Chain-Rule - 01

Explanation:-

Cost of 21 pencils and 9 clippers = Rs. 819
Cost of 7 pencils and 3 clippers = $$\frac{{819}}{3}$$ = Rs. 273

Explanation:-

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

Explanation:-

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

Explanation:-

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

Explanation:-

Let, L = large ships, M = medium ships, S = small ships
Now from the question,
4L ≡ 7S 15L = $$\left( {\frac{7}{4} \times 15} \right)$$ S = $$\frac{{105}}{4}$$ S
Also, 2L = $$\left( {\frac{7}{4} \times 2} \right)$$ S = $$\frac{7}{2}$$ S
3M ≡ 2L + 1S ≡ $$\left( {\frac{7}{2} + 1} \right)$$ S = $$\frac{9}{2}$$ S
⇔ 7M = $$\left( {\frac{9}{2} \times \frac{1}{3} \times 7} \right)$$ S = $$\frac{{21}}{2}$$ S
$$\eqalign{ & \therefore \,\left( {{\text{15L}} + {\text{7M}} + {\text{4S}}} \right){\text{ ships}} \equiv \left( {\frac{{105}}{4} + \frac{{21}}{2} + 14} \right){\text{S}} = \frac{{203}}{4}{\text{S}} \cr & \left( {{\text{12 large}} + 14{\text{ medium}} + 2{\text{1 small}}} \right){\text{ ships}} \equiv \left[ {\left( {\frac{7}{4} \times 12} \right) + \left( {\frac{{21}}{2} \times 2} \right) + 21} \right]{\text{S}} = \left( {21 + 21 + 21} \right){\text{S}} = 63{\text{S}} \cr}$$
Let the required number of journeys be x
More ships, Less journeys (Indirect proportion)
$$\eqalign{ & \therefore \,63:\frac{{203}}{4}::36:x \cr & \Leftrightarrow 63x = \frac{{203}}{4} \times 36 = 1827 \cr & \Leftrightarrow x = \frac{{1827}}{{63}} \cr & \Leftrightarrow x = 29 \cr}$$

Alternate:

Now from the question,
$$\eqalign{ & 4L = 7S \cr & \Rightarrow \frac{L}{S} = \frac{7}{4} \cr & \Rightarrow L = 7x;S = 4x \cr & 3M = 2L + S \cr & \Rightarrow M = \frac{{2 \times 7x + 4x}}{3} \cr & \Rightarrow M = 6x \cr}$$
Thus, ratio of large, medium and small = 7 : 6 : 4
So, numbers of journeys:
$$\eqalign{ & = \frac{{(15 \times 7 + 7 \times 6 + 14 \times 4)36}}{{12 \times 7 + 14 \times 6 + 21 \times 4}} \cr & = \frac{{7308}}{{252}} \cr & = 29 \cr}$$

Explanation:-

1 man = 2 boys
⇒ (12 men + 18 boys) ≡ (12 × 2 ×18) boys = 42 boys
Let required number of boys = x
⇒ (21 men + x boys) ≡ (21 × 2 × x) boys = (42 + x) boys
Less days, More boys (Indirect proportion)
More hours per day, Less boys (Indirect proportion)
More work, More boys (Direct proportion)
$$\left. \begin{gathered} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\text{Days 50}}:60 \hfill \\ {\text{Hours per day 9}}:\frac{{15}}{2} \hfill \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\text{Work }}1:2 \hfill \\ \end{gathered} \right\}::42:\left( {42 + x} \right)$$
$$\eqalign{ & \therefore \,\left[ {50 \times 98 \times 1 \times \left( {42 + x} \right)} \right] = \left( {60 \times \frac{{15}}{2} \times 2 \times 42} \right) \cr & \Leftrightarrow \left( {42 + x} \right) = \frac{{37800}}{{450}} \cr & \Leftrightarrow 42 + x = 84 \cr & \Leftrightarrow x = 42 \cr}$$

Explanation:-

Ratio of time taken by a woman, a man and a boy
$$\eqalign{ & = 8:6:12 \cr & = 4:3:6 \cr}$$
So, 4 women ≡ 3 men ≡ 6 boy
(12 mens + 12 womens + 12 boys)
$$\eqalign{ & = \left[ {12 + \left( {\frac{3}{4} \times 12} \right) + \left( {\frac{3}{6} \times 12} \right)} \right]{\text{men}} \cr & {\text{ = }}\left( {12 + 9 + 6} \right){\text{men}} \cr & = 27{\text{ men}} \cr}$$
Let the required number of days be x
More men, Less days (Indirect proportion)
More working hours, Less days (Indirect proportion)
$$\left. \begin{gathered} {\text{Working hours 8}}:6 \hfill \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\text{Men 27}}:9 \hfill \\ \end{gathered} \right\}::6:x$$
$$\eqalign{ & \therefore \,27 \times 8 \times x = 9 \times 6 \times 6 \cr & \Leftrightarrow x = \frac{{\left( {9 \times 6 \times 6} \right)}}{{\left( {27 \times 8} \right)}} \cr & \Leftrightarrow x = \frac{3}{2} \cr & \Leftrightarrow x = 1\frac{1}{2} \cr}$$

Explanation:-

Let the team take x days to finish 360 pieces
Then, number of pieces made each day = $$\frac{{360}}{x}$$
More number of pieces per day, Less days (Indirect proportion)
$$\eqalign{ & \therefore \,\left( {\frac{{360}}{x} + 4} \right):\frac{{360}}{x}::x:\left( {x - 1} \right) \cr & \Leftrightarrow \left( {\frac{{360}}{x} + 4} \right):\left( {x - 1} \right) = \frac{{360}}{x}:x \cr & \Leftrightarrow 360 - \frac{{60}}{x} + 4x - 4 = 360 \cr & \Leftrightarrow 4x - \frac{{360}}{x} - 4 = 0 \cr & \Leftrightarrow x - \frac{{90}}{x} - 1 = 0 \cr & \Leftrightarrow {x^2} - x - 90 = 0 \cr & \Leftrightarrow \left( {x - 10} \right)\left( {x + 9} \right) = 0 \cr & \Leftrightarrow x = 10 \cr}$$

Explanation:-

Initially, Let there be x men having food for y days
After 10 days, x men had food for days
Also, $$\left( {x - \frac{x}{5}} \right)$$ men had food for y days
$$\eqalign{ & \therefore \,x\left( {y - 10} \right) = \frac{{4x}}{5} \times y \cr & \Leftrightarrow 5xy - 50x = 4xy \cr & \Leftrightarrow xy - 50x = 0 \cr & \Leftrightarrow x\left( {y - 50} \right) = 0 \cr & \Leftrightarrow y - 50 = 0 \cr & \Leftrightarrow y = 50 \cr}$$

Explanation:-

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