(a) Simplify, without using tables or calculator : 
(b) Given that log102 = 0.3010 and log103 = 0.4771 . Evaluate correct to 2 significant figures and without using calculator and table
.log10 1.125
ANSWER
(a)
(b) Given log102 = 0.3010 and log103 = 0.4771
Log 10 1. 125 = Log 10 1125/ 1000 = 45/40;
= Log 10 9/8 ; Log 109 – log 810 = Log 1032 – Log 1023
= 2log 103 – 3log 102 = (0.4771) – (0.3010)
= 0.9542 – 0.903
= 0.051
(a) Solve : 7x+4 < 1/2(4x+3) .
(b) Salem, Sunday and Shaka shared a sum of N1,100.00. For every N2.00 that Salem gets, Sunday gets 50 kobo and for every N4.00 Sunday gets, Shaka gets N2.00. Find Shaka’s share.
ANSWER
(a) 7x+4 < 1 /2(4x+3)
7x+4 < 2x+ 3 / 2 collect like terms
7x−2x < 3 /2−4
5x < −5/2 divide both sides by 5
x < −1/2.
(b) When Salem gets N2 ——> Sunday gets 50 kobo
(Recall N1 = 100 kobo)
Salem gets 200 kobo ——> Sunday gets 50 kobo.
Sunday gets N4 —–> Shaka gets N2.
====> Sunday gets 400 kobo ——> Shaka gets 200 kobo.
When Sunday gets 400 kobo ——> Salem gets 1600 kobo.
∴∴ Salem : Sunday : Shaka = 1600 : 400 : 200
= 8 : 2 : 1
∴ Shaka’s share=1/11×N1,100 = N100
(a) The present ages of a father and his son are in the ratio 10 : 3. If the son is 15 years old now, in how many years will the ratio of their ages be 2 : 1?
(b) The arithmetic mean of x, y and z is 6 while that of x, y, z, l, u, v and w is 9. Calculate the arithmetic mean of l, u, v and w.
ANSWER
Sum of the ration = 10+3 = 13
Son’s present age = 15yrs
Father’s present age = let it be k
Sum of their ages = y
3———-15
13———-y
cross multiply 3 x y = 13 x 15
then y = 65 = sum of their ages .
then fathers age is 65-15 = 50
In z years time, the ratio of their ages = 2 : 1
2/1 = 50+z / 15+z
cross multiply 2(15+z) = 50+z
30 + 2z = 50+z
z = 20
Therefore, in 20 years’ time, the ratio of their age will be 2 : 1.
(b) x + y + z / 3 = 6
x + y + z = 18 ——– (1)
x + y + z + l + u + v + w / 7 = 9
x + y + z + l + u + v + w = 63 ———- (2)
18 + l + u + v + w = 63
∴l+u+v+w=63−18=45
Mean of l, u, v and w = 45/4 = 11.25
The area of a circle is 154cm2. It is divided into three sectors such that two of the sectors are equal in size and the third sector is three times the size of the other two put together. Calculate the perimeter of the third sector. [Take π=22/7]
Answer
A boy 1.2m tall, stands 6m away from the foot of a vertical lamp pole 4.2m long. If the lamp is at the tip of the pole,
(a) represent this information in a diagram ;
(b) calculate the (i) length of the shadow of the boy cast by the lamp ; (ii) angle of elevation of the lamp from the boy, correct to the nearest degree.
Answer
(a) 
From triangle TPQ
Tan x = opp/adj = 3/6 = 0.5
x = tan-1 0.5 ; x = 26.570
x = 270
to get y = 180-90-27 = 630
(bi) The boy’s shadow z
Tan 630 = z / 1.2 then 1.2 x tan 630
therefore z = 2.4m
(bii) Angle of elevation of the boy
From triangle TPQ
Tan x = opp/adj = 3/6 = 0.5
x = tan-1 0.5 ; x = 26.570
x = 270
(a) Two positive whole numbers p and q are such that p is greater than q and their sum is equal to three times their difference;
(i) Express p in terms of q ; (ii) Hence, evaluate p2+q2 / pq.
(b) A man sold 100 articles at 25 for N66.00 and made a gain of 32%. Calculate his gain or loss percent if he sold them at 20 for N50.00.
Answer
(a)(i) p>q….(1)
p+q=3(p−q)……(2)
From (2), p+q=3p−3q collect like terms
p−3p=−3q−q⟹−2p=−4q
p=2q
(ii)
(b)
Selling price = 100×66 / 25
= N264.00
Using SP−CP / CP= 264−CP / CP = 32 / 100
100(264−CP)=32CP
26400=32CP+100CP=132CP
CP=26400 / 132
= N200
When he sells 20 for N50, Selling price = 100×50 / 20=N250.00
Hence, he made gain.
1 / 4 ×100
= N25
(a) Copy and complete the table of values for the relation y=3x2−5x−7.
(b) Using scales of 2 cm to 1 unit on the x- axis and 2 cm to 5 units on the y- axis, draw the graph of y=3x2−5x−7,−3≤x≤4.
(c) From the graph : (i) find the roots of the equation 3x2−5x−7=0 ; (ii) estimate the minimum value of y ; (iii) calculate the gradient of the curve at the point x = 2.
Answer
(a) 
(c)
(1) Roos of equation are x = -09 and 2.5
(ii) Minimum value of y = -9
(iii) gradient of the curve at the point = 2 is
gradient = y2 – y1 /x2 – x1
= 10 – -0 /4 – 1 = 20/2.3 = 6.7
(a) If (3 – x), 6, (7 – 5x) are consecutive terms of a geometric progression (GP) with constant ratio r > 0, find the :
(i) values of x ; (ii) constant ratio.
(b) 
In the diagram, |AB| = 3 cm, |BC| = 4 cm, |CD| = 6 cm and |DA| = 7 cm. Calculate <ADC, correct to the nearest degree.
Answer
(3 – x) = T1 (1st term – a)
6 = T2 (2nd term – ar )
(7 – 5x) = T3 (3rd term – ar2 )
Common ratio = 6/3-x = 7-5x /6 cross multiply 36 = (3-x)(7-5x)
36=21−15x−7x+5x2
36=21−22x+5x2
5x2−22x+21−36=0
5x2−22x−15=0
5x2−25x+3x−15=0⟹5x(x−5)+3(x−5)=0
(x−5)(5x+3)=0
x=−3 / 5 or = 5
(ii) constant ratio.
r = 6/3-x ; where x = 5 then r = 6/3-5 = -3
Where x = -3/5 ; r = 6/3-x = 6/3-(-3/5) = 5/3
(b) 
Considering ∆ AEC
b2 = 42 + 32
b2 = 16 + 9 = 25 ;
b = √25 = 5cm
Considering ∆ ACD
using cosine rule ;
(a) Using ruler and a pair of compasses only, construct : (i) a trapezium WXYZ such that |WX| = 10.2 cm, |XY| = 5.6 cm, |YZ| = 5.8 cm, < WXY = 60° and WX is parallel to YZ (ii) a perpendicular from Z to meet WX at N.
(b) Measure :
(i) |WZ| ; (ii) |ZN| .
Answer
(a)
(b)
(b) From the diagram above,
(i) |WZ| = 5.25 cm ; (ii) |ZN| = 4.86 cm.
(a) 
A segment of a circle is cut off from a rectangular board as shown in the diagram. If the radius of the circle is 1 1/2 times the length of the chord; calculate, correct to 2 decimal places, the perimeter of the remaining portion. [Take π=22/7]
(b) (b) Evaluate without using calculators or tables, 3 /√3(2 /√3 – √12 /6 )
Answer
(a) 
Length of chord is 14cm
Radius = 3/2 X 14cm = 21 cm .
the chord AB = 2r Sin θ/2
14 = 2(21) sin θ/2
sin θ/2 = 14 / 42 = 0.333
θ/2 = sin-1 0.333 = 19.469
θ = 19.469 x 2 = 38. 938
Length of the arc = θ/360 x 2nr
38.938 /360 x 2 x 22/7 x 21
= 14.277cm
Perimeter of the remaining portion = 22 + 12 + 12 + 5 + 3 + 14.277
= 68.277 cm = 68.28
The frequency distribution table shows the marks obtained by 100 students in a Mathematics test.
(a) Draw the cumulative curve for the distribution.
(b) Use the graph to find the : (i) 60th percentile ; (ii) probability that a student passed the test if the pass mark was fixed at 35%.
Answer
(a) image
b(i) 60th percentile = 60/100 x 100 = 60
@ 60th perc 55.5
(ii) If the pass mark is 35% = 35/100 x 100 = 35
@35 = 47.5
An aeroplane flies due North from a town T on the equator at a speed of 950km per hour for 4 hours to another town P. It then flies eastwards to town Q on longitude 65°E. If the longitude of T is 15°E,
(a) represent this information in a diagram ;
(b) calculate the : (i) latitude of P, correct to the nearest degree ; (ii) distance between P and Q, correct to four significant figures. [Take π=22 / 7; Radius of the earth = 6400km].
Answer
(a) image
b(i) Length of arc PT = speed x time = 950 x 4 = 3800km
distance = 3800km
distance = θ / 360 x 2Πr
3800 = θ /3600 x 2 x 22/7 x 6400
θ = 34.010
(ii) Distance between P and Q, correct to four significant figures.
Longitude difference = 65° – 15° = 50°
Using d = θ /360 x 2Πr
where r = R cos θ
d = θ /360 x 2Π R cos θ
= 50/360 x 2 x 22/7 6400 x cos 34.010
= 4631.9km = 4632(to four significant figure)
When one end of a ladder, LM, is placed against a vertical wall at a point 5 metres above the ground, the ladder makes an angle of 37° with the horizontal ground.
(a) Represent this information in a diagram ;
(b) Calculate, correct to 3 significant figures, the length of the ladder ;
(c) If the foot of the ladder is pushed towards the wall by 2 metres, calculate, correct to the nearest degree, the angle which the ladder now makes with the ground.
Answer
(a) 
(b) Length of ladder = xm
sin 37 = 5 /x ∴ x sin 37 = 5
x = 5/ sin 37 = 5/ 0.6018 = 8.31m
x = 8.31m
(c) 
First lets get z
z2 = 8.312 – 52 = 69.06 – 25
= 69.06 – 25 = 44.06
z = √44.06 = 6.64m
z = 6.64m
since the ladder was moved 2m towards the wall then z = 6.64 – 2 = 4.64m
Let yo be the new angle the ladder makes
Then cos y = 4.64 / 8.31 = 0.5584
yo = cos-1 0.5584 = 56.089° = 56°(to the nearest degree)
