Chain Rules

Let, L = large ships, M = medium ships, S = small ships
Now from the question,
4L ≡ 7S
$${\text{15L}} = \left( {\frac{7}{4} \times 15} \right){\text{S}} = \frac{{105}}{4}{\text{S}}$$
$${\text{Also, 2L}} = \left( {\frac{7}{4} \times 2} \right){\text{S}} = \frac{7}{2}{\text{S}}$$
$${\text{3M}} \equiv 2{\text{L}} + {\text{1S }} \equiv \left( {\frac{7}{2} + 1} \right){\text{S}} = \frac{9}{2}{\text{S}}$$
$$ ⇒ {\text{ 7M}} = \left( {\frac{9}{2} \times \frac{1}{3} \times 7} \right)\,\,{\text{S}} = \frac{{21}}{2}{\text{S}}$$
$$\therefore \,\left( {{\text{15L}} + {\text{7M}} + {\text{14S}}} \right){\text{ ships}}$$
$$ \equiv \left( {\frac{{105}}{4} + \frac{{21}}{2} + 14} \right){\text{S}}$$
$$ = \frac{{203}}{4}{\text{S}}$$
$$\left( {{\text{12 large}} + 14{\text{ medium}} + 2{\text{1 small}}} \right){\text{ ships}}$$
$$ \equiv \left[ {\left( {\frac{7}{4} \times 12} \right) + \left( {\frac{{21}}{2} \times 2} \right) + 21} \right]{\text{S}}$$
$$ = \left( {21 + 21 + 21} \right){\text{S}}$$
$$ = 63\,{\text{S}}$$
Let the required number of journeys be x
More ships, Less journeys (Indirect proportion)
$$\therefore \,63:\frac{{203}}{4}::36:x$$
$$ ⇒ 63x = \frac{{203}}{4} \times 36 = 1827$$
$$ ⇒ x = \frac{{1827}}{{63}}$$
$$ ⇒ x = 29$$
Alternate:
Now from the question,
$$4L = 7S$$
$$ ⇒ \frac{L}{S} = \frac{7}{4}$$
$$ ⇒ L = 7x;S = 4x$$
$$3M = 2L + S$$
$$ ⇒ M = \frac{{2 \times 7x + 4x}}{3}$$
$$ ⇒ M = 6x$$
Thus, ratio of large, medium and small = 7 : 6 : 4
So, numbers of journeys:
$$ = \frac{{(15 \times 7 + 7 \times 6 + 14 \times 4)36}}{{12 \times 7 + 14 \times 6 + 21 \times 4}}$$
$$ = \frac{{7308}}{{252}}$$
$$ = 29$$

Cloth is required for 1 shirt
= 2 m, 60 cm or 260 cm
Cloth is required for 7 shirt
= 260 × 7
= 1820 cm or 18 m 20 cm

1 score of papers = 20 papers
Cost of 4 dozen papers = ₹24
Cost of 20 papers
$$ = \frac{{24}}{{4 \times 12}} \times 20$$
$$ = {\text{₹}}{\text{.10}}$$

Cost of 8 fan's and 14 oven's is ₹ 36520
Cost of 4 fan's and 7 oven's is ₹ $$\frac{{{\text{36520}}}}{2}$$  = ₹18260
Cost of 12 fan's and 21 oven's is
= 18260 × 3
= ₹ 54780

Cost of 5 kgs apples ₹ 450
Cost of 1 kg apples ₹ $$\frac{{{\text{450}}}}{5}$$
Cost of 8 kgs apples = $$\frac{{450}}{5} \times 8$$   = ₹720
Cost of 12 dozens mangoes ₹ 4320
Cost of 1 dozen mangoes ₹ $$\frac{{{\text{4320}}}}{{12}}$$
Cost of 8 dozen mangoes = $$\frac{{4320}}{{12}} \times 8$$   = ₹ 2880
Cost of 4 kgs oranges = ₹ 240
Cost of 1 kg orange = ₹ $$\frac{{{\text{240}}}}{4}$$
Cost of 8 kgs oranges = $$\frac{{240}}{4} \times 8$$   = ₹ 480
Total cost = 720 + 2880 + 480 = ₹ 4080

Let the required quantity of coal be×metric tonnes
More engines, More coal (Direct proportion)
More hours per day, More coal (Direct proportion)
More rate, More coal (Direct proportion)
\[\left. \begin{gathered} \,\,\,\,\,\,\,\,\,\,\,\,\,\,{\text{Engines 5}}:8 \hfill \\ {\text{Hours per day 9}}:10 \hfill \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\text{Rate }}\frac{1}{3}:\frac{1}{4} \hfill \\ \end{gathered} \right\}::6:x\]
$$\therefore \,\,\left( {5 \times 9 \times \frac{1}{3} \times x} \right) = \left( {8 \times 10 \times \frac{1}{4} \times 6} \right)$$
$$ ⇒ 15x = 120$$
$$ ⇒ x = 8$$

Let the required number of days be×
2 men of latter type = 3 men of former type
12 men of latter type
= $$\left( {\frac{3}{2} \times 12} \right)$$
= 18 men of former type
More men, Less days (Indirect proportion)
Less working hours, More days (Indirect proportion)
\[\left. \begin{gathered} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\text{Men 18}}:9 \hfill \\ {\text{Working hrs 6}}:\frac{{15}}{2} \hfill \\ \end{gathered} \right\}::20:x\]
$$\therefore \,18 \times 6 \times×= 9 \times \frac{{15}}{2} \times 20$$
$$ ⇒ 108x = 1350$$
$$ ⇒ x = \frac{{25}}{2}$$
$$ ⇒ x = 12\frac{1}{2}$$

3 women ≡ 2 men
So, 21 women ≡ 14 men
Less men, More days (Indirect proportion)
Less hours per day, More days (Indirect proportion)
\[\left. \begin{gathered} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\text{Men 14}}:15 \hfill \\ {\text{Hours per day 6}}:8 \hfill \\ \end{gathered} \right\}::21:x\]
$$\therefore \,\left( {14 \times 6 \times x} \right) = \left( {15 \times 8 \times 21} \right)$$
$$ ⇒ x = \frac{{\left( {15 \times 8 \times 21} \right)}}{{\left( {14 \times 6} \right)}}$$
$$ ⇒ x = 30$$
∴ Required number of days = 30

Let the remaining food last for×days
4000 soldiers had provision for 160 days
3200 soldiers had provision for×days
Less men, More days (Indirect proportion)
$$\therefore \,3200:4000::160:x$$
$$ ⇒ 3200x = 4000 \times 160$$
$$ ⇒ x = \frac{{\left( {4000 \times 160} \right)}}{{3200}}$$
$$ ⇒ x = 200$$

Let the remaining food last for×days
500 men had provision for = (27 - 3) = 24 days
(500 + 300) men had provision for×days
More men, Less days (Indirect proportion)
$$\therefore \,800:500::24:x$$
$$ ⇒ \left( {800 \times x} \right) = \left( {500 \times 24} \right)$$
$$ ⇒ x = \frac{{\left( {500 \times 24} \right)}}{{800}}$$
$$ ⇒ x = 15$$