Method I

Let t = total hours both worked at the same rate.

The fraction of time taken to do the work for 1 hour =$\frac{1}{\mathrm{t}}$

Time taken for Kojo to complete the work is 5 hours

Kojo's fraction of time in 1 hour = $\frac{1}{5}$

Time taken for Edna to complete the work is 4 hours

Edna's fraction of time in 1 hour = $\frac{1}{4}$

If you sum each of their fractions, you must get the fraction of total time taken by both in 1 hour

$\frac{1}{5}$ + $\frac{1}{4}$ = $\frac{1}{\mathrm{t}}$

$\frac{\mathrm{4\; +\; 5}}{20}$ = $\frac{1}{\mathrm{t}}$

$\frac{9}{20}$ = $\frac{1}{\mathrm{t}}$

Cross multiply.

9 x t = 20 x 1

9t = 20

Divide both sides by 9

t = $\frac{20}{9}$ hours or 2$\frac{2}{9}$ hours or 2.22 hours

Method II

Rate of doing work = $\frac{\mathrm{Work\; done}}{\mathrm{Time}}$

Let w = the work to be done.

Time taken for Kojo to do the w work is 5 hours

Kojo's rate = $\frac{\mathrm{w}}{5}$

Time taken for Edna to do the w work is 4 hours

Edna's rate = $\frac{\mathrm{w}}{4}$

Let t = the time taken when both worked together at the same rate.

Rate when both worked together = $\frac{\mathrm{w}}{\mathrm{t}}$

Note: the work (w) is the same.

When you add both of their rates, it should give you the rate when they both worked together.

$\frac{\mathrm{w}}{5}$ + $\frac{\mathrm{w}}{4}$ = $\frac{\mathrm{w}}{\mathrm{t}}$

Factorize w out.

w($\frac{1}{5}$ + $\frac{1}{4}$) = w($\frac{1}{\mathrm{t}}$)

The w's cancel each other.

$\frac{1}{5}$ + $\frac{1}{4}$ = $\frac{1}{\mathrm{t}}$

$\frac{\mathrm{4\; +\; 5}}{20}$ = $\frac{1}{\mathrm{t}}$

$\frac{9}{20}$ = $\frac{1}{\mathrm{t}}$

Cross multiply.

9 x t = 1 x 20

9t = 20

Divide both sides by 9

t = $\frac{20}{9}$ hours or 2$\frac{2}{9}$ hours or 2.22 hours

Method I

The difference (5 - 4) of the ratios represents how much more Ansah received than Boadu.

More ratio = 5 - 4 = 1

More amount = GH₵ 5,000

Boadu's ratio = 4

If 1 = GH₵ 5000

4 = $\frac{\mathrm{4\; x\; GH₵\; 5000}}{1}$ = GH₵ 20,000

∴ Boadu received GH₵ 20,000.00

Method II

Using a variable to represent total profit shared.

Let y = total profit shared.

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

Ansah's ratio = 5

Boadu's ratio = 4

Sum of ratios = 5 + 4 = 9

Ansah's share = $\frac{5}{9}$ x y = $\frac{\mathrm{5y}}{9}$

Boadu's share = $\frac{4}{9}$ x y = $\frac{\mathrm{4y}}{9}$

If you subtract Boadu's share from Ansah's share, you will get how much more Ansah received than Boadu.

$\frac{\mathrm{5y}}{9}$ - $\frac{\mathrm{4y}}{9}$ = 5000

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

5y - 4y = 5000 x 9

y = 45000

Total amount shared = GH₵ 45,000

Boadu's share = $\frac{4}{9}$ x y = $\frac{4}{9}$ x GH₵ 45000 = GH₵ 20,000.00

Method I

Using proportionality

The difference of the ratios (subtracting the smaller from the larger) represents how many more is the larger than the smaller.

Difference of ratios = 7 - 5 = 2

The dealer sold 142 more motorcycles than cars means the ratio 2 represents 142

Ratio for cars = 5 and that of motorcycles = 7

Calculating cars sold

If 2 = 142

5 = $\frac{\mathrm{5\; x\; 142}}{2}$ = 355

∴ number of cars sold = 355

Calculating motorcycles sold

If 2 = 142

7 = $\frac{\mathrm{7\; x\; 142}}{2}$ = 497

∴ number of motorcycles sold = 497

Method II

Using variables to represent each items sold

Let m = number of cars sold and n = number of motorcycles sold.

A dealer sold 142 more motorcycles than cars means if you add 142 to cars sold, you will get the number of motorcycles sold.

n = m + 142 ------- eqn 1

Total items sold = m + n

Number of items sold = $\frac{\mathrm{The\; ratio\; of\; the\; item}}{\mathrm{Sum\; of\; ratios}}$ x Total items sold

Ratio for car = 5 and that of motorcycle = 7

Sum of ratios = 5 + 7 = 12

Number of cars sold = $\frac{5}{12}$ x (m + n)

But number of cars sold = m

m = $\frac{5}{12}$ x (m + n)

Get rid of the fraction by multiplying both sides by the denominator (12)

12 x m = 5(m + n)

12m = 5m + 5n

12m - 5m = 5n

7m = 5n

Divide both sides by 7

m = $\frac{\mathrm{5n}}{7}$

Substitute m = $\frac{\mathrm{5n}}{7}$ into ---- eqn 1

n = 142 + $\frac{\mathrm{5n}}{7}$

Get rid of the fraction by multiplying both sides by the denominator (7)

7 x n = 142 x 7 + 5n

7n = 994 + 5n

7n - 5n = 994

2n = 994

Divide both sides by 2

n = $\frac{994}{2}$ = 497

Substitute n = 497 into ---- eqn 1

497 = 142 + m

497 - 142 = m

355 = m

∴ cars sold = 355 and motorcycles sold = 497

Method I

The sum of the ratios represents the total cost of producing the article.

The difference of the ratios of materials and labour represents how much more of materials than that of labour.

Sum of ratios = 5 + 7 + 13 = 25

Difference of the ratios of cost of materials and labour = 13 - 7 = 6

If 6 = 840

25 = $\frac{\mathrm{25\; x\; 840}}{6}$ = Le 3,500

Method II

Let y = the total cost of producing the article.

Value of a ratio = $\frac{\mathrm{Ratio}}{\mathrm{Sum\; of\; ratios}}$ x Total

Sum of ratios = 5 + 7 + 13 = 25

Ratio for labour = 7

Labour = $\frac{7}{25}$ x y = $\frac{\mathrm{7y}}{25}$

Ratio for cost of materials = 13

Cost of materials = $\frac{13}{25}$ x y = $\frac{\mathrm{13y}}{25}$

If you subtract the cost of labour from the cost of materials, you will get how much more of cost of materials than labour.

$\frac{\mathrm{13y}}{25}$ - $\frac{\mathrm{7y}}{25}$ = 840

$\frac{\mathrm{13y\; -\; 7y}}{25}$ = 840

$\frac{\mathrm{6y}}{25}$ = 840

Get rid of the fraction by multiplying both sides by the denominator 25

6y = 840 x 25

Divide both sides by 6

y = $\frac{\mathrm{840\; x\; 25}}{6}$ = Le 3,500

Total cost of producing the article = Le 3,500.00

Z varies directly as x

Z ∝ x

Z varies inversely as twice the cube root of y

Z ∝ $\frac{1}{2\left(\sqrt[3]{y}\right)}$

Combining the two.

Z ∝ $\frac{\mathrm{x}}{2\left(\sqrt[3]{y}\right)}$

Remove the proportionality sign by introducing a constant.

Z = k x $\frac{\mathrm{x}}{2\left(\sqrt[3]{y}\right)}$

Z = $\frac{\mathrm{kx}}{2\left(\sqrt[3]{y}\right)}$

$\sqrt[3]{y}$ = ${\mathrm{y}}^{\frac{1}{3}}$

Z = $\frac{\mathrm{kx}}{2\left(y^{\frac{1}{3}}\right)}$

Z = 8

x = 4

y = $\frac{1}{8}$

8 = 2 x 2 x 2 = 2³

y = $\frac{1}{2^3}$

a^-b = $\frac{1}{{\mathrm{a}}^{\mathrm{b}}}$

y = $\frac{1}{2^3}$ = 2^-3

8 = $\frac{\mathrm{4k}}{2\left({\left(2^{-3}\right)}^{\frac{1}{3}}\right)}$

8 = $\frac{\mathrm{2k}}{2^{\mathrm{-3\; x}\frac{1}{3}}}$

8 = $\frac{\mathrm{2k}}{2^{-1}}$

2k = k x 2

2 = 2¹

a^m ÷ aⁿ = $\frac{{\mathrm{a}}^{\mathrm{m}}}{{\mathrm{a}}^{\mathrm{n}}}$ = a^{m - n}

8 = k x 2^1-(-1)

Note: -- = + or -(-) = +

8 = k x 2^{1 + 1}

8 = k x 2²

8 = k x 4

Divide both sides by 4

k = $\frac{8}{4}$ = 2

Z = $\frac{\mathrm{2x}}{2\left(\sqrt[3]{y}\right)}$

The 2 at both numerator and denominator cancels each other.

Z = $\frac{\mathrm{x}}{\sqrt[3]{y}}$

Get rid of the fraction by multiplying both sides by the denominator $\sqrt[3]{y}$

Z x $\sqrt[3]{y}$ = x

Divide both sides by Z

$\sqrt[3]{y}$ = $\frac{\mathrm{x}}{\mathrm{Z}}$

Get rid of the cube root by cubing both sides.

y = ${\left(\frac{\mathrm{x}}{\mathrm{Z}}\right)}^3$