# 2024/2025 Waec SSCE Further Mathematics Expo Questions and Answers

Answers Loading….. …..

Keep Refreshing this Page.

FURTHER MATHS ANSWERS

(1)

5^4^(3x/4 -1) + 5^3(x-1)/5^(3x – 2)

5^3x/4 – 4 + 5^3x – 3/5^3x – 2

5^3x – 4 + 5^3x – 3/5^3x – 2

(5^3x ÷ 5^4) + 5^3x ÷ 5^3÷5^3x – 2

5^3x/625 + 5^3x/125÷5^3x- 2

5^3x + 5(5^3x) ÷ 5^3x-3

5^3x + 5(5^3x)/625 ÷ 5^3x/25

Let 5^3x = y

y + 5(y)/625 ÷ y/25

y + 5y/625 × 25/y

=6y/625 × 25/y

=6/25

1(b)

F(x+2)-6x^2+5x-8 f(5

X+2=5z

X=5-2

X=3

6x^2 +5x-8

=6(3)^2 + 5(3)-8

=54+15-8

=6!

(2a)

Using y2 – y1/x2 – x1

Where y2 = 7, y1 = -5,

X2 = -2, and X1 = 7

7 – -5/-2 – 7

=7+5/-9

=12/-9

=4/-3

Coordinate points :

-4/3(3 : 2)

=-12/3 : -8/3

= -4 : -2⅔

X = (-4, 2⅔)

(2b)

2/1-√2 – 2/2+√2

=2(2+√2)-2(1-√2)/(1-√2)(2+√2)

=4+2√2 – 2+2√2/2+√2-2√2 – 2

=2 + 4√2/-√2

=(2+4√2)(-√2)/-√2(-√2)

= -2√2 – 4(2)/2

= -8 – 2√2/2

= -4 – √2

(3)

Sn = A/2[2n+(n-1)d]

Where Sn = 165

a = -3, d = 2

165 = A/2[2(-3)+(n-1)2]

165 = n[-6+2n-2]/2

165×2 = n[2n – 8]

330 = 2n² – 8n

2n² – 8n – 330 = 0

n²-4n-165 = 0

Using -b±√b²-4ac/2a

4±√-4²-4(1)(-165)/2(1)

4±√16 + 660/2

4±√676 = 4±26/2

4+26/2 = 30/2

= 15 terms

(4)

Draw the right angled triangle

Using Pythagoras theorem

Third side = √(p+q)² – (p-q)²

=√(p+q+p-q)(p+q-p+q)

Difference of two squares.

=√(2p)(2q)

=√4pq

Adjacent side = 2√pq

Tanx = opp/adj = p – q/2√pq

1 – tan²X = 1-(p-q)²/4pq

=(4pq)-(p²-2pq+q²)/4pq

= -p²+6pq-q²/4pq

= -(p² – 6pq + q²)/4pq

(5)

Draw the diagram

Using cosine law

Cos∅ = 16²+10²-14²/2(16)(10)

Cos∅ = 256 + 100 – 196/320

Cos∅ = 160/320

Cos∅ = 0.5

∅ = cos-¹(0.5)

∅ = 60°

Angle between 10N and 16N

= 180 – ∅ (sum of angles on a straight line)

= 180 – 60

=120°

(6)

Draw the diagram

Taking moment about the pivot,

(T × 25)=(50×10)+(20×45)

25T = 500 + 900

25T = 1400

T = 1400/25

T = 56N

(7)

In a tabular form

Under class interval:

1-5, 6-10, 11-15, 16-20, 21-25, 26-30

Under class mark (X):

3, 8, 13, 18, 23, 28

Under X-Xbar:

-10, -5, 0, 5, 10, 15

Under frequency:

18, 12, 25, 15, 20, 10

Ef = 100

Under f(X – XbarA):

-180, -60, 0, 75, 200, 150

f(X – XbarA) = 185

Where xA = 13

Mean = xA + Ef(X – Xbar)/Ef

=13 + 185/100

=13 + 1.85

=14.85years

PLS NOTE THAT XBAR LOOKS LIKE X WITH MINUS SIGN ON TOP.

(10a) given 4x² – px +1 = 0

For real roots: b² – 4ac >0

(-p) ² – 4(4) (1) > 0

p² – 16 > 0

p² >16

p > ± 4

(10bi) Given: (1 +3x)⁶

Using pascal’s triangle: 1, 6, 15, 20, 15, 6, 1

(1)⁶(3x)º + 6(1)⁵ (3x)¹ + 15 (1)⁴ (3x)₂ + 20(1)³ (3x)³ + (15) (1)²(3x)⁴+6(1)¹(3x)⁵ + 1(1)º(3x)⁶

1 + 6(3x) + 15 (9x²) + 20 (27x³) + 15 (81x⁴) + 6(243 x⁵) 729x⁶

1 + 18x + 135x² + 540x³ + 1215x⁴ 1458x⁵ + 729x⁶

(ii) (1.03)⁶ = (1 + 3(0.01)]

Therefore (1.03)⁶ = 1 + 18(0.01) + 135 (0.01)² + 540(0.01)³ + 1215(0.01)⁴

+ 1458 (0.01)⁵ + 729 (0.01)⁶

+ 1 + 0.18 + 0.0135 + 0.005 + 0.00001215

+ 0.0000001458 + 0.000000000729

= 1.1940523

= 1.194 (4s.f)

(12) prob (pass) = 60% = 60/100 =3/5

Prob (fail) 1-3/5 = 2/5

(a) Prob (atleast two failed) = 1 – prob (ome pass)

= 1 – 10Ci (3/5)¹ (2/5)⁹

1 – (10!/9!) (3/5) (2/5)⁹

= 1 – 10 (3/5) (2/5)⁹

= 1 – 10 (3/5) (0.000262144))

= 1 – 0.001572864

= 0.9984

(12b) prob (exastly half passed)

= 10C5 (3/5)⁵ (2/5)⁵

= 10!/5!5! (6/25)⁵

= 252 (6/25)⁵

= 252 x 0.0007962624

= 0.2007

(12c) prob (at most two failed)

= prob (zero/fail) + prob (one/fail) +prob (two/fail)

= 10C (2/5)º (3/5)10 + 10C (2/5)¹ (3/5)⁹ + 10 C2(2/5)² (3/5)⁸

= (3/5)10 + 10 (2/5) (3/5)⁹ + 45 (2/5)²(3/5)⁸

= 0.060466176 + 0.040310784 + 0.120932352

= 0.1673

(14a)

Draw the diagram

(14b)

From the diagram

a = v – 20/4

2.5 = v – 20/4

V – 20 = 10

V = 10 + 20 = 30m/s

acceleration/retardation = 3/4

2.5/30/T-12 = 3/4

2.5(T – 12)/30 = 3/4

(T – 12) = 30×3/2.5×4

= 90/10 = 9

T – 12 = 9

T = 9 + 12 = 21

t = T – 12

t = 21 – 12

t = 9secs

(14c)

Total distance of the journey

= Area of BCDI + Area of AFEO + Area of DFI

= 1/2(12+8)10 + 1/2(9×10) + 1/2(21+21)20

= (20/2)10 + 90/2 +(42/2)20

=10(10) + 45 + 21(20)

=100 + 45 + 420

= 565m