**Mathematics OBJ: 1-10: CABDBADCBC 11-20: BCBCBBAACD 21-30: DCCBCCABBA 31-40: AABDDCDDCC 41-50: BBCDCCBCCD**

Section A:

Section B:

(1a)

110ₓ = 40₅

change all to base ten.

110ₓ = 40₅

1x²+1x ¹+0x⁰ = 45¹+0*5⁰

x² + x + 0 = 20

x² + x – 20 = 0

Solve quadratically

x² + 5x – 4x – 20 = 0

x(x+5) – 4(x+5) = 0

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

Either x-4 = 0 or x + 5 = 0

x = -5 or 4

therefore, x= 4

(2a)

y2-y1/x2-x1 = y-y1/x-x1

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

-3-7/2+2 = y-7/x+2

-10/4=y-7/x+2

4(y-7)= -10(x+2)

4y-28= -10x-20

4y= -10x+28-20

4y= -10x+8

2y= -5x+4

OR

5x+2y=4

(3a)

Total ratio =3+5+8=16

Ali’s share =3/16 * 420,000/1

=210,000

Son’s of Ali and Yusuf’s share = N78,750+ N210,000

= N288,750

(4)

Since <PQR = <PRS = 90°

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

(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

(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

(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

(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

(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

(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

(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°

Section A:

*1,2,3,4,5*Section B:

*6,8,11,12,13*(1a)

110ₓ = 40₅

change all to base ten.

110ₓ = 40₅

1x²+1x ¹+0x⁰ = 45¹+0*5⁰

x² + x + 0 = 20

x² + x – 20 = 0

Solve quadratically

x² + 5x – 4x – 20 = 0

x(x+5) – 4(x+5) = 0

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

Either x-4 = 0 or x + 5 = 0

x = -5 or 4

therefore, x= 4

(1b)

15/√75 + √108 + √432

= 15/√253 + √363 + √144*3

= 15/5√3 + 6√3 + 12√3

= 3/√3 + 18√3

= 3√3/3 = 18√3

= √3 + 18√3

=19√3

CLICK HERE FOR THE IMAGE

(2a)

y2-y1/x2-x1 = y-y1/x-x1

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

-3-7/2+2 = y-7/x+2

-10/4=y-7/x+2

4(y-7)= -10(x+2)

4y-28= -10x-20

4y= -10x+28-20

4y= -10x+8

2y= -5x+4

OR

5x+2y=4

(2b)

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

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

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

8+3(a/b)= 5 [ 5-(a/b)]

8+3(a/b)=25-5(a/b)

3(a/b) + 5(a/b)=25-8

8(a/b)=17

a/b=17/8

CLICK HERE FOR THE IMAGE

(3a)

Total ratio =3+5+8=16

Ali’s share =3/16 * 420,000/1

=210,000

Son’s of Ali and Yusuf’s share = N78,750+ N210,000

= N288,750

(3b)

2(1/8)^x=32^x-1

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

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

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

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

-3x+1=5x-5

-5x-3x= -5-1

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

x=3/4

CLICK HERE FOR THE IMAGE

(4)

Since <PQR = <PRS = 90°

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

CLICK HERE FOR THE IMAGE

(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

CLICK HERE FOR THE IMAGE

(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

CLICK HERE FOR THE IMAGE

(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

(8bii)

|AC|/66 = Cos35°

|AC| = 66 x cos35°

= 66 x 0.8192

= 54.0672

Tan = 41.86°

Angle of elevation of top of tower from c = 41.85°

CLICK HERE FOR THE IMAGE

(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%

CLICK HERE FOR THE IMAGE

(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

CLICK HERE FOR THE IMAGE

(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

CLICK HERE FOR THE IMAGE

(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

CLICK HERE FOR THE IMAGE

Nice one

my name is Muhammad Abdullahi

ESSAY MATHS

How to subscribe

thanks for the maths

pls obj

Maths objetive

We need the obj now plz gidi

Obj pleas… Maths

mathematics obj answers

thanks what of the objective we have started

tomorrow economics answers