P is inversely proportional to Q₂

P ∝ $\frac{1}{{\mathrm{Q}}^2}$

Get rid of the proportionality sign (∝) by introducing a constant.

P = k x $\frac{1}{{\mathrm{Q}}^2}$

P = $\frac{\mathrm{k}}{{\mathrm{Q}}^2}$

When Q = 6, P = 2.25

$\frac{\mathrm{k}}{6^2}$ = 2.25

$\frac{\mathrm{k}}{36}$ = 2.25

Multiply both sides by 36

k = 2.25 x 36 = 81

P = $\frac{\mathrm{k}}{{\mathrm{Q}}^2}$

P = $\frac{81}{{\mathrm{Q}}^2}$

When Q = 3

P = $\frac{81}{3^2}$

P = $\frac{81}{9}$ = 9

The more the workers, the lesser the time.

Hence time (t) is inversely proportional to the number of workers (w).

t ∝ $\frac{1}{\mathrm{w}}$

Remove the proportionality by introducing a constant (k).

t = k x $\frac{1}{\mathrm{w}}$

Use the known information to solve for k

7 workers used 6 days.

6 = k x $\frac{1}{7}$

k = 6 x 7

The general equation is:

t = 6 x 7 x $\frac{1}{\mathrm{w}}$

t = $\frac{\mathrm{6\; x\; 7}}{\mathrm{w}}$

Calculate the time (t) for 2 workers.

w = 2

t = $\frac{\mathrm{6\; x\; 7}}{2}$ = 3 x 7 = 21

∴ It will take the two men 21 days to complete the work.

Ratios = 2 : 4 : 10

2 divides all.

Ratios = 1 : 2 : 5

Sum of ratios represents the total oranges shared.

The smallest ratio represents the least share.

Sum of ratios = 1 + 2 + 5 = 8

The smallest ratio = 1

If 8 = 32

1 = $\frac{\mathrm{1\; x\; 32}}{8}$ = 4

∴ The least share is 4

x : y : z = 2 : 3 : 4

x = 2
y = 3
z = 4

Substitute the values into the equation $\frac{\mathrm{9x\; +\; 3y}}{\mathrm{6z\; -\; 2y}}$

$\frac{\mathrm{9x\; +\; 3y}}{\mathrm{6z\; -\; 2y}}$ = $\frac{\mathrm{9\; x\; 2\; +\; 3\; x\; 3}}{\mathrm{6\; x\; 4\; -\; 2\; x\; 3}}$

$\frac{\mathrm{9x\; +\; 3y}}{\mathrm{6z\; -\; 2y}}$ = $\frac{\mathrm{18\; +\; 9}}{\mathrm{24\; -\; 6}}$

$\frac{\mathrm{9x\; +\; 3y}}{\mathrm{6z\; -\; 2y}}$ = $\frac{27}{18}$ = 1.5

If $0.045 = ₦1.00

$18,000.00 = $\frac{\mathrm{₦1.00\; x\; \$18,000.00}}{\mathrm{\$0.045}}$ = ₦ 400,000

$\frac{\mathrm{4m\; +\; 3n}}{\mathrm{4m\; -\; 3n}}$ = $\frac{5}{2}$

Method I

The numerator at the left equals the numerator at the right and the denominator at the left equals the denominator at the right.

4m + 3n = 5 ---- eq. 1

4m - 3n = 2 ---- eq. 2

Let's get rid of n by adding --- eq. 1 and 2

8m = 7

Divide both sides by 8

m =$\frac{7}{8}$

Get rid of m by subtracting -- eq. 2 from eq. 1

3n--3n = 3

-- = +

3n + 3n = 3

6n = 3

Divide both sides by 6

n = $\frac{3}{6}$ = $\frac{1}{2}$

Note: alternatively, you can substitute the value of m into any of the equations and solve for n.

m : n = $\frac{7}{8}$ : $\frac{1}{2}$

m : n = $\frac{7}{8}$ ÷ $\frac{1}{2}$

Reciprocate the fraction at the right of ÷ and change the ÷ to multiplication (x)

m : n = $\frac{7}{8}$ x $\frac{2}{1}$

Note: 2 divides itself 1 time and 8, 4 times.

m : n = $\frac{7}{4}$

⇒ m : n = 7 : 4

Method II

Cross multiply.

$\frac{\mathrm{4m\; +\; 3n}}{\mathrm{4m\; -\; 3n}}$ = $\frac{5}{2}$

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

2 x 4m + 2 x 3n = 5 x 4m - 5 x 3n

8m + 6n = 20m - 15n

6n + 15n = 20m - 8m

21n = 12m

Divide both sides by n in order to get m : n

21 = $\frac{\mathrm{12m}}{\mathrm{n}}$

Divide both sides by 12

$\frac{21}{12}$ = $\frac{\mathrm{m}}{\mathrm{n}}$

Note: 3 divides 21, 7 times and 12, 4 times.

$\frac{7}{4}$ = $\frac{\mathrm{m}}{\mathrm{n}}$

∴ m : n = 7 : 4

The total sales is 100%

Change the mixed fraction to improper fraction

12$\frac{1}{2}$% = $\frac{\mathrm{2\; x\; 12\; +\; 1}}{2}$% = $\frac{\mathrm{24\; +\; 1}}{2}$% = $\frac{25}{2}$%

If $\frac{25}{2}$% = $35,000.00

100% = $35,000.00 x 100% ÷ $\frac{25}{2}$%

% cancels each other.

Reciprocate the fraction at the right of the ÷ and change ÷ to multiplication (x)

Note: reciprocate means the numerator becomes the denominator and the denominator becomes the numerator.

100% = $35,000.00 x 100 x $\frac{2}{25}$ = $280,000

The sum of ratios represents the total members

5 + 2 represents Total members

7 represents the total members of the committee

The ratio for girls is 5 and that of boys is 2.

Total members is 35

If 7 = 35

5 = $\frac{\mathrm{35\; x\; 5}}{7}$ = 25

∴ there are 25 girls

If 7 = 35

2 = $\frac{\mathrm{35\; x\; 2}}{7}$ = 10

∴ there are 10 boys

The number of girls remain the same but we now wish to increase the number of boys.

The new ratio for boys is 4

We know the number of girls.

If 5 = 25

4 = $\frac{\mathrm{25\; x\; 4}}{5}$ = 20

The boys must be 20 in order to get 5 : 4 ratio.

∴ 10 more boys must be added to the current 10 boys to get 20 boys in the committee.

P ∝ $\frac{{\mathrm{v}}^2}{\mathrm{x}}$

Replace x by vt

P ∝ $\frac{{\mathrm{v}}^2}{\mathrm{vt}}$

v² = v x v

P ∝ $\frac{\mathrm{v\; x\; v}}{\mathrm{vt}}$

The v cancel each other.

P ∝ $\frac{\mathrm{v}}{\mathrm{t}}$

The sum of the ratio represents the entire money shared.

Sum of ratios = 3 + 2
Sum of ratios = 5

Rose's ratio is 2.

If 5 = GH₵ 875.08
2 = $\frac{\mathrm{GH₵\; 875.08\; x\; 2}}{5}$ = GH₵ 350.032

Rose's share = GH₵ 350.032

Sum of Efua's and Ama's ratios = 1 + 2
Sum of Efua's and Ama's ratios = 3

Ama's ratio = 2

The sum of ratios represents the entire money both shared.

If 3 = GH₵ 350.032
2 = $\frac{\mathrm{GH₵\; 350.032\; x\; 2}}{3}$ = GH₵ 233.3546666666667

Ama's share = GH₵ 233.35

y ∝ x²

Remove the proportionality by introducing a constant.

y = kx²

Subtitute the known values to solve for the constant k

y = 3 when x = $\sqrt{2}$

3 = k x ${\left(\sqrt{2}\right)2}^{}$

When you square a square root, the square cancels out the square root.

3 = k x 2

Divide both sides by 2

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

Subtitute the value of k into the equation

y = $\frac{3}{2}$ x x²

When x = 2

y = $\frac{3}{2}$ x 2²

2² = 2 x 2

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

The 2 dividing 3 cancels out one of the 2s

y = 3 x 1 x 2

y = 6

Directly proportion → R ∝ L

Inversely proportional → R ∝ $\frac{1}{\mathrm{P}}$

When you join the two relations,

R ∝ $\frac{\mathrm{L}}{\mathrm{P}}$

Remove the proportionality and introduce a constant k.

R = k x $\frac{\mathrm{L}}{\mathrm{P}}$

Now use the given information to calculate for k.

R = 3
L = 9
P = 0.8

3 = k x $\frac{9}{0.8}$

Get rid of the fraction by multiplying both sides by the denominator 0.8.

3 x 0.8 = k x 9

Divide both sides by 9

k = $\frac{\mathrm{3\; x\; 0.8}}{9}$ = $\frac{0.8}{3}$

Substitute the value of k into the equation.

R = $\frac{0.8}{3}$ x $\frac{\mathrm{L}}{\mathrm{P}}$

R = $\frac{\mathrm{0.8L}}{\mathrm{3P}}$

When L = 15 and P = 1.8

R = $\frac{\mathrm{0.8\; x\; 15}}{\mathrm{3\; x\; 1.8}}$ = $\frac{12}{5.4}$ = 2.2222 ≈ 2.2

x is directly proportional to y

x ∝ y

x is inversely proportional to z

x ∝ $\frac{1}{\mathrm{z}}$

Combining the two, x ∝ $\frac{\mathrm{y}}{\mathrm{z}}$

Introduce a constant k to remove the proportionality constant.

x = k x $\frac{\mathrm{y}}{\mathrm{z}}$

Substitute the known values and solve for the constant k

x = 15
y = 10
z = 4

15 = k x $\frac{10}{4}$

Get rid of the fraction by multiplying both sides by the denominator, 4

4 x 15 = k x 10

60 = 10k

Divide both sides by 10

k = $\frac{60}{10}$ = 6

Substitute the value of k into the relation.

x = k x $\frac{\mathrm{y}}{\mathrm{z}}$

k = 6

x = 6 x $\frac{\mathrm{y}}{\mathrm{z}}$

x = $\frac{\mathrm{6\; x\; y}}{\mathrm{z}}$

x = $\frac{\mathrm{6y}}{\mathrm{z}}$

m : n = 2 : 1

Comparing the left hand and right hand sides, m = 2 and n = 1

$\frac{3{\mathrm{m}}^2-2{\mathrm{n}}^2}{{\mathrm{m}}^2+\mathrm{mn}}$ = $\frac{3{\mathrm{m}}^2-2{\mathrm{n}}^2}{{\mathrm{m}}^2+\mathrm{mn}}$

Substitute the value of m and n into the given relation.

$\frac{3{\mathrm{m}}^2-2{\mathrm{n}}^2}{{\mathrm{m}}^2+\mathrm{mn}}$ = $\frac{\mathrm{3\; x}{2}^2-\mathrm{2\; x}{1}^2}{2^2+\mathrm{2\; x\; 1}}$

$\frac{3{\mathrm{m}}^2-2{\mathrm{n}}^2}{{\mathrm{m}}^2+\mathrm{mn}}$ = $\frac{\mathrm{3\; x\; 4}-\mathrm{2\; x\; 1}}{4+2}$

$\frac{3{\mathrm{m}}^2-2{\mathrm{n}}^2}{{\mathrm{m}}^2+\mathrm{mn}}$ = $\frac{12-2}{4+2}$

$\frac{3{\mathrm{m}}^2-2{\mathrm{n}}^2}{{\mathrm{m}}^2+\mathrm{mn}}$ = $\frac{10}{6}$

$\frac{3{\mathrm{m}}^2-2{\mathrm{n}}^2}{{\mathrm{m}}^2+\mathrm{mn}}$ = $\frac{5}{3}$

H varies directly as p

H ∝ p

H varies inversely as the square of y

H ∝ $\frac{1}{{\mathrm{y}}^2}$

Combining the two

H ∝ $\frac{\mathrm{p}}{{\mathrm{y}}^2}$

Introduce a constant k to remove the proportionality sign.

H = k x $\frac{\mathrm{p}}{{\mathrm{y}}^2}$

Use the given values to find k.

H = 1

p = 8

y = 2

1 = k x $\frac{8}{2^2}$

1 = k x $\frac{8}{4}$

1 = k x 2

Divide both sides by 2

k = $\frac{1}{2}$

Substitute the value of k into the general relation.

H = k x $\frac{\mathrm{p}}{{\mathrm{y}}^2}$

H = $\frac{1}{2}$ x $\frac{\mathrm{p}}{{\mathrm{y}}^2}$

H = $\frac{\mathrm{1\; x\; p}}{\mathrm{2\; x}{\mathrm{y}}^2}$

H = $\frac{\mathrm{p}}{2{\mathrm{y}}^2}$

Mary : Charity = 4 : 5

The difference in their ratios represent how much more or less one gets.

The sum of their ratios represents the total profit made.

Note: difference means subtraction.

Difference in ratios = 5 - 4 = 1

Sum of ratios = 4 + 5 = 9

Mary received GH₵ 5,000.00 less than Charity

The difference in ratios (1) represents how much less Mary received.

If 1 = GH₵ 5,000

9 = $\frac{\mathrm{9\; x\; GH₵\; 5,000}}{1}$ = GH₵ 45,000

∴ The profit made is GH₵ 45,000

y ∝ $\frac{1}{{\mathrm{x}}^2}$

NB: ∝ is the symbol for proportional
inversely proportional is 1 over the expression

Introduce a constant to remove the proportionality sign.

y = k x $\frac{1}{{\mathrm{x}}^2}$

Use the known values to calculate the constant(k) to get the general expression

y = 100 and x = 3

100 = k x $\frac{1}{3^2}$

100 = k x $\frac{1}{9}$

Get rid of the fraction by multiplying both sides of the equation by the denominator 9

100 x 9 = k x $\frac{1}{9}$ x 9

100 x 9 = k x 1

k = 900

Substitute the value of k to get the general equation.

y = 900 x $\frac{1}{{\mathrm{x}}^2}$ = $\frac{900}{{\mathrm{x}}^2}$

When y = 25

25 = $\frac{900}{{\mathrm{x}}^2}$

Get rid of the fraction by multiplying both sides of the equation by the denominator x²

25 x x² = $\frac{900}{{\mathrm{x}}^2}$ x x²

25 x x² = 900

Divide both sides by 25

x² = $\frac{900}{25}$

x² = 36

Take square root of both sides to get rid of the square.

x = $\sqrt{\mathrm{36}}$ = 6

y varies inversely as x ⇒ y ∝ $\frac{1}{\mathrm{x}}$

Remove the proportionality sign (∝) by introducing a constant (k)

y = k x $\frac{1}{\mathrm{x}}$ = $\frac{\mathrm{k}}{\mathrm{x}}$

When x = 3, y = 6.

6 = $\frac{\mathrm{k}}{3}$

Solve for k.

k = 6 x 3 = 18

y = $\frac{18}{\mathrm{x}}$

When x = 9

y = $\frac{18}{9}$ = 2