Waec gce mathematics Question & Answer Expo 2022

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Waec Gce 2022 General Mathematics Questions & Answers Now Available

 

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Waec gce mathematics Question 2022

Waec gce mathematics Question 2022
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Waec gce mathematics Answer 2022

(1a)

√75 − √3 (2√3−3)−5√27

= √(5×5×3) − (2×3)+3(√3) −5√(9×3)

= 5(√3) −6 +3(√3) −(5×3)√3

= 5√(3) + 3(√3) −15(√3) −6

= −7(√3) −6

 

(1b)

10(x−4)=4(2x−1)+5

10x−40=8x−4+5

10x−8x=1+40

2x=41

x=41/2

x= 20.5

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

 

(2ai)

8C3

= (8!)/(3!×(8−3)!)

= (8×7×6×5!)/(3!×(8−3)!)

=(8×7×6×5!)/(3!×5!)

= (8×7×6)/(3×2×1)

= (336)/6

= 56

 

(2aii)

7P4

=7!/(7−4)!

= (7×6×5×4×3!)/(3!)

= 7×6×5×4

= 840

 

(2b)

25=(1/5)(125^x)

5^2 =5^(−1) *5^(3x)

2= −1 + 3x

3x = 2+1

3x = 3

x = 1

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

 

(3i)

C = a + bN

80,000 = a + 4b……..(i)

110,000 = a + 6b…….. (ii)

Subtracting equ(i) from (ii)

30,000= 3b

b = 30000/3

b = 10,000

Substituting b in equ(i)

80,000 = a + 4b

80,000 = a + 40,000

a = 80,000 – 40,000

a = 40,000

The relationship between C and N is given by;

C = a + bN

C = 40,000 + 10,000N

 

(3ii)

When N = 9

C = 40,000 + 10,000(9)

C = 40,000 + 90,000

C = ₦130,000

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

 

(4i)

Mean = Σx/n

x̄ = (12+19+20+21+22+22+16+8)/8

x̄ = 140/8

x̄ = 17.5

Read: WAEC GCE Government Answers 2022

(4ii)

8, 12, 16, 19, 20, 21, 22, 22

Median = 19+20/2

= 39/2

= 19.5

 

(4iii)

Mode = 22

 

(4iv)

TABULATE

 

x | 12 | 19 | 20 | 21 | 22 | 22| 16 | 8

 

(x – x̄) | – 5.5 | 1.5 | 2.5 | 3.5 | 4.5 | 4.5 | -1.5 | -9.5

 

(x – x̄)² | 30.25 | 2.25 | 6.25 | 12.25 | 20.25 | 20.25 | 2.25 | 90.25

 

Σ(x – x̄)² = 184

= √(Σ(x – x̄)²/N

= √184/8

= √23

= 4.796

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

 

(5a)

= (8x²+5x-3)/x³

= ∫(8x²+5x-3)/x³ dx

= ∫(8x²/x³) dx + ∫(5x/x³) dx – ∫3/x³ dx

= ∫(8/x)dx + ∫(5/x²)dx – ∫(3/x³)dx

= 8 ∫(1/x)dx + 5 ∫(1/x²)dx – 3 ∫(1/x³)dx

= 8 ∫ (1/x)dx + 5∫(x-²)dx – 3∫(x-³)dx

= 8Inx + 5(x-²+¹/-2+1) – 3(x-³+¹/-3+1)+c

= 8Inx + 5(x-¹/-1) – 3(x-²/-2)+c

= 8Inx – 5(1/x) – 3(1/x²)(-1/2) + c

= 8Inx – 5/x + (3/2x²) + c

 

(5b)

x² + y² – 6x – 2y – 15 =0

Grouping;

x² – 6x + y² – 2y = 15

Adding half the square of the coefficient of x and y to both sides

(x² – 6x + 3²) + (y² – 2y +1²) = 15 + (3)² + (1)²

(x – 3)² + (y – 1)² = 15+9+1

(x – 3)² + (y – 1)² = 25

h = -3, k = -1

Center = C (-3, – 1)

R² = 25

R = √25

R = 5

Center = C (-3, – 1), Radius = 5

 

(5c)

Check the image

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

Check: WAEC GCE Agric Science Answer 2022

(6bi)

f(x) = 2x – 8

f(-2) = 2(-2) – 8

= – 4 – 8

= – 12

(6bii)

f(6) = 2(6) – 6

= 12 – 6

= 6

(6biii)

f(9) = 2(9) – 8

= 18 – 8

= 10

(6biv)

f(0) = 2(0) – 8

= 0 – 8

= – 8

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

 

(7c)

x³ + tx² – x – 8 is divisible by x-1

x – 1 =0

x = 1

Substituting x= 1

(1)³ + t(1)² – 1 – 8 =0

1 + t – 1 = 8

t = 8

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

 

(8a)

(6x – 10)/(x² – 2x – 3)

= (6x – 10)/(x+1)(x-3)

A/(x+1) + B/(x-3) = 6x-10

A(x-3) +B(x+1) = 6x-10

Substituting x= -1

A(-1 – 3) + B(-1 +1) =6(-1) – 10

A(-4) = – 6 – 10

-4A = -16

A = 4

Substituting x= 3

A(3-3) +B(3+1) = 6(3) – 10

B(4) = 18 – 10

4B = 8

B = 2

= A/(x+1) + B/(x-3)

= 4/(x+1) + 2/(x-3)

 

(8b)

(5^2y) – (5^1+y) + 6 =0

(5^2y) – (5^1)(5^y) + 6 =0

Let 5^y = x

x² – 5x + 6 =0

x² – 3x – 2x + 6 =0

(x² – 3x) (-2x + 6) =0

x(x – 3) – 2(x – 3) =0

(x – 2) (x – 3) =0

x – 2 =0

x = 2

x – 3 =0

x = 3

Recall:

5^y = x

Substituting

5^y = 2

y = Log₅(2)

y₁= 0.4307

5^y = 3

y = Log₅(3)

y₂ = 0.6826

y₁= 0.4307, y₂ = 0.6826

 

(8ci)

Check the image

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

 

(12)

TABULATE

X | 7 | 3 | 5 | 5 | 7 | 4 | 6 | 4 | 6 | 7

 

Y | 5 | 7 | 6 | 8 | 5 | 6 | 5 | 7 | 4 | 7

 

xy | 35 | 21 | 30 | 40 | 35 | 24 | 30 | 28 | 24 | 49

 

x² | 49 | 9 | 25 | 25 | 49 | 16 | 36 | 16 | 36 | 49

 

y² | 25 | 49 | 36 | 64 | 25 | 36 | 25 | 49 | 16 | 49

 

Σx² = 310, Σx = 54, Σxy = 316

Σy² = 374, Σy = 60, n = 10

 

(12ai)

x̄ = Σx/n

x̄ = 54/10

x̄ = 5.4

 

(12aii)

ȳ = Σy/n

ȳ = 60/10

ȳ = 6

 

(12b)

Check the image

 

(12c)

The negative coefficient -12.06 suggests that as the independent variable increases, the dependent variable tends to decrease.

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

 

(13a)

Statistics is the discipline that concerns the collection, organization, analysis, interpretation, and presentation of data.

 

(13b)

(i) Statistics helps in providing a better understanding and accurate description of nature’s phenomena.

(ii) Statistics helps in the proper and efficient planning of a statistical inquiry in any field of study.

(iii) Statistics helps in collecting appropriate quantitative data.

(iv) Statistics helps in presenting complex data in a suitable tabular, diagrammatic and graphic form for an easy and clear comprehension of the data.

 

(13c)

-Measure of Central Tendency-

(i) Mean

(ii) Median

 

-Measure of Dispersion-

(i) Standard deviation

(ii) Variance

 

(13d)

The mean is more important and the mo*2021 NABTEB GCE ADVANCED MATHEMATICS ANSWERS*

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

 

(1a)

√75 − √3 (2√3−3)−5√27

= √(5×5×3) − (2×3)+3(√3) −5√(9×3)

= 5(√3) −6 +3(√3) −(5×3)√3

= 5√(3) + 3(√3) −15(√3) −6

= −7(√3) −6

 

(1b)

10(x−4)=4(2x−1)+5

10x−40=8x−4+5

10x−8x=1+40

2x=41

x=41/2

x= 20.5

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

 

(2ai)

8C3

= (8!)/(3!×(8−3)!)

= (8×7×6×5!)/(3!×(8−3)!)

=(8×7×6×5!)/(3!×5!)

= (8×7×6)/(3×2×1)

= (336)/6

= 56

 

(2aii)

7P4

=7!/(7−4)!

= (7×6×5×4×3!)/(3!)

= 7×6×5×4

= 840

 

(2b)

25=(1/5)(125^x)

5^2 =5^(−1) *5^(3x)

2= −1 + 3x

3x = 2+1

3x = 3

x = 1

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

 

(3i)

C = a + bN

80,000 = a + 4b……..(i)

110,000 = a + 6b…….. (ii)

Subtracting equ(i) from (ii)

30,000= 3b

b = 30000/3

b = 10,000

Substituting b in equ(i)

80,000 = a + 4b

80,000 = a + 40,000

a = 80,000 – 40,000

a = 40,000

The relationship between C and N is given by;

C = a + bN

C = 40,000 + 10,000N

 

(3ii)

When N = 9

C = 40,000 + 10,000(9)

C = 40,000 + 90,000

C = ₦130,000

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

 

(4i)

Mean = Σx/n

x̄ = (12+19+20+21+22+22+16+8)/8

x̄ = 140/8

x̄ = 17.5

 

(4ii)

8, 12, 16, 19, 20, 21, 22, 22

Median = 19+20/2

= 39/2

= 19.5

 

(4iii)

Mode = 22

 

(4iv)

TABULATE

 

x | 12 | 19 | 20 | 21 | 22 | 22| 16 | 8

 

(x – x̄) | – 5.5 | 1.5 | 2.5 | 3.5 | 4.5 | 4.5 | -1.5 | -9.5

 

(x – x̄)² | 30.25 | 2.25 | 6.25 | 12.25 | 20.25 | 20.25 | 2.25 | 90.25

 

Σ(x – x̄)² = 184

= √(Σ(x – x̄)²/N

= √184/8

= √23

= 4.796

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

 

(5a)

= (8x²+5x-3)/x³

= ∫(8x²+5x-3)/x³ dx

= ∫(8x²/x³) dx + ∫(5x/x³) dx – ∫3/x³ dx

= ∫(8/x)dx + ∫(5/x²)dx – ∫(3/x³)dx

= 8 ∫(1/x)dx + 5 ∫(1/x²)dx – 3 ∫(1/x³)dx

= 8 ∫ (1/x)dx + 5∫(x-²)dx – 3∫(x-³)dx

= 8Inx + 5(x-²+¹/-2+1) – 3(x-³+¹/-3+1)+c

= 8Inx + 5(x-¹/-1) – 3(x-²/-2)+c

= 8Inx – 5(1/x) – 3(1/x²)(-1/2) + c

= 8Inx – 5/x + (3/2x²) + c

 

(5b)

x² + y² – 6x – 2y – 15 =0

Grouping;

x² – 6x + y² – 2y = 15

Adding half the square of the coefficient of x and y to both sides

(x² – 6x + 3²) + (y² – 2y +1²) = 15 + (3)² + (1)²

(x – 3)² + (y – 1)² = 15+9+1

(x – 3)² + (y – 1)² = 25

h = -3, k = -1

Center = C (-3, – 1)

R² = 25

R = √25

R = 5

Center = C (-3, – 1), Radius = 5

 

(5c)

Check the image

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

 

(6bi)

f(x) = 2x – 8

f(-2) = 2(-2) – 8

= – 4 – 8

= – 12

(6bii)

f(6) = 2(6) – 6

= 12 – 6

= 6

(6biii)

f(9) = 2(9) – 8

= 18 – 8

= 10

(6biv)

f(0) = 2(0) – 8

= 0 – 8

= – 8

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

 

(7c)

x³ + tx² – x – 8 is divisible by x-1

x – 1 =0

x = 1

Substituting x= 1

(1)³ + t(1)² – 1 – 8 =0

1 + t – 1 = 8

t = 8

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

 

(8a)

(6x – 10)/(x² – 2x – 3)

= (6x – 10)/(x+1)(x-3)

A/(x+1) + B/(x-3) = 6x-10

A(x-3) +B(x+1) = 6x-10

Substituting x= -1

A(-1 – 3) + B(-1 +1) =6(-1) – 10

A(-4) = – 6 – 10

-4A = -16

A = 4

Substituting x= 3

A(3-3) +B(3+1) = 6(3) – 10

B(4) = 18 – 10

4B = 8

B = 2

= A/(x+1) + B/(x-3)

= 4/(x+1) + 2/(x-3)

 

(8b)

(5^2y) – (5^1+y) + 6 =0

(5^2y) – (5^1)(5^y) + 6 =0

Let 5^y = x

x² – 5x + 6 =0

x² – 3x – 2x + 6 =0

(x² – 3x) (-2x + 6) =0

x(x – 3) – 2(x – 3) =0

(x – 2) (x – 3) =0

x – 2 =0

x = 2

x – 3 =0

x = 3

Recall:

5^y = x

Substituting

5^y = 2

y = Log₅(2)

y₁= 0.4307

5^y = 3

y = Log₅(3)

y₂ = 0.6826

y₁= 0.4307, y₂ = 0.6826

 

(8ci)

Check the image

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

 

(12)

TABULATE

X | 7 | 3 | 5 | 5 | 7 | 4 | 6 | 4 | 6 | 7

 

Y | 5 | 7 | 6 | 8 | 5 | 6 | 5 | 7 | 4 | 7

 

xy | 35 | 21 | 30 | 40 | 35 | 24 | 30 | 28 | 24 | 49

 

x² | 49 | 9 | 25 | 25 | 49 | 16 | 36 | 16 | 36 | 49

 

y² | 25 | 49 | 36 | 64 | 25 | 36 | 25 | 49 | 16 | 49

 

Σx² = 310, Σx = 54, Σxy = 316

Σy² = 374, Σy = 60, n = 10

 

(12ai)

x̄ = Σx/n

x̄ = 54/10

x̄ = 5.4

 

(12aii)

ȳ = Σy/n

ȳ = 60/10

ȳ = 6

 

(12b)

Check the image

 

(12c)

The negative coefficient -12.06 suggests that as the independent variable increases, the dependent variable tends to decrease.

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

 

(13a)

Statistics is the discipline that concerns the collection, organization, analysis, interpretation, and presentation of data.

 

(13b)

(i) Statistics helps in providing a better understanding and accurate description of nature’s phenomena.

(ii) Statistics helps in the proper and efficient planning of a statistical inquiry in any field of study.

(iii) Statistics helps in collecting appropriate quantitative data.

(iv) Statistics helps in presenting complex data in a suitable tabular, diagrammatic and graphic form for an easy and clear comprehension of the data.

 

(13c)

-Measure of Central Tendency-

(i) Mean

(ii) Median

 

-Measure of Dispersion-

(i) Standard deviation

(ii) Variance

 

(13d)

The mean is more important and the most used measure of central tendency because it uses all values in the data set to give you an average. For data from skewed distributions, the median is better than the mean because it isn’t influenced by extremely large values.

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

st used measure of central tendency because it uses all values in the data set to give you an average. For data from skewed distributions, the median is better than the mean because it isn’t influenced by extremely large values.

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

How to pass waec gce 2022 mathematics exam?

The mathematics paper can be one of the most terrifying and stressful assessments of your life so far, especially if you’re afraid that you won’t pass it. It doesn’t matter whether you failed the last test or just didn’t pay attention in class – it’s important to stay calm and realize that there are many ways to ace your exam! Here are five tips on how to pass your mathematics paper.

 

Assess your current knowledge

Look at your past papers and assess how you performed. Have you gotten better over time? Do you tend to score better on certain topics or in certain parts of a test? If so, focus on what worked for you in these previous papers, as well as on areas where you’ve struggled. This will give you a good idea of where your strengths and weaknesses lie, which is important information when it comes time to study.

 

Set a plan with goals

Setting goals helps you plan out how you’re going to get there. You can make as detailed or vague a plan as you like, but setting a goal helps provide a target date for your success and provides you with direction. Make sure it’s challenging enough so that you’ll actually reach it—you don’t want to set a goal that will be too easy (or boring) and not feel accomplished when you do complete it. Go ahead and get out your calculator; I promise we’ll keep things super-simple. Set three goals: one short-term, one medium-term, and one long-term goal. This way, once you’ve reached one goal, there’s something else right around the corner! Ready? Go!

 

Show you care about it

A common mistake is not actually caring about your topic. If you are passionate about it, show that in your content. By showing a passion for whatever you’re writing about, people will be more willing to read and share it because they can tell you are vested in what you’re saying. And if you don’t care about whatever it is, why should anyone else?

 

Practice, practice, practice!

In order to learn how to pass mathematics paper, you first need to practice. Once you’ve got some problems under your belt, read and understand each problem before moving on. If a question gives you trouble, don’t rush; look it over carefully and try working it out with a partner if possible. If all else fails, call your teacher and ask for help! (If possible, do so after class or during office hours.) It’s better for a professor or TA to tell you No, that answer is wrong than for you to submit an incorrect one on your own—and potentially receive a zero for an assignment. Remember: Practice makes perfect!

 

Know where to go when you need help

Whether you’re preparing for an exam or working on homework, it’s easy to find yourself in a bind. Fortunately, there are hundreds of online resources that can help you with math questions. There are dozens of forums and websites where you can seek help from other students and many textbooks offer glossaries or guides that answer basic questions. What’s more, there are sites like Khan Academy and TutorVista that offer comprehensive courses on various mathematical concepts. You can search by topic or difficulty level and get one-on-one instruction from teachers who have been tested by their peers. If you still feel lost in your class, consider talking with your professor during office hours or signing up for an extra credit tutoring session after class.

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