Enroll to XAT 2023 Mocks & Sectionals <\/a><\/p>\n Question 1:\u00a0<\/b>A glass contains 500 cc of milk and a cup contains 500 cc of water. From the glass, 150 cc of milk is transferred to the cup and mixed thoroughly. Next, 150 cc of this mixture is transferred from the cup to the glass. Now, the amount of water in the glass and the amount of milk in the cup are in the ratio<\/p>\n a)\u00a01 : 1<\/p>\n b)\u00a010 : 13<\/p>\n c)\u00a03 : 10<\/p>\n d)\u00a010 : 3<\/p>\n 1)\u00a0Answer\u00a0(A)<\/strong><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n Initially: a glass 500cc milk and a cup\u00a0500cc water<\/p>\n Step 1:\u00a0150 cc of milk is transferred to the cup from glass<\/p>\n After step 1: Glass – 350 cc milk, Cup – 150 cc milk and 500 cc water<\/p>\n Step 2:\u00a0150 cc of this mixture is transferred from the cup to the glass<\/p>\n After step 2:<\/p>\n Glass – 350 cc milk + 150 cc mixture with milk:water ratio 3:10<\/p>\n Cup – 500 cc mixture with milk:water ratio 3:10<\/p>\n water in glass :\u00a0milk in cup =\u00a0$\\frac{10}{13}\\times150\\ :\\ \\frac{3}{13}\\times500=1:1$<\/p>\n The answer is option A.<\/p>\n Question 2:\u00a0<\/b>There are two containers of the same volume, first container half-filled with sugar syrup and the second container half-filled with milk. Half the content of the first container is transferred to the second container, and then the half of this mixture is transferred back to the first container. Next, half the content of the first container is transferred back to the second container. Then the ratio of sugar syrup and milk in the second container is<\/p>\n a)\u00a04 : 5<\/p>\n b)\u00a06 : 5<\/p>\n c)\u00a05 : 4<\/p>\n d)\u00a05 : 6<\/p>\n 2)\u00a0Answer\u00a0(D)<\/strong><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b> Step 1:\u00a0Half the content of the first container is transferred to the second container<\/p>\n Step 2:\u00a0Half of the mixture of second container\u00a0is transferred back to the first container<\/p>\n Step 3:\u00a0\u00a0Half the content of the first container is transferred back to the second container<\/p>\n Sugar syrup\u00a0: Milk in second container = 62.5 : 75 = 5 : 6<\/p>\n The answer is option D.<\/p>\n Question 3:\u00a0<\/b>A mixture contains lemon juice and sugar syrup in equal proportion. If a new mixture is created by adding this mixture and sugar syrup in the ratio 1 : 3, then the ratio of lemon juice and sugar syrup in the new mixture is<\/p>\n a)\u00a01 : 4<\/p>\n b)\u00a01 : 5<\/p>\n c)\u00a01 : 6<\/p>\n d)\u00a01 : 7<\/p>\n 3)\u00a0Answer\u00a0(D)<\/strong><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n Lemon juice : sugar syrup in the mixture is 1:1, i.e. 50% Lemon juice and 50% sugar syrup.<\/p>\n In sugar syrup, 100% is sugar syrup.<\/p>\n These two are mixed in the ratio 1:3.<\/p>\n Lemon juice =\u00a0$\\ \\frac{\\ 1\\left(50\\%\\right)}{1+3}$<\/p>\n Sugar syrup =\u00a0$\\ \\frac{\\ 1\\left(50\\%\\right)+3\\left(100\\%\\right)}{1+3}=\\frac{350}{4}$<\/p>\n Required ratio = 50:350 = 1:7<\/p>\n The answer is option D.<\/p>\n Question 4:\u00a0<\/b>Ankita buys 4 kg cashews, 14 kg peanuts and 6 kg almonds when the cost of 7 kg cashews is the same as that of 30 kg peanuts or 9 kg almonds. She mixes all the three nuts and\u00a0marks a price for the mixture in order to make a profit of \u20b91752. She sells 4 kg of the mixture at this marked price and the remaining at a 20% discount on the marked price, thus making a total profit of \u20b9744. Then the amount, in rupees, that she had spent in buying almonds is<\/p>\n a)\u00a01680<\/p>\n b)\u00a01176<\/p>\n c)\u00a02520<\/p>\n d)\u00a01440<\/p>\n 4)\u00a0Answer\u00a0(A)<\/strong><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n It is given,<\/p>\n 7C = 30P = 9A and Ankita bought 4C, 14P and 6A.<\/p>\n Let 7C = 30P = 9A = 630k<\/p>\n C = 90k, P = 21k, and A = 70k<\/p>\n Cost price of 4C, 14P and 6A = 4(90k)+14(21k)+6(70k) = 1074k<\/p>\n Marked up price = 1074k + 1752<\/p>\n S.P =\u00a0$\\frac{1}{6}\\left(1074k+1752\\right)+\\left(\\frac{4}{5}\\right)\\left(\\frac{5}{6}\\right)\\left(1074k+1752\\right)$ =\u00a0$\\frac{5}{6}\\left(1074k+1752\\right)$<\/p>\n S.P – C.P = profit<\/p>\n $1460-\\frac{1074k}{6}=744$<\/p>\n $\\frac{1074k}{6}=716$<\/p>\n k = 4<\/p>\n Money spent on buying almonds = 420k = 420*4 = Rs 1680<\/p>\n The answer is option A.<\/p>\n Question 5:\u00a0<\/b>If a certain weight of an alloy of silver and copper is mixed with 3 kg of pure silver, the resulting alloy will have 90% silver by weight. If the same weight of the initial alloy is mixed with 2 kg of another alloy which has 90% silver by weight, the resulting alloy will have 84% silver by weight. Then, the weight of the initial alloy, in kg, is<\/p>\n a)\u00a03.5<\/p>\n b)\u00a02.5<\/p>\n c)\u00a03<\/p>\n d)\u00a04<\/p>\n 5)\u00a0Answer\u00a0(C)<\/strong><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n Let the alloy contain x Kg silver and y kg copper Take XAT 2023 Mock Tests<\/a><\/p>\n Question 6:\u00a0<\/b>A person buys tea of three different qualities at \u20b9 800, \u20b9 500, and \u20b9 300 per kg, respectively, and the amounts bought are in the proportion 2 : 3 : 5. She mixes all the tea and sells one-sixth of the mixture at \u20b9 700 per kg. The price, in INR per kg, at which she should sell the remaining tea, to make an overall profit of 50%, is<\/p>\n a)\u00a0653<\/p>\n b)\u00a0688<\/p>\n c)\u00a0692<\/p>\n d)\u00a0675<\/p>\n 6)\u00a0Answer\u00a0(B)<\/strong><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n Considering the three kinds of tea are A, B, and C.<\/p>\n The price of kind A = Rs 800 per kg.<\/p>\n The price of kind B = Rs 500 per kg.<\/p>\n The price of kind C = Rs 300 per kg.<\/p>\n They were mixed in the ratio of 2 : 3: 5.<\/p>\n 1\/6 of the total mixture is sold for Rs 700 per kg.<\/p>\n Assuming the ratio of mixture to A =\u00a012kg, B =\u00a018kg, C =30 kg.<\/p>\n The total cost price is 800*12+500*18+300*30 = Rs 27600.<\/p>\n Selling 1\/6 which is 10kg for Rs 700\/kg the revenue earned is Rs 7000.<\/p>\n In order to have an overall profit of 50 percent on Rs 27600.<\/p>\n Thes selling price of the 60 kg is Rs 27600*1.5 = Rs 41400.<\/p>\n Hence he must sell the remaining 50 kg mixture for Rs 41400 – Rs 7000 = 34400.<\/p>\n Hence the price per kg is Rs 34400\/50 = Rs 688<\/p>\n Question 7:\u00a0<\/b>Two alcohol solutions, A and B, are mixed in the proportion 1:3 by volume. The volume\u00a0of the mixture is then doubled by adding solution A such that the resulting mixture\u00a0has 72% alcohol. If solution A has 60% alcohol, then the percentage of alcohol in\u00a0solution B is<\/p>\n a)\u00a090%<\/p>\n b)\u00a094%<\/p>\n c)\u00a092%<\/p>\n d)\u00a089%<\/p>\n 7)\u00a0Answer\u00a0(C)<\/strong><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n Initially let’s consider A and B as one component<\/p>\n The volume of the mixture is doubled by\u00a0adding A(60% alcohol) i.e they are mixed in 1:1 ratio and the resultant mixture has 72% alcohol.<\/p>\n Let the percentage of alcohol in component 1 be ‘x’.<\/p>\n Using allegations ,\u00a0$\\frac{\\left(72-60\\right)}{x-72}=\\frac{1}{1}$ => x= 84<\/p>\n Percentage of alcohol in\u00a0A = 60% => Let’s percentage of alcohol in B = x%<\/p>\n The resultant mixture has 84% alcohol. ratio = 1:3<\/p>\n Using allegations ,\u00a0$\\frac{\\left(x-84\\right)}{84-60}=\\frac{1}{3}$<\/p>\n => x= 92%<\/p>\n Question 8:\u00a0<\/b>An alloy is prepared by mixing three metals A, B and C in the proportion 3 : 4 : 7 by volume. Weights of the same volume of the metals A. B and C are in the ratio 5 : 2 : 6. In 130 kg of the alloy, the weight, in kg. of the metal C is<\/p>\n a)\u00a048<\/p>\n b)\u00a084<\/p>\n c)\u00a070<\/p>\n d)\u00a096<\/p>\n 8)\u00a0Answer\u00a0(B)<\/strong><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n Let the volume of Metals A,B,C we 3x, 4x, 7x<\/p>\n Ratio weights of given volume be 5y,2y,6y<\/p>\n .’. 15xy+8xy+42xy=130 => 65xy=130 => xy=2.<\/p>\n .’.`The weight, in kg. of the metal C is 42xy=84.<\/p>\n Question 9:\u00a0<\/b>Imagine you have two large pitchers, A and B. A contains 10 litres of wine and B contains 10 litres of water. One litre of water is removed from B and poured into A. The liquid is mixed very well. Then one litre of the mixture from A is poured into B. Which of the following statements is true?<\/p>\n a)\u00a0The water contents in A is more than the wine contents in B<\/p>\n b)\u00a0The water contents in A is less than the wine contents in B<\/p>\n c)\u00a0The water contents in A is same as the wine contents in B<\/p>\n d)\u00a0None of these<\/p>\n 9)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/p>\n Initially,\u00a0A has\u00a0 10 L wine and B has 10L water<\/p>\n Process 1 :\u00a0When 1 L of water is poured from B to A<\/p>\n A now has 11L ( 10 L wine and 1L water) and B has 9L water<\/p>\n Process 2:\u00a0 1L from A is put into B<\/p>\n 1L of liquid from A will have the composition of\u00a010\/11L Wine and 1\/11L Water<\/p>\n Composition of A(10L)\u00a0 = $\\frac{100}{11}$ L wine and\u00a0$\\frac{10}{11}$ L water<\/p>\n Composition of B(10L) = $\\frac{10}{11}$ L wine and $\\frac{100}{11}$ L water<\/p>\n Question 10:\u00a0<\/b>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 plate vessels should be mixed to obtain a new mixture in vessel C, only five maintaining half milk and half water?<\/p>\n a)\u00a01:1<\/p>\n b)\u00a07:5<\/p>\n c)\u00a02:4<\/p>\n d)\u00a01:3<\/p>\n 10)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n The fraction of milk in A = 4\/(4+3) = 4\/7<\/p>\n The fraction of milk in B = 2\/(2+3) = 2\/5<\/p>\n Let them be mixed in the ratio x:y such that their fraction is 1\/2 $\\frac{4x}{7}$ +\u00a0$\\frac{2y}{5}$ =\u00a0$\\frac{x+y}{2}$<\/p>\n Upon solving,<\/p>\n $\\frac{4x}{7}$ – $\\frac{x}{2}$ = $\\frac{y}{2}$ –\u00a0$\\frac{2y}{5}$<\/p>\n $\\frac{x}{14}$ = $\\frac{y}{10}$<\/p>\n $\\frac{x}{y}$ = $\\frac{7}{5}$<\/p>\n Required ratio is 7:5<\/p>\n
\n![](\"https:\/\/cracku.in\/media\/uploads\/image_2bk81Ma.png\")
<\/p>\n
\nNow when mixed with 3Kg Pure silver
\nwe get\u00a0$\\frac{\\left(x+3\\right)}{x+y+3}=\\frac{9}{10}$
\nwe get 10x+30 =9x+9y+27
\n9y-x=3\u00a0\u00a0\u00a0 (1)
\nNow as per condition 2
\nsilver in 2nd alloy = 2(0.9) =1.8
\nso we get$\\frac{\\left(x+1.8\\right)}{x+y+2}=\\frac{21}{25}$
\nwe get 21y-4x =3 \u00a0 (2)
\nsolving (1) and (2) we get y= 0.6 and x =2.4
\nso x+y = 3<\/p>\n
\nIt can be written as<\/p>\n