Download Ratio and Proportion Questions for CAT<\/a><\/p>\n CAT Online Coaching<\/a><\/p>\n Question 1:\u00a0<\/b>A student took five papers in an examination, where the full marks were the same for each paper. His marks in these papers were in the proportion of 6 : 7 : 8 : 9 : 10. In all papers together, the candidate obtained 60% of the total marks. Then the number of papers in which he got more than 50% marks is<\/p>\n a)\u00a02<\/p>\n b)\u00a03<\/p>\n c)\u00a04<\/p>\n d)\u00a05<\/p>\n 1)\u00a0Answer\u00a0(C)<\/strong><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n Let the marks in the five papers be 6k, 7k, 8k, 9k and 10k respectively. Question 2:\u00a0<\/b>Fresh grapes contain 90% water by weight while dried grapes contain 20% water by weight and the remaining proportion being pulp. What is the weight of dry grapes available from 20 kg of fresh grapes?<\/p>\n a)\u00a02 kg<\/p>\n b)\u00a02.4 kg<\/p>\n c)\u00a02.5 kg<\/p>\n d)\u00a0None of these<\/p>\n 2)\u00a0Answer\u00a0(C)<\/strong><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n Fresh grapes contain 90% water so water in 20kg of fresh pulp = (90\/100)x20= 18kg.<\/p>\n In 20kg fresh grapes, the weight of water is 18kg and the weight of pulp is 2kg.<\/p>\n The concept that we apply in this question is that the weight of pulp will remain the same in both dry and fresh grapes. If this grape is dried, the water content will change but pulp content will remain the same.<\/p>\n Suppose the weight of the dry grapes be D.<\/p>\n 80% of the weight of dry grapes = weight of the pulp = 2 kg<\/p>\n (80\/100) x D =2 kg.<\/p>\n D = 2.5 kg<\/p>\n Question 3:\u00a0<\/b>Two liquids A and B are in the ratio 5 : 1 in container 1 and 1 : 3 in container 2. In what ratio should the contents of the two containers be mixed so as to obtain a mixture of A and B in the ratio 1 : 1?<\/p>\n a)\u00a02 : 3<\/p>\n b)\u00a04 : 3<\/p>\n c)\u00a03 : 2<\/p>\n d)\u00a03 : 4<\/p>\n 3)\u00a0Answer\u00a0(D)<\/strong><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n Fraction of A in contained 1 = $\\frac{5}{6}$<\/p>\n Fraction of A in contained 2\u00a0= $\\frac{1}{4}$<\/p>\n Let the ratio of liquid required from containers 1 and 2 be x:1-x<\/p>\n x($\\frac{5}{6}$) + (1-x)($\\frac{1}{4}$) =\u00a0$\\frac{1}{2}$<\/p>\n $\\frac{7x}{12}$ = $\\frac{1}{4}$<\/p>\n => x = $\\frac{3}{7}$<\/p>\n => Ratio = 3:4<\/p>\n Question 4:\u00a0<\/b>The value of each of a set of coins varies as the square of its diameter, if its thickness remains constant, and it varies as the thickness, if the diameter remains constant. If the diameter of two coins are in the ratio 4 : 3, what should be the ratio of their thickness’ be if the value of the first is four times that of the second?<\/p>\n a)\u00a016 : 9<\/p>\n b)\u00a09 : 4<\/p>\n c)\u00a09 : 16<\/p>\n d)\u00a04 : 9<\/p>\n 4)\u00a0Answer\u00a0(B)<\/strong><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n Value of coin = $k (2r)^2 t$ (where k is proportionality constant, 2r is diameter and t is thickness) $k (2r_1)^2 t_1 = 4 \\times (k(2r_2)^2 t_2)$<\/p>\n or $\\frac{t_1}{t_2} = \\frac{9}{4}$ \u00a0(As ratio of diameters 2r will be 9:4)<\/p>\n Question 5:\u00a0<\/b>There are two containers: the first contains 500 ml of alcohol, while the second contains 500 ml of water. Three cups of alcohol from the first container is taken out and is mixed well in the second container. Then three cups of this mixture is taken out and is mixed in the first container. Let A denote the proportion of water in the first container and B denote the proportion of alcohol in the second container. Then,<\/p>\n a)\u00a0A > B<\/p>\n b)\u00a0A < B<\/p>\n c)\u00a0A = B<\/p>\n d)\u00a0Cannot be determined<\/p>\n 5)\u00a0Answer\u00a0(C)<\/strong><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n Let the volume of the cup be V. Volume of alcohol in the first container is 500-3V Hence, after adding back 3 cups of the mixture, amount of water in the first container is $0+\\frac{1500V}{500+3V} $ So, the required proportion of water in the first container and alcohol in the second container are equal.<\/p>\n Checkout: CAT Free Practice Questions and Videos<\/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 6:\u00a0<\/b>What is the maximum amount of sucrose (to the nearest gram) that can be added to one-gram of saccharin such that the final mixture obtained is atleast 100 times as sweet as glucose?<\/p>\n a)\u00a07<\/p>\n b)\u00a08<\/p>\n c)\u00a09<\/p>\n d)\u00a0100<\/p>\n 6)\u00a0Answer\u00a0(B)<\/strong><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n For the mixture to be 100 times as sweet as glucose, its sweetness relative to the mixture should be at least 74.<\/p>\n 1 gm of saccharin = 675<\/p>\n Let the number of grams of sucrose to be added be N. Thus, the total weight of the mixture = N + 1.<\/p>\n So, (675 + N) \/ (N+1) = 74<\/p>\n => 675 + N = 74N + 74<\/p>\n => 601 = 73N => N = 8.23<\/p>\n When N=9, sweetness will be S = (675+9)\/10 = 684\/10 = 68.4<\/p>\n When N=8, sweetness will be S = (675+8)\/9 = 683\/9 = 75.8<\/p>\n So, option b) is the correct answer.<\/p>\n Question 7:\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 7)\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 8:\u00a0<\/b>A stall sells popcorn and chips in packets of three sizes: large, super, and jumbo. The numbers of large, super, and jumbo packets in its stock are in the ratio 7 : 17 : 16 for popcorn and 6 : 15 : 14 for chips. If the total number of popcorn packets in its stock is the same as that of chips packets, then the numbers of jumbo popcorn packets and jumbo chips packets are in the ratio<\/p>\n a)\u00a01 : 1<\/p>\n b)\u00a08 : 7<\/p>\n c)\u00a04 : 3<\/p>\n d)\u00a06 : 5<\/p>\n 8)\u00a0Answer\u00a0(A)<\/strong><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n The ratio of L, S, J for popcorn = 7 : 17 : 16<\/p>\n Let them be 7$x$, 17$x$ and 16$x$<\/p>\n The ratio of L, S, J for chips = 6 : 15 : 14<\/p>\n Let them 6$y$, 15$y$ and 14$y$<\/p>\n Given, 40$x$ = 35$y$, $x = \\frac{7y}{8}$<\/p>\n Jumbo popcor = 16$x$ = 16 *\u00a0$\\frac{7y}{8}$= 14$y$<\/p>\n Hence, the ratio of jumbo popcorn and jumbo chips = 1 : 1<\/p>\n Question 9:\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
\nSo, the total marks in all the 5 papers put together is 40k. This is equal to 60% of the total maximum marks. So, the total maximum marks is 5\/3 * 40k
\nSo, the maximum marks in each paper is 5\/3 * 40k \/ 5 = 40k\/3 = 13.33k
\n50% of the maximum marks is 6.67k
\nSo, the number of papers in which the student scored more than 50% is 4<\/p>\n
\nSo (value of first coin) = 4 (value of second coin)<\/p>\n
\nHence, after removing three cups of alcohol from the first container,<\/p>\n
\nVolume of water in the second container is 500 and volume of alcohol in the second container is 3V.
\nSo, in each cup, the amount of water contained is $\\frac{500}{500+3V}*V$<\/p>\n
\nAmount of alcohol contained in the second container is $3V – \\frac{9V^2}{500+3V} = \\frac{1500V}{500+3V}$<\/p>\n![\"\"](\"https:\/\/cracku.in\/media\/uploads\/2015\/11\/18\/capture.PNG\")
<\/p>\n