Download Mixture and Alligation Questions for CAT<\/a><\/p>\n CAT Online Coaching<\/a><\/p>\n Question 1:\u00a0<\/b>A milkman mixes 20 litres of water with 80 litres of milk. After selling one-fourth of this mixture, he adds water to replenish the quantity that he had sold. What is the current proportion of water to milk? a)\u00a02 : 3<\/p>\n b)\u00a01 : 2<\/p>\n c)\u00a01 : 3<\/p>\n d)\u00a03 : 4<\/p>\n 1)\u00a0Answer\u00a0(A)<\/strong><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n After selling 1\/4th of the mixture, the remaining quantity of water is 15 liters and milk is 60 liters. So the milkman would add 25 liters of water to the mixture. The total amount of water now is 40 liters and milk is 60 liters. Therefore, the required ratio is 2:3.<\/p>\n Instructions<\/b><\/p>\n DIRECTIONS for the following two questions: The following table presents the sweetness of different items relative to sucrose, whose sweetness is taken to be 1.00.<\/p>\n Question 2:\u00a0<\/b>Approximately how many times sweeter than sucrose is a mixture consisting of glucose, sucrose and fructose in the ratio of 1: 2: 3?<\/p>\n a)\u00a01.3<\/p>\n b)\u00a01<\/p>\n c)\u00a00.6<\/p>\n d)\u00a02.3<\/p>\n 2)\u00a0Answer\u00a0(A)<\/strong><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n The relative sweetness of the mixture is (1*0.74 + 2*1 + 3*1.7) \/ (1+2+3) = 7.84\/6 = 1.30<\/p>\n Option a) is the correct answer.<\/p>\n Question 3:\u00a0<\/b>Bottle 1 contains a mixture of milk and water in 7: 2 ratio and Bottle 2 contains a mixture of milk and water in 9: 4 ratio. In what ratio of volumes should the liquids in Bottle 1 and Bottle 2 be combined to obtain a mixture of milk and water in 3:1 ratio?<\/p>\n a)\u00a027:14<\/p>\n b)\u00a027:13<\/p>\n c)\u00a027:16<\/p>\n d)\u00a027:18<\/p>\n 3)\u00a0Answer\u00a0(B)<\/strong><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n The ratio of milk and water in Bottle 1 is 7:2 and the ratio of milk and water in Bottle 2 is 9:4 Let the ratio in which they should be mixed be equal to X:1.<\/p>\n Hence, the total volume of milk is $\\frac{7X}{9}+\\frac{9}{13}$ Hence,\u00a0$\\frac{7X}{9}+\\frac{9}{13} = 3*(\\frac{2X}{9}+\\frac{4}{13})$ Therefore $X = \\frac{27}{13}$<\/p>\n Question 4:\u00a0<\/b>Consider three mixtures \u2014 the first having water and liquid A in the ratio 1:2, the second having water and liquid B in the ratio 1:3, and the third having water and liquid C in the ratio 1:4. These three mixtures of A, B, and C, respectively, are further mixed in the proportion 4: 3: 2. Then the resulting mixture has<\/p>\n a)\u00a0The same amount of water and liquid B<\/p>\n b)\u00a0The same amount of liquids B and C<\/p>\n c)\u00a0More water than liquid B<\/p>\n d)\u00a0More water than liquid A<\/p>\n 4)\u00a0Answer\u00a0(C)<\/strong><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n The proportion of water in the first mixture is $\\frac{1}{3}$ The proportion of water in the second mixture is $\\frac{1}{4}$ The proportion of water in the third mixture is $\\frac{1}{5}$ As they are mixed in the ratio 4:3:2, the final amount of water is $4 \\times \\frac{1}{3} + 3 \\times \\frac{1}{4} + 2 \\times \\frac{1}{5} = \\frac{149}{60}$ Hence, the ratio of Water : A : B : C in the final mixture is $\\frac{149}{60}:\\frac{8}{3}:\\frac{9}{4}:\\frac{8}{5} = 149:160:135:96$<\/p>\n From the given choices, only option C \u00a0is correct.<\/p>\n Question 5:\u00a0<\/b>Two types of tea, A and B, are mixed and then sold at Rs. 40 per kg. The profit is 10% if A and B are mixed in the ratio 3 : 2, and 5% if this ratio is 2 : 3. The cost prices, per kg, of A and B are in the ratio<\/p>\n a)\u00a017 : 25<\/p>\n b)\u00a018 : 25<\/p>\n c)\u00a019 : 24<\/p>\n d)\u00a021 : 25<\/p>\n 5)\u00a0Answer\u00a0(C)<\/strong><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n The selling price of the mixture is Rs.40\/kg. The profit is 5% if the 2 varieties are mixed in the ratio 2:3. Equating (1) and (2), we get, Therefore, option C is the right answer.<\/p>\n Question 6:\u00a0<\/b>A trader sells 10 litres of a mixture of paints A and B, where the amount of B in the mixture does not exceed that of A. The cost of paint A per litre is Rs. 8 more than that of paint B. If the trader sells the entire mixture for Rs. 264 and makes a profit of 10%, then the highest possible cost of paint B, in Rs. per litre, is<\/p>\n a)\u00a016<\/p>\n b)\u00a026<\/p>\n c)\u00a020<\/p>\n d)\u00a022<\/p>\n 6)\u00a0Answer\u00a0(C)<\/strong><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n Let the price of paint B be x. We know that the amount of paint B in the mixture does not exceed the amount of paint A. Therefore, paint B can at the maximum compose 50% of the mixture.<\/p>\n The seller sells 10 litres of paint at Rs.264 earning a profit of 10%. Therefore, the cost of 1 litre of the mixture = Rs.24<\/p>\n We have to find the highest possible cost of paint B. => (x+x+8)\/2 = 24 Therefore, option\u00a0C is the right answer.<\/p>\n Question 7:\u00a0<\/b>A 20% ethanol solution is mixed with another ethanol solution, say, S of unknown concentration in the proportion 1:3 by volume. This mixture is then mixed with an equal volume of 20% ethanol solution. If the resultant mixture is a 31.25% ethanol solution, then the unknown concentration of S is<\/p>\n a)\u00a030%<\/p>\n b)\u00a040%<\/p>\n c)\u00a050%<\/p>\n d)\u00a060%<\/p>\n 7)\u00a0Answer\u00a0(C)<\/strong><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n Let the volume of the first and the second solution be 100 and 300. Question 8:\u00a0<\/b>A jar contains a mixture of 175 ml water and 700 ml alcohol. Gopal takes out 10% of the mixture and substitutes it by water of the same amount. The process is repeated once again. The percentage of water in the mixture is now<\/p>\n a)\u00a030.3<\/p>\n b)\u00a035.2<\/p>\n c)\u00a025.4<\/p>\n d)\u00a020.5<\/p>\n 8)\u00a0Answer\u00a0(B)<\/strong><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n Final quantity of\u00a0alcohol in the mixture = $\\dfrac{700}{700+175}*(\\dfrac{90}{100})^2*[700+175]$ = 567 ml<\/p>\n Therefore, final quantity of water in the mixture = 875 – 567 = 308 ml<\/p>\n Hence, we can say that the percentage of water in the mixture = $\\dfrac{308}{875}\\times 100$ = 35.2 %<\/p>\n Question 9:\u00a0<\/b>There are two drums, each containing a mixture of paints A and B. In drum 1, A and B are in the ratio 18 : 7. The mixtures from drums 1 and 2 are mixed in the ratio 3 : 4 and in this final mixture, A and B are in the ratio 13 : 7. In drum 2, then A and B were in the ratio<\/p>\n a)\u00a0251 : 163<\/p>\n b)\u00a0239 : 161<\/p>\n c)\u00a0220 : 149<\/p>\n d)\u00a0229 : 141<\/p>\n
\n[CAT 2004]<\/p>\n![\"\"](\"https:\/\/cracku.in\/media\/uploads\/2015\/11\/18\/capture.PNG\")
<\/p>\n
\nTherefore, the proportion of milk in Bottle 1 is $\\frac{7}{9}$ and the proportion of milk in Bottle 2 is $\\frac{9}{13}$<\/p>\n
\nThe total volume of water is $\\frac{2X}{9}+\\frac{4}{13}$
\nThey are in the ratio 3:1<\/p>\n
\nTherefore, $91X+81=78X+108$<\/p>\n
\nThe proportion of Liquid A in the first mixture is $\\frac{2}{3}$<\/p>\n
\nThe proportion of Liquid B in the second mixture is $\\frac{3}{4}$<\/p>\n
\nThe proportion of Liquid C in the third mixture is $\\frac{4}{5}$<\/p>\n
\nThe final amount of Liquid A in the mixture is $4\\times\\frac{2}{3} = \\frac{8}{3}$
\nThe final amount of Liquid B in the mixture is $3\\times\\frac{3}{4} = \\frac{9}{4}$
\nThe final amount of Liquid C in the mixture is $2\\times\\frac{4}{5} = \\frac{8}{5}$<\/p>\n
\nLet a be the price of 1 kg of tea A in the mixture and b be the price per kg of tea B.
\nIt has been given that the profit is 10% if the 2 varieties are mixed in the ratio 3:2
\nLet the cost price of the mixture be x.
\nIt has been given that 1.1x = 40
\nx = 40\/1.1
\nPrice per kg of the mixture in ratio 3:2 =\u00a0$\\frac{3a+2b}{5}\u00a0$
\n$\\frac{3a+2b}{5} = \\frac{40}{1.1}$
\n$3.3a+2.2b=200$ ——–(1)<\/p>\n
\nPrice per kg of the mixture in ratio 2:3 =\u00a0$\\frac{2a+3b}{5}$
\n$\\frac{2a+3b}{5} = \\frac{40}{1.05}$
\n$2.1a+3.15b=200$\u00a0——(2)<\/p>\n
\n$3.3a+2.2b = 2.1a+3.15b$
\n$1.2a=0.95b$
\n$\\frac{a}{b} = \\frac{0.95}{1.2}$
\n$\\frac{a}{b} = \\frac{19}{24}$<\/p>\n
\nPrice of paint A = x+8<\/p>\n
\n=> The cost price of 10 litres of the paint mixture = Rs. 240<\/p>\n
\nWhen we increase the cost of paint B, the cost of paint A will increase too. If the cost price of the mixture is closer to the cost of paint B, then the amount of paint B present in the mixture should be greater than the amount of paint A present in the mixture.
\nThe highest possible cost of paint B will be obtained when the volumes of paint A and paint B in the mixture are equal.<\/p>\n
\n2x = 40
\nx = Rs. 20<\/p>\n
\nWhen they are mixed, quantity of ethanol in the mixture
\n= (20 + 300S)
\nLet this solution be mixed with equal volume i.e. 400 of third solution in which the strength of ethanol is 20%.
\nSo, the quantity of ethanol in the final solution
\n= (20 + 300S + 80) = (300S + 100)
\nIt is given that, 31.25% of 800 = (300S + 100)
\nor, 300S + 100 = 250
\nor S = $\\frac{1}{2}$ = 50%
\nHence, 50 is the correct answer.<\/p>\n