Common difference (d) = 11 - 7 = 15 - 11 = 4

U_n = 115

U₁ = 7

115 = 7 + (n - 1) x 4

115 = 7 + 4n - 4

115 = 7 - 4 + 4n

115 = 3 + 4n

115 - 3 = 4n

112 = 4n

Divide both sides by 4

n = $\frac{112}{4}$

n = 28

Σfx = 2 + 5 + 7 + 9 + 15 = 38

Σf = 1 + 1 + 1 + 1 + 1 = 5

Note: f is 1 since each number occurred only once.

Mean (µ) = $\frac{38}{5}$ = 7.6

Mean Deviation = $\frac{\mathrm{\Sigma f|(x\; -\; \mu )|}}{\mathrm{\Sigma f}}$

Σf|(x - µ)| = 17.6

Σf = 5

Mean Deviation = $\frac{17.6}{5}$

Mean Deviation = 3.52

Definition of locus: A locus is the set of all points that satisfy a given condition. In this case, the condition is that the distance from a point P to M is equal to the distance from P to N.

Perpendicular bisector property: The perpendicular bisector of a line segment passes through the midpoint of that segment and is perpendicular to it. Any point on the perpendicular bisector is equidistant from the two endpoints of the line segment.

Speed = 60 miles per hour

3600 s = 1 hour

Notes:
1. 60 minutes = 1 hour
2. 60 seconds = 1 minutes
3. 60 x 60 seconds = 3600 seconds = 1 hour

If 3600 seconds = 1 hour

15 seconds = $\frac{\mathrm{15\; seconds\; x\; 1\; hour}}{\mathrm{3600\; seconds}}$ = 0.00417 hour

Distance = Length of train

Distance = Speed x Time

Distance = 60 miles per hour x 0.00417 hour

Note: per hour cancels hour

Length of train ≈ 0.25 miles

If 1 mile = 1760 yards

0.25 mile = $\frac{\mathrm{0.25\; mile\; x\; 1760\; yards}}{\mathrm{1\; mile}}$ = 440.352 yards ≈ 440 yards

Note: the mile cancel each other

Let k = Kwaku's age

Kwaku is 2 years older than Atanga, Atanga's age = k - 2

Kwaku is 6 years younger than Esi, Esi's age = k + 6

Kwaku's age + Atanga's age + Esi's age = 70

k + (k - 2) + (k + 6) = 70

k + k - 2 + k + 6 = 70

k + k + k + 6 - 2 = 70

3k + 4 = 70

3k = 70 - 4

3k = 66

Divide both sides by 3

k = $\frac{66}{3}$

k = 22

Kwaku is 22 years

Atanga's age = 22 - 2 = 20 years

Esi's age = 22 + 6 = 28 years

Ratios of ages = 20 : 22 : 28

Ratios of ages = 10 : 11 : 14

Total ratio = 10 + 11 + 14 = 35

Amount received = $\frac{\mathrm{Ratio}}{\mathrm{Sum\; of\; ratios}}$ x Total Amount

Ratio for the eldest = 14

Sum of ratios = 35

Total amount shared = $153.00

Eldest's share = $\frac{14}{35}$ x $153.00

Eldest's share = $61.20

$\frac{log{3}^{(2x\; -\; 1)}}{log{3}^{243}}$ = $\frac{2}{5}$

243 = 3⁵

log_a^ab = b x log_a^a

But log_a^a = 1

log_a^ab = b x 1

log_a^ab = b

log₃²⁴³ = log₃³⁵

log₃³⁵ = 5

$\frac{log{3}^{(2x\; -\; 1)}}{5}$ = $\frac{2}{5}$

Since the denominator at the left equals to the denominator at the right, the numerators are also the same.

log₃^{(2x - 1)} = 2

If log_b^a = c, b^c = a

3² = 2x - 1

9 = 2x - 1

9 + 1 = 2x

10 = 2x

Divide both sides by 2

x = $\frac{10}{2}$

x = 5

Since the difference (subtraction) is 6 years, when you add 6 years to the younger age, you should get the eldest's age.

Let y = Esi's age

Amina's age = Esi's age + 6

Amina's age = y + 6

Amina's age + Esi's age = 24

(y + 6) + y = 24

y + 6 + y = 24

y + y + 6 = 24

2y = 24 - 6

2y = 18

Divide both sides by 2

y = $\frac{18}{2}$

y = 9

Esi's age is 9 years

Note: modulo is the remainder after division

3 (mod 11) = 1 x 11 + 3 = 14

3 (mod 11) = 2 x 11 + 3 = 25

3 (mod 11) = 3 x 11 + 3 = 36

Using 14 = 3 (mod 11)

4x + 1 = 14

4x = 14 - 1

4x = 13

4 cannot divide 13 without a remainder. Hence we cannot use 14 = 3 (mod 11)

Using 25 = 3 (mod 11)

4x + 1 = 25

4x = 25 - 1

4x = 24

Divide both sides by 4

x = $\frac{24}{4}$

x = 6

Both methods of finding the area requires that we solve for |QT|.

Considering the triangle

Apply pythagoras theorem

|QT|² + |RT|² = |QR|²

|QT|² + 3.3² = 5.5²

|QT|² + 10.89 = 30.25

|QT|² = 30.25 - 10.89

|QT|² = 19.36

|QT| = √19.36

|QT| = 4.4 cm

Method I

Find area of trapezium

Area of trapezium = $\frac{1}{2}$ x (Sum of parallel sides) x Height

Area of trapezium = $\frac{1}{2}$ x (|PQ| + [|PQ| + |RT|]) x |QT|

Area of trapezium = $\frac{1}{2}$ x (6.7 cm + [6.7 cm + 3.3 cm]) x 4.4 cm

Area of trapezium = $\frac{1}{2}$ x (6.7 cm + 10.0 cm) x 4.4 cm

Area of trapezium = $\frac{1}{2}$ x 16.7 cm x 4.4 cm

Area of trapezium = 36.74 cm²

Method II

Find area of rectangle and triangle

Area of rectangle = |PQ| x |QT|

Area of rectangle = 6.7 cm x 4.4 cm = 29.48 cm²

Area of triangle = $\frac{1}{2}$ x Base x Height

Area of triangle = $\frac{1}{2}$ x |RT| x |QT|

Area of triangle = $\frac{1}{2}$ x 3.3 cm x 4.4 cm

Area of triangle = 7.26 cm²

Total area = Area of rectangle + Area of triangle

Total area = 29.48 cm² + 7.26 cm²

Total area = 36.74 cm²

Let |MQ| = y

Since ∆OMN is similar to ∆OPQ, the corresponding sides are proportional.

Scale factor = $\frac{\mathrm{|OP|}}{\mathrm{|ON|}}$ = $\frac{\mathrm{|OQ|}}{\mathrm{|OM|}}$ = $\frac{\mathrm{|PQ|}}{\mathrm{|MN|}}$

|OM| = 10 - |MQ|

|OM| = 10 - y

$\frac{\mathrm{8\; cm}}{\mathrm{6\; cm}}$ = $\frac{\mathrm{10\; cm}}{\mathrm{10\; -\; y}}$

$\frac{4}{3}$ = $\frac{10}{\mathrm{10\; -\; y}}$

Cross-multiply

4 (10 - y) = 3 x 10

40 - 4y = 30

-4y = 30 - 40

-4y = -10

Divide both sides by -4

y = $\frac{-10}{-4}$

y = 2.5 cm

|MQ| = 2.5 cm

2120₃ = 2 x 3³ + 1 x 3² + 2 x 3¹ + 0 x 3⁰

2120₃ = 2 x 27 + 1 x 9 + 2 x 3 + 0 x 1

2120₃ = 54 + 9 + 6 + 0

2120₃ = 69

5p² + 4pq - 15pr - 12qr = 5p² + 4pq - 15pr - 12qr

5p² + 4pq - 15pr - 12qr = p(5p + 4q) - 3r(5p + 4q)

5p² + 4pq - 15pr - 12qr = (5p + 4q)(p - 3r)

Method I

Calculating fraction for SHS 3 and later calculating the number

Number of students in form = Fraction x Total Population

The total fraction is 1 which represents the 1800 students

Fraction in SHS 1 and 2 = $\frac{4}{9}$ + $\frac{13}{36}$

Fraction in SHS 1 and 2 = $\frac{\mathrm{4\; x\; 4\; +\; 1\; x\; 13}}{36}$

Fraction in SHS 1 and 2 = $\frac{\mathrm{16\; +\; 13}}{36}$

Fraction in SHS 1 and 2 = $\frac{29}{36}$

Fraction in SHS 3 = 1 - $\frac{29}{36}$

Fraction in SHS 3 = $\frac{\mathrm{36\; -\; 29}}{36}$

Fraction in SHS 3 = $\frac{7}{36}$

Number of students in SHS 3 = $\frac{7}{36}$ x 1800

Number of students in SHS 3 = 350

Method II

Calculating number of students in SHS 1 and 2

Number of students in SHS 1 = $\frac{4}{9}$ x 1800

Number of students in SHS 1 = 800

Number of students in SHS 2 = $\frac{13}{36}$ x 1800

Number of students in SHS 2 = 650

Total number of students in SHS 1 and 2 = 800 + 650 = 1450

Number of students in SHS 3 = Total Population - Number of students in SHS 1 and 2

Number of students in SHS 3 = 1800 - 1450

Number of students in SHS 3 = 350

3y + bx - 6 = 0

Since the line passes through (3, 4), when you substitute the x value, 3 and y value 4 into the equation, the result is supposed to be 0. Then you can find the b value.

3(4) + b(3) - 6 = 0

12 + 3b - 6 = 0

3b + 12 - 6 = 0

3b + 6 = 0

3b = -6

Divide both sides by 3

b = $\frac{-6}{3}$

b = -2

g:x → $\frac{\mathrm{2x\; -\; 1}}{\mathrm{3x\; -\; 2}}$

For g not defined, the denominator must be 0

3x - 2 = 0

3x = 2

Divide both sides by 3

x = $\frac{2}{3}$

In ∆PQR, |QR| is the hypothenuse since ∠ QPR = 90°

From pythagoras theorem, p² = r² + q²

Substituting q² = 25 - r² into the equation

p² = r² + (25 - r²)

p² = r² + 25 - r²

p² = 25

Take square root of both sides

p = √25

p = 5

Note: sum of all central angles is 360°

360 - 2x = x (Opposite angles)

360 = x + 2x

360 = 3x

Divide both sides by 3

x = $\frac{360}{3}$

x = 120°

Sum of the angles of the rhombus is 360°

∠ OKL + ∠ KLM + ∠ LMO + ∠ MOK = 360°

∠ KLM = x = 120°

∠ MOK = 360 - 2x = 360 - 2(120) = 120°

∠ OKL + 120° + ∠ LMO + 120° = 360°

∠ OKL + ∠ LMO + 240° = 360°

∠ OKL + ∠ LMO = 360° - 240°

∠ OKL + ∠ LMO = 120°

Opposite angles of a rhombus are equal. Hence ∠ OKL = ∠ LMO

∠ OKL = ∠ LMO = $\frac{120}{2}$ = 60°

Area of field = Area of the rectangle - Area of the semicircle

Area of rectangle = |PQ| x |QR|

Area of rectangle = 8 cm x 4 cm = 32 cm²

Area of semicircle = $\frac{1}{2}$ x π x r²

Radius = $\frac{1}{2}$ x (8 cm - [2 cm + 2 cm])

Radius = $\frac{1}{2}$ x (8 cm - 4 cm)

Radius = $\frac{1}{2}$ x 4 cm

Radius = 2 cm

Area of semicircle = $\frac{1}{2}$ x $\frac{22}{7}$ x (2 cm)²

Area of semicircle = $\frac{1}{2}$ x $\frac{22}{7}$ x 4 cm²

Area of semicircle = $\frac{88}{14}$ cm²

Area of semicircle = 6.29 cm²

Area of field = Area of rectangle - Area of semicircle

Area of field = 32 cm² - 6.29 cm²

Area of field = 25.71 cm²

Area of field ≈ 25.7 cm²

Total pebbles = 12 + 8 + 10

Total pebbles = 30

Probability (First pebble is green) = $\frac{\mathrm{Number\; of\; green\; pebbles}}{\mathrm{Total\; number\; of\; pebbles}}$

Probability (First pebble is green) = $\frac{10}{30}$

Probability (First pebble is green) = $\frac{1}{3}$

Probability (Second pebble is not red) = $\frac{\mathrm{Number\; of\; non-red\; pebbles}}{\mathrm{Total\; number\; of\; pebbles}}$

Number of non-red pebbles = 8 + 10

Number of non-red pebbles = 18

Probability (Second pebble is not red) = $\frac{18}{30}$

Probability (Second pebble is not red) = $\frac{3}{5}$

Since the selection is with replacement, the two events are independent.

Note: and in probability means multiplication

Probability = Probability (First pebble is green) x Probability (Second pebble is not red)

Probability = $\frac{1}{3}$ x $\frac{3}{5}$

Probability = $\frac{\mathrm{1\; x\; 3}}{\mathrm{3\; x\; 5}}$

Probability = $\frac{3}{15}$

Probability = $\frac{1}{5}$

(4x + 5)(3 - 2x) = 4x(3 - 2x) + 5(3 - 2x)

(4x + 5)(3 - 2x) = 12x - 8x² + 15 - 10x

(4x + 5)(3 - 2x) = -8x² + (12x - 10x) + 15

(4x + 5)(3 - 2x) = -8x² + 2x + 15

Coefficient of x² is -8

Let g = Gifty's current age

Gifty is 19 years younger than her mother means if you add 19 years to Gifty's age, you will get that of the mother's age.

Mother's age = g + 19

In three years time,

Gifty's age = g + 3

Mother's age = (g + 19) + 3 = g + 22

The sum of their ages in three years will be:

(g + 3) + (g + 22) = 67

Simplifying:

g + 3 + g + 22 = 67

2g + 25 = 67

2g= 42

Divide both sides by 2

$\frac{\mathrm{2g}}{2}$ = $\frac{42}{2}$

g= 21.

Gifty is currently 21 years old.

Mother's current age = g + 19 = 21 + 19 = 40

The sum of their ages now is: 21 + 40 = 61

Mean = $\frac{\mathrm{Sum\; of\; values}}{\mathrm{Number\; of\; values}}$

Given that the mean of a, b, c, d and e is 15

$\frac{\mathrm{a\; +\; b\; +\; c\; +\; d\; +\; e}}{5}$ = 15

Get rid of the denominator by multiplying both sides by the denominator, 5

5 x $\frac{\mathrm{a\; +\; b\; +\; c\; +\; d\; +\; e}}{5}$ = 15 x 5

a + b + c + d + e = 75

Mean of (a + 1), (b + 3), (c + 5), (d + 7) and (e + 9)

Mean = $\frac{\mathrm{(a\; +\; 1)\; +\; (b\; +\; 3)\; +\; (c\; +\; 5)\; +\; (d\; +\; 7)\; +\; (e\; +\; 9)}}{5}$

(a + 1) + (b + 3) + (c + 5) + (d + 7) + (e + 9) = a + b + c + d + e + (1 + 3 + 5 + 7 + 9)

(a + 1) + (b + 3) + (c + 5) + (d + 7) + (e + 9) = a + b + c + d + e + 25

But a + b + c + d + e = 75

(a + 1) + (b + 3) + (c + 5) + (d + 7) + (e + 9) = 75 + 25 = 100

Mean = $\frac{100}{5}$ = 20

The figures are 5, 4, 0, 5, 1, 4 and the third falls on 0.

The next figure after the third figure is 5.

Since 5 is greater than 4, we add 1 to the third figure (0) to get 1.

Therefore, 5.40514 rounded to three significant figures is 5.41.

Let y = amount invested at 10%

The amount invested at 11% is 140% of the amount invested at 10%

Amount invested at 11% = $\frac{140}{100}$ x y = 1.4y

3$\frac{1}{2}$ years = 3.5 years

Time (T) = 3.5

Interest at 11%

Principal = 1.4y

Rate = 11%

Interest at 11% = $\frac{\mathrm{1.4y\; x\; 3.5\; x\; 11}}{100}$

Interest at 11% = $\frac{\mathrm{53.9y}}{100}$

Interest at 11% = 0.539y

Interest at 10%

Principal = y

Rate = 10%

Interest at 10% = $\frac{\mathrm{y\; x\; 3.5\; x\; 10}}{100}$

Interest at 10% = $\frac{\mathrm{35y}}{100}$

Interest at 10% = 0.35y

Total interest = Interest at 11% + Interest at 10%

Total interest = GH₵ 22,580.60

22580.60 = 0.539y + 0.35y

22580.60 = 0.889y

Divide both sides by 0.889

y = $\frac{22580.60}{0.889}$

y = 25400

Amount invested at 10% = GH₵ 25,400

Amount invested at 11% = 1.4y = 1.4 x y

Amount invested at 11% = 1.4 x 25400 = GH₵ 35,560

Note: 1 mark is deducted for answers without unit.

Radius = 2 m

Volume of the hemisphere = $\frac{1}{2}$ x $\frac{4}{3}$ x $\frac{22}{7}$ x 2³

Volume of the hemisphere = 16.76 m³

1 m³ = 1000 litres

Volume of the hemisphere = 16.76 x 1000 = 16760 litres

Area of farm = 500 m²

Water required per square metre = 50 litres

Total water required = 500 x 50 litres

Total water required = 25,000 litres

Since 16,760 litres < 25,000 litres, a full tank of water would not be sufficient to irrigate the farm.