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**Question 1**
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Solve the inequality x - 1 > 4(x + 2)

**Answer Details**

x - 1 > 4(x + 2) = x - 1 > 4x + 8

4x + 8 < x - 1 = 4x - x < -1 -8

= 3x < -9

∴ x < -3

**Question 2**
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If P = 23 $\frac{2}{3}$ (1−r2n2 $\frac{1-{r}^{2}}{{n}^{2}}$), find n when r = 13 $\frac{1}{3}$ and p = 1

**Answer Details**

If P = 23
$\frac{2}{3}$ (1−r2n2
$\frac{1-{r}^{2}}{{n}^{2}}$), find n when r = 13
$\frac{1}{3}$ and p = 1

p = 2(1−r2)3n2
$\frac{2(1-{r}^{2})}{3{n}^{2}}$ when r = 13
$\frac{1}{3}$ and p = 1

1 = 23
$\frac{2}{3}$ (1−(13)2)n2
$\frac{(1-(\frac{1}{3}{)}^{2})}{{n}^{2}}$

n2 = 2(3−1)3×3
$\frac{2(3-1)}{3\times 3}$

n2 = 2×23×3
$\frac{2\times 2}{3\times 3}$

= 49
$\frac{4}{9}$

n = 49
$\frac{4}{9}$

= 23

**Question 3**
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What is the locus or the mid-point of all the chords of length 6cm with circle of radius 5cm and with center 0?

**Answer Details**

x2 + 33 = 52

x2 + 9 = 25

x2 = 25 – 9

x2 = 16

x = √ 16

= 4 cm

∴ The locus is a circle of radius 4 cm with the center 0

**Question 4**
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The goals scored by 40 football teams from three league divisions are recorded below:

Number of goals0123456frequency431516101 $\begin{array}{cccccccc}\text{Number of goals}& 0& 1& 2& 3& 4& 5& 6\\ \text{frequency}& 4& 3& 15& 16& 1& 0& 1\end{array}$

What is the total number of goals scored by all the terms?

**Answer Details**

Let x rep. number of goals

"f"frequencyxffx0401332153031648414500616
$\begin{array}{ccc}"f& "& frequency\\ x& f& fx\\ 0& 4& 0\\ 1& 3& 3\\ 2& 15& 30\\ 3& 16& 48\\ 4& 1& 4\\ 5& 0& 0\\ 6& 1& 6\end{array}$

∑fx
$\sum fx$ = 91

Total number of goals scored is ∑fx
$\sum fx$= 91

**Question 5**
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Two chords QR and NP of a circle intersect inside the circle at x. If RQP = 37o, RQN = 49o and QPN = 35o, find PRQ

**Answer Details**

In PNO, ONP

= 180 - (35 + 86)

= 180 - 121

= 59

PRQ = QNP = 59(angles in the same segment of a circle are equal)

**Question 6**
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If P varies inversely as V and V varies directly as R2, find the relationship between P and R given that R = 7 when P = 2.

**Answer Details**

P = 1v
$\frac{1}{v}$ and vR2 = P = kv
$\frac{k}{v}$......(i)

and v KR2 .......(ii)

(where k is constant)

Subst. for v in equation (i) = p = 12KR $\frac{{1}^{2}}{KR}$.....(ii)

when r = 7, p = 2

2 = k72 $\frac{k}{{7}^{2}}$

k = 2 x 49

= 98

Subt. foe k in ....(iii)

P = 98R2 $\frac{98}{{R}^{2}}$

PR2 = 98

and v KR2 .......(ii)

(where k is constant)

Subst. for v in equation (i) = p = 12KR $\frac{{1}^{2}}{KR}$.....(ii)

when r = 7, p = 2

2 = k72 $\frac{k}{{7}^{2}}$

k = 2 x 49

= 98

Subt. foe k in ....(iii)

P = 98R2 $\frac{98}{{R}^{2}}$

PR2 = 98

**Question 7**
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Simplify without using tables log26log28 $\frac{lo{g}_{2}6}{lo{g}_{2}8}$ - log23log212

**Answer Details**

log26 - log23 = log2 (63
$\frac{6}{3}$) = log22

log28 - 212
$\frac{1}{2}$ log28 - log212
$\frac{1}{2}$ log28 - log214
$\frac{1}{4}$

log22log232=log225log22
$\frac{lo{g}_{2}2}{lo{g}_{2}32}=\frac{lo{g}_{2}2}{5lo{g}_{2}2}$

= 15
$\frac{1}{5}$

N.B. log22 = 1

**Question 8**
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Four boys and ten girls can cut a field in 5 hours if the boys work at 54 $\frac{5}{4}$ the rate at which the girls work. How many boys will be needed to cut the field in 3 hours?

**Answer Details**

Let x represents number of boys that can work at 54
$\frac{5}{4}$ the rate at which the 10 girls work

For 1hr. x boys will work for 154
$\frac{\frac{1}{5}}{4}$ x 10

x = 45
$\frac{4}{5}$ x 10 = 8 boys

8 boys will do the work of ten girls at the same rate 4 + 8 = 12 boys cut the field in 5 hrs

For 3 hrs, 12×53
$\frac{12\times 5}{3}$ boys will be needed = 20 boys

**Question 9**
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A rectangular lawn has an area of 1815 square yards. if its length is 50 meters, find its width in meters given that 1 metre equals 1.1 yards

**Answer Details**

1m = 1.1 yard, length(L)= 50m

= (50 x 1.1)yards

= 55 yards

Area(A) = length(L) y width (W)

1815 = 55y width (W)

width (w) = 181555
$\frac{1815}{55}$

= 33 years

**Question 10**
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In the figure, find the value of x

**Answer Details**

Z = X = Y = 180o .........(i)

2Z + Y + Y = 190 = 2Z + 2Y = 180

Z + Y = 90o........(ii)

hence x + (z + y) = 180

x + 90o = 180o

x = 180o - 90o

= 90o

**Question 11**
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From two points x and y, 8m apart, and in line with a pole, the angle of elevation of the top of the pole are 30o and 60o respectively. Fins the height of the pole assuming that x, y and the foot of the pole are the same horizontal plane and x and y are on the same side of the pole.

**Answer Details**

From the diagram, WYZ = 60o, XYW = 180o - 60o

= 120o

LX = 30o

XWY = 180o - 120o + 30o

XWY = 30o

WXY = XYW = 30o

Side XY = YW

YW = 8m, sin 60o = 32
$\frac{3}{2}$

? sin 60o = hYW
$\frac{h}{YW}$, sin 60o = h8
$\frac{h}{8}$

h = 8 x 32
$\frac{3}{2}$

= 4?3

**Question 12**
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Find the eleventh term of the progression 4, 8, 16.....

**Answer Details**

a = 4, r = 42
$\frac{4}{2}$

84
$\frac{8}{4}$ = 2

n = 11

Tn = arn - 1

T11 = 4(2)11 - 1

4(2)10 = 212

since 4 = 22

= 212

**Question 13**
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In the diagram above, QR // TS, QR : TS = 2.3, Find the ratio of the area of triangle PQR to the area of the trapezium QRST.

**Answer Details**

h1 = h1+h21.5 $\frac{{h}_{1}+{h}_{2}}{1.5}$ = h1h2=2 $\frac{{h}_{1}}{{h}_{2}}=2$

A△PQRATP(QRST)=12×2×h112×h2(2+3) $\frac{{A}_{\u25b3PQR}}{{A}_{TP(QRST)}}=\frac{\frac{1}{2}\times 2\times {h}_{1}}{\frac{1}{2}\times {h}_{2}(2+3)}$

25×h1h2=25×2 $\frac{2}{5}\times \frac{{h}_{1}}{{h}_{2}}=\frac{2}{5}\times 2$

= 45

**Question 14**
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Taking the period of day light on a certain day to be from 5.30a.m to 7.00p.m. Calculate the angle of a pie chart designed to show the periods of the day light and of darkness on that day

**Answer Details**

Period of day light is from 5.30 to 7.00p.m - 13hrs 30min. interval

24hrs make 1 day

∴ period of darkness = 24hrs - 13hrs 30mins = 10hrs 30mins.

Angle of sector of daylight = 13×60+30×360(24×60)
$\frac{13\times 60+30\times 360}{(24\times 60)}$

= 8101440
$\frac{810}{1440}$ x 360o

= 2916001440
$\frac{291600}{1440}$

= 202.5o

202.5o = 202.30o

darkness period = 360o - 202o 30'

= 157o30'

= 202o30' 157o 30'

**Question 15**
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If tan θ $\theta $ = m2−n22mn $\frac{{m}^{2}-{n}^{2}}{2mn}$ find secθ

**Answer Details**

Tan θ
$\theta $ = m2−n22mn
$\frac{{m}^{2}-{n}^{2}}{2mn}$

OppAdj
$\frac{\text{Opp}}{\text{Adj}}$ by pathagoras theorem

= Hyp2 = Opp2 + Adj2

Hyp2 = (m2 - n2) + (2mn)2

Hyp2 = m4 - 2m2n4 - 4m2 - n2

Hyp2 = m4 + 2m2 + n2n

Hyp2 = (m2 - n2)2

Hyp2 = m2+n22mn

**Question 16**
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Two brothers invested a total of ₦5,000.00 on a farm project, the farm yield was sold for ₦15,000.00 at the end of the season. If the profit was shared in the ratio 2 : 3, what is the difference in the amount to profit received by the brothers?

**Answer Details**

Total amount invested by A and B = ₦5,000

farm yield was sold for ₦15,000.00

profit = 15,000.00 - 5,000.00

= ₦10,000.00

Profit was shared in ratio 2 : 3

2 + 3 = 5

A received 25
$\frac{2}{5}$ of profit = 25
$\frac{2}{5}$ x 10,000 = ₦4,000.00

A receive 35
$\frac{3}{5}$ of profit = 35
$\frac{3}{5}$ x 10,000 = ₦6,000.00

Difference in profit received = ₦6,0000 - ₦4,000.00

= ₦2,000.00

**Question 17**
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Convert 241 in base 5 to base 8,

**Answer Details**

2415 = 2 x 52 + 4 x 5 + 1 x 5 + 1 x 5o

50 + 20 + 1 = 7110

Convert 7110 to base 8

87188R781R080R1
$\begin{array}{cc}8& 71\\ 8& 8R7\\ 8& 1R0\\ 8& 0R1\end{array}$

= 1078

**Question 18**
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Reduce each number to two significant figures and then evaluate 0.021741×1.20470.023789

**Answer Details**

0.021741×1.20470.023789
$\frac{0.021741\times 1.2047}{0.023789}$ = 0.022×1.20.024
$\frac{0.022\times 1.2}{0.024}$ (to 216)

= 0.02640.024
$\frac{0.0264}{0.024}$

= 1.1

**Question 19**
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The numbers 3, 2, 8, 5, 7, 12, 9 and 14 are the marks scored by a group of students in a class test. If P is scored by an group of students in a class test. If P is the mean and Q the median, the P + Q is

**Answer Details**

Re-arranging them 2, 3, 5, 7, 8, 9, 12, 14

Mean = P and median = Q

P = 2+2+5+7+8+9+12+148
$\frac{2+2+5+7+8+9+12+14}{8}$

608
$\frac{60}{8}$ = 7.5

median (Q) = 7+82
$\frac{7+8}{2}$

152
$\frac{15}{2}$ = 7.5

P + Q = 7.5 + 7.5 = 15

**Question 20**
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Find the probability of selecting a figure which is a parallelogram from a square, a rectangle, a rhombus, a kite and a trapezium

**Answer Details**

A rectangle, a square and a rhombus are the only parallelogram

∴ Probability of a parallelogram = 35

**Question 21**
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The base of a pyramid is a square of side 8cm. If its vertex is directly above the centre, find the height, given that the edge is 4.3cm

**Answer Details**

Base of pyramid of a square of side 8cm vertex directly above the centre edge = 4√3
$4\sqrt{3}$cm

From the diagram, the diagonal of one base is AC2 = 82 + 82

Ac2 = 64 + 64 = 128

AC = 8√2
$8\sqrt{2}$

but OC = 12
$\frac{1}{2}$AC = 8√22
$\sqrt{\frac{2}{2}}$ = 4√2
$4\sqrt{2}$cm

OE = h = height

h2 = (4√3
$4\sqrt{3}$)2

16 x 2 - 16 x 2

48 - 32 = 16

h = √16
$16$

= 4

**Question 22**
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An (n−2)2 $(n-2{)}^{2}$ sided figure has n diagonals, find the number n of diagonals for a 25 sided figure

**Question 23**
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The minimum value of y $y$ in the equation y=x2−6x+8

**Question 24**
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By selling 20 oranges for ₦1.35 a trader makes a profit of 8%. What is his percentage gain or loss if he sells the same 20 oranges for ₦1.10?

**Answer Details**

profit 8% of ₦1.35 = 8100
$\frac{8}{100}$ x ₦1.35 = ₦0.08

Cost price = ₦1.35 - ₦0.10 = ₦1.25

If he sells the 20 oranges for ₦1.10 now

%loss = actual lossCost price
$\frac{\text{actual loss}}{\text{Cost price}}$ x 100

125−1.101.25
$\frac{125-1.10}{1.25}$ x 100

= 0.15×1001.25
$\frac{0.15\times 100}{1.25}$

= 151.25
$\frac{15}{1.25}$

= 12%

**Question 25**
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If the length of a square is increased by 20% while while its width is decreased by 20% to form a rectangle, what is the ratio of the area of the rectangle to the area of the square?

**Answer Details**

Length and width of a square is 100%

Length increased by 20% and

Width decreased by 20% to form a rectangle

Length of rectangle = 120% to form a rectangle

Length of rectangle = 120% and

Width of rectangle = 80%

Area of rectangle = L x W

Area of square = W

Ratio of the area of the rectangle to the area of the square

A = Area of rectangleArea of square
$\frac{\text{Area of rectangle}}{\text{Area of square}}$

120×30100×100
$\frac{120\times 30}{100\times 100}$ = 96100
$\frac{96}{100}$

= 24 : 25

**Question 26**
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A train moves from P to Q at an average speed of 90km/h and immediately returns from Q to P through the same route at an average speed of 45km/h. Find the average speed for the entire journey

**Answer Details**

Average speed from P to Q = 90km\h

Average speed from O to P = 45km/h

Average for the entire journey = 90 + 45

1352
$\frac{135}{2}$ = 67.50 km/h

**Question 27**
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The figure is an example of the construction of a

**Answer Details**

QR is a given line and P is a given point. The construction is the perpendicular from a given point P to a given line QR

**Question 28**
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Evaluate without using tables (0.008) -13 $\frac{1}{3}$ x (0.16) - 32

**Answer Details**

(0.008) -13
$\frac{1}{3}$ x (0.16) - 32
$\frac{3}{2}$ = (8 x 10-3)10-3 x (16 x 10-2)-32
$\frac{3}{2}$

= (23)103
$\frac{({2}^{3})}{{10}^{3}}$ - 3 x (24)10
$\frac{({2}^{4})}{10}$ - −32
$\frac{-3}{2}$

= 6258

**Question 29**
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A room is 12m long, 9m wide and 8m high. Find the cosine of the angle which a diagonal of the room makes with the floor of the room

**Answer Details**

ABCD is the floor, by pythagoras

AC2 = 144 + 81 = √225
$225$

AC = 15cm

Height of room is 8m, diagonal of floor is 15m

∴ The cosine of the angle which a diagonal of the room makes with the floor is EC2 = 152

cosine = adjhyp
$\frac{\text{adj}}{\text{hyp}}$ = 1517

**Question 30**
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Find the least length of a rod which can be cut into exactly equal strips, each of either 40cm or 48cm in length

**Answer Details**

The least length is 4048
$\frac{40}{48}$ = 58
$\frac{5}{8}$

for the rod to be cut in exactly equal trips

Ratio 56
$\frac{5}{6}$ : 4840
$\frac{48}{40}$

564048
$\frac{\frac{5}{6}}{\frac{40}{48}}$ = 1

56
$\frac{5}{6}$ x 4840
$\frac{48}{40}$ = 240240
$\frac{240}{240}$ = 1

The least length = 240cm

**Question 31**
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