Three equal jars are filled with a mixture of spirit and water. The proportion of spirit to water in each jar is 2:3, 3:4 and4:5. The contents of these jars are emptied into a single vessel. What is the proportion of spirit and water in it?
Solution
Correct Answer: Option A
For convenience in calculation, we’ll have to suppose the capacity of each vessel to be the L.C.M. of (2 + 3), (3 + 4) and (4 + 5) i.e., 315 litres. Because it hardly matters whether the capacity of each vessel is 10 litres, or 315 litres or 1000 litres. The only thing is that they should have equal quantity of mixtures.
Quantity of spirit = \(\left( {\frac{2}{5} \times 315} \right) + \left( {\frac{3}{7} \times 315} \right) + \left( {\frac{4}{9} \times 315} \right)\)
= [126 + 135 + 140] = 401 litres
Total quantity of mixtures = (3 × 315 litres) = 945 litres
Quantity of water in mixture = (945 - 401) = 544 litres
Thus, \(\frac{{quantity\;of\;spirit\;in\;mixture}}{{quantity\;of\;water\;in\;mixture}} = \frac{{401}}{{544}}\)
Theorem: if x glasses of equal size are filled with a mixture of spirit and water. The ratio of spirit and water in each glass are as follows a1:b1, a2:b2,…………….ax:bx. If the contents of all the x glasses are emptied into a single vessel, then proportion of spirit and water in it is given by \(\left( {\frac{{{a_1}}}{{{a_1} + {b_1}}} + \frac{{{a_2}}}{{{a_2} + {b_2}}} + \ldots + \frac{{{a_x}}}{{{a_x} + {b_x}}}} \right):\left( {\frac{{{b_1}}}{{{a_1} + {b_1}}} + \frac{{{b_2}}}{{{a_2} + {b_2}}} + \ldots + \frac{{{b_x}}}{{{a_x} + {b_x}}}} \right)\)
\(\therefore \frac{{quantity\;of\;spirit\;in\;mixture}}{{quantity\;of\;water\;in\;mixture}} = \left( {\frac{2}{{2 + 1}} + \frac{3}{{3 + 4}} + \frac{4}{{4 + 5}}} \right):\left( {\frac{1}{{2 + 1}} + \frac{4}{{3 + 7}} + \frac{5}{{4 + 5}}} \right) = \frac{{401}}{{544}}\)
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