- Without using tables or calculator, simplify:
leaving the answer in standard form (scientific notation).
leaving the answer in standard form (scientific notation).
(NOT DRAWN TO SCALE)
ANSWER
- (a) 6 x 32 x 0.004 ; = 0.6 x 16 = 16
1.2x 0.008 x 0.16 1.2 x 0.16 2 x 0.16
100/2 = 50
(b) <AEB= 1800 – 3X0= 1800 ———— (i)
<BHC= 1200 – 7X0= 1800 ———— (ii)
By Elimination (eqtn 1 n eqtn 2)
60 – 4x= 0
60 15 = 4x
4 1 4
:. x = 150
Let <GHB = yo
7xo = yo = 1800 (sum of angle on straight line)
7(24) + yo = 1800 ; 1680 + yo = 1800
Y0 = 1800 n 1680 ; y0 = 120 ; <GHB = 120
- Simply
, leaving the answer in surd form (radicals)
, leaving the answer in surd form (radicals)- If 124n= 232five, find
.
.
ANSWER
- (a) 3√75 n √12 + √108
3√25 x 3 n √4 x 3 + √36 x 3
3 x 5√3 n 2√3 + 6√3
= 15√3 n 2√3 + 6√3 = 19√3
(b) 124n = 232five
1 x n2 + 2 x n1 + 4 x n0 = 2 x 52 + 3 x 51 + 2 x 50
n2 + 2n + 4 = 50 + 15 + 2 ; n2 + 2n + 4 = 67
n2 + 2n + 4n 67 = 0 ; n2 + 2n n 63 = 0
n2 + 9n n 7n n 63 =0 ; n(n + 9) n 7(n + 9) =0
(n + 9)(n n 7) = 0 ; n = n9 or n = 7
the base n cannot be negative :. n = 7
- Solve the simultaneous equations:
- A man drives from Ibadan to Oyo, a distance of 48
in 45 minutes. If he drives at 72 
where the surface is good and 48 where it is bad, find the number of kilometers of good surface.
in 45 minutes. If he drives at 72 
where the surface is good and 48 
where it is bad, find the number of kilometers of good surface.ANSWER
- (a) 1 + 1 = 5 ————- (i)
X y
1 – 1 = 1 ————– (ii)
X y
Make the equation (i) and (ii) to be linear from equation (i)
1 + 1 = 5
X y
Y + x = 5
Xy 1
y + x = 5xy ——————- (iii)
1 – 1 = 1
x y
x – y = 1
xy 1
x n y = yx ——————— (iv)
y +x= 5xyn————– (iii)
x n y = yx ————— (iv)
add equation (iii) and (iv)
2x=6xy
Divide both sides by 2x
2x = 6xy
2x 2x
1 = 3y ; 3y =1 ; y = 1/3
Substitute y 1/3 into equation (iii)
Equation (iii) is
Y+x=5xy : 1/3 + x =5x (1/3)
1 + x =5x
3 1 3
Multiply through by 3
3 (1/3) + 3(x) = 3 (5x/3)
1+3x =5x : 1=5x -3x : 1=2x: 2x=1: x ½
:. X =1/2 , y =1/3
(b) Total distance = 48km
Total time =45 mins = (45/60)hr = 0.75hr
:. Let the distance cover on a good surface =xkm and the
Time = t1 hr
:. Average speed on good surface = distance
Time
i.e 72 = x
t1 :. X =72t1 …………………… (i)
speed on bad surface = 48km/hr
let the time spent on bad surface = t2 h
:. Average speed on bad surface = distance
Time
i.e 48 = x
t2
y=48t2……………… (ii)
From above:
Total distance = 48km
i.e x +y = 48km ……………(iii)
Total time = 0.75h ; i.e t1 +t2 =0.75 ………….(iv)
Substitute equations (i) and (ii) into equation (iii)
72 t1 + 48t2 = 48km
Dividing through by 24
We have: 3t1 +2t2 = 2 ……………. (v)
Solving equation (iv) and (v) simultaneously from above
t1 + t2 = 0.75 ………….. (iv) ; 3t1 +2t2= 2……………. (v)
Multiply equation (iv) by 3
Multiply equation (v) by 1
We have
3t1 + 3t2 = 2.25
N3t1 + 2t2 = 2
t2 = 0.25
-
(a)
In the diagram, O is the centre of the circle radius
and ∠
If the area of the shaded part is 504 
2, calculate the value of 
(b) Two isosceles triangles
are drawn on opposite sides of a common base PQ. If ∠
, calculate the value of ∠
.
and ∠
If the area of the shaded part is 504 
2, calculate the value of 
are drawn on opposite sides of a common base PQ. If ∠
, calculate the value of ∠
.ANSWER
- (a) Shaded part is a segment
Area of segment = Area of sector n Area of triangle
504 cm2 = θ – 1r2 sinθ
360 2
=90 x 22 x r2 –1 x r2sin900
360 7 2
504=1 x 22r2 – 1r2
4 77 2
=22r2 – 1r2
28 2
504 = 22r2 – 14r2 ; 504 =8r2 ; 8r2 = 504 x28
28 28
r2 = 504 x 28 = r2 = 1764 =√1764 = 42cm
8
(b) from ∆PQS
Let the base angle = a and b
Where a = b
109 + a + a = 180 (sum of < sin ∆ )
2a + 109= 180
2a = 180 -109 : 2a = 71
A =71/2 = a = 35.5
:. 0
The value of
+PQS (660+35.50)
<RQS=101.50
A building contractor tendered for two independent contracts, X and Y. The probabilities that he will win contract X is 0.5 and not win contract Y is 0.3. What is the probability that he will win:
- both contract;
- exactly one of the contracts;
- neitherof the contracts?
ANSWER
- Probability of win x = 0.5
Probability of win y = 1-0.3=0.7
Probability of not win y = 0.3
- Probability he wins both contract = probability (x and y) = 0.5 x 0.7 = 0.35
- Probability he wins exactly one = (prob win x) x
(prob not win y) + prob (win y) x prob (not win y)
=(0.5 x 0.3) + (0.7 x 0.5) = 0.15 + 0.35 = 0.50
- Probability neither x nor y
Probability (not x and not y)
0.5 x 0.3 = 0.15
- If
- A television set was marked for sale at
in order to make a profit of 20
. The television set was actually sold at a discount of 5
Calculate, correct to 2 significant figures, the actual percentage profit.
in order to make a profit of 20
. The television set was actually sold at a discount of 5
Calculate, correct to 2 significant figures, the actual percentage profit.
ANSWER
3 = 1/3
2p – ½ ¼p + 1
3 = 1/3 ; 3 = 1
2p/1 –1/2 p/4 + 1/1 4p-1 p+ 4
2 4
4p-1 p+ 4 3 x 2 = 4
2 4 4p-1 3 (P+4)
= 6 = 4
4p – 1 3p +12
Cross multiply; 6 (3p +12) = 4 (4p-1) ; 18p + 72 = 16p – 4
Collect like terms
18p – 16p = -4-72; 2p =-76
:.p = -76 = -38
2
(b) Market price = GH€ 760.00
Percentage marked price = 100 + 20 = 120%
Selling price = 760.00 –(5/100 x 760.00)
= 760 – 38 = Gh€ 722
Let the cost price = GH€ x
% cost price = 100
:. 760 – 120 ; x =100
By simple proportion
120 x = 760 x 100
X = 760 x 100
120
GH€ 633.33
Profit = selling price – cost price
722.00 – 633.33 = GH€ 88.67
:.% profit = profit x 100
Cost price
= 88.67 x 1000 = 14%
633.337
- Copy and compete the table of values for the relation
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| 1.0 |
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| 2.7 |
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| 0.0 |
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- Using scales of
on the 
axis and 2 
unit on the 
axis, draw the graph of 
- Use the graph to find the values of
for which 
on the 
axis and 2 
unit on the 
axis, draw the graph of 
for which 
Answer
(a)
X | 00 | 300 | 600 | 900 | 1200 | 1500 | 1800 | 2100 | 2400 | 2700 |
2Sin x | 0.00 | 1.00 | 1.73 | 2.00 | 1.73 | 1.00 | 0.00 | 1.00 | 1.73 | 2.00 |
+1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 |
Y | 1.00 | 2.00 | 2.73 | 3.00 | 2.73 | 2.00 | 1.00 | 0.00 | 2.73 | 1.00 |
- Copy and complete the following table for multiplication modulo 11.
 | 1 | 5 | 9 | 10 |
| 1 | 1 | 5 | 9 | 10 |
| 5 | 5 | |||
| 9 | 9 | |||
| 10 | 10 |
Use the table to:
evaluate (9 5 ) (10 10)- find the truth set of
- 10 m=2,
- n n = 4
- When a fraction is reduced to its lowest term, it is equal to
. The numerator of the fraction when doubled would be 34 greater than the denominator. Find the fraction.
evaluate (9 5 ) (10 10)
10 m=2,
n n = 4
. The numerator of the fraction when doubled would be 34 greater than the denominator. Find the fraction.Answer
(a) (i) (9 ⊗ 5) ⊗ (10 ⊗ 10)
9 x 5 = 45/11 =4 Rem (1)
10 x 10 = 100/11 = 9 Rem (1)
:. (1) ⊗ (1) = 1
- 10 ⊗ m = 2
m = 10 ⊗ 2
m = 20 (mod 11)
m = 1 Rem 9
m = 9
ii. n ⊗ n = 4
9 ⊗ 9 = 4
(b) Let the fraction be x/y
x/y = ¾ ………… (i)
2x = 34 + y ……… (ii)
From equation (i)
X = 3y/4
Substituting 3y/4 for x in equation (ii)
2x = 3x + y ; 2(3y/4) = 34 +y
3y = 2 (34 +y) ; 3y = 68 +2y
3y – 2y = 68 ; y = 68
X = 3(y)/4 = 3 x 68/4
x = 51
Therefore the fraction is 51/68
(a)
U
Inthe Venn diagram, P, Q and R are subsets of the universal set U.
If n(U) = 125, find:
- the value of x;
- n(PÈQÇR’)
In the diagram, O is the centre of the circle. If WX is parallel to YZ and ∠WXY = 50o, find the value of:
- ∠WXZ;
- ∠YEZ.
Answer
(a) 16-2x +5x +4x +8x + 7x +6 +x+19-3x+4=125
Collect like terms
16+6+19+4+5x+4x+8x+7x+x-3x-2x =125
45+25x-5x=125 ; 45 +20x=125
20x = 125-45 ; 20x = 80 ; x 80/20 ; x = 4
(ii) To find n(PᵕQᵔRI)
(PᵕQᵔRI) = (16-2x,5x,6+x,4)
= (16-2 (4), 5 (4), 6 +4,4)
=(8,20,10,4) ; (PᵕQᵔRI) = 4
(b) (i) <XYW =900 (angle in a semi-circle)
In ∆XWY, <x + <w =900 (rem ps of a∆ )
i.e 50 +W =900
:. W =900 -500
<WYZ = <W = 400 (Alternative angles)
:. <WYZ = 400
(ii) <WOZ = 2<WYZ (<Substende by the WZ
at the centre is twice that in the circumference
:.<WOZ = 2 x 400 = 800
<WEO + <W +<WOZ = 1800
(sum of interior <s of a ∆ )
i.e <WEO + 400+800=1800
:.<WEO = 1800 – 1200 = 600
<YEZ=<WEO(Vertically opp. <s) :.<YES=600
- Solve: ( x – 2)(x – 3) = 12
(b)
Answer
. (a) (x-2)(x-3)=12
X(x-3)-2(x-3)=12 ; x2-3x-2x+6=12
X2-5x+6-12=0 ; x2-5x-6=0
X2-6x+x-6=0 ; x(x-6)+1(x-6)=0
(x+1)(x-6)=0 ; x+1=0 or x-6 =0
X=-1 or x=6
- (b) Area of shaded portion = Area of sector – Area of triangle
Θ x – ½ Sinθ
360
60 x 22 x 7 x 7 – (1/2 x 7 x 7 Sin 60)
360 7
=25.67 – 21.22=4.45
:. Area of the shaded portion = 2×4.45=8.9cm
Scores | 1 | 2 | 3 | 4 | 5 | 6 |
Frequency | 2 | 5 | 13 | 11 | 9 | 10 |
The table shows the distribution of outcomes when a die is thrown 50 times. Calculate the:
(a) mean deviation of the distribution;
(b) probability that a score selected at random is at least a 4.
Answer
Scores (x) | f | fx | x-m=d | d | fd |
1 | 2 | 2 | -3 | 3 | 6 |
2 | 5 | 10 | -2 | 2 | 10 |
3 | 13 | 39 | -1 | 1 | 13 |
4 | 11 | 44 | 0 | 0 | 0 |
5 | 9 | 45 | 1 | 1 | 9 |
6 | 10 | 60 | 2 | 2 | 20 |
Mean x ∑f/ ∑f = 200/50
(a) Mean deviation ∑f/∑f = 58/50 = 1.16
(b) Probability (at least 4 score) = 11 + 9 +10/50 = 30/50
= 0.6
(a) Given that 5cos (x+8.5) 0 – 1 = 0, 00 ≤ x ≤ 90 0, calculate, correct to the nearest degree, the value of 
.
(b) The bearing of Q from P is 0150 0 and the bearing of P from R is 0150 0. If Q and R are
24 km and 32 km respectively from P:
- represent this information in a diagram;
- calculate the distance between Q and R, correct to two decimal places;
- find the bearing of R from Q, correct to the nearest degree.
Answer
(a) Given that:
5cosx(x + 8.5) 0 -1= 0 ; cos (x+8.5) 0=1/5
Cos (x+8.5) 0 = 0.2 ; x +8.50=cos-10.2
X + 8.50 = 78.460
Collect like terms
X=78.460-8.50 ; x0=69.960 ; x0=700
(b)(1)
(ii) By the cosine rule,
|QR|2 = 322 + 242 − 2 × 32 × 24 × cos45
|QR|2 = 1024 + 576 −1536 cos45
= 1600 − 1086.1056
|QR|2 =
|QR|= √513.8944 = 22.669km
≊22.67km ( 2 decimal place)
(iii) By the sine rule,
32 / sinα = 22.67/sin45
sinα = 32 × sin45 / 22.67
= 0.9981
α = sin −1(0.9981) = 86.4787°
The diagram below shows all the angles at Q;
reflex < NQR = 360° – (86.47° + 30°) = 360° – 116.47°
= 243.53°
Hence, the bearing of R from Q = 244° (to the nearest degree).
(a) Two functions, f and g, are defined by f : x→2x2 − 1 and g : x→ 3x + 2 where x is a real number.
(i) If f(x−1) − 7=0, find the values of x.
(ii) Evaluate : f(−1/2).g(3) / f(4)−g(5).
(b) An operation, (∗) is defined on the set R, of real numbers, by m ∗ n = −n / m2 +1, where m, n ∈ R. If −3, −10 ∈ R, show whether or not ∗ is commutative.
Answer
(a) f : x → 2x2−1; g: x →3x+2
(i) f(x−1)−7 = 0
f(x−1) = 2(x−1)2−1 = 2(x2−2x+1)−1
= 2x2−4x+2−1
f(x−1)−7 = 2x2−4x+1−7 = 2x2−4x−6=0
2x2− 6x + 2x − 6=0 ⟹ 2x(x−3) + 2(x−3) = 0
(2x+2)(x−3) = 0 ⟹2x = −2; x=3
x = −1;3
(ii) f(−1/2).g(3) / f(4)−g(5)
f(−1/2) = 2(−1/2)2 −1 = 1/2 − 1 = − 1/2
g(3) = 3 (3) + 2 = 9 + 2 = 11
f(4) = 2 (42) − 1 = 32−1= 31
g(5) = 3(5) + 2 = 15 + 2 = 17
f(−1/2).g(3) / f(4)−g(5) = (−1/2).(11) / 31−17
= −11/2 / 14
= −11 / 28
