷ 7/8 = 2/3 x 8/7= 16/21

2. a)
3q² x 2q = 6q³ =6q²
  q         q

b)
(2p²)² x 3p = 4p⁴ x 3p = 12 p⁵ = 2 p⁴
     6p           6p      6p


3.
a)
2x + 5y = 20 (1)
3x + 3y = 21 (2)

6x +15y = 60 (1)x3=(3)
6x + 6y = 42 (2)x2=(4)
     9y = 18 (1)-(2)=(5)
Hence y=2
Substitute into (1) to find x=5
Check in (2)

b)
5x - 2y = 13
2x + 3y = 9
10x - 4y = 26
10x +15y = 45
19y=19
y=1
x=3

c)
3x - 2y = 7
2x - y = 6
4x - 2y =12
x=5
y=4


4
a)i)
(x+2)(3x+1)=198
3x²+x+6x+2=198
3x²+7x+2=198
3x²+7x-196=0
(3x+28)(x-7)=0
Hence x-7=0 or 3x+28=0
x=7

ii)
2(x+2) + 2(3x+1)=62
2x+4+6x+2=62
8x+6=62
8x=56
x=7

b)i)
(x+2)(2x-1)=150
2x²-x+4x-2=150
3x²+3x-2=150
3x²+3x-152=0
(3x+19)(x-8)=0
Hence x-7=0 or 3x+28=0
x=7

ii)
2(x+2) + 2(2x-1) = 56
2x+4 + 4x-2 = 56
6x + 2 = 56
6x = 54
x = 9m

5)
a)
x²-2x-2=0
(x-1)²-3=0
(x-1)²=3
(x-1)= ± root3
x=1+ root3 and 1-root3

b)
4x²+3x-2=0
x²+(3/4)-(1/2)=0
(x+³/₈)²-⁹/₆₄-¹/₂=0
(x+³/₈)²=⁴¹/₆₄
x+³/₈x=±root(41/16)
x=-(3/8)+ root(41/64) and -(3/8)-root(41/64)
c)
5x²-6x-4=0
x²-(6/5)-(4/5)=0
(x-³/₅)²-⁹/₂₅-⁴/₅=0
(x-³/₅)²=²⁹/₂₅
x-³/₅x=±root(29/25)
x=(3/5)+ root(29/25) and (3/5)-root(29/25)

6)
a)
Using Difference of 2 squares...
25x²y²-100=
(5xy-10)(5xy+10)

b)
81p²-36q²=
9(9p²-4q²)= Using Difference of 2 squares...
=9(3p-2q)(3p+2q)


7)
a)
15a² x b² = 5ab²
  6    a     2


b)
x³y² x 15xy = 3x³y
 10x   10y²    20

c)
(2x+5)(x-1) = (x-1)
(2x+5)(2x-5)  (2x-5)

d)
3(x+3)(x-2) = 3(x-2)
2(x+3)(x-3)   2(x-3)

e)
(x+1)(x+1) = (x+1)
(x+1)(x-1)   (x-1)


f)
3(x²+5x+4) = 3(x+4)(x+1) = 3(x+1)
 2x²+7x-4    (2x-1)(x+4)   (2x-1)


8)
The greater than or equal to symbol does not work in some versions of Netscape, so I've used the greater to symbol instead
a)
3x²-4>11
3x²>15
x²>5
±x>±root5
x>root5 or x<-root5

2x²-7<23
2x²<30
x²<15
±x<±root15
-root15<x<root15

c)
3x-3>2x+7
x-3>7
x>10

d)
5x-1<14
5x<15
x<3


9)
a)
pq - 2p = 5p + 3
pq = 7p + 3
q = 7p+3
     p

b)
a = b + c
       d+1
a(d+1) = b(d+1) + c
ad + a = bd + b + c
ad - bd = b + c - a
d(a-b) = b + c - a
d = b + c - a
     (a-b)

c)
Mb = 100(a-b)
Mb = 100a - 100b
Mb + 100b = 100a
b(M+100) = 100a
b = 100a
    M+100

d)
S(u+v) = uv
Su + Sv = uv
Su = uv - Sv
Su = v(u-S)
  Su  = v
(u-S)

e)
 a  =   b 
x+1   2x-1
a(2x-1) = b(x+1)
2ax - a = bx + b
2ax - bx = a + b
x(2a - b) = a + b
x =  a+b 
    2a+b

f)
The PI symbol is not supported by some versions of netscape, so I've used a 'mu'(µ) instead!
T² = 4 µ²(^L/_g)
 Tg  = L
4µ²


10
a)
16^-1/2 x 5⁰= 1/4 x 1 = 1/4
b)
(¹/₂₇)^-1/3 = 27^1/3 = 3

c)
49^3/2 x 81 ^-1/4 = 7³ x 3^-1 = 343/3


d)
125^2/3 x 8^2/3 = 5² x 2² = 25 x 4 = 100

e)
12^1/2 x 9 ^-1/4 = root12 x root(1/3) = root4 = 2


11
a)

x	x3-2x	=0?
1	-1	too small
2	4	too big
1.5	0.375	too big
1.4	-0.056	too small
1.45	0.148625	too big
1.41	-0.016779	too small
1.42	0.023288	too big
1.415	0.003148375	too big

So x=1.41 to 2dp (there are other solutions at 0 and -1.41)

b)There are no real solutions to this equation!
c)

x	2x2+x+1	=60?
4	38	too small
5	63	too big
4.5	49.75	too small
4.9	60.23	too big
4.8	57.52	too small
4.88	59.68	too small
4.89	59.9563	too small
4.895	60.093075	too big

So x=4.89 to 2dp (there is another solution at -4.22)

12)
a)
y=2x-3
y² = (2x-3)(2x-3) = 4x² - 6x - 6x + 9 = 4x² - 12x + 9
x² + y² = 5
x² + 4x² - 12x + 9 = 5
5x² - 12x + 4 = 0
Solve by whatever method you want (complete the square, quadratic formula etc)
x = 2 and x = 0.4
Substitute these values into the linear equation
y = 1 and y = -2.2
Check that x² + y² = 5
Coordinates are (2,1) and (0.4,-2.2)

13)
y=2x+1
y² = (2x+1)(2x+1) = 4x² + 2x + 2x + 1 = 4x² + 4x + 1
x² + y² = 12
x² + 4x² + 4x + 1 = 12
5x² + 4x - 11 = 0
Solve by whatever method you want (complete the square, quadratic formula etc)
x = 1.14 and x = -1.94
Substitute these values into the linear equation
y = 3.27 and y = -2.87
Check that x² + y² = 12
Coordinates are (1.14,3.27) and (-1.94,-2.87)


14)
£217000 x 1.24^-1=
£175000
(1.24 because it's a 24% increase, ^-1 because you're going back in time)

15)
£931.72 x 1.075^-3=
£750
(1.075 because it's a 7.5% increase, ^-3 because you're going back in time 3 years)

16)
a)
100 b)
100 x 1.2⁵ =
248.832
c)
By trial and error (or logarithms?)
t=12.6 (1 dp)


17)
Has anybody actually read this far? a)

d	frequency	width	fd	midpoint	mp x f
0<d<5	3	5	3÷5=0.6	2.5	7.5
5<d<10	9	5	9÷5=1.8	7.5	67.5
10<d<15	34	5	34÷5=6.8	12.5	85
15<d<20	49	5	49÷5=9.8	17.5	171.5
20<d<30	17	10	17÷10=1.7	25	425
30<d<40	41	10	41÷10=4.1	35	1435
40<d<60	27	20	27÷20=1.35	50	1350
60<d<100	20	40	20÷40=0.5	80	1600

b)Mean = sum of (mp x f) = 5141.5 = 25.7075
         sum of f           200

18)
a)
b)
Add the first 7 probabilities = 510/720 = 17/24
c)BBR+BRB+RBB = 378/720 = 21/40

19)


640 x (¹/₄)³ = 10 cm³

20)
a)
Using similarity...
the base of the big cone is three times that of the small cone
So... The big cone is three times the height of the small one
So... x = 24 cm

b) By Pythagoras
h² = 36² - 12²
h² = 1296 - 144
h² = 1152
h = 24 root2 = 33.94 cm


c) The large cone has a volume of 1/3 x 12 x 12 x µ x 33.94 = 5118.0cm³
The smaller cone has a volume (1/64) of this (see Q19!) = 80.0cm³
The difference is 5038.1cm³


21)
Volume = µ x Radius² x h
Hence Radius = root(V/(µ x h))
Maximum value will be when V is maximum and h is minimum (and use smallest value of pi allowable!)
Radius = root(505/(3.14 x 12.85))
=3.538 cm

22)
a)
Linear equation y = mx + c
m is the gradient change in y = 19-5 = 14 = 7
                  change in x    6-0    6   3
c is the intercept with the y axis = 5
Equation is therefore y= 7/3 x + 5

b)
By Pythagoras
d² = 14² + 6²
d² = 196 + 36
d² = 232
d = root232 = 15.2 units

23)
A is (0,4) becasue the amplitude is 4
B is (90,0) (because the period is 180)
C is (180,-4) (for the smae reasons)


24)
a)
AC = AD + AC = b + a+b = 2b+a
BC = -BA + AD + CD = -a + b + a+b = 2b
BE = (1/2)BC= b
AE = AB + BE = a + b = a+b

b)
AD is parallel to BC hence it is a trapezium

25) a) AB = DC and AD = BC (it is a parallelogram)
Hence AC = AB + BC = p + q
AE = (1/2)AC = 1/2(p+q)= 1/2p + 1/2q
BD = BA + AD = -p + q = q-p
BE = BA+AE = -p + 1/2p+ 1/2q= 1/2q - 1/2p
b)
2(BE) = BD hence E is the midpoint


26)
a)

Using Trig: SOH CAH TOA
Tan Q = 3.2/4.8 = 2/3
Q = Tan^-1(2/3)
= 33.7^o

b)

Using Trig: SOH CAH TOA
H = 3.2/ Sin 60 = 3.2/ 0.866=
3.7 m

27)

x	1	2	4	a
Z	100	12.5	b	0.1

When x is multiplied by two, Z is multiplied by 1/8, hence the relationship is inverse cubic.
n = 3
K = Z x³
Then substituting a pair of values from the table (eg Z=100, x =1)
K = 100
a) Sustituting x = 4 into Z=100/x³ gives 1.5625
Rearranging the formula gives x=cuberoot(K/Z) b) Sustituting Z = 0.1 x=cuberoot(100/Z) gives x = 10


28)

v	1	4	36	a
y	12	6	b	3/25

When v is multiplied by four, Z is multiplied by 1/2, hence the relationship is inverse square root.
n = n
K = y root(v)
Then substituting a pair of values from the table (eg y=12, v=1)
K = 12
a) Sustituting v = 36 into y=12/root(v) gives a=2
Rearranging the formula gives V=y²/K²
b) Sustituting y = 3/25 into V=y²/12² gives V = 10000

29)
a)

x-1 + 2x+1 =
 2     3  

3(x-1) + 2(2x+1) =
       6

3x-3 + 4x+2 =
     6

7x-1
  6

b)

3x+4 - 2x+5 =
  4     5

5(3x+4) - 4(2x+5) =
       4x5

15x+20 - 8x-20 =
      20

7x
20

c)
  1  +  2   =
2x+1   3x-2

1(3x-2) + 2(2x+1) =
  (2x+1)(3x-2)   

3x-2+4x+2    =
(2x+1)(3x-2)

    7x      
(2x+1)(3x-2)


d)
2x+1 - x+3 =
2x-1   x+4

(2x+1)(x+4) - (x+3)(2x-1) =
         (2x-1)(x+4)

2x²+5x+4 - (2x²+5x-3) =
         (2x-1)(x+4)

     7      =
(2x-1)(x+4)


30)
a)
 1   = 4  
x-1   3x-2

1(3x-2) = 4(x-1)

3x-2 = 4x-4

x=2

b)
 2  = 3 -  1 
x-2   x   x-4

2x(x-4) = 3(x-2)(x-4) - 1x(x-2)

2x²-8x = 3x²-18x+24 - x²+2x

8x=24

x=3

c)
2 - x = 1
        x

2x - x² = 1

x²-2x+1 = 0

(x-1)(x-1)=0
x=1

d)
x = 2x-5 + 3
     x-1

x(x-1) = 2x-5 + 3(x-1)

x²-x = 2x-5 + 3x-3

x²-6x+8=0
By factorising
(x-2)(x+4)=0
Hence
x= 2 or 4

e)
12(2x-4) + 16(x-2) + 12(9)=0

24x-48+16x-32+108=0
40x+28=0
x=-7/10


31)




32
a)

Using Trig: SOH CAH TOA
Tan Q = 3/8
Q = Tan^-1(3/8)
= 20.6^o
b)

First find DB
By Pythagoras
DB² = AB² + AB²
DB² = 16 + 64
DB² = 80
DB = root80 = 8.944
Using Trig: SOH CAH TOA
Tan Q = 3/8.944...
Q = Tan^-1(3/8.944...)
= 18.5^o
c)

First find FB or FA By Pythagoras
FB² = CB² + FC²
FB² = 64 + 9
FB² = 73
FB = root73 = 8.544

By Pythagoras
FA² = FB² + AB²
FB² = 73 + 16
FB² = 89
FB = root89 = 9.434

Using Trig: SOH CAH TOA
Tan Q = 4/8.544...
Q = Tan^-1(4/8.544...)
= 7.3^o