Mixture and Alligation is an important topic in the Quant section of the SNAP Exam. You can also download this Free Mixture and Alligation Questions for SNAP PDF (with answers) by Cracku. These questions will help you to practice and solve the Mixture and Alligation questions in the SNAP exam. Utilize this PDF practice set, <\/strong>which is one of the best sources for practising.<\/p>\n Download Mixture and Alligation Questions for SNAP<\/a><\/p>\n Enroll to SNAP 2022 Crash Course<\/a><\/p>\n Question 1:\u00a0<\/b>A vessel contains a mixture of milk and water in the respective ratio of 14 : 3. 25.5 litres of the mixture is taken out from the vessel and 2.5 litres of pure water and 5 litres of pure milk is added to the mixture. If the resultant mixture contains 20% water, what was the initial quantity of mixture in the vessel before the replacement? (in litres)<\/p>\n a)\u00a051<\/p>\n b)\u00a0102<\/p>\n c)\u00a068<\/p>\n d)\u00a085<\/p>\n e)\u00a034<\/p>\n 1)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let the total quantity of mixture in the vessel initially = $17x$ litres<\/p>\n => Quantity of milk = $\\frac{14}{17} \\times 17x = 14x$ litres<\/p>\n Quantity of water = $17x – 14x = 3x$ litres<\/p>\n Acc. to ques,<\/p>\n => $\\frac{14x – (\\frac{14}{17} \\times 25.5) + 5}{3x – (\\frac{3}{17} \\times 25.5) + 2.5} = \\frac{80}{20}$<\/p>\n => $\\frac{14x – 21 + 5}{3x – 4.5 + 2.5} = \\frac{4}{1}$<\/p>\n => $\\frac{14x – 16}{3x – 2} = \\frac{4}{1}$<\/p>\n => $14x – 16 = 12x – 8$<\/p>\n => $14x – 12x = 16 – 8$<\/p>\n => $x = \\frac{8}{2} = 4$<\/p>\n $\\therefore$ Initial quantity of mixture in the vessel before the replacement = $17 \\times 4 = 68$ litres<\/p>\n Question 2:\u00a0<\/b>A vessel contains 100 litres mixture of milk and water in the respective ratio of 22 : 3. 40 litres of the mixture is taken out from the vessel and 4.8 litres of pure milk and pure water each is added to the mixture. By what percent is the quantity of water in the final mixture less than the quantity of milk?<\/p>\n a)\u00a078${1 \\over 2}$<\/p>\n b)\u00a079${1 \\over 6}$<\/p>\n c)\u00a072${5 \\over 6}$<\/p>\n d)\u00a076<\/p>\n e)\u00a077${1 \\over 2}$<\/p>\n 2)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n Quantity of milk in vessel = $\\frac{22}{25} \\times 100 = 88$ litres<\/p>\n => Quantity of water = $100 – 88 = 12$ litres<\/p>\n 40 litres of the mixture is taken out, i.e., $\\frac{40}{100} = (\\frac{2}{5})^{th}$<\/p>\n => Milk left = $88 – \\frac{2}{5} \\times 88 = 52.8$ litres<\/p>\n Water left =\u00a0$12 – \\frac{2}{5} \\times 12 = 7.2$ litres<\/p>\n Now, 4.8 lires of milk and water are added.<\/p>\n => Quantity of milk in the vessel = 52.8 + 4.8 = 57.6 litres<\/p>\n Quantity of water in the vessel = 7.2 + 4.8 = 12 litres<\/p>\n $\\therefore$ Required % = $\\frac{57.6 – 12}{57.6} \\times 100$<\/p>\n = $\\frac{475}{6} = 79 \\frac{1}{6} \\%$<\/p>\n Question 3:\u00a0<\/b>Jar A has 60 litres of mixture of milk and water in the respective ratio of 2 : 1. Jar B which had 40 litres of mixture of milk and water was emptied into jar A, as a result in jar A, the respective ratio of milk and water became 13 : 7. What was the quantity of water in jar B?<\/p>\n a)\u00a08 litres<\/p>\n b)\u00a015 litres<\/p>\n c)\u00a022 litres<\/p>\n d)\u00a07 litres<\/p>\n e)\u00a01 litre<\/p>\n 3)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n Jar A has 60 litres of mixture of milk and water in the respective ratio of 2 : 1<\/p>\n => Quantity of milk in Jar A = $\\frac{2}{3} \\times 60 = 40$ litres<\/p>\n Quantity of water in Jar A = $60 – 40 = 20$ itres<\/p>\n Let quantity of water in Jar B = $x$ litres<\/p>\n => Quantity of milk in Jar B = $(40 – x)$ litres<\/p>\n Acc. to ques, => $\\frac{40 + (40 – x)}{20 + x} = \\frac{13}{7}$<\/p>\n => $560 – 7x = 260 + 13x$<\/p>\n => $13x + 7x = 560 – 260$<\/p>\n => $20x = 300$<\/p>\n => $x = \\frac{300}{20} = 15$ litres<\/p>\n Question 4:\u00a0<\/b>Jar A has 36 litres of mixture of milk and water in the respective ratio of 5 : 4. Jar B which had 20 litres of mixture of milk and water, was emptied into jar A, and as a result in jar A, the respective ratio of milk and water becomes 5: 3. What was the quantity of water in jar B?<\/p>\n a)\u00a05 litres<\/p>\n b)\u00a03 litres<\/p>\n c)\u00a08 litres<\/p>\n d)\u00a02 litres<\/p>\n e)\u00a01 litre<\/p>\n 4)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n Jar A has 36 litres of mixture of milk and water in the respective ratio of 5 : 4<\/p>\n => Quantity of milk in Jar A = $\\frac{5}{9} \\times 36 = 20$ litres<\/p>\n Quantity of water in Jar A = $36 – 20 = 16$ itres<\/p>\n Let quantity of water in Jar B = $x$ litres<\/p>\n => Quantity of milk in Jar B = $(20 – x)$ litres<\/p>\n Acc. to ques, => $\\frac{20 + (20 – x)}{16 + x} = \\frac{5}{3}$<\/p>\n => $120 – 3x = 80 + 5x$<\/p>\n => $5x + 3x = 120 – 80$<\/p>\n => $8x = 40$<\/p>\n => $x = \\frac{40}{8} = 5$ litres<\/p>\n Question 5:\u00a0<\/b>A vessel contains 100 litres mixture of milk and water in the respective ratio of 22 : 3. 40 litres of the mixture is taken out from the vessel and 4.8 litres of pure milk and pure water each is added to the mixture. By what percent is the quantity of water in the final mixture less than the quantity of milk?<\/p>\n a)\u00a078${1 \\over 2}$<\/p>\n b)\u00a079${1 \\over 6}$<\/p>\n c)\u00a072${5 \\over 6}$<\/p>\n d)\u00a076<\/p>\n e)\u00a077${1 \\over 2}$<\/p>\n 5)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n Quantity of milk in vessel = $\\frac{22}{25} \\times 100 = 88$ litres<\/p>\n => Quantity of water = $100 – 88 = 12$ litres<\/p>\n 40 litres of the mixture is taken out, i.e., $\\frac{40}{100} = (\\frac{2}{5})^{th}$<\/p>\n => Milk left = $88 – \\frac{2}{5} \\times 88 = 52.8$ litres<\/p>\n Water left =\u00a0$12 – \\frac{2}{5} \\times 12 = 7.2$ litres<\/p>\n Now, 4.8 lires of milk and water are added.<\/p>\n => Quantity of milk in the vessel = 52.8 + 4.8 = 57.6 litres<\/p>\n Quantity of water in the vessel = 7.2 + 4.8 = 12 litres<\/p>\n $\\therefore$ Required % = $\\frac{57.6 – 12}{57.6} \\times 100$<\/p>\n = $\\frac{475}{6} = 79 \\frac{1}{6} \\%$<\/p>\n Question 6:\u00a0<\/b>Jar A has 60 litres of mixture of milk and water in the respective ratio of 2 : 1. Jar B which had 40 litres of mixture of milk and water was emptied into jar A, as a result in jar A, the respective ratio of milk and water became 13 : 7. What was the quantity of water in jar B?<\/p>\n a)\u00a08 litres<\/p>\n b)\u00a015 litres<\/p>\n c)\u00a022 litres<\/p>\n d)\u00a07 litres<\/p>\n e)\u00a01 litre<\/p>\n 6)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n Jar A has 60 litres of mixture of milk and water in the respective ratio of 2 : 1<\/p>\n => Quantity of milk in Jar A = $\\frac{2}{3} \\times 60 = 40$ litres<\/p>\n Quantity of water in Jar A = $60 – 40 = 20$ itres<\/p>\n Let quantity of water in Jar B = $x$ litres<\/p>\n => Quantity of milk in Jar B = $(40 – x)$ litres<\/p>\n Acc. to ques, => $\\frac{40 + (40 – x)}{20 + x} = \\frac{13}{7}$<\/p>\n => $560 – 7x = 260 + 13x$<\/p>\n => $13x + 7x = 560 – 260$<\/p>\n => $20x = 300$<\/p>\n => $x = \\frac{300}{20} = 15$ litres<\/p>\n Take\u00a0 SNAP mock tests here<\/strong><\/span><\/a><\/p>\n Enrol to 10 SNAP Latest Mocks For Just Rs. 499<\/a><\/strong><\/span><\/p>\n Question 7:\u00a0<\/b>Jar A has 36 litres of mixture of milk and water in the respective ratio of 5 : 4. Jar B which had 20 litres of mixture of milk and water, was emptied into jar A, and as a result in jar A, the respective ratio of milk and water becomes 5: 3. What was the quantity of water in jar B?<\/p>\n a)\u00a05 litres<\/p>\n b)\u00a03 litres<\/p>\n c)\u00a08 litres<\/p>\n d)\u00a02 litres<\/p>\n e)\u00a01 litre<\/p>\n 7)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n Jar A has 36 litres of mixture of milk and water in the respective ratio of 5 : 4<\/p>\n => Quantity of milk in Jar A = $\\frac{5}{9} \\times 36 = 20$ litres<\/p>\n Quantity of water in Jar A = $36 – 20 = 16$ itres<\/p>\n Let quantity of water in Jar B = $x$ litres<\/p>\n => Quantity of milk in Jar B = $(20 – x)$ litres<\/p>\n Acc. to ques, => $\\frac{20 + (20 – x)}{16 + x} = \\frac{5}{3}$<\/p>\n => $120 – 3x = 80 + 5x$<\/p>\n => $5x + 3x = 120 – 80$<\/p>\n => $8x = 40$<\/p>\n => $x = \\frac{40}{8} = 5$ litres<\/p>\n Question 8:\u00a0<\/b>Jar A contains 78 litres of milk and water in the respective ratio of 6 : 7. 26 litres of the mixture was taken out from Jar A. What quantity of milk should be added to jarA, so that water constitutes 40% of the resultant mixture in jar A?<\/p>\n a)\u00a08 litres<\/p>\n b)\u00a036 litres<\/p>\n c)\u00a012 litres<\/p>\n d)\u00a014 litres<\/p>\n e)\u00a018 litres<\/p>\n 8)\u00a0Answer\u00a0(E)<\/strong><\/p>\n Solution:<\/b><\/p>\n Jar A has 78 litres of mixture of milk and water in the respective ratio of 6 : 7<\/p>\n => Quantity of milk in Jar A = $\\frac{6}{13} \\times 78 = 36$ litres<\/p>\n Quantity of water in Jar A = $78 – 36 = 42$ litres<\/p>\n 26 litres of the mixture was taken out from Jar A, i.e., $\\frac{26}{78} = (\\frac{1}{3})^{rd}$<\/p>\n => Milk left = $36 – \\frac{1}{3} \\times 36 = 24$<\/p>\n Water left\u00a0= $42 – \\frac{1}{3} \\times 42 = 28$<\/p>\n Let milk added to jar A = $x$ litres<\/p>\n Acc. to ques, => $\\frac{24 + x}{28} = \\frac{60}{40}$<\/p>\n => $\\frac{24 + x}{28} = \\frac{3}{2}$<\/p>\n => $48 + 2x = 84$<\/p>\n => $2x = 84 – 48 = 36$<\/p>\n => $x = \\frac{36}{2} = 18$ litres<\/p>\n Question 9:\u00a0<\/b>A vessel contains 60 litres of milk. 6 litres of milk is taken out and 6 litres of water is added to the vessel. Again 6 litres of mixture from the vessel is withdrawn and 6 litres of water is added to the vessel. The ratio of milk and water in the resulting mixture in the vessel is<\/p>\n a)\u00a081 : 19<\/p>\n b)\u00a071 : 29<\/p>\n c)\u00a061 : 39<\/p>\n d)\u00a061 : 29<\/p>\n e)\u00a0None of these<\/p>\n 9)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n If we are taking out 6 litres out of a 60 litre solution and replacing it with water,<\/p>\n => We are replacing $\\frac{1}{10}$th of the solution with water<\/p>\n => Fraction of milk will become $\\frac{9}{10}$th of original.<\/p>\n If the process is repeated ‘n’ times, fraction of milk will become $(\\frac{9}{10})^n$ of the original.<\/p>\n Here, the process is done twice.<\/p>\n => Final quantity of milk = $(\\frac{9}{10})^2 \\times 60$ = 48.6 litres<\/p>\n and Final quantity of water = 60 – 48.6 = 11.4 litres<\/p>\n $\\therefore$ Required ratio = $\\frac{48.6}{11.4}$ = 81\u00a0: 19<\/strong><\/p>\n Question 10:\u00a0<\/b>The milk and water in two vessels A and B are in the ratio 4 : 3 and 2 : 3 respectively. In what ratio the liquids in both the vessels be mixed to obtain a new mixture in vessel C consisting half milk and half water?<\/p>\n a)\u00a08 : 3<\/p>\n b)\u00a07 : 5<\/p>\n c)\u00a04 : 3<\/p>\n d)\u00a02 : 3<\/p>\n e)\u00a0None of these<\/p>\n 10)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let mixture in vessel A = $x$ ml<\/p>\n and mixture in vessel B = $y$ ml<\/p>\n => Milk in vessel\u00a0A = $\\frac{4x}{7}$<\/p>\n Milk in vessel B = $\\frac{2y}{5}$<\/p>\n Acc to ques,<\/p>\n => $\\frac{4x}{7} + \\frac{2y}{5} = \\frac{1}{2} (x + y)$<\/p>\n => $\\frac{4x}{7} – \\frac{x}{2} = \\frac{y}{2} – \\frac{2y}{5}$<\/p>\n => $\\frac{x}{14} = \\frac{y}{10}$<\/p>\n => $\\frac{x}{y} = \\frac{14}{10} = \\frac{7}{5}$<\/p>\n Question 11:\u00a0<\/b>Jar A contains ‘X’ litre of pure milk only. A 27 litre mixture of milk and water in the respective ratio of 4 : 5, is added to jar A. The new mixture thus formed in jar A contains 70% milk, what is the value of X ?<\/p>\n a)\u00a023<\/p>\n b)\u00a030<\/p>\n c)\u00a027<\/p>\n d)\u00a048<\/p>\n e)\u00a028<\/p>\n 11)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n Quantity of milk in 27 litre mixture = $\\frac{4}{4 + 5} \\times 27 = 12$ litre<\/p>\n Quantity of water = $27 – 12 = 15$ litre<\/p>\n Ratio of milk and water in the new mixture = $70 : 30 = 7 : 3$<\/p>\n Acc to ques,<\/p>\n => $\\frac{X + 12}{15} = \\frac{7}{3}$<\/p>\n => $3X + 36 = 15 \\times 7 = 105$<\/p>\n => $3X = 105 – 36 = 69$<\/p>\n => $X = \\frac{69}{3} = 23$ litre<\/p>\n Question 12:\u00a0<\/b>In a vessel there is 40 litres mixture of milk and water. There is 15% water in the mixture. The milkman sells 10 litres of mixture to a customer and thereafter adds 12.5 litres of water to the remaining mixture. What is the respective ratio of milk and water in the new mixture ?<\/p>\n a)\u00a02 : 3<\/p>\n b)\u00a03 : 2<\/p>\n c)\u00a03 : 4<\/p>\n d)\u00a04 : 3<\/p>\n e)\u00a0None of these<\/p>\n 12)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n Mixture remaining after selling 10 litres = 40 – 10 = 30 litres<\/p>\n Now, quantity of water in 30 litres of mixture = $\\frac{15}{100} * 30$ = 4.5 litres<\/p>\n Milk = 30 – 4.5 = 25.5 litres<\/p>\n After adding 12.5 litres of water, total quantity of water = 12.5 + 4.5 = 17 litres<\/p>\n $\\therefore$ Required ratio of milk and water = 25.5 : 17<\/p>\n = 1.5 : 1 = 3 : 2<\/p>\n Question 13:\u00a0<\/b>18 litres of pure water was added to a vessel containing 80 litres of pure milk. 49 litres of the resultant mixture was then sold and some more quantity of pure milk and pure water was added to the vessel in the respective ratio of 2 : 1. If the resultant respective ratio of milk and water in the vessel was 4 : 1, what was the quantity of pure milk added in the vessel ? (in litres)<\/p>\n a)\u00a04<\/p>\n b)\u00a08<\/p>\n c)\u00a010<\/p>\n d)\u00a012<\/p>\n e)\u00a02<\/p>\n 13)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n 18 litres of pure water was added to a vessel containing 80 litres of pure milk.<\/p>\n Total mixture = 80 + 18 = 98 litres<\/p>\n Now, 49 litres i.e., $\\frac{1}{2}$is removed, => Milk left = $\\frac{80}{2}$ = 40 litres<\/p>\n Water\u00a0left = $\\frac{18}{2}$ =\u00a09 litres<\/p>\n Let milk added be $2x$ litres and water added is $x$ litres<\/p>\n => $\\frac{40 + 2x}{9 + x} = \\frac{4}{1}$<\/p>\n => $40 + 2x = 36 + 4x$<\/p>\n => $2x = 40 – 36 = 4$<\/p>\n => $x = \\frac{4}{2} = 2$<\/p>\n $\\therefore$ Quantity of milk added = $2 \\times 2 = 4$ litres<\/p>\n Question 14:\u00a0<\/b>A vessel contains a mixture of milk and water in the respective ratio of 10 : 3. Twenty-six litre of this mixture was taken out and replaced with 8 litre of water. If the resultant respective ratio of milk and water in the mixture was 5 : 2, what was the initial quantity of mixture in the vessel ? (in litre)<\/p>\n a)\u00a0143<\/p>\n b)\u00a0182<\/p>\n c)\u00a0169<\/p>\n d)\u00a0156<\/p>\n e)\u00a0130<\/p>\n 14)\u00a0Answer\u00a0(E)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let quantity of Milk and water be M and W respectively. Question 15:\u00a0<\/b>In a 90 litres mixture of milk and water, percentage of water is only 30%. The milkman gave 18 litres of this mixture to a customer and then added 18 litres of water to the remaining mixture. What is the percentage of milk in the final mixture ?<\/p>\n a)\u00a064<\/p>\n b)\u00a048<\/p>\n c)\u00a052<\/p>\n d)\u00a068<\/p>\n e)\u00a056<\/p>\n 15)\u00a0Answer\u00a0(E)<\/strong><\/p>\n Solution:<\/b><\/p>\n In 90 liters of mixture,
\nM : W =10 : 3
\n3M=10W
\nIn 26 litre of mixture
\nM =26(10\/13) = 20 litre and
\nW =26(3\/13) = 6 litre
\n8 litre of water is added.
\nResulting ratio of M and W is
\nM-20 : W-6+8 = 5 : 2
\n2(M-20) = 5(W+2)
\n2M – 40 = 5W + 10
\nMultiplyin all the terms by 2.
\n4M – 80 = 10 W + 20
\nReplacing 10W with 3M.
\n4M – 80 = 3M + 20
\nM = 100
\nHence W would be 30.
\nTotal quantity = 100+30 = 130.
\nOption E is the correect answer.<\/p>\n
\nAmount of water=$\\frac{90\\times30}{100}$.
\n=27 liters.
\nAmount of milk=$\\frac{90\\times70}{100}$.
\n=63 liters.
\nSimilarly, in 18 liters of mixture,
\nAmount of water=$\\frac{18\\times30}{100}$.
\n=5.4 liters.
\nAmount of milk=$\\frac{18\\times70}{100}$.
\n=12.6 liters.
\nAfter removing 18 liters of solution,
\nAmount of water=27-5.4=21.6 liters.
\nAmount of milk=63-12.6=50.4 liters.
\nAfter adding 18 liters of water,
\nAmount of water in the solution=21.6+18=39.6 liters.
\nHence, Percentage of milk in solution=$\\frac{50.4}{50.4+39.6}\\times 100$.
\n=56%.
\nHence, Option E is correct.<\/span><\/span><\/span><\/p>\n