Alligation — Part 2 MCQs with Explanations

Total milk = 60 litres Drawn off = 12 litres $$\frac{{{\text{Final quantity}}}}{{{\text{Initial quantity}}}}$$ $$ = {\left( {1 - \frac{x}{c}} \right)^t}$$ X = Replaced quantity C = Capacity T = Number of process $$\eqalign{ & \frac{{{\text{Final quantity}}}}{{{\text{Initial quantity}}}} = {\left( {1 - \frac{{12}}{{60}}} \right)^2} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = {\left( {\frac{4}{5}} \right)^2} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{16}}{{25}} \cr} $$ Ratio of milk and water in the resultant mixture : = 16 : 9

Water content in 40 litres of mixture. $$\eqalign{ & = 40 \times \frac{{10}}{{100}} \cr & = 4{\text{ litres}} \cr} $$ ∴ Milk in the mixture. $$\eqalign{ & {\text{ = 40}} - 4 \cr & = 36{\text{ litres}} \cr} $$ Let x litres of water is mixed $$\eqalign{ & \Rightarrow \frac{{4 + x}}{{40 + x}} = \frac{{20}}{{100}} \cr & \Rightarrow x = 5 {\text{ litres}} \cr} $$

According to the question, Mixture of copper and aluminium = 2000 gms 30% is copper means = $$\frac{30}{100}$$ × 2000 = 600 gms copper $$\eqalign{ & \Rightarrow \frac{{600}}{{1400 + x}} = \frac{{20\% }}{{80\% }} \cr & \Rightarrow 1400 + x = 2400 \cr} $$ ⇒ x = 1000 gms

Percentage of sugar in 4 litres mixture. $$\eqalign{ & = \frac{9}{{5 \times 4}} \times 100 \cr & = 45\% \cr} $$

According to the question, Selling price of the mixture = Rs. 320, Gain = 20% ∴ Cost price of the mixture $$\eqalign{ & = 320 \times \frac{{100}}{{120}} \cr & = {\text{Rs}}{\text{. }}\frac{{800}}{3} \cr} $$ Now using alligation method $$\eqalign{ & {\text{Ratio of quantity }} \cr & \to 40:260 \cr & \,\,\,\,\,\,\,\,\,\,\,2:13 \cr} $$

Let the quantity of liquid drawn out = x $$\eqalign{ & \Rightarrow \frac{{3 - \frac{3}{4}x}}{{1 - \frac{1}{4}x + x}} = \frac{1}{1} \cr & \Rightarrow 12 - 3x = 4 - x + 4x \cr & \Rightarrow 8 = 6x \cr & \Rightarrow x = \frac{4}{3} \cr} $$ Hence, required part of quantity $$\eqalign{ & {\text{ = }}\frac{{\frac{4}{3}}}{4} \cr & {\text{ = }}\frac{1}{3} \cr} $$

According to the question, \[\left. \begin{gathered} {\text{Alloy}}\,{\text{A}} \to 5{ \times _2}:3{ \times _2} = 8{ \times _2} \hfill \\ {\text{Alloy}}\,{\text{B}} \to 5\,\,\,\,\,\,\,\,:11\,\,\,\,\, = 16 \hfill \\ \end{gathered} \right]\] Equal quantity are mixed $$\eqalign{ & {\text{Alloy A }} \to 10\,\,\,:\,\,\,6\,\,\,\, = 16 \cr & {\text{Alloy B }} \to {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} \,5\,\,\,\,\,:\,\,\,11\, = 16 \cr & {\text{Alloy C }} \to {\bf{15}}\,\,\,:\,\,{\bf{17}} \cr} $$

Let first blend is 2 kg and second blend is 3 kg. $$\eqalign{ & {\text{Total cost price }}{\text{ }} \cr & = \left( {35 \times 2} \right) + \left( {40 \times 3} \right) \cr & = 70 + 120 \cr & = {\text{Rs}}{\text{. 190}} \cr & {\text{Total selling price }} \cr & {\text{ = }}\left( {{\text{ }}1 \times 46} \right) + \left( {4 \times 55} \right) \cr & = 266\left[ {\frac{1}{5}{\text{ of 5kg = 1kg}}} \right] \cr & \therefore {\text{Profit percent }} \cr & \Rightarrow \frac{{{\text{Total profit}}}}{{{\text{Total cost price}}}} \times 100 \cr & \Rightarrow \frac{{\left( {266 - 190} \right)}}{{190}} \times 100 \cr & \Rightarrow \frac{{76}}{{190}} \times 100 \cr & \Rightarrow 40\% \cr} $$

$$\eqalign{ & {\text{Number of girls}} \cr & = \frac{1}{2} \times 12500 \cr & = 6250 \cr} $$

Quantity of tin in 60 kg of A $$\eqalign{ & = \left( {60 \times \frac{2}{5}} \right)\,{\text{kg}} \cr & = 24\,{\text{kg}} \cr} $$ Quantity of tin in 100 kg of B $$\eqalign{ & = \left( {100 \times \frac{1}{5}} \right)\,{\text{kg}} \cr & = 20\,{\text{kg}} \cr} $$ ∴ Quantity of tin in the new alloy = (24 + 20) kg = 44 kg

According to the question, $$\eqalign{ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\text{Milk : Water}} \cr & {\text{Initial Ratio}}\,\,\,\,\,{{\text{7}}_{ \times 2}}:{3_{ \times 2}} \cr & {\text{Final Ratio}}\,\,\,\,\,\,\,{{\text{2}}_{ \times 7}}:{1_{ \times 7}} \cr} $$ ∴ Remember water is added not milk, so make milk equal \[\begin{gathered} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\text{M}}\,\,\,\,{\text{:}}\,\,\,\,{\text{W}} \hfill \\ \left. \begin{gathered} {\text{Initial}}\,\,\,\,14\,\,\,\,:\,\,\,\,6 \hfill \\ {\text{Final}}\,\,\,\,\,\,14\,\,\,\,:\,\,\,\,7 \hfill \\ \end{gathered} \right) \hfill \\ \end{gathered} \] $$\eqalign{ & \cr & = \,20\,{\text{unit}} \cr & 1 \cr} $$ $$\eqalign{ & {\text{20 units = 80 litres}} \cr & {\text{1 unit = 4 litres}} \cr & {\text{Water added = 4 litres}} \cr} $$

In 300 kg of solution, Percentage of sugar = 40% ∴ Sugar = $$\frac{{300 \times 40}}{{100}}$$ = 120 kg Let x kg of sugar be added. Now, according to the question $$\eqalign{ & \frac{{120 + x}}{{300 + x}} = \frac{1}{2} \cr & \Rightarrow 240 + 2x = 300 + x \cr & \Rightarrow 2x - x = 300 - 240 \cr & \Rightarrow x = 60\,{\text{kg}} \cr} $$

According to the question, $${\text{Milk}}\,\,\,\,:\,\,\,\,{\text{Water}}$$ \[\left. \begin{gathered} \,\,\,\,\,7\,\,\,\,\,\,\,\,:\,\,\,\,\,\,\,5 \hfill \\ \,\,\,\,\,7\,\,\,\,\,\,\,\,:\,\,\,\,\,\,\,8 \hfill \\ \end{gathered} \right)\,3\,{\text{unit}}\] ∴ Remember water is added and not milk, so make milk equal but here milk is already equal $$\eqalign{ & {\text{3 units = 15 litres}} \cr & {\text{1 unit = 5 litres}} \cr & {\text{8 units = 40 litres}} \cr} $$ Total quantity of water in the new mixture = 40 litres

According to the question, By using alligation method : Ratio of quantity $$ \to \boxed{\,\,1\,\,\,\,\,\,:\,\,\,\,\,\,3\,\,}$$

According to the question, Average price of mixed tea $$\eqalign{ & {\text{ = }}\frac{{280 \times 7 + 240 \times 9}}{{16}} \cr & {\text{ = }}\frac{{1960 + 2160}}{{16}} \cr & {\text{ = }}\frac{{4120}}{{16}} \cr & {\text{ = Rs}}{\text{. 257}}{\text{.50}} \cr} $$

Let the price of water = Rs. 0 According to the question, Cost price of pure milk = Rs. 1.80 Cost price of the mixture = Rs. 0.90 Now using Alligation method. Therefore the amount of water is equal to the amount of Milk in mixture. i.e. 16 litter Milk in Given Mixture

According to the question, Cost price of the mixture = 30 × 9.50 + 40 × 8.50 = 285 + 340 = Rs. 625 Selling price of the mixture = 8.90 × 70 = Rs. 623 Loss = Cost price - Selling price Loss = 625 - 623 = Rs. 2

Capacities of vessels $$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,3:2:1$$ $$\eqalign{ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\text{M}}\,{\text{:}}\,{\text{W}}\,\,\,\,\,\,\,\,\,\,\,\,\,{\text{T}}\,\,{\text{Mixture}} \cr & {\text{V - 1}} \to {{\text{(}}5\,\,\,:2\,\,\, = \,\,\,\,\,7{\text{)}}_{ \times 5}} \cr & {\text{V - 2}} \to {{\text{(4}}\,\,\,:1\,\,\, = \,\,\,\,\,5{\text{)}}_{ \times 7}} \cr & {\text{V - 3}} \to {{\text{(4}}\,\,\,:1\,\,\, = \,\,\,\,\,5{\text{)}}_{ \times 7}} \cr} $$ Equate the mixture $$\eqalign{ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\text{M}}\,{\text{:}}\,{\text{W}}\,\,\,\,\,\,\,\,\,\,{\text{T}}\,\,{\text{Mixture}} \cr & \left( {{\text{V - 1}}} \right) \to 25:10\,\,\,\,\,\,\, = 35 \cr & \left( {{\text{V - 2}}} \right) \to 28:7\,\,\,\,\,\,\,\, = 35 \cr & \left( {{\text{V - 3}}} \right) \to 28:7\,\,\,\,\,\,\,\, = 35 \cr} $$ $$\eqalign{ & {\text{Capacities}}\,\,\,\,\,\,\,\,\,\,\,\,\,{\text{M}}:{\text{W}}\,\,\, = {\text{Total Mix}}{\text{.}} \cr & \left( {{\text{V}} - 1} \right) \times 3\, \to 75:30\,\,\, = 105 \cr & \left( {{\text{V}} - 2} \right) \times 2 \to 56:14\,\,\, = 70 \cr & \left( {{\text{V}} - 1} \right) \times 1\, \to 28:7\,\,\,\, = 35 \cr} $$ Water taken out $$ \Rightarrow \frac{1}{3}{\text{ of water in (V - 1)}}$$ $$ + \frac{1}{2}{\text{ of water in (V - 2)}}$$ $$ + \frac{1}{7}{\text{ of water in (V - 3)}}$$ $$\eqalign{ & \Rightarrow \frac{1}{3} \times 30 + \frac{1}{2} \times 14 + \frac{1}{7} \times 7 \cr & \Rightarrow 10 + 7 + 1 \cr & \Rightarrow 18 \cr} $$ Similarly mixture will be $$ \Rightarrow \frac{1}{3} \times 105 + \frac{1}{2} \times 70 + \frac{1}{7}$$ × 35 ⇒ 75 $$\eqalign{ & \therefore \% {\text{ of water = }}\frac{{18}}{{75}} \times 100 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 24\% \cr} $$

Let them be mixed in the ratio x : y Then, in 1 st alloy, Zinc = $$\frac{x}{3}$$ and Copper = $$\frac{{2x}}{3}$$ 2 nd alloy, Zinc = $$\frac{{2y}}{5}$$ and Copper = $$\frac{{3y}}{5}$$ Now, we have $$\eqalign{ & \frac{x}{3} + \frac{{2y}}{5}:\frac{{2x}}{3} + \frac{{3y}}{5} = 5:8 \cr & {\text{or,}}\frac{{5x + 6y}}{{10x + 9y}} = \frac{5}{8} \cr & {\text{or,}}40x + 48y = 50x + 45y \cr & {\text{or,}}10x = 3y \cr & \therefore \frac{x}{y} = \frac{3}{{10}} \cr} $$ Thus, the required ratio = 3 : 10 Alligation Method : You must know that we can apply this rule over the fractional value of either zinc or copper. Let us consider the fractional value of zinc. Therefore, they should be mixed in the ratio $$\eqalign{ & = \frac{1}{{65}}:\frac{2}{{39}} \cr & {\text{or,}}\frac{1}{{65}} \times \frac{{39}}{2} \cr & = \frac{3}{{10}} \cr & {\text{or,}}\,\,3:10 \cr} $$

Let the quantity of alcohol and water be 4x litres and 3x litres respectively $$\eqalign{ & \frac{{4x}}{{\left( {3x + 5} \right)}} = \frac{4}{5} \cr & \Rightarrow 20x = 4\left( {3x + 5} \right) \cr & \Rightarrow 8x = 20 \cr & \Rightarrow x = 2.5 \cr} $$ Quantity of alcohol = (4 × 2.5) litres = 10 litres.

According to the question, When there is no profit no loss, Here, Cost price = Selling price ∴ Cost price of mixed sugar = Rs. 16/kg Apply alligation $$ \Rightarrow {\text{Ratio }} \to 1:4$$

According to the question, \[\begin{gathered} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\text{Acid}}\,\,\,\,\,\,\,\,\,\,\,\,\,{\text{Water}}\,\,\,\,\,\,{\text{Total}} \hfill \\ \left. \begin{gathered} {\text{Sample I}}\,\,\,\,\,\,\,\,{{\text{2}}_{ \times 40}} = 80\,\,\,\,:\,\,\,\,{1_{ \times 40}} = 40\,\,\,\,\,\,{3_{ \times 40}} \hfill \\ \hfill \\ {\text{Sample II}}\,\,\,\,\,\,{3_{ \times 24}} = 72\,\,\,\,\,:\,\,\,\,{2_{ \times 24}} = 48\,\,\,\,{5_{ \times 24}} \hfill \\ \hfill \\ {\text{Sample III}}\,\,\,\,\,\underline {{5_{ \times 15}} = 75} \,\,\,\,:\,\,\,\,\,\underline {{3_{ \times 15}} = 45} \,\,\,\,{8_{ \times 15}} \hfill \\ \end{gathered} \right]120 \hfill \\ \end{gathered} \] \[{\text{Ratio of quantity }}\boxed{\,\,\,227\,\,\,\,\,:\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,133\,\,\,}\] $${\text{Required ratio = }}133:227$$

According to the question, Cost price of the mixture = 15 × 29 + 25 × 20 = Rs. 935 Selling price of the mixture = 27 × 40 = Rs. 1080 Profit = Selling price - Cost price Profit = 1080 - 935 = Rs. 145

By alligation method - Then, Lala give loan in ratio 2 : 1, to Arun and Bhatia

∴According to the question, Container : $$\eqalign{ & \,\,\,\,\,\,\,\,{\text{I}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\text{II}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\text{III}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\text{IV}} \cr & {\text{M}}\,:\,{\text{W}}\,\,\,{\text{M}}\,:\,{\text{W}}\,\,\,{\text{M}}\,:\,{\text{W}}\,\,\,{\text{M}}\,:\,{\text{W}} \cr & \,\,\,\,5:3\,\,\,\,\,\,\,\,\,\,\,2:1\,\,\,\,\,\,\,\,\,\,\,3:2\,\,\,\,\,\,\,\,\,\,7:4 \cr} $$ $$\eqalign{ & {\bf{Container}}\,\,({\text{I}}): \cr & \frac{{{\text{Milk}}}}{{{\text{Water}}}} = \frac{5}{3} = 1.67 \cr & \cr & {\bf{Container}}\,\,({\text{II}}): \cr & \frac{{{\text{Milk}}}}{{{\text{Water}}}} = \frac{2}{1} = 2 \cr & \cr & {\bf{Container}}\,\,({\text{III}}): \cr & \frac{{{\text{Milk}}}}{{{\text{Water}}}} = \frac{3}{2} = 1.5 \cr & \cr & {\bf{Container}}\,\,({\text{IV}}): \cr & \frac{{{\text{Milk}}}}{{{\text{Water}}}} = \frac{7}{4} = 1.75 \cr} $$ ∴ The quantity of milk relative to water minimum in container third.

According to the question, \[\begin{gathered} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\text{Chromium}}\,\,{\text{Steel}} \hfill \\ {\text{Type 1}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,2\,\,\,\,:\,\,\,\,\,11 \hfill \\ {\text{Type 2}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,5\,\,\,\,{\text{:}}\,\,\,\,\,\,\,21 \hfill \\ \end{gathered} \] Now using alligation, \[{\text{Ratio of quantity }} \to \boxed{\,\,1\,\,\,\,\,\,:\,\,\,\,\,\,2\,\,}\]

In 2 litres of first container, Spirit = $$2 \times \frac{4}{5} = \frac{8}{5}$$ water = $$2 \times \frac{1}{5} = \frac{2}{5}$$ In 3 litres of first container, Spirit =$$3 \times \frac{11}{15} = \frac{11}{5}$$ water = $$3 \times \frac{4}{15} = \frac{4}{5}$$ In 4 litres of first container, Spirit =$$4 \times \frac{7}{10} = \frac{14}{5}$$ water = $$4 \times \frac{3}{10} = \frac{6}{5}$$ ∴ Required ratio $$ = \left( {\frac{8}{5} + \frac{{11}}{5} + \frac{{14}}{5}} \right) : \left( {\frac{2}{5} + \frac{4}{5} + \frac{6}{5}} \right) $$ $$ = \frac{{33}}{5} : \frac{{12}}{5} $$ = 33 : 12 = 11 : 4

According to the question, $$\eqalign{ & \,\,\,\,\,\,\,\,\,\,\,\,\,{\text{Alcohol}}\,\,\,\,\,\,{\text{Water}} \cr & {\text{Vessel A}}\,\,\,\,\,\,\,\,5\,\,\,\,:\,\,\,\,\,{\text{3}} \cr & {\text{Vessel B}}\,\,\,\,\,\,\,\,5\,\,\,\,{\text{:}}\,\,\,\,\,4 \cr} $$ Now using alligation,

According to the question, $$\eqalign{ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\text{Acid}}\,\,\,\,\,\,\,{\text{Water}} \cr & {\text{Vessel A}}\,\,\,\,\,{\text{4}}\,\,\,\,:\,\,\,\,{\text{3}} \cr & {\text{Vessel B}}\,\,\,\,\,\,{\text{5}}\,\,\,\,{\text{:}}\,\,\,\,\,{\text{3}} \cr} $$ Now using alligation,

$$\eqalign{ & {\text{According to the question,}} \cr & {\text{Mixture }} - {\text{1 = 20 litres}} \cr & {\text{Mixture }} - {\text{2 = 36 litres}} \cr & {\text{In Mixture }} - {\text{1 ratio of }} \cr & {\text{ }}\frac{{{\text{Spirit}}}}{{{\text{Water}}}} = \left. {\frac{3}{7}} \right\rangle 10{\text{ units}} \cr & {\text{In Mixture }} - 2{\text{ ratio of}} \cr & {\text{ }}\frac{{{\text{Spirit}}}}{{{\text{Water}}}} = \left. {\frac{7}{5}} \right\rangle 12{\text{ units}} \cr & \cr & {\text{10 units }} \to {\text{20 litres}} \cr & \,\,\,{\text{1 unit }}\,\,\, \to {\text{2 litres}} \cr & {\text{12 units }} \to {\text{36 litres}} \cr & \,\,\,{\text{1 unit }}\,\,\, \to {\text{3 litres}} \cr & \cr & \therefore {\text{In Mixture }} - {\text{1 ratio of }} \cr & \frac{{{\text{Spirit}}}}{{{\text{Water}}}} = \frac{{3 \times 2}}{{7 \times 2}} = \frac{6}{{14}} \cr & \cr & \therefore {\text{In Mixture }} - 2{\text{ ratio of }} \cr & \frac{{{\text{Spirit}}}}{{{\text{Water}}}} = \frac{{7 \times 3}}{{5 \times 3}} = \frac{{21}}{{15}} \cr & \cr & {\text{Ratio of spirits and water}} \cr & {\text{ = }}\frac{{6 + 21}}{{14 + 15}} \cr & {\text{ = }}\frac{{27}}{{29}} \cr & {\text{ = 27:29}} \cr} $$

According to the question, $$\eqalign{ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\text{Acid}}\,\,\,\,\,\,{\text{Water}} \cr & {\text{Vessel A}}\,\,{\text{4}}\,\,\,:\,\,\,\,{\text{3}} \cr & {\text{Vessel B}}\,\,\,2\,\,\,\,{\text{:}}\,\,\,\,\,{\text{3}} \cr} $$ Now using alligation,

\[\begin{array}{*{20}{c}} {}&{{\text{Acid}}}&{{\text{Water}}} \\ {{\text{Old}} \to }&{36\% }&{64\% } \\ {{\text{New}} \to }&{30\% }&{70\% } \end{array}\] $$\eqalign{ & {\text{Old mixture}} = {\text{Remaining}} + {\text{Taken out}} \cr & = 20{\text{L}} + 4{\text{L}} \cr & = 24{\text{L}} \cr & {\text{Water}} = 24 \times \frac{{64}}{{100}} \Rightarrow 15.36{\text{L}} \cr} $$

The ratio of copper and tin = 3x : 2x According the question, 3x + 250 = 2(2x) 3x + 250 = 4x x = 250 ∴ Quantity of tin in the alloy = 2x = 2 × 250 = 500 gm

$$\eqalign{ & {\text{After removed 12L}} \cr & \,\,\,\,\,\,\,{\text{Acid}}\,\,\,\,\,\,\,\,:\,\,\,\,\,\,\,{\text{Water}} \cr & \Rightarrow 12 \times \frac{5}{8},\,12\, \times \frac{3}{8} \cr & \Rightarrow - 7.5,\, - 4.5 \cr & {\text{Acid}}\,\,\,\,\,\,\,\,\,\,{\text{Water}} \cr & \,\,\,{\text{20L }}\,\,\,\,\,\,\,\,\,\,{\text{ 12L}} \cr & - 7.5{\text{ }}\,\,\,\,{\text{ }} - 4.5 \cr & {\text{ }}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\text{ }} + 7.5 \cr & \overline {\,12.5\,\,\,\,\,:\,\,\,\,\,\,15\,\,} \cr & \,\,\,\,\,\,5\,\,\,\,\,\,\,\,:\,\,\,\,\,\,\,\,6 \cr} $$

2 : 3 is the correct answer to “The price of a variety of a commodity is Rs. 7 per kg and that of another is Rs. 12 per kg. Find the ratio in which two varieties should be mixed so that the price of the mixture is Rs. 10 per kg.”. This question is standard in competitive examination syllabi for Alligation.

\[\begin{array}{*{20}{c}} {{\text{Total capacity}}}&{{\text{Milk}}}&{}&{{\text{Water}}} \\ {15}&9&:&6 \\ {}&3&:&2 \end{array}\] Half liter is taken out Milk = 500 ml × $$\frac{3}{5}$$ = 300 ml

3 : 1 Answer

After replacing 20 litre of liquid for first time, ratio of \[\begin{array}{*{20}{c}} {{\text{Ethanol}}}&:&{{\text{Water}}} \\ {60}&:&{20} \end{array}\] Now, again removing 20 litre of the mixture in the given proportion. \[\begin{array}{*{20}{c}} {{\text{Ethanol}}}&:&{{\text{Water}}} \\ {45}&:&{15} \end{array}\] Now, an adding 20 litre water Total water in the drum = 15 + 20 = 35 litre $$\eqalign{ & {\bf{Alternate}}\,{\bf{Solution:}} \cr & {\text{Final quantity}} = {\text{Initial quantity}} \times {\left( {1 - \frac{x}{c}} \right)^n} \cr & x \to {\text{ quantity taken out}} \cr & c \to {\text{ total capacity}} \cr & n \to {\text{ number of process}} \cr & = 80 \times {\left( {1 - \frac{{20}}{{80}}} \right)^2} \cr & = 80 \times {\left( {\frac{3}{4}} \right)^2} \cr & = 80 \times \frac{3}{4} \times \frac{3}{4} \cr & {\text{Final quantity of external}} = 45{\text{ litre}} \cr & {\text{and Water}} = 80 - 45 = 35{\text{ litre}} \cr} $$

$$\eqalign{ & {\text{I}} \to \frac{{40 \times 60}}{{100}} = 24 \cr & {\text{II}} \to \frac{{35 \times 80}}{{100}} = 28 \cr & {\text{Acid }}\% = \frac{{52}}{{75}} \times 100 = 69\frac{1}{3}\% \cr} $$

$$\eqalign{ & {\text{Let third variety}} = x \cr & 80\,\,:\,\,50\,\,:\,\,x \cr & \,\,1\,\,\,:\,\,\,\,2\,\,\,:\,\,\,3 \cr & \overline {\,80 + 100 + 3x\,} = \left( {3 + 2 + 1} \right) \times 75 \cr & 180 + 3x = 6 \times 75 \cr & 3x = 270 \cr & x = 90{\text{ per kg}} \cr} $$