WAEC Further Mathematics Questions and Answers – 2019 May/June Expo

West Africa Examination Council, WAEC Further Mathematics Questions and Answers for 2019 May/June Examination is now available.
WAEC Further Mathematics Questions and Answers

ATTENTION:- PLEASE ALWAYS SUBSCRIBE FOR EARLIER ANSWERS, SUBSCRIBERS GET ANSWERS EARLIER THAN WE POSTED.

👑 2019 WAEC GCE FURTHER MATHEMATICS ANSWERS👑

100% VERIFIED FURTHER MATHEMATICS OBJ ANSWERS

1-10: AACACBDABC
11-20: DAACCDBDAA
21-30: BCCBABABCC
31-40: BCCBCCDDDD

100% VERIFIED FURTHER MATHEMATICS THEORY ANSWERS

There two sections in this question ie Section A and Section B.
Section A (1-8)
Section B (9-15) divided into THREE PARTS ie I,II,III.

Candidates are required to answer All in Section A and Any FOUR in section B but atleast ONE from each Part in Section B.

1-8 (must)
Choose one or two from either these parts
Part I (9,10,11)
Part II (12,13)
Part III (14,15)

Total questions to be answered is 12
==================================

(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
[CLICK HERE]
==================================

(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
[CLICK HERE]
==================================

(4)
Draw the right angled triangle [CLICK HERE]

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 [CLICK HERE]

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 [CLICK HERE]
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

Reader Also Read:  2020 WAEC Practical And Physical Geography Questions And Answers

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.
[CLICK HERE]
==================================

(8)
[CLICK HERE]
==================================

(9a)
Y = 7 – 6/x and y + 2x – 3 = 0
Substitute eqn (1) into eqn(2)
7 – 6/x + 2x – 3 = 0
Multiply through by X
7x – 6 + 2x² – 3x = 0
2x² + 4x – 6 = 0
X² + 2x – 3 = 0
(x² + 3x – x – 3) = 0
X(X+3)-1(x+3)=0
(x-1)(x+3)=0
X – 1 = 0 or X + 3 = 0
X = 1 or X = -3
But y = 7 – 6/x
When X = 1
y = 7 – 6/1
y = 7 – 6
y = 1

When X = -3
y = 7 – 6/-3
y = 7 + 2
y = 9
Coordinates are (1, 1) and (-3, 9)

(9b)
Draw the diagram [ CLICK HERE ]

Gradient of AB = 9 – 1/-3 -1
=8/-4 = -2
Midpoint of AB = (1+3/2, 1-9/2) = (2, -4)
Gradient of perpendicular = -1/-2 = 1/2

equation of perpendicular is
y-(-4)/x-2 = 1/2
y + 4/x – 2 = 1/2
y = 1/2x – 1 – 4
y = 1/2x – 5
OR
2y = X – 10

==================================

(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

(10bii)
(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)

[CLICK HERE]
==================================
(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

12. [CLICK HERE]
==================================

(14a)
Draw the diagram [CLICK HERE]

(14b)
From the diagram [CLICK HERE]
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

GRAPH. [CLICK HERE]
4a & b. [CLICK HERE]
4c. [CLICK HERE]

=======COMPLETED=======

Don’t Just Copy Alone, Drop your Comments, Share with your friends on Facebook & Twitter Using the below share buttons and like our Facebook fan pages Here
BEST OF LUCK

We Update Our Social Media Platform With Latest Updates, Don’t Miss Out Join US On Facebook and For Latest Updates.

1 Comment

Leave a Reply

Your Phone Number will not be published.




This site uses Akismet to reduce spam. Learn how your comment data is processed.