Question

Two vessels P and Q contain wine and water in the ratios of 5: 2 and 8: 5, respectively. Find the ratio in which these mixtures are to be mixed to get a new mixture containing wine and water in the ratio of 9: 4.
(A) 2: 7 (B) 7: 2 (C) 7: 3 (D) 3: 7 (E) None of these

Answer 1

Hint: As given mixtures have wine and water in ratio therefore they have some common factor between them, we will create equations and solve them to find the ratio of wine and water in the new mixture.

Complete step-by-step answer:
Wine: Water in vessel $ = 5:2$
Wine: Water in vessel $ = 8:5$
Final ratio of Water: Water $ = 9:4$
Let, x will be the common factor of the first vessel and y is the common factor of the second vessel.
So, wine and water amount from first vessel = $\dfrac{{5x}}{7}$ and $\dfrac{{2x}}{7}$ [5:2$ \Rightarrow 5 + 2 = 7$]
Wine and water amount from second vessel = $\dfrac{{8y}}{{13}}$ and $\dfrac{{5y}}{{13}}$ [8:5$ \Rightarrow 8 + 5 = 13$]
So, in new mixture:
Total wine $ = (\dfrac{{5x}}{7} + \dfrac{{8y}}{{13}})$
Total water $ = (\dfrac{{2x}}{7} + \dfrac{{5y}}{{13}})$
Now, new mixture are in ratio 9:4
$(\dfrac{{5x}}{7} + \dfrac{{8y}}{{13}}):(\dfrac{{2x}}{7} + \dfrac{{5y}}{{13}}) = 9:4$ (Given)
$ \Rightarrow \dfrac{{\dfrac{{5x}}{7} + \dfrac{{8y}}{{13}}}}{{\dfrac{{2x}}{7} + \dfrac{{5y}}{{13}}}} = \dfrac{9}{4}$
Do Cross Multiplication:

$ \Rightarrow \dfrac{{65x + 56y}}{{26x + 35y}} = \dfrac{9}{4}$
Do Cross Multiplication:
$ \Rightarrow 65x \cdot 4 + 56y \cdot 4 = 26x \cdot 9 + 35y \cdot 9$
$ \Rightarrow 260x + 224y = 234x + 315y$
$ \Rightarrow 260x - 234x = 315y - 224y$
$ \Rightarrow 26x = 91y$
$ \Rightarrow \dfrac{x}{y} = \dfrac{{91}}{{26}}$
$ \Rightarrow \dfrac{x}{y} = \dfrac{7}{2}$
Therefore, the new mixture is in ratio - 7:2
Hence, Option (B) is correct.

Note: This question can also be answered with the help of alligation. Alligation is a rule that enables us to find the ratio in which two or more ingredients at the given price must be mixed to produce a mixture of desired price. Alligation is an old and practical method of solving arithmetic problems related to mixtures of ingredients. There are two types of alligation: alligation medial, used to find the quantity of a mixture given the quantities of its ingredients, and alligation alternate, used to find the amount of each ingredient needed to make a mixture of a given quantity. Alligation media is merely a matter of finding a weighted mean. Alligation alternate is more complicated and involves organizing the ingredients into high and low pairs which are then traded off.