 WAEC Mathematics Questions and Answers for free to all Weac candidates In Ghana, Liberia, Nigeria, Sierra Leone, and The Gambia. On this page, all the Waec Mathematics questions and answers for 2023 and the most common questions and answers are released here.

Waec Candidates that applied for the West African Examination Council (WAEC) SSCE Examination will write their Waec Mathematics For general students. All details you need for you to be successful and pass this 2023 Waec Exam will also be given and make sure you read all through.

### 2023 Waec Mathematics Exam Papers

2023 Waec Mathematics Exam Papers Are

• Waec Mathematics Essay Questions
• Waec Mathematics Objectives Questions,

You are writing the 2 papers in only one day. In this post, the previous Year’s Waec questions and answers for Mathematics are released and the 2023 Waec Mathematics Exam Questions will also be released for those participating in the 2023 Waec examination.

### 2023 Waec Mathematics Questions And Answers Objective Questions (paper 1)

Previous Year WAEC Mathematics Obj Answers.

11-CCBCABABBA
21-AACCDCDABC
31-BCDCDCBABA

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

### 2023 Waec Mathematics Questions and Answers THEORY (paper 2)

2)
ii Area if cross-section
=Area of rectangle PQRU-Area of semi-circle
=(PUxPQ) – pie ITSI/4
=(4×11)-22/7×7/4
=44-22/4
=44-5.5
=38.5m^2

(3a)
(3b)
h=7cm
T=462cm^2
T=pie sqr + 2 pie rH
462= pie( r ^2 +27r)
462/3.142
=r sqr= =14r
r sqr+ 14r=147
r sqr +14r- 147=0
r= 14+-( (sqr(14))-41147)/2*1
r=-(14 +-(196-4))/2
r=(-14+_12)/2
r=(-14+12)/2 or
r=2/2
=1cm

4a)
Pr(3)=x/total
total=25+30+x+28+40+32
=155+x
0.225=x/155+ x
=34.875+0.225x=x
x-0.225x=34.875
=0.775x=34.875 degree
x=34.875/0.775
=45

(4b)
Pr(even)=30+28+40+32
=90
Pr(even)=90/200=9/20
Pr(prime no)=25=30+45+40
=140
Pr(prime no)=140/200=14/20

Pr(even or prime)=9/20 + 14/20

5a)
Area of triangle=1/2bh where h=6
:. 1/26/1b/1=36
6b/2=36/1
6b=72
B=72/6
Base of angle PSR=12cm
Since |TS| //PQ
QR=PR-PQ
QR=12-8
QR=4cm

5b)
Draw d triangle
From angle ABC, to get |BC|
Tan60/1=10cm/|BC|
|BC|=10.65/tan60
|BC|=10.65/1.7321
=6.1486 =6.15m
Hence |BC|=|DE|
From angle AED, to get |AE|
Draw d diagram
|AB|=6.15tan45
6.15*1
=6.15m
There4, th height of the tree =10.65-6.15
=4.50m

9a)
x 62 63 64 65 66 67 68
tally 1 iiiii iiiii iiiii ii iiii iiii iiii iiiii iiiii iiiii i ii
freq 1 5 12 14 10 6 2 total = 50
fx 62 315 768 910 660 402 136 tot = 3253
x-x- -3.06 -2.06 -1.06 0.06 0.94 1.94 2.94
(x-x-) 9.3636 4.2436 1.1236 0.0036 0.8836 3.7636 8.6436
f(x-x)2 9.3636 21.218 13.4832 0.0504 8.836 22.5816 17.2872 tot
= 92.82
bi) mean = EFX/EF = 3253/50 =65.06
bii) Standard deviation = Square root of EF(x-x)2/Ef
= Square root of 92.82/50
= square root of 1.8564
= 1.36

7a)
x-2/4 :x+2/2x
Cross multiply
2x(x-2)=4(x+2)
Opening the bracket
2x^2-4x = 4x+8
Collect like terms together
2x^2-4x-4x-8=0
2x^2-8x-8=0
Solving with completing the square method
2x^2-8x=8
Divide through with the coefficient of x^2
2x^2/2-8x/2=8/2
X^2-4x=4
Half of coefficient of x
(-4 x 1/2)^2 = (-2)^2
X^2-4x+(-2)^2=4+(-2)^2
X^2-4x+(-2)^2=4+4
(x-2)^2=8
X-2:square root 8
X=2 + or – squar root 8
X=2+ square root 8 or x=2- square root 8
X=2+2.828=4.828
X=2-2.828= -0.828
X=4:83 or – 0.83.

13a)
(x1,y1) =(2,-3)
for 2x + y =6
y=-2x +6
compare y = mx +6
m2 = -2
for parallel lines, m1=m2
y-y1/x-x1 = m
y–3/x-2 =-2/1
y + 3 = -2(x -2)
y + 3 = -2x + 4
y = -2x + 4 -3
y = -2x +1

13b)
T={2,3,5,7}
I. X angle Y= (x+y+xy) mod 8
– |2|3|5|7
2|4|5|7|1
3|5|6|0|2
5|7|0|2|4
7|1|2|4|6

13bii)
I)
2angle(D angle 7)
2 angle 4, since (5 angle 7)=4
2 angle 4=6
II)
2 angle n =5 angle 7
Since 5 angle 7=4
:. 2 angle n=4
And n=2 because
2angle2=4 5 angle 7

11a)
3p+4q/3p-4q* 2/1 find p:q
1(rp+4q)= 2(3p-4q)
3p+4q=6p-8q
Collect like terms
3p-6p=-8q-4q
-3p=-12q
Ther4 p:q = -3p/-3=-12q/-3
P=4q, p=4, q=1
P:q= 4:1
11b)
Ts=pq-(2+2)
Ts=pq-4
Circumference of Ts= 2pie(TS)

11bi)
Ts=pq-(2+2)
Ts=pq-4
Circumference of Ts= 2pie(TS)
=2pie(pq-4)
Perimeter=pq+4+4+2+2+2pie(pq-4)
34=pq+12+2pie(pq-4)
34-12=pq+2piePQ-8pie
22=PQ(1+2pie)-8pie
22=pq(1(222/7)-822/7
22=pq(1+44/7)-176/7
22=PQ(7+44/7)-176/7
22+176/7 =pq(51/7)
154+176/7=Pq(51/7)
330/7=pq(51/7)
330/51=pq
Pq=6.4706

11bii)
TS=PQ-4=6.4706-4=2.4706m
Area of semi-circle TS=pieD
=7.7626m^2
Area of rectangle PqRU=|pu|*|Pq|
4**6.4706=25.8824
Area of the cross section=area of rectangle-area of semi-circle
=25.8824-7.7626
=18.1198m^2

6a)
Sn=n/2(2a+(n-1)d)
A=1, d=2, n=n
Sn=n/2(21+(n-1)2)
N/2(2+2,-2)
=n/22=n^2

6b
n(E)=95, n(BnT)=7
N(U)=95, n(BnTnC’)=7n(B’nTnC)=3,n(BnTnC)=8
n(BnT’nC’)=x, n(BnT’nC),n(b)=47, n(T)=30
i) Draw d venn diagram
ii)To find x,
X+x+7+8=47
2x=47-15=32
X=32/2 =16
iii)No of those who traveled by at least two means
7+3+16+8=34

8a)
KGF = 110
2x = r = y
if KGF = 110,

Before you leave this page kindly make sure you understand and know how WASSCE grades your WAEC Subject. The reason why we are doing this, is not to frighten or scare you but to put you in check and see reasons why you need to be serious with your Mathematics studies for the 2023 Agric Exam.

### Waec Grading System For all Waec Candidates

Do you know that the West African Examination Council (WAEC) Board has published the Waec grading system of results? Kindly check below to see the meaning of the Waec Grading Result.

The table below shows candidates’ positioning of the Waec grading of results ranging from A1, B2, B3, C4, C5, C6, D7, E8, F9 are the complete list of Waec Grading Result. All Waec candidates must fall into one of the Waec Grading Systems.

 GRADE NUMERIC VALUE INTERPRETATION A1 1 EXCELLENT B2 2 VERY GOOD B3 3 GOOD C4 4 CREDIT C5 5 CREDIT C6 6 CREDIT D7 7 PASS E8 8 PASS F9 9 FAIL

1. A1 Excellent 75% – 100%
2. B2 Very good 70% – 74%
3. B3 Good 65%  69%
4. C4 Credit 60% – 64%
5. C5 Credit 55% – 59%
6. C6 Credit 50% – 54%
7. D7 Pass 45% – 49%
8. E8 Pass 40% – 45%
9. F9 Failure 0% – 44%

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