Ratio and Proportion is one of the most important topics in the CAT Quantitative Ability Section. You can check out these Ratio and Proportion questions in the<\/span> CAT Previous year’s papers<\/a>. <\/strong>If you want to learn the basics, you can watch these videos on Ratio and Proportion basics<\/strong><\/a>. This article will look into some important Ratio and Proportion Questions for CAT. These are good sources for practice; If you want to practice these questions, you can download this CAT Ratio and Proportion Most Important Questions PDF below, which is completely Free.<\/p>\n Download Averages, Ratio and Proportion Questions for cat<\/a><\/p>\n Enroll for CAT 2023 Crash Course<\/a><\/p>\n Question 1:\u00a0<\/b>Ten years ago, the ages of the members of a joint family of eight people added up to 231 years. Three years later, one member died at the age of 60 years and a child was born during the same year. After another three years, one more member died, again at 60, and a child was born during the same year. The current average age of this eight-member joint family is nearest to a)\u00a023 years<\/p>\n b)\u00a022 years<\/p>\n c)\u00a021 years<\/p>\n d)\u00a025 years<\/p>\n e)\u00a024 years<\/p>\n 1)\u00a0Answer\u00a0(E)<\/strong><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n Ten years ago, the total age of the family is 231 years.<\/p>\n Seven years ago, (Just before the death of the first person), the total age of the family would have been 231+8*3 = 231+24 = 255. After the death of one member, the total age is 255-60 =\u00a0195 years.<\/p>\n Since a child takes birth in the same year, the number of members remain the same i.e. (7+1) = 8<\/p>\n Four years ago, (i.e. 6 years after start date)\u00a0one of the member of age 60 dies, Since a child takes birth in the same year, the number of members remain the same i.e. (7+1) = 8<\/p>\n After 4 more years, the current total age of the family is = 8×4 + 159 = 191 years<\/p>\n The average age is 191\/8 = 23.875 years = 24 years (approx)<\/p>\n Alternatively, Question 2:\u00a0<\/b>A shipping clerk has five boxes of different but unknown weights each weighing less than 100 kgs. The clerk weighs the boxes in pairs. The weights obtained are 110, 112, 113, 114, 115, 116, 117, 118, 120 and 121 kgs. What is the weight, in kgs, of the heaviest box?<\/p>\n a)\u00a060<\/p>\n b)\u00a062<\/p>\n c)\u00a064<\/p>\n d)\u00a0cannot be determined<\/p>\n Enroll for CAT 2023 Crash Course<\/a><\/p>2)\u00a0Answer\u00a0(B)<\/strong><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n Let the individual weights be a,b,c,d,e in increasing order such that e is max and a is min. Adding all the addition of weight together we get 4*(a+b+c+d+e) = 1156 so a+b+c+d+e = 289 . Out of these a+b will be lowest sum and d+e will be the max . so a+b=110 and d+e=121 so we get value of c as 58 . now c have the 3rd highest weight so addition of c and e must give the second largest total i.e 120 . hence e = 120-58 = 62<\/p>\n Question 3:\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 3)\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 4:\u00a0<\/b>I have one-rupee coins, 50-paisa coins and 25-paisa coins. The number of coins are in the ratio 2.5 : 3 : 4. If the total amount with me is Rs. 210, find the number of one-rupee coins.<\/p>\n a)\u00a090<\/p>\n b)\u00a085<\/p>\n c)\u00a0100<\/p>\n d)\u00a0105<\/p>\n 4)\u00a0Answer\u00a0(D)<\/strong><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n Let’s say number of coins are 2.5x , 3x and 4x Question 5:\u00a0<\/b>An elevator has a weight limit of 630 kg. It is carrying a group of people of whom the heaviest weighs 57 kg and the lightest weighs 53 kg. What is the maximum possible number of people in the group?<\/p>\n 5)\u00a0Answer:\u00a011<\/b><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n It is given that the maximum weight limit is 630. The lightest person’s weight is 53 Kg and the heaviest person’s weight is 57 Kg.<\/p>\n In order to have maximum people in the lift, all the remaining people should be of the lightest weight possible, which is 53 Kg.<\/p>\n Let there be n people.<\/p>\n 53 + n(53) + 57 = 630<\/p>\n n is approximately equal to 9.8. Hence, 9 people are possible.<\/p>\n Therefore, a total of 9 + 2 = 11 people can use the elevator.<\/p>\n Checkout: CAT Free Practice Questions and Videos<\/strong><\/a><\/em><\/p>\n Question 6:\u00a0<\/b>Out of the shirts produced in a factory, 15% are defective, while 20% of the rest are sold in the domestic market. If the remaining 8840 shirts are left for export, then the number of shirts produced in the factory is<\/p>\n a)\u00a013600<\/p>\n b)\u00a013000<\/p>\n c)\u00a013400<\/p>\n d)\u00a014000<\/p>\n 6)\u00a0Answer\u00a0(B)<\/strong><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n Let the total number of shirts be x. Hence 8840 = 0.8 * (0.85x) . Question 7:\u00a0<\/b>In an apartment complex, the number of people aged 51 years and above is 30 and there are at most 39 people whose ages are below 51 years. The average age of all the people in the apartment complex is 38 years. What is the largest possible average age, in years, of the people whose ages are below 51 years?<\/p>\n a)\u00a027<\/p>\n b)\u00a025<\/p>\n c)\u00a026<\/p>\n d)\u00a028<\/p>\n 7)\u00a0Answer\u00a0(D)<\/strong><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n The possible average age of people\u00a0whose ages are below 51 years will be maximum if the average age of the number of people aged\u00a051 years and above is minimum. Hence, we can say that that there are 30 people having same age 51 years.<\/p>\n Let ‘x’ be the maximum average age of people whose ages are below 51.<\/p>\n Then we can say that,<\/p>\n $\\dfrac{51*30+39*x}{30+39} = 38$<\/p>\n $\\Rightarrow$ $1530+39x = 2622$<\/p>\n $\\Rightarrow$ $x = 1092\/39 = 28$<\/p>\n Hence, we can say that option D is the correct answer.<\/p>\n 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>Ramesh and Gautam are among 22 students who write an examination. Ramesh scores 82.5. The average score of the 21 students other than Gautam is 62. The average score of all the 22 students is one more than the average score of the 21 students other than Ramesh. The score of Gautam is<\/p>\n a)\u00a053<\/p>\n b)\u00a051<\/p>\n c)\u00a048<\/p>\n d)\u00a049<\/p>\n 9)\u00a0Answer\u00a0(B)<\/strong><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n Assume the average of 21 students other than Ramesh = a<\/p>\n Sum of the scores of 21 students other than Ramesh = 21a<\/p>\n Hence the average of 22 students = a+1<\/p>\n Sum of the scores of all 22 students = 22(a+1)<\/p>\n The score of Ramesh = Sum of scores of all 22 students – Sum of the scores of\u00a021 students other than Ramesh = 22(a+1)-21a=a+22\u00a0 = 82.5\u00a0 \u00a0(Given)<\/p>\n => a = 60.5<\/p>\n Hence,\u00a0sum of the scores of all 22 students = 22(a+1) = 22*61.5 = 1353<\/p>\n Now the sum of the scores of students other than Gautam = 21*62 = 1302<\/p>\n Hence the score of Gautam = 1353-1302=51<\/p>\n Question 10:\u00a0<\/b>Identical chocolate pieces are sold in boxes of two sizes, small and large. The large box is sold for twice the price of the small box. If the selling price per gram of chocolate in the large box is 12% less than that in the small box, then the percentage by which the weight of chocolate in the large box exceeds that in the small box is nearest to<\/p>\n a)\u00a0144<\/p>\n b)\u00a0127<\/p>\n c)\u00a0135<\/p>\n d)\u00a0124<\/p>\n
\n[CAT 2007]<\/p>\n
\nThis is because, in 3 years, every person in the family would have aged by 3 years,
\nTotal change in age = 231+24 = 255<\/p>\n
\ntherefore,\u00a0total age of the family is 195+24-60 = 159 years.<\/p>\n
\nSince the number of members is always the same throughout
\nThe 2 older members dropped their age by 60
\nSo, after 10yrs, total age = 231 + 8*10 – 2*60\u00a0= 191
\nAverage age = 191\/8 = 23.875 $\\simeq$ 24<\/p>\n
\nSo (value of first coin) = 4 (value of second coin)<\/p>\n
\nSo total amount will be = 2.5x + 3x(0.5) + 4x(0.25) = 210
\nSo x = 42
\nAnd number of 1 rs. coins = 2.5x = 105<\/p>\n
\nHence number of non defective shirts = x – 15% of x = 0.85x
\nNumber
\nof shirts left for export = No of non defective shirts – number of
\nshirts sold in domestic market
\n= No of non defective shirts – 20% of No of non defective shirts
\n= 80% of No of non defective shirts<\/p>\n
\nSolving for x we get, x = 13000<\/p>\n