Here you can download a free Ratio and Proportion questions PDF with answers for MAH MBA CET 2022 by Cracku. These are some tricky questions in the MAH MBA CET 2022 exam that you need to find the Ratio and Proportion of answers for the given questions. These questions will help you to make practice and solve the Ratio and Proportion\u00a0questions in the MAH MBA CET exam. Utilize this best PDF practice set<\/strong> which is included answers in detail. Click on the below link to download the Ratio and Proportion MCQ<\/strong> PDF for MBA-CET 2022 for free.<\/p>\n Download Ratio and Proportion Questions for MAH-CET<\/a><\/p>\n Enroll to MAH-CET 2022 Crash Course<\/a><\/p>\n Question 1:\u00a0<\/b>Six years ago Jagannath was twice as old as Badri if the ratio of their present age is 9:5 respectively .What is the difference between their present ages?<\/p>\n a)\u00a024<\/p>\n b)\u00a030<\/p>\n c)\u00a050<\/p>\n d)\u00a0Cannot determined<\/p>\n e)\u00a0None of these<\/p>\n 1)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let present age of Jagannath = $9x$ years<\/p>\n => Badri’s present age = $5x$ years<\/p>\n According to ques, => $(9x-6)=2 \\times (5x-6)$<\/p>\n => $9x-6=10x-12$<\/p>\n => $10x-9x=12-6$<\/p>\n => $x=6$<\/p>\n $\\therefore$ Difference between their present ages = $9x-5x=4x$<\/p>\n = $4 \\times 6=24$<\/p>\n => Ans – (A)<\/p>\n Question 2:\u00a0<\/b>An amount of Rs 1,25,000 is to be distributed among the Sudhir,Soni,Shakunthala in the respective ratio of 2 : 3 : 5.What will be the difference between Soni’s and Sudhir’s \u00a0Share?<\/p>\n a)\u00a025000<\/p>\n b)\u00a012500<\/p>\n c)\u00a018750<\/p>\n d)\u00a02500<\/p>\n e)\u00a0None of these<\/p>\n 2)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let amount received by\u00a0Sudhir,Soni and Shakunthala be $2x,3x$ and $5x$ respectively.<\/p>\n => Total amount = $(2x+3x+5x)=125,000$<\/p>\n => $10x=125,000$<\/p>\n => $x=\\frac{125,000}{10}=12500$<\/p>\n $\\therefore$ Difference between Soni’s and Sudhir’s \u00a0Share = $3x-2x=x = Rs.$ $12,500$<\/p>\n => Ans – (B)<\/p>\n Question 3:\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 3)\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 4:\u00a0<\/b>The perimeter of a rectangular field is 240 metre. The ratio between the length and breadth of the field is 8:7. Find the area of the field.<\/p>\n a)\u00a03854 sq. m.<\/p>\n b)\u00a03584 sq. m.<\/p>\n c)\u00a03684 sq. m.<\/p>\n d)\u00a03666 sq. m.<\/p>\n e)\u00a0None of these<\/p>\n 4)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let the length and breadth of the rectangular field be $8x$ m and $7x$ m respectively.<\/p>\n Perimeter = $2 (8x + 7x) = 240$<\/p>\n => $15x = \\frac{240}{2} = 120$<\/p>\n => $x = \\frac{120}{15} = 8$<\/p>\n $\\therefore$ Area of field = $8x \\times 7x = 56 x^2$<\/p>\n = $56 \\times (8)^2 = 3584 m^2$<\/p>\n Question 5:\u00a0<\/b>If A’s salary is Rs. 10,000 less than B’s salary, B’s salary is 15,000 less than C’s salary and the sum of A, B and C’s salary Is Rs. 65,000, find the salary of A.<\/p>\n a)\u00a0Rs. 10000<\/p>\n b)\u00a0Rs. 12000<\/p>\n c)\u00a0Rs. 15000<\/p>\n d)\u00a0Rs. 25000<\/p>\n e)\u00a0None of these<\/p>\n 5)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let C’s salary = $Rs.$ $x$<\/p>\n => B’s salary = $Rs.$ $(x-15,000)$<\/p>\n and A’s salary = $Rs.$ $(x-15,000-10,000)$ = $Rs.$ $(x-25,000)$<\/p>\n Sum\u00a0of A, B and C’s salary = Rs. 65,000<\/p>\n => $(x)+(x-15,000)+(x-25,000)=65,000$<\/p>\n => $3x-40,000=65,000$<\/p>\n => $3x=65,000+40,000=1,05,000$<\/p>\n => $x=\\frac{1,05,000}{3}=35,000$<\/p>\n $\\therefore$ A’s salary = $(x-25,000)=35,000-25,000 = Rs.$ $10,000$<\/p>\n => Ans – (A)<\/p>\n Take Free MAH-CET mock tests here<\/a><\/strong><\/span><\/p>\n Enroll to 5 MAH CET Latest Mocks For Just Rs. 299<\/a><\/strong><\/span><\/p>\n Question 6:\u00a0<\/b>An article is sold at a loss of 10%. Its cost price is Rs.800. A discount of 20% was offered on the labelled price while selling. What is the loss per cent at the labelled price ?<\/p>\n a)\u00a010%<\/p>\n b)\u00a015%<\/p>\n c)\u00a020%<\/p>\n d)\u00a025%<\/p>\n e)\u00a0None of these<\/p>\n 6)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let labelled price of the item = $Rs. 100x$<\/p>\n Discount % = 20%<\/p>\n => Selling price = $100x-\\frac{20}{100} \\times 100x = Rs. 80x$<\/p>\n Cost price = Rs. 800<\/p>\n Loss % = $\\frac{800-80x}{800} \\times 100=10$<\/p>\n => $\\frac{800-80x}{8}=10$<\/p>\n => $100-10x=10$<\/p>\n => $10x=100-10=90$<\/p>\n => $x=\\frac{90}{10}=9$<\/p>\n Thu, labelled price = $100 \\times 9 = Rs. 900$<\/p>\n and Selling price = $80 \\times 9 = Rs. 720$<\/p>\n $\\therefore$ Loss % at labelled price = $\\frac{900-720}{900} \\times 100$<\/p>\n = $\\frac{180}{9}=20\\%$<\/p>\n => Ans – (C)<\/p>\n Question 7:\u00a0<\/b>One fourth of two-fifth of 30% of a number x is equal to 15. Find 20% of the same number.<\/p>\n a)\u00a0100<\/p>\n b)\u00a0120<\/p>\n c)\u00a0105<\/p>\n d)\u00a080<\/p>\n e)\u00a0None of these<\/p>\n 7)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n According to ques,<\/p>\n => $\\frac{1}{4} \\times \\frac{2}{5} \\times \\frac{30}{100} \\times x=15$<\/p>\n => $\\frac{3x}{100}=15$<\/p>\n => $x=15 \\times \\frac{100}{3}=500$<\/p>\n $\\therefore$ 20% of $x$ = $\\frac{20}{100} \\times 500$<\/p>\n = $20 \\times 5=100$<\/p>\n => Ans – (A)<\/p>\n Question 8:\u00a0<\/b>Three girls start jogging from the same point around a circular track and each one completes one round in 24 seconds, 36 seconds and 48 seconds respectively. After how much time will they meet at one point?<\/p>\n a)\u00a02 minutes, 20 seconds<\/p>\n b)\u00a02 minutes, 24 seconds<\/p>\n c)\u00a04 minutes, 12 seconds<\/p>\n d)\u00a03 minutes, 36 seconds<\/p>\n e)\u00a0None of these<\/p>\n 8)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n Time after which they will meet at one point is the L.C.M. of the time taken by each girl<\/p>\n Prime factorization of\u00a0:<\/p>\n 24 = $2^3 \\times 3$<\/p>\n 36 = $2^2 \\times 3^2$<\/p>\n 48 = $2^4 \\times 3$<\/p>\n L.C.M. (24,36,48) = $2^4 \\times 3^2 = 144$<\/p>\n $\\therefore$ They will meet after 144 seconds = 2 min 24 sec<\/p>\n => Ans – (B)<\/p>\n Question 9:\u00a0<\/b>If the Numerator of the fraction is increased by the 600% and Denominator is increased by 200 % \u00a0The resulting fraction is\u00a0$2\\frac{4}{5}$ . Then what is the original Fraction ?<\/p>\n a)\u00a0$\\frac{4}{7}$<\/p>\n b)\u00a0$\\frac{13}{12}$<\/p>\n c)\u00a0$\\frac{11}{12}$<\/p>\n d)\u00a0$\\frac{6}{5}$<\/p>\n e)\u00a0None of these<\/p>\n 9)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let the numerator = $x$ and denominator = $y$<\/p>\n If numerator is increased by 600%, => New numerator = $7x$<\/p>\n Similarly, new denominator = $3y$<\/p>\n => Fraction = $\\frac{7x}{3y} = 2\\frac{4}{5}$<\/p>\n => $\\frac{7x}{3y}=\\frac{14}{5}$<\/p>\n => $\\frac{x}{y} = \\frac{14}{5} \\times \\frac{3}{7}$<\/p>\n => $\\frac{x}{y}=\\frac{6}{5}$<\/p>\n => Ans – (D)<\/p>\n Question 10:\u00a0<\/b>Prerna decided to donate 15% of her salary to an orphanage. On the day of donation she changed her mind and donated Rs. 1,896 which was 80% of what she had decided earlier. How much is Prerna’s salary?<\/p>\n a)\u00a0Rs. 18,500<\/p>\n b)\u00a0Rs. 10,250<\/p>\n c)\u00a0Rs. 15,800<\/p>\n d)\u00a0Cannot be determined<\/p>\n e)\u00a0None of these<\/p>\n 10)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let Prerna’s salary = $Rs.100x$<\/p>\n Amount which Prerna decided to donate = $\\frac{15}{100} \\times 100x=Rs.15x$<\/p>\n Amount which she actually donated = Rs. 1896<\/p>\n According to ques,<\/p>\n => $\\frac{80}{100} \\times 15x=1896$<\/p>\n => $\\frac{4}{5} \\times 15x=1896$<\/p>\n => $12x=1896$<\/p>\n => $x=\\frac{1896}{12}=158$<\/p>\n $\\therefore$ Prerna’s salary = $100 \\times 158=Rs.$ $15,800$<\/p>\n => Ans – (C)<\/p>\n Question 11:\u00a0<\/b>The respective ratio between the monthly salaries of Rene and Som is 5 : 3. Out of her monthly salary Rene gives ${1 \\over 6}$th as rent, ${1 \\over 5}$th to her mother, 30% as her education loan and keeps 25% aside for miscellaneous expenditure. Remaining Rs. 5000 she keeps as savings. What is Som\u2019s monthly salary?<\/p>\n a)\u00a0Rs. 21000<\/p>\n b)\u00a0Rs. 24000<\/p>\n c)\u00a0Rs. 27000<\/p>\n d)\u00a0Rs. 36000<\/p>\n e)\u00a0Rs. 18000<\/p>\n 11)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let monthly salary of Rene = $Rs. 1500x$<\/p>\n => Monthly salary of Som = $Rs. 900x$<\/p>\n Amount given as rent by Rene = $\\frac{1}{6} \\times 1500x = 250x$<\/p>\n Amount given by Rene to her mother = $\\frac{1}{5} \\times 1500x = 300x$<\/p>\n Amount for loan = $\\frac{30}{100} \\times 1500x = 450x$<\/p>\n Amount kept aside = $\\frac{25}{100} \\times 1500x = 375x$<\/p>\n => Amount left = $1500x – (250x + 300x + 450x + 375x) = 5000$<\/p>\n => $1500x – 1375x = 125x = 5000$<\/p>\n => $x = \\frac{5000}{125} = 40$<\/p>\n $\\therefore$ Som’s salary = $900 \\times 40 = Rs. 36,000$<\/p>\n Question 12:\u00a0<\/b>A and B started a business with initial investments in the respective ratio of 18 : 7. After four months from the start of the business, A invested Rs. 2000 more and B invested Rs. 7000 more. At the end of one year, if the profit was distributed among them in the ratio of 2 : 1 respectively, what was the total initial investment with which A and B started the business?<\/p>\n a)\u00a0Rs. 50,000<\/p>\n b)\u00a0Rs. 25,000<\/p>\n c)\u00a0Rs. 1,50,000<\/p>\n d)\u00a0Rs. 75,000<\/p>\n e)\u00a0Rs. 1,25,000<\/p>\n 12)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let amount invested by A = $Rs. 18x$<\/p>\n => Amount invested by B = $Rs. 7x$<\/p>\n After four months from the start of the business, A invested Rs. 2000 more and B invested Rs. 7000 more<\/p>\n Thus, ratio of profit received by A : B<\/p>\n = $[(18x \\times 4) + (18x + 2000) \\times 8] : [(7x \\times 4) + (7x + 7000) \\times 8]$<\/p>\n = $(72x + 144x + 16000) : (28x + 56x + 56000)$<\/p>\n = $(216x + 16000) : (84x + 56000) = (54x + 4000) : (21x + 14000)$<\/p>\n Acc. to ques, => $\\frac{54x + 4000}{21x + 14000} = \\frac{2}{1}$<\/p>\n => $54x + 4000 = 42x + 28000$<\/p>\n => $54x – 42x = 12x = 28000 – 4000 = 24000$<\/p>\n => $x = \\frac{24000}{12} = 2000$<\/p>\n $\\therefore$ Total initial investment = $18x + 7x = 25x$<\/p>\n = $25 \\times 2000 = Rs. 50,000$<\/p>\n Question 13:\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 13)\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 14:\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 14)\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 15:\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 15)\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