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Large, medium and small ships are used to bring water. 4 large ships carry as much water as 7 small ships, 3 medium ships carry the same amount of water as 2 large ships and 1 small ship. 15 large, 7 medium and 14 small ships, each made 36 journeys and brought a certain quantity of water. In how many journeys would 12 large, 14 medium and 21 small ships bring the same quantity ?
Answer & Solution
Answer: Option
B
Solution:
Let, L = large ships, M = medium ships, S = small ships
Now from the question,
4L ≡ 7S
$$\eqalign{ & {\text{15L}} = \left( {\frac{7}{4} \times 15} \right){\text{S}} = \frac{{105}}{4}{\text{S}} \cr & {\text{Also, 2L}} = \left( {\frac{7}{4} \times 2} \right){\text{S}} = \frac{7}{2}{\text{S}} \cr & {\text{3M}} \equiv 2{\text{L}} + {\text{1S }} \equiv \left( {\frac{7}{2} + 1} \right){\text{S}} = \frac{9}{2}{\text{S}} \cr & \Leftrightarrow {\text{ 7M}} = \left( {\frac{9}{2} \times \frac{1}{3} \times 7} \right)\,\,{\text{S}} = \frac{{21}}{2}{\text{S}} \cr & \therefore \,\left( {{\text{15L}} + {\text{7M}} + {\text{14S}}} \right){\text{ ships}} \cr & \equiv \left( {\frac{{105}}{4} + \frac{{21}}{2} + 14} \right){\text{S}} \cr & = \frac{{203}}{4}{\text{S}} \cr & \left( {{\text{12 large}} + 14{\text{ medium}} + 2{\text{1 small}}} \right){\text{ ships}} \cr & \equiv \left[ {\left( {\frac{7}{4} \times 12} \right) + \left( {\frac{{21}}{2} \times 2} \right) + 21} \right]{\text{S}} \cr & = \left( {21 + 21 + 21} \right){\text{S}} \cr & = 63\,{\text{S}} \cr} $$
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:
$$\eqalign{ & = \frac{{(15 \times 7 + 7 \times 6 + 14 \times 4)36}}{{12 \times 7 + 14 \times 6 + 21 \times 4}} \cr & = \frac{{7308}}{{252}} \cr & = 29 \cr} $$
Now from the question,
4L ≡ 7S
$$\eqalign{ & {\text{15L}} = \left( {\frac{7}{4} \times 15} \right){\text{S}} = \frac{{105}}{4}{\text{S}} \cr & {\text{Also, 2L}} = \left( {\frac{7}{4} \times 2} \right){\text{S}} = \frac{7}{2}{\text{S}} \cr & {\text{3M}} \equiv 2{\text{L}} + {\text{1S }} \equiv \left( {\frac{7}{2} + 1} \right){\text{S}} = \frac{9}{2}{\text{S}} \cr & \Leftrightarrow {\text{ 7M}} = \left( {\frac{9}{2} \times \frac{1}{3} \times 7} \right)\,\,{\text{S}} = \frac{{21}}{2}{\text{S}} \cr & \therefore \,\left( {{\text{15L}} + {\text{7M}} + {\text{14S}}} \right){\text{ ships}} \cr & \equiv \left( {\frac{{105}}{4} + \frac{{21}}{2} + 14} \right){\text{S}} \cr & = \frac{{203}}{4}{\text{S}} \cr & \left( {{\text{12 large}} + 14{\text{ medium}} + 2{\text{1 small}}} \right){\text{ ships}} \cr & \equiv \left[ {\left( {\frac{7}{4} \times 12} \right) + \left( {\frac{{21}}{2} \times 2} \right) + 21} \right]{\text{S}} \cr & = \left( {21 + 21 + 21} \right){\text{S}} \cr & = 63\,{\text{S}} \cr} $$
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:
$$\eqalign{ & = \frac{{(15 \times 7 + 7 \times 6 + 14 \times 4)36}}{{12 \times 7 + 14 \times 6 + 21 \times 4}} \cr & = \frac{{7308}}{{252}} \cr & = 29 \cr} $$