WAEC SSCE - General Mathematics - 2022 (Essay)
Frage 1 Bericht
(a) Given that (7 -2x), 9, (5x + 17) are consecutive terms of a Geometric Progression (G. P) with common ratio, r>0, find the values of x.
(b) Two positive numbers are in the ratio 3:4. The sum of thrice the first number and twice the second is 68. Find the smaller number.
(a) r =97−2x
r = 5x+17−2x
97−2x = 5x+17−2x
9x9=(7-2x) (5x +17)
81 = 35x + 119 - 10x2 - 34x
10x2 - x - 38
(x -2)(10x+19) = 0
x = 2, x = 1910
x = 2
(b) Let the integers be x and y
x:y= 3:4
xy = 34
x = 3y4
3x + 2y=68
3( 34 y)+ 2y = 68
94 y + 2y = 68
9y + 8y = 272
17y = 272
y= 16
x = 34 * 16 →12
Antwortdetails
(a) r =97−2x
r = 5x+17−2x
97−2x = 5x+17−2x
9x9=(7-2x) (5x +17)
81 = 35x + 119 - 10x2 - 34x
10x2 - x - 38
(x -2)(10x+19) = 0
x = 2, x = 1910
x = 2
(b) Let the integers be x and y
x:y= 3:4
xy = 34
x = 3y4
3x + 2y=68
3( 34 y)+ 2y = 68
94 y + 2y = 68
9y + 8y = 272
17y = 272
y= 16
x = 34 * 16 →12
Frage 2 Bericht
(a) The probability that an athlete will not win any of three races is 1/4.If the athlete runs in all the races, what is the probability that the athlete will win;
(i) only the second race; (ii) all the three races; (ii) only two of the races?
(b) A cone with perpendicular height 24cm has a volume of 1200cm\(^{3}\). Find the volume of a cone with same base radius and height 84cm. [Take pi = \(\frac{22}{7}\)]
9(a)(i)
p = 14 ,
p = 1 - 14
p = 34
= 364
= 0.04687 ≈ 0.05
(ii) p(winning all) = 34 * 34 * 34
= 2764 = 0.422
(iii) p(winning two)
= (34 * 34 * 14 ) + (34 * 14 * 34 ) + (14 * 34 * 34 )
= 964 + 964 + 964
= 2764 = 0.4218
= 0.42
(b) Volume of a cone = 1/3πr2 h
1200 = 13
* 227
* r2
* 24
r2 = 1200∗3∗722∗24
r2 = 25200528
r2 = 47.727
r = 4–√7.727
r = 6.9085cm
Volume of a cone with same radius and height 84cm:
v = 13 * 227 * (6.9085)2 * 84
= 13
* 227
* 47.727 * 84
volume = 4200cm3
Antwortdetails
9(a)(i)
p = 14 ,
p = 1 - 14
p = 34
= 364
= 0.04687 ≈ 0.05
(ii) p(winning all) = 34 * 34 * 34
= 2764 = 0.422
(iii) p(winning two)
= (34 * 34 * 14 ) + (34 * 14 * 34 ) + (14 * 34 * 34 )
= 964 + 964 + 964
= 2764 = 0.4218
= 0.42
(b) Volume of a cone = 1/3πr2 h
1200 = 13
* 227
* r2
* 24
r2 = 1200∗3∗722∗24
r2 = 25200528
r2 = 47.727
r = 4–√7.727
r = 6.9085cm
Volume of a cone with same radius and height 84cm:
v = 13 * 227 * (6.9085)2 * 84
= 13
* 227
* 47.727 * 84
volume = 4200cm3
Frage 3 Bericht
| Item | food & drinks | fuel | rent | building project | education | savings |
| Percentage% | 35 | 7.5 | 1.0 | 15 | 17.5 | x |
The table shows the monthly expenditure (in percentages) of Mr. Okafor's salary.
(a) Calculate the percentage of Mr. Okafor's salary that was. put into salary.
(b) Illustrate the information on a pie chart.
(c) If Mr. Okafor's annual gross salary is $28,800.00 and he pays tax of 12%.
Calculate: (i) his monthly tax; (ii) amount saved each month.
(a) 35 + 7.5 + 10 + 15 + 17.5 + x = 100
85 + x = 100.
x = 100 - 85
x = 15
| items | % | degree |
| food & drinks | 35 | 35100∗360 = 126º |
| fuel | 7.5 | 7.5100∗360 = 27º |
| rent | 10 | 10100∗360 = 36º |
| building project | 15 | 15100∗360 = 54º |
| education | 17.5 | 17.5100∗360 = 63º |
| savings | 15 | 15100∗360 = 54º |
(c) income tax = 12100∗28,800 = $3,456.00
Monthly tax = 345612 = $288.00
(ii) Monthly net salary = 112(288−3456) = $2,112.00
Amount saved each month = 15100∗2112 = $316.80
Antwortdetails
(a) 35 + 7.5 + 10 + 15 + 17.5 + x = 100
85 + x = 100.
x = 100 - 85
x = 15
| items | % | degree |
| food & drinks | 35 | 35100∗360 = 126º |
| fuel | 7.5 | 7.5100∗360 = 27º |
| rent | 10 | 10100∗360 = 36º |
| building project | 15 | 15100∗360 = 54º |
| education | 17.5 | 17.5100∗360 = 63º |
| savings | 15 | 15100∗360 = 54º |
(c) income tax = 12100∗28,800 = $3,456.00
Monthly tax = 345612 = $288.00
(ii) Monthly net salary = 112(288−3456) = $2,112.00
Amount saved each month = 15100∗2112 = $316.80
Frage 4 Bericht
In the diagram, BCDE is a circle with centre
A. ∠BCD = (2x + 40)°, ∠BAD = (5x - 35)º, ∠BED = (2y+ 10)° and ∠ADC=40". Find:
(a) the values of x and y;
(a) 2x + 40 + 2y + 10 = 180°...(1)
2x + 2y = 130
x + y = 65º
5x - 35 = 2(2y+ 10) ... (2)
5x - 35 = 4y +20
5x - 4y = 55
5x - 4(5-x) = 55
5x - 260 + 4x =55
9x = 315
x = 35°
y= 65 - 35
y=30°
∠ABC + [2 x º35 + 40] + [5(35)-350)] + 40° =360º
∠ABC + 110 + 140°+ 40º = 360
∠ABC + 290° = 360°
∠ABC = 70°
Antwortdetails
(a) 2x + 40 + 2y + 10 = 180°...(1)
2x + 2y = 130
x + y = 65º
5x - 35 = 2(2y+ 10) ... (2)
5x - 35 = 4y +20
5x - 4y = 55
5x - 4(5-x) = 55
5x - 260 + 4x =55
9x = 315
x = 35°
y= 65 - 35
y=30°
∠ABC + [2 x º35 + 40] + [5(35)-350)] + 40° =360º
∠ABC + 110 + 140°+ 40º = 360
∠ABC + 290° = 360°
∠ABC = 70°
Frage 5 Bericht
\[ \frac{\sqrt{2}/3 - 1}{(\sqrt{3}/3)(1)} = \frac{\sqrt{3} - 3}{\sqrt{3}} \]
(√3−3√3 ) = √3√3 \
3−√33
= 1 - √3
20 = 15 + 10 - x
x = 25 - 20 = 5
P(both) = 520
= 14 or 0.25
Antwortdetails
(√3−3√3 ) = √3√3 \
3−√33
= 1 - √3
20 = 15 + 10 - x
x = 25 - 20 = 5
P(both) = 520
= 14 or 0.25
Frage 6 Bericht
(a) Copy and complete the table of values for \(y = 3\mathrm{Sin}x + 7\mathrm{Cos}x\) for 0°
| xº | 0 | 20 | 40 | 60 | 80 | 100 | 120 | 140 | 160 | 180 |
| y | 7.0 | 4.2 | -0.9 |
(b) Using a scale of 2cm to 20° on the x-axis and 2cm to 2 units on the y-axis, draw the graph of \(y = 3\mathrm{Sin}x + 7\mathrm{Cos}x\) for 0°
(c) Using the graph, find the;
(i) value of y when x= 150°,
(i) range of values of x for which y > 0.
| xº | 0 | 20 | 40 | 60 | 80 | 100 | 120 | 140 | 160 | 180 |
| y | 7.0 | 7.6 | 7.3 | 6.1 | 4.2 | 1.7 | -0.9 | -3.4 | -5.6 | 7.0 |
(b)
(C)(i) From the graph, When x is 150° y = -4.6 ± 0.2
(ii) Range of value of x for which y > is x: 0º ≤ x ≤ 113
Antwortdetails
| xº | 0 | 20 | 40 | 60 | 80 | 100 | 120 | 140 | 160 | 180 |
| y | 7.0 | 7.6 | 7.3 | 6.1 | 4.2 | 1.7 | -0.9 | -3.4 | -5.6 | 7.0 |
(b)
(C)(i) From the graph, When x is 150° y = -4.6 ± 0.2
(ii) Range of value of x for which y > is x: 0º ≤ x ≤ 113
Frage 7 Bericht
The graph shows the relation of the form \(y = mx^2 + nx + r\), where m, n and r are constants.
Using the graph:
(a) state the scale used on both axes; (b) find the values of m, n and r; (c) find the gradient of the line through P and Q; (d) state the range of values of x for which y > Q.
(a) Scale
- x-axis: 2 cm represents 2 units, i.e. 1 cm represents 1 unit.
- y-axis: 2 cm represents 10 units, i.e. 1 cm represents 5 units.
(b) Values of \(m\), \(n\) and \(r\)
From the graph, the curve cuts the \(x\)-axis at \(x=-2\) and \(x=4\). Hence,
\[y=-(x+2)(x-4)\]
\[y=-\left(x^2-2x-8\right)=-x^2+2x+8.\]
Comparing with \(y=mx^2+nx+r\),
\[\boxed{m=-1,\quad n=2,\quad r=8.}\]
(c) Gradient of \(PQ\)
From the graph, \(P(-5,-27)\) and \(Q(3,5)\).
\[\text{Gradient of }PQ=\frac{5-(-27)}{3-(-5)}=\frac{32}{8}=\boxed{4}.\]
(d) Range of \(x\) for which \(y>0\)
The graph is above the \(x\)-axis between its roots \(-2\) and \(4\). Therefore,
\[\boxed{-2<x<4}.\]
Antwortdetails
(a) Scale
- x-axis: 2 cm represents 2 units, i.e. 1 cm represents 1 unit.
- y-axis: 2 cm represents 10 units, i.e. 1 cm represents 5 units.
(b) Values of \(m\), \(n\) and \(r\)
From the graph, the curve cuts the \(x\)-axis at \(x=-2\) and \(x=4\). Hence,
\[y=-(x+2)(x-4)\]
\[y=-\left(x^2-2x-8\right)=-x^2+2x+8.\]
Comparing with \(y=mx^2+nx+r\),
\[\boxed{m=-1,\quad n=2,\quad r=8.}\]
(c) Gradient of \(PQ\)
From the graph, \(P(-5,-27)\) and \(Q(3,5)\).
\[\text{Gradient of }PQ=\frac{5-(-27)}{3-(-5)}=\frac{32}{8}=\boxed{4}.\]
(d) Range of \(x\) for which \(y>0\)
The graph is above the \(x\)-axis between its roots \(-2\) and \(4\). Therefore,
\[\boxed{-2<x<4}.\]
Frage 8 Bericht
Given that \( y = \left(\frac{pr}{m} - p^2r\right)^{-\frac{3}{2}} \)
(a) make r the subject;
(b) find the value of r when y = -8, m = 1 and p = 3.
(a) y =(prm−p2r )−32
y−23
=(prm−pr2
)
my−23 =(pr−pr2mm )
my−23 = pr - p2 rm
my−23 = r(p - p2 m)
r = (my−2/3m−p2m )
(b) r = (1−8−2/33−32∗1
)
= 14 ÷ -6
= - 1−6
= - 124
Antwortdetails
(a) y =(prm−p2r )−32
y−23
=(prm−pr2
)
my−23 =(pr−pr2mm )
my−23 = pr - p2 rm
my−23 = r(p - p2 m)
r = (my−2/3m−p2m )
(b) r = (1−8−2/33−32∗1
)
= 14 ÷ -6
= - 1−6
= - 124
Frage 9 Bericht
| Age | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
| No. of children | 2 | 6 | 5 | 4 | 6 | 9 | 8 | 5 |
The table shows the distribution of ages of a number of children in a school. If the mean of the distribution is 7, find the;
(a) value of x, (b) standard deviation of their ages.
Mean = (3x2 +4x6 + 5x5 + 6x + 7x6 + 8x9 + 9x8 + 10x5) ÷ (2 + 6 + 5 + x + 6 + 9 + 8 + 5)
7 = (291 + 6x) ÷ (41+x)
7(41+x) = 291 + 6x
287 + 7x = 291 + 6x
x = 4
(b)
| Age(x in yrs) | No.of child | Fx2 |
| 3 | 2 | 18 |
| 4 | 6 | 96 |
| 5 | 5 | 125 |
| 6 | 4 | 144 |
| 7 | 6 | 294 |
| 8 | 9 | 576 |
| 9 | 8 | 648 |
| 10 | 5 | 500 |
| Σf = 45 | Σfx2 = 2401 |
Standard deviation = 19645−−−√=4–√.3555
= 2.087
Antwortdetails
Mean = (3x2 +4x6 + 5x5 + 6x + 7x6 + 8x9 + 9x8 + 10x5) ÷ (2 + 6 + 5 + x + 6 + 9 + 8 + 5)
7 = (291 + 6x) ÷ (41+x)
7(41+x) = 291 + 6x
287 + 7x = 291 + 6x
x = 4
(b)
| Age(x in yrs) | No.of child | Fx2 |
| 3 | 2 | 18 |
| 4 | 6 | 96 |
| 5 | 5 | 125 |
| 6 | 4 | 144 |
| 7 | 6 | 294 |
| 8 | 9 | 576 |
| 9 | 8 | 648 |
| 10 | 5 | 500 |
| Σf = 45 | Σfx2 = 2401 |
Standard deviation = 19645−−−√=4–√.3555
= 2.087
Frage 10 Bericht
(a) The diameter of a cylinder closed at both ends is 7cm. If the total surface area is 209cm\(^{2}\), calculate the height. [Take pi = 22/7].
(b) The points X and Y, 19m apart are on the same side of a tree. The angles of elevation of the top, T, of the tree from X and Y on the horizontal ground with the foot of the tree are 43º and 38° respectively.
(i) Illustrate the information in a diagram. (ii) Find, correct to one decimal place, the height of the tree.
209 = 2 * 227 * (72)2 + 2 * 227 * 72 * h
209 = 77 + 22h
132 = 22h
h = 6cm
(b) tan 43º = hy
0.9125y = h ...[i]
tan 38º = hy+19
0.7813(y+19) = h ...[ii]
0.7813(y+19) = 0.9125y
0.7813y + 14.8447 = 0.93257
14.8447 = 0.1512y
y= 98.1792
= 98.2m
h = 0.9325 x 98.1792
= 91.55
=91.6m
Antwortdetails
209 = 2 * 227 * (72)2 + 2 * 227 * 72 * h
209 = 77 + 22h
132 = 22h
h = 6cm
(b) tan 43º = hy
0.9125y = h ...[i]
tan 38º = hy+19
0.7813(y+19) = h ...[ii]
0.7813(y+19) = 0.9125y
0.7813y + 14.8447 = 0.93257
14.8447 = 0.1512y
y= 98.1792
= 98.2m
h = 0.9325 x 98.1792
= 91.55
=91.6m
Frage 11 Bericht
A chord subtends an angle of 72º at the centre of a circle of radius 24.5m. Calculate the perimeter of the minor segment. [Take π = \( \frac{22}{7} \)
From ΔOKB
sin 36° = |KB|24.5
|KB| = 0.5878 x 24.5
=14.4011m
Or
Length of arc = 72360∗2∗227∗24.7
=30.8m
Perimeter = 2(14.4011) + 30.8
= 59.6022
= 59.6m
Antwortdetails
From ΔOKB
sin 36° = |KB|24.5
|KB| = 0.5878 x 24.5
=14.4011m
Or
Length of arc = 72360∗2∗227∗24.7
=30.8m
Perimeter = 2(14.4011) + 30.8
= 59.6022
= 59.6m
Frage 12 Bericht
(a) In the diagram, O is the centre of the circle XYZ. angle ZXO=34 ° and angle XOY=146 ° . Find angle OYZ.
(b) The exterior angles of a polygon are 42, 38,57, x, (x+ y)°. (2x- 15)° and (3x -y)º? If x is 7 less than y, find the values of x and y.
(a) ∠XYZ = 12 (146) = 73º
∠OZX = 34º
∠OZY = 73 - 34
∠OYZ = ∠OZY = 39
(b) x = y - 7
42 + 38 + 57 + x + x + y + 2x - 15 + 3x - y = 360
122°+7x = 360° 7x 238°
x = 34
: 34 = y - 7
34 + 7 = y
41 = y
Antwortdetails
(a) ∠XYZ = 12 (146) = 73º
∠OZX = 34º
∠OZY = 73 - 34
∠OYZ = ∠OZY = 39
(b) x = y - 7
42 + 38 + 57 + x + x + y + 2x - 15 + 3x - y = 360
122°+7x = 360° 7x 238°
x = 34
: 34 = y - 7
34 + 7 = y
41 = y
Frage 13 Bericht
(a) A man purchased 180 copies of a book at N250.00 each. He sold y copies at N300.00 each and the rest at a discount of 5 kobo in the Naira of the cost price.
If he made a profit of N7,125.00, find the value of y.
(b) A trader bought x bags of rice at a cost \(C = 24x + 103\) and sold them at a price, \(S = \frac{x^2}{20} - 33x\).
Find the expression for the profit (i) If 20 bags of rice were sold,
(ii) calculate the percentage profit.
(a) Total cost 180 x 250
=N45,000.00
Total selling price = 300y + (180-y) x 250 x 95100
300y + 42750 -237.5y
= 62.5y + 42750
Profit = 62.5y + 42750 -4500
7125 = 62.5y - 2250
9375 = 62.5y
y = 150
(b)) Profit = x220−33x (24x+ 103)
9x - x220 - 103
(ii) Cost = 24(20) +103 = 583.00
Selling Price =33(20) - 20220 = 640
Percentage profit = 640−583583 x 100%
= 9.78%
Antwortdetails
(a) Total cost 180 x 250
=N45,000.00
Total selling price = 300y + (180-y) x 250 x 95100
300y + 42750 -237.5y
= 62.5y + 42750
Profit = 62.5y + 42750 -4500
7125 = 62.5y - 2250
9375 = 62.5y
y = 150
(b)) Profit = x220−33x (24x+ 103)
9x - x220 - 103
(ii) Cost = 24(20) +103 = 583.00
Selling Price =33(20) - 20220 = 640
Percentage profit = 640−583583 x 100%
= 9.78%
