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(1a)

110x = 40s

Converting both sides to base 10, we have;

(1xX²)+(1xX¹)+(0×X^0) = (4×5¹) + (0×5^0)

X² + X + 0 = 20 + 0

X² + X - 20 = 0

X² + 5x - 4x - 20 = 0

X(x+5)-4(x+5) = 0

(x-4)(x+5) = 0

Since X must be positive

X - 4 = 0

X = 4

(1b)

15/√75 + √108 + √432

= 15/√25×3 + √36×3 + √144×3

=15/5√3 + 6√3 + 12√3

= 3/√3 + 18√3

=3√3/3 + 18√3

= √3 + 18√3

= 19√3

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(2a)

The equation of the line through the points

A(-2,7) and B(2,-3)

Using the equation Y=mx+b

Where m=slope of the gradient, b=the intercept at the vertical axis

Hence slope=Change in Y/Change in X

=>Y2-Y1/X2-X1=Y-y1/X-x1

=(-3-7)/(2--2)=Y-7/X+2

=-10/4=(Y-7)/(X+2)

=4(y-2)=-10(x+2)

4y-28=-10x-20

4y=-10x+8

5x+2y=4

(2b)

(5b-a)/(8b+3a)=1/5

5(5b-a)=1(8b+3a)

25b-5a=8b+3a

25b-8b=3a+5a

17b=8a

Therefore a/b=17/8

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(3a)

Ali : Masah : Yusuf = #420,000

3 : 5 : 8

Sum of ratio shared :

3 + 5 + 8 = 16

Therefore

Ali share = 3/16 × 420,000

= #78,750

Musah Share = 5/16 × 420,000

=#131,250

Yusuf share = 8/16 × 420,000

=#210,000

Therefore

Sum of Ali + Yusuf

=#78,750 + #210,000

= #288,750

name

(3b)

Solve : 2(1/8)² = 32^x-1

2¹ × (8^-1)^x = 32^x-1

2¹ × (2^-3(-1))^x = 2^5(x - 1)

2¹ × (2^-3)^x = 2^5x - 5

2¹ × 2^-3x = 2^5x - 5

2^1+(-3x) = 2^5x - 5

2^1-3x = 2^5x - 5

1 - 3x = 5x - 5

-3x - 5x = -5 - 1

-8x/-8 = -6/-8

X = 3/4

-=-=-=-=-=-=-=-=-=-=-=-=-=-=-==-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-

(4)

Since Using Pythagoras theorem

|PR|² = |PQ|² + |QR|²

|PR|² = 3² + 4²

|PR|² = 9 + 16

|PR|² = 25 PR = √25

|PR| = 5cm

Considering PRS

|PS|² = |PR|²+|SR|²

13² = 5² + |SR|²

169 = 25 + |SR|²

|SR|² = 169 - 25

|SR|² = 144

|SR| = √144 = 12cm

Hence the area of the quadrilateral = Area of triangle PQR + area of PRS

= 1/2bh + 1/2bh

= 1/2×4×3 + 1/2×12×5

= 6+30 = 36cm

name

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(5a)

No of red balls = 3

No of green balls = 5

No of blue balls = x

Prob.(red ball) = no of total outcome/no of possible outcome

Pr(red) = 3/3+5+x = 1/6

3/8+x = 1/6

6(3) = 1(8+x)

18 = 8 + x

X = 18 - 8 = 10

Therefore the no of blue ball = 10

(5b)

Probability of picking a green ball

P(g) = no of green balls/no of possible outcome

P(g) = 5/3+5+10 = 5/18

=5/18

name

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(6ai)

F α M1M2/d²

F = KM1M2/d²

Given F = 20N, M1= 25kg, M2 = 10kg and d = 5m

20 = k(25)(10)/5²

250k = 500

k = 500/250 = 2

Expression is

F = 2M1M2/d²

(6aii)

Making d subject

d = √2M1M2/F

d = √2 ×7.5×4/30

d = √60/30 = √2

d = √2m or 1.41m

(6b)

Draw the diagram

X+X+60+X+80+X+40+X+20 = 540(sum of angles in a Pentagon)

5x + 200 = 540

5x = 540 - 200

5x = 340

X = 340/5

X = 68

name

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(8a)

1/3x - 1/4(x+2)>_ 3x -1⅓

1/3x - 1/4(x+2)>_3x - 4/3

Multiply through by the L. C. M(12), we have

4x - 3(x + 2)>_36x - 16

4x - 3x - 6 >_ 36x - 16

-6+16 >_36x + 3x - 4x

10 >_ 35x

35x _< 10

X = 10/35

X = 2/7

(8bi)

Draw the triangle

|AB|/66 = sin35

|AB| = 66sin35 = 66×0.5736 = 37.8576

Draw the right angled triangle

|AD|/|AB| = Tan52

|AD| = 37.8576 × Tan52° = 37.8576 × 1.2799 = 48.45m

Height of tower = 48.45m

name

name

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(10)

130kg of tomatoes for #52,000

Half of the tomatoes

130/2 = 65kg sold at 30%

Profit = #52,000/2 = 26,000

#26,000 = 100%

X = 130%

X = 26000 × 130/100

= #33,800

Then 65kg was then sold at reduction of 12% per kg

Recall that the initial cost price = 52000/130

=400kg

65kg sold at = 33,000/65

=#520/kg

Then for 12% reduction

520 × 88/100 = 457.6/kg

(a)

The new selling price per kg = #457.6/kg

(b) 65kg - 5kg = 60kg

(60kg×457.6kg)+33,800

= #61,256.00

#profit = selling price /cost price × 1000/1

=61256/52000×100/1= 117.8

= 17.8%

name

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(11ai)

ar² = 1/4 ......(1)

ar^5= 1/32 .....(2)

Divide eqn (2) by eqn(1)

ar^5/ar² = 1/32÷1/4

r³ = 1/32 × 4/1

r³= 1/8

r³ = 2-³

r = 2-¹

r = 1/2

Common ratio = 1/2

Put this into eqn (1)

a(1/2)² = 1/4

a(1/4) = 1/4

a = (1/4)/(1/4) = 1

First term, a = 1

(11aii)

Seventh term, T7 = ar^6

=(1)(1/2)^6

=1/64

(11b)

Given : X = 2 and X = -3

(X - 2)(X + 3) = 0

X² + 3x - 2x - 6 , 0

X² + x - 6 = 0

Comparing with ax²+bx+c = 0

a = 1

b = 1

C = -6

name

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(12a)

Given : siny = 8/17

Draw the right angle

From Pythagorean triple, third side is 15

Draw the right angle triangle

tan y = 8/15

tan y/1+2tany = 8/15/1+2(8/15) = 8/15/1+16/15

tany /1+2tan y = 8/31

(12b)

Amount shared = #300,000

Otobo's share = #60,000

Ade's share = 5/12 × #(300,000-60,000)

= 5/12 × #240,000

=#100,000

Adeobi's share = #300,000 - (#60,000 + #100,000)

= 300,000 - 160,000

=#140,000

Ratio : Otobo : Ade : Adeola

60,000 : 100,000 : 140,000

60 : 100 : 140

6 : 10 : 14

3 : 5 : 7

(13a)

RT^0=RS^0=90° (radius o targets)

T^0S=2TU^0 (angle at centre = 2 time angle at circumstance)

TOs=2*68°=136°

Now RT^0+RS^0+T^0S+SRT^0=360°(sum of angle in a quadrilateral)

90° + 90° + 136° + x = 360°

X+316°=360°

X=360°-316

X=44°

(13b)

Let tank B hold x litres

; Tank A hold (x+600)literally

3(x-100)=(x+600-100)

3x-300=x+500

3x-x=500+300

2x=800

X=800/2=400

Tank B holds 400 litres

Tank A holds (400+600)=1000litres

(10)

130kg of tomatoes for #52,000

Half of the tomatoes

130/2 = 65kg sold at 30%

Profit = #52,000/2 = 26,000

#26,000 = 100%

X = 130%

X = 26000 × 130/100

= #33,800

Then 65kg was then sold at reduction of 12% per kg

Recall that the initial cost price = 52000/130

=400kg

65kg sold at = 33,000/65

=#520/kg

Then for 12% reduction

520 × 88/100 = 457.6/kg

(a)

The new selling price per kg = #457.6/kg

(b) 65kg - 5kg = 60kg

(60kg×457.6kg)+33,800

= #61,256.00

#profit = selling price /cost price × 1000/1

=61256/52000×100/1= 117.8

= 17.8%

(12a)

Given : siny = 8/17

Draw the right angle

From Pythagorean triple, third side is 15

Draw the right angle triangle

tan y = 8/15

tan y/1+2tany = 8/15/1+2(8/15) = 8/15/1+16/15

tany /1+2tan y = 8/31

(12b)

Amount shared = #300,000

Otobo's share = #60,000

Ade's share = 5/12 × #(300,000-60,000)

= 5/12 × #240,000

=#100,000

Adeobi's share = #300,000 - (#60,000 + #100,000)

= 300,000 - 160,000

=#140,000

Ratio : Otobo : Ade : Adeola

60,000 : 100,000 : 140,000

60 : 100 : 140

6 : 10 : 14

3 : 5 : 7

(11ai)

ar² = 1/4 ......(1)

ar^5= 1/32 .....(2)

Divide eqn (2) by eqn(1)

ar^5/ar² = 1/32÷1/4

r³ = 1/32 × 4/1

r³= 1/8

r³ = 2-³

r = 2-¹

r = 1/2

Common ratio = 1/2

Put this into eqn (1)

a(1/2)² = 1/4

a(1/4) = 1/4

a = (1/4)/(1/4) = 1

First term, a = 1

(11aii)

Seventh term, T7 = ar^6

=(1)(1/2)^6

=1/64

(11b)

Given : X = 2 and X = -3

(X - 2)(X + 3) = 0

X² + 3x - 2x - 6 , 0

X² + x - 6 = 0

Comparing with ax²+bx+c = 0

a = 1

b = 1

C = -6

110x = 40s

Converting both sides to base 10, we have;

(1xX²)+(1xX¹)+(0×X^0) = (4×5¹) + (0×5^0)

X² + X + 0 = 20 + 0

X² + X - 20 = 0

X² + 5x - 4x - 20 = 0

X(x+5)-4(x+5) = 0

(x-4)(x+5) = 0

Since X must be positive

X - 4 = 0

X = 4

(1b)

15/√75 + √108 + √432

= 15/√25×3 + √36×3 + √144×3

=15/5√3 + 6√3 + 12√3

= 3/√3 + 18√3

=3√3/3 + 18√3

= √3 + 18√3

= 19√3

-=-=-=-=-=-=-=-=-=-=-=-=-=-=-==-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-

(2a)

The equation of the line through the points

A(-2,7) and B(2,-3)

Using the equation Y=mx+b

Where m=slope of the gradient, b=the intercept at the vertical axis

Hence slope=Change in Y/Change in X

=>Y2-Y1/X2-X1=Y-y1/X-x1

=(-3-7)/(2--2)=Y-7/X+2

=-10/4=(Y-7)/(X+2)

=4(y-2)=-10(x+2)

4y-28=-10x-20

4y=-10x+8

5x+2y=4

(2b)

(5b-a)/(8b+3a)=1/5

5(5b-a)=1(8b+3a)

25b-5a=8b+3a

25b-8b=3a+5a

17b=8a

Therefore a/b=17/8

=-=-=-=-=-=-=-=-=-=-=-=-=-=-==-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-

(3a)

Ali : Masah : Yusuf = #420,000

3 : 5 : 8

Sum of ratio shared :

3 + 5 + 8 = 16

Therefore

Ali share = 3/16 × 420,000

= #78,750

Musah Share = 5/16 × 420,000

=#131,250

Yusuf share = 8/16 × 420,000

=#210,000

Therefore

Sum of Ali + Yusuf

=#78,750 + #210,000

= #288,750

name

(3b)

Solve : 2(1/8)² = 32^x-1

2¹ × (8^-1)^x = 32^x-1

2¹ × (2^-3(-1))^x = 2^5(x - 1)

2¹ × (2^-3)^x = 2^5x - 5

2¹ × 2^-3x = 2^5x - 5

2^1+(-3x) = 2^5x - 5

2^1-3x = 2^5x - 5

1 - 3x = 5x - 5

-3x - 5x = -5 - 1

-8x/-8 = -6/-8

X = 3/4

-=-=-=-=-=-=-=-=-=-=-=-=-=-=-==-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-

(4)

Since Using Pythagoras theorem

|PR|² = |PQ|² + |QR|²

|PR|² = 3² + 4²

|PR|² = 9 + 16

|PR|² = 25 PR = √25

|PR| = 5cm

Considering PRS

|PS|² = |PR|²+|SR|²

13² = 5² + |SR|²

169 = 25 + |SR|²

|SR|² = 169 - 25

|SR|² = 144

|SR| = √144 = 12cm

Hence the area of the quadrilateral = Area of triangle PQR + area of PRS

= 1/2bh + 1/2bh

= 1/2×4×3 + 1/2×12×5

= 6+30 = 36cm

name

-=-=-=-=-=-=-=-=-=-=-=-=-=-=-==-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-

(5a)

No of red balls = 3

No of green balls = 5

No of blue balls = x

Prob.(red ball) = no of total outcome/no of possible outcome

Pr(red) = 3/3+5+x = 1/6

3/8+x = 1/6

6(3) = 1(8+x)

18 = 8 + x

X = 18 - 8 = 10

Therefore the no of blue ball = 10

(5b)

Probability of picking a green ball

P(g) = no of green balls/no of possible outcome

P(g) = 5/3+5+10 = 5/18

=5/18

name

-=-=-=-=-=-=-=-=-=-=-=-=-=-=-==-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-

(6ai)

F α M1M2/d²

F = KM1M2/d²

Given F = 20N, M1= 25kg, M2 = 10kg and d = 5m

20 = k(25)(10)/5²

250k = 500

k = 500/250 = 2

Expression is

F = 2M1M2/d²

(6aii)

Making d subject

d = √2M1M2/F

d = √2 ×7.5×4/30

d = √60/30 = √2

d = √2m or 1.41m

(6b)

Draw the diagram

X+X+60+X+80+X+40+X+20 = 540(sum of angles in a Pentagon)

5x + 200 = 540

5x = 540 - 200

5x = 340

X = 340/5

X = 68

name

-=-=-=-=-=-=-=-=-=-=-=-=-=-=-==-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-

(8a)

1/3x - 1/4(x+2)>_ 3x -1⅓

1/3x - 1/4(x+2)>_3x - 4/3

Multiply through by the L. C. M(12), we have

4x - 3(x + 2)>_36x - 16

4x - 3x - 6 >_ 36x - 16

-6+16 >_36x + 3x - 4x

10 >_ 35x

35x _< 10

X = 10/35

X = 2/7

(8bi)

Draw the triangle

|AB|/66 = sin35

|AB| = 66sin35 = 66×0.5736 = 37.8576

Draw the right angled triangle

|AD|/|AB| = Tan52

|AD| = 37.8576 × Tan52° = 37.8576 × 1.2799 = 48.45m

Height of tower = 48.45m

name

name

-=-=-=-=-=-=-=-=-=-=-=-=-=-=-==-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-

(10)

130kg of tomatoes for #52,000

Half of the tomatoes

130/2 = 65kg sold at 30%

Profit = #52,000/2 = 26,000

#26,000 = 100%

X = 130%

X = 26000 × 130/100

= #33,800

Then 65kg was then sold at reduction of 12% per kg

Recall that the initial cost price = 52000/130

=400kg

65kg sold at = 33,000/65

=#520/kg

Then for 12% reduction

520 × 88/100 = 457.6/kg

(a)

The new selling price per kg = #457.6/kg

(b) 65kg - 5kg = 60kg

(60kg×457.6kg)+33,800

= #61,256.00

#profit = selling price /cost price × 1000/1

=61256/52000×100/1= 117.8

= 17.8%

name

-=-=-=-=-=-=-=-=-=-=-=-=-=-=-==-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-

(11ai)

ar² = 1/4 ......(1)

ar^5= 1/32 .....(2)

Divide eqn (2) by eqn(1)

ar^5/ar² = 1/32÷1/4

r³ = 1/32 × 4/1

r³= 1/8

r³ = 2-³

r = 2-¹

r = 1/2

Common ratio = 1/2

Put this into eqn (1)

a(1/2)² = 1/4

a(1/4) = 1/4

a = (1/4)/(1/4) = 1

First term, a = 1

(11aii)

Seventh term, T7 = ar^6

=(1)(1/2)^6

=1/64

(11b)

Given : X = 2 and X = -3

(X - 2)(X + 3) = 0

X² + 3x - 2x - 6 , 0

X² + x - 6 = 0

Comparing with ax²+bx+c = 0

a = 1

b = 1

C = -6

name

-=-=-=-=-=-=-=-=-=-=-=-=-=-=-==-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-

(12a)

Given : siny = 8/17

Draw the right angle

From Pythagorean triple, third side is 15

Draw the right angle triangle

tan y = 8/15

tan y/1+2tany = 8/15/1+2(8/15) = 8/15/1+16/15

tany /1+2tan y = 8/31

(12b)

Amount shared = #300,000

Otobo's share = #60,000

Ade's share = 5/12 × #(300,000-60,000)

= 5/12 × #240,000

=#100,000

Adeobi's share = #300,000 - (#60,000 + #100,000)

= 300,000 - 160,000

=#140,000

Ratio : Otobo : Ade : Adeola

60,000 : 100,000 : 140,000

60 : 100 : 140

6 : 10 : 14

3 : 5 : 7

(13a)

RT^0=RS^0=90° (radius o targets)

T^0S=2TU^0 (angle at centre = 2 time angle at circumstance)

TOs=2*68°=136°

Now RT^0+RS^0+T^0S+SRT^0=360°(sum of angle in a quadrilateral)

90° + 90° + 136° + x = 360°

X+316°=360°

X=360°-316

X=44°

(13b)

Let tank B hold x litres

; Tank A hold (x+600)literally

3(x-100)=(x+600-100)

3x-300=x+500

3x-x=500+300

2x=800

X=800/2=400

Tank B holds 400 litres

Tank A holds (400+600)=1000litres

(10)

130kg of tomatoes for #52,000

Half of the tomatoes

130/2 = 65kg sold at 30%

Profit = #52,000/2 = 26,000

#26,000 = 100%

X = 130%

X = 26000 × 130/100

= #33,800

Then 65kg was then sold at reduction of 12% per kg

Recall that the initial cost price = 52000/130

=400kg

65kg sold at = 33,000/65

=#520/kg

Then for 12% reduction

520 × 88/100 = 457.6/kg

(a)

The new selling price per kg = #457.6/kg

(b) 65kg - 5kg = 60kg

(60kg×457.6kg)+33,800

= #61,256.00

#profit = selling price /cost price × 1000/1

=61256/52000×100/1= 117.8

= 17.8%

(12a)

Given : siny = 8/17

Draw the right angle

From Pythagorean triple, third side is 15

Draw the right angle triangle

tan y = 8/15

tan y/1+2tany = 8/15/1+2(8/15) = 8/15/1+16/15

tany /1+2tan y = 8/31

(12b)

Amount shared = #300,000

Otobo's share = #60,000

Ade's share = 5/12 × #(300,000-60,000)

= 5/12 × #240,000

=#100,000

Adeobi's share = #300,000 - (#60,000 + #100,000)

= 300,000 - 160,000

=#140,000

Ratio : Otobo : Ade : Adeola

60,000 : 100,000 : 140,000

60 : 100 : 140

6 : 10 : 14

3 : 5 : 7

(11ai)

ar² = 1/4 ......(1)

ar^5= 1/32 .....(2)

Divide eqn (2) by eqn(1)

ar^5/ar² = 1/32÷1/4

r³ = 1/32 × 4/1

r³= 1/8

r³ = 2-³

r = 2-¹

r = 1/2

Common ratio = 1/2

Put this into eqn (1)

a(1/2)² = 1/4

a(1/4) = 1/4

a = (1/4)/(1/4) = 1

First term, a = 1

(11aii)

Seventh term, T7 = ar^6

=(1)(1/2)^6

=1/64

(11b)

Given : X = 2 and X = -3

(X - 2)(X + 3) = 0

X² + 3x - 2x - 6 , 0

X² + x - 6 = 0

Comparing with ax²+bx+c = 0

a = 1

b = 1

C = -6

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