Download Top-20 SSC MTS Ratio and Proportion Questions PDF. Ration and Proportion questions based on asked questions in previous year exam papers very important for the SSC MTS exam.<\/p>\n
Download Ratio And Proportion Questions For SSC MTS<\/a><\/p>\n Last Day – 10 SSC MTS Mocks for just Rs. 117<\/a><\/p>\n Take a free mock test for SSC MTS<\/a><\/p>\n SSC MTS Previous Papers<\/a> (Download PDF)<\/p>\n Question 1:\u00a0<\/b>The ratio of two numbers is 5 : 4. If the sum of both the numbers is 180, then what is the smaller number among both the numbers?<\/p>\n a)\u00a0100<\/p>\n b)\u00a080<\/p>\n c)\u00a060<\/p>\n d)\u00a075<\/p>\n Question 2:\u00a0<\/b>Rs 6300 is divided among A, B and C in the ratio of 1\/2: 1: 3\/5. What is the share (in Rs) of B?<\/p>\n a)\u00a03300<\/p>\n b)\u00a02700<\/p>\n c)\u00a03000<\/p>\n d)\u00a04200<\/p>\n Question 3:\u00a0<\/b>S, T and U together can complete a work in 30 days. If the ratio of efficiency of S, T and U is 20: 15: 12 respectively, then in how many days U alone can complete the same work?<\/p>\n a)\u00a0195\/2<\/p>\n b)\u00a0235\/2<\/p>\n c)\u00a0225\/2<\/p>\n d)\u00a0215\/2<\/p>\n Question 4:\u00a0<\/b>In what ratio must a mixture of 20% milk strength be mixed with that of 60% milk strength so as to get a new mixture of 25% milk strength ?<\/p>\n a)\u00a07: 1<\/p>\n b)\u00a04: 1<\/p>\n c)\u00a05: 2<\/p>\n d)\u00a09: 2<\/p>\n Question 5:\u00a0<\/b>Rs 18200 is divided among X, Y and Z in the ratio of 1\/3: 1\/4: 1\/2. What is the share (in Rs) of X ?<\/p>\n a)\u00a07000<\/p>\n b)\u00a04400<\/p>\n c)\u00a04200<\/p>\n d)\u00a05600<\/p>\n SSC MTS Previous Papers PDF<\/a><\/p>\n Take a free mock test for SSC MTS<\/a><\/p>\n Question 6:\u00a0<\/b>The ratio of milk and water in three samples is 1: 3, 3: 5 and 11: 5. A mixture containing equal quantities of all three samples is made. What will be the ratio of milk and water in the new mixture?<\/p>\n a)\u00a07: 9<\/p>\n b)\u00a05: 7<\/p>\n c)\u00a015: 13<\/p>\n d)\u00a09: 11<\/p>\n Question 7:\u00a0<\/b>Of three positive numbers, the ratio of first and second is 5: 2, ratio of second and third is 5: 4. The product of first and third is 1800. What is the sum of the three numbers?<\/p>\n a)\u00a043<\/p>\n b)\u00a0133<\/p>\n c)\u00a0119<\/p>\n d)\u00a0129<\/p>\n Question 8:\u00a0<\/b>Two whole numbers are such that the square of first number exceeds the second by 112 and the ratio of the numbers is 4:3. What is the value of smaller number?<\/p>\n a)\u00a03<\/p>\n b)\u00a04<\/p>\n c)\u00a012<\/p>\n d)\u00a036<\/p>\n Question 9:\u00a0<\/b>The sum of three numbers is 162. If the ratio of the first number to the second number is 5: 7 and that of the second number to the third number is 5: 3, then what is the second number?<\/p>\n a)\u00a035<\/p>\n b)\u00a070<\/p>\n c)\u00a080<\/p>\n d)\u00a040<\/p>\n Question 10:\u00a0<\/b>The milk and water in two vessels are in the ratio of 3: 1 and 7: 11 respectively. In what ratio should the liquid in both the vessels be mixed to obtain a new mixture containing half milk and half water?<\/p>\n a)\u00a05:7<\/p>\n b)\u00a04:9<\/p>\n c)\u00a01:1<\/p>\n d)\u00a04:7<\/p>\n SSC Study Material (18000 Solved Questions)<\/a><\/p>\n General Science Notes for SSC Exams<\/a><\/p>\n Question 11:\u00a0<\/b>The ratio of three numbers is 5: 6: 8. If the sum of the three numbers is 380, then a)\u00a075<\/p>\n b)\u00a080<\/p>\n c)\u00a0120<\/p>\n d)\u00a0100<\/p>\n Question 12:\u00a0<\/b>If in the population of a town there is an increase of 44% in the first year and 75% in the second year, then what is the average rate of increase in the population?<\/p>\n a)\u00a045%<\/p>\n b)\u00a059.5%<\/p>\n c)\u00a074%<\/p>\n d)\u00a076%<\/p>\n Question 13:\u00a0<\/b>In a mixture of milk and water, there is 19% milk. If the quantity of milk in the mixture is 437 mililitres, then what is the total quantity (in millilitres) of mixture?<\/p>\n a)\u00a02300<\/p>\n b)\u00a02200<\/p>\n c)\u00a02350<\/p>\n d)\u00a02400<\/p>\n Question 14:\u00a0<\/b>The ratio of two numbers is 6 : 11 and their difference is 65. What will be the smaller number among both the numbers?<\/p>\n a)\u00a058<\/p>\n b)\u00a065<\/p>\n c)\u00a052<\/p>\n d)\u00a078<\/p>\n Question 15:\u00a0<\/b>If 35% of (x + y) = 40% of (x – y), then x is what percentage of y?<\/p>\n a)\u00a01500<\/p>\n b)\u00a0150<\/p>\n c)\u00a015<\/p>\n d)\u00a0105<\/p>\n Question 16:\u00a0<\/b>The ratio of three numbers is 2 : 3 : 5. If the sum of the three numbers is 275, then what is the largest among the three numbers?<\/p>\n a)\u00a0142<\/p>\n b)\u00a082.5<\/p>\n c)\u00a0137.5<\/p>\n d)\u00a0152<\/p>\n Question 17:\u00a0<\/b>The sum of three numbers is 67. If the ratio of the first number to the second number is 3 : 5 and that of the second number to the third number is 4 : 7, then what is the second number?<\/p>\n a)\u00a020<\/p>\n b)\u00a024<\/p>\n c)\u00a018<\/p>\n d)\u00a016<\/p>\n Question 18:\u00a0<\/b>A 90-litre mixture of milk and water contains 80% milk. How much water (in litres) must be added to make the amount of water in the mixture 40%?<\/p>\n a)\u00a030<\/p>\n b)\u00a040<\/p>\n c)\u00a015<\/p>\n d)\u00a025<\/p>\n Question 19:\u00a0<\/b>A dishonest shopkeeper mixes vegetable oil with ghee in the ratio 1:3. It is also known that the ratio of the cost price of vegetable oil and ghee is 2:5. If the shopkeeper sells the mixture at 10% more then cost price of pure ghee, find out his profit percentage in this transaction.<\/p>\n a)\u00a0$29\\frac{7}{17}$%<\/p>\n b)\u00a0$27\\frac{3}{17}$%<\/p>\n c)\u00a0$27\\frac{7}{17}$%<\/p>\n d)\u00a0$27\\frac{9}{17}$%<\/p>\n Question 20:\u00a0<\/b>Price of a diamond is directly proportional to the cube of its weight. A man broke the diamond accidentally in three pieces in the ratio of 1 : 2 : 3 and thus, the net valuation of all 3 diamonds now is Rs. 21600. What was the original price (in Rs) of the diamond?<\/p>\n a)\u00a010000<\/p>\n b)\u00a015000<\/p>\n c)\u00a0169200<\/p>\n d)\u00a0129600<\/p>\n 200+ SSC Important Practice Sets<\/a><\/p>\n GK Q&A for Competitive Exams (Download PDF)<\/a><\/p>\n Answers & Solutions:<\/strong><\/span><\/p>\n 1)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Let the numbers be $5x$ and $4x$<\/p>\n Sum = $5x+4x=180$<\/p>\n => $9x=180$<\/p>\n => $x=\\frac{180}{9}=20$<\/p>\n $\\therefore$ Smaller number = $4\\times20=80$<\/p>\n => Ans – (B)<\/p>\n 2)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Total amount = Rs. 6300<\/p>\n Ratio of amount with A : B : C = $\\frac{1}{2}:1:\\frac{3}{5}$<\/p>\n => Share of B = $\\frac{1}{(\\frac{1}{2}+1+\\frac{3}{5})}\\times6300$<\/p>\n = $\\frac{10}{5+10+6}\\times6300$<\/p>\n = $10\\times300=Rs.$ $3000$<\/p>\n => Ans – (C)<\/p>\n 3)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Let efficiencies of S, T and U be $20x,15x$ and $12x$ units\/day respectively.<\/p>\n S, T and U together can complete the work in 30 days<\/p>\n => Total work to be done = $30\\times(20x+15x+12x)=1410x$ units<\/p>\n $\\therefore$ Time taken by U alone to complete the work = $\\frac{1410x}{12x}=\\frac{235}{2}$ days<\/p>\n => Ans – (B)<\/p>\n 4)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Let the ratio in which\u00a020% milk be mixed with that of 60% milk = $x:y$<\/p>\n According to ques,<\/p>\n => $20x+60y=25(x+y)$<\/p>\n => $20x+60y=25x+25y$<\/p>\n => $25x-20x=60y-25y$<\/p>\n => $5x=35y$<\/p>\n => $\\frac{x}{y}=\\frac{35}{5}=7$<\/p>\n $\\therefore$ Required ratio =\u00a07\u00a0: 1<\/strong><\/p>\n => Ans – (A)<\/p>\n 5)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Total amount = Rs. 18,200<\/p>\n Ratio of amount with X : Y : Z = $\\frac{1}{3}:\\frac{1}{4}:\\frac{1}{2}$<\/p>\n => Share of X = $\\frac{\\frac{1}{3}}{(\\frac{1}{3}+\\frac{1}{4}+\\frac{1}{2})}\\times18,200$<\/p>\n = $(\\frac{1}{3}\\div\\frac{4+3+6}{12})\\times18,200$<\/p>\n = $\\frac{1}{3}\\times\\frac{12}{13}\\times18,200$<\/p>\n = $4\\times1400=Rs.$ $5600$<\/p>\n => Ans – (D)<\/p>\n 6)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Let quantity of each sample is L.C.M. (4,8,16) = 16 unit<\/p>\n => Milk in first sample = $\\frac{1}{(1+3)}\\times16=4$ unit<\/p>\n and water in first sample = $16-4=12$ units<\/p>\n Similarly, milk in second sample =\u00a0$\\frac{3}{(3+5)}\\times16=6$ units and water = $16-6=10$ units<\/p>\n Milk in third sample =\u00a0$\\frac{11}{(11+5)}\\times16=11$ units and water = $16-11=5$ units<\/p>\n $\\therefore$ Ratio of milk and water in the new mixture = $\\frac{(4+6+11)}{(12+10+5)}$<\/p>\n = $\\frac{21}{27}=\\frac{7}{9}$<\/p>\n => Ans – (A)<\/p>\n 7)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Let the numbers be $x,y,z$<\/p>\n Thus,\u00a0$x:y=5:2$ ———–(i)<\/p>\n and $y:z=5:4$\u00a0————-(ii)<\/p>\n Multiplying equation (i) by 5 and equation (ii) by 2<\/p>\n => $x:y:z=25:10:8$<\/p>\n Let $x=25k$, $y=10k$ and $z=8k$<\/p>\n => Product = $25k\\times8k=1800$<\/p>\n => $k^2=\\frac{1800}{200}=9$<\/p>\n => $k=\\sqrt9=3$<\/p>\n $\\therefore$ Sum of three numbers = $25k+10k+8k=43k$<\/p>\n = $43\\times3=129$<\/p>\n => Ans – (D)<\/p>\n 8)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Let the numbers be $4x$ and $3x$<\/p>\n According to ques,<\/p>\n => $(4x)^2-(3x)^2=112$<\/p>\n => $16x^2-9x^2=7x^2=112$<\/p>\n => $x^2=\\frac{112}{7}=16$<\/p>\n => $x=\\sqrt{16}=4$<\/p>\n $\\therefore$ Smaller number = $3\\times4=12$<\/p>\n => Ans – (C)<\/p>\n 9)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Let the numbers be $x,y,z$<\/p>\n Thus,\u00a0$x:y=5:7$ ———–(i)<\/p>\n and $y:z=5:3$\u00a0————-(ii)<\/p>\n Multiplying equation (i) by 5 and equation (ii) by 7<\/p>\n => $x:y:z=25:35:21$<\/p>\n Let $x=25k$, $y=35k$ and $z=21k$<\/p>\n =>\u00a0Sum of three numbers = $25k+35k+21k=81k=162$<\/p>\n => $k=\\frac{162}{81}=2$<\/p>\n $\\therefore$ Second number = $35\\times2=70$<\/p>\n => Ans – (B)<\/p>\n 10)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Milk in 1 litre mixture of vessel 1 = $\\frac{3}{4}$<\/p>\n Milk\u00a0in 1 litre mixture of vessel 2 = $\\frac{7}{18}$<\/p>\n Milk\u00a0in 1 litre of the new mixture = $\\frac{1}{2}$<\/p>\n Let ratio of liquid from both vessels be $x:y$<\/p>\n => $(\\frac{3}{4}x)+(\\frac{7}{18}y)=\\frac{1}{2}(x+y)$<\/p>\n Multiply both sides by L.C.M. (4,18,2) = 36<\/p>\n => $27x+14y=18x+18y$<\/p>\n => $27x-18x=18y-14y$<\/p>\n => $9x=4y$<\/p>\n => $\\frac{x}{y}=\\frac{4}{9}=$<\/p>\n $\\therefore$ Liquid of both the vessels should be mixed in the ratio\u00a04\u00a0: 9\u00a0<\/strong>such that a new mixture containing half milk and half water is obtained.<\/p>\n => Ans – (B)<\/p>\n 11)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Let three numbers be $5x,6x$ and $8x$ respectively.<\/p>\n Sum = $5x+6x+8x=380$<\/p>\n => $x=\\frac{380}{19}=20$<\/p>\n $\\therefore$ Smallest number = $5\\times20=100$<\/p>\n => Ans – (D)<\/p>\n 12)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Let population of town initially = $100x$<\/p>\n Population after 44% increase in 1st year = $100x+\\frac{44}{100}\\times100x=144x$<\/p>\n Population after 75% increase in 2nd year = $144x+\\frac{75}{100}\\times144x=252x$<\/p>\n Thus, rate of increase in population = $\\frac{(252x-100x)}{100x}\\times100=152\\%$<\/p>\n => Average rate of increase in the population in 2 years = $\\frac{152}{2}=76\\%$<\/p>\n => Ans – (D)<\/p>\n 13)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Let total quantity of mixture (milk+water) = $100x$ ml<\/p>\n => Quantity of milk = $\\frac{19}{100}\\times100x=19x$ ml<\/p>\n Also, it is given that quantity of milk = 437 ml<\/p>\n => $19x=437$<\/p>\n => $x=\\frac{437}{19}=23$<\/p>\n $\\therefore$ Total quantity of mixture = $100\\times23=2300$ ml<\/p>\n => Ans – (A)<\/p>\n 14)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Let the numbers be $6x$ and $11x$<\/p>\n Difference =\u00a0$11x-6x=65$<\/p>\n => $x=\\frac{65}{5}=13$<\/p>\n $\\therefore$ Smaller number = $6\\times13=78$<\/p>\n => Ans – (D)<\/p>\n 15)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Expression\u00a0:\u00a035% of (x + y) = 40% of (x – y)<\/p>\n => $\\frac{35}{100}\\times(x+y)=\\frac{40}{100}\\times(x-y)$<\/p>\n => $35x+35y=40x-40y$<\/p>\n => $40x-35x=35y+40y$<\/p>\n => $5x=75y$<\/p>\n => $\\frac{x}{y}=\\frac{75}{5}=15$<\/p>\n $\\therefore$ Required % = $15\\times100=1500\\%$<\/p>\n => Ans – (A)<\/p>\n 16)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Let three numbers be $2x,3x$ and $5x$ respectively.<\/p>\n Sum = $2x+3x+5x=275$<\/p>\n => $x=\\frac{275}{10}=27.5$<\/p>\n $\\therefore$ Largest number = $5\\times27.5=137.5$<\/p>\n => Ans – (C)<\/p>\n 17)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Ratio of first number to second number = 3\u00a0: 5 and ratio\u00a0of the second number to the third number = 4\u00a0: 7<\/p>\n Multiplying first equation by 4 and second equation by 5 to make the second (common) term equal.<\/p>\n Let first number be = $12x$, second number = $20x$ and third number = $35x$<\/p>\n => Sum = $12x+20x+35x=67x=67$<\/p>\n => $x=\\frac{67}{67}=1$<\/p>\n $\\therefore$ Second number = $20\\times1=20$<\/p>\n => Ans – (A)<\/p>\n 18)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Quantity of mixture = 90 litres 19)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Let us assume that the cost of pure ghee is Rs. 5X\/kg.<\/p>\n So, the cost of vegetable oil = $\\frac{2}{5} \\times 5X = Rs. 2X\/kg$<\/p>\n Let us assume that the shopkeeper used 3Y kg of pure ghee and Y kg of vegetable oil to prepare the mixture.<\/p>\n Hence, the total cost incurred by the shopkeeper = 5X * 3Y + 2X * Y = 17XY<\/p>\n Total weight of mixture = 3Y + Y = 4Y kg<\/p>\n Selling price of mixture = $\\frac{100+10}{100} \\times 5X = Rs 5.5X\/kg$<\/p>\n Total revenue generated by selling the mixture = 5.5X * 4Y = 22XY.<\/p>\n Hence, we can see that the shopkeeper made profit of 22XY – 17XY = 5XY.<\/p>\n Profit percentage of the shopkeeper in the transaction = $\\frac{5XY}{17XY} \\times 100 = 29\\frac{7}{17}$%.<\/p>\n Therefore, option\u00a0A is the right answer.<\/p>\n 20)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Let weight of original diamond = $6x$ and price of diamond = Rs. $P$.
\nwhat is the smallest among the three numbers?<\/p>\n
\nQuantity of milk = $\\frac{80}{100}\\times90=72$ litres
\n$\\Rightarrow$ Quantity of Water = $90-72=18$ litres
\nLet us assume that $x$ litres of water is to be added.
\nAccording to problem statement,
\n$\\Rightarrow$ $\\frac{40}{100}(x+90)=18+x$
\n$\\Rightarrow$ $2*(x+90) = 5*(18+x)$
\n$\\Rightarrow$ $5x – 2x = 180 – 90$
\n$\\Rightarrow$ $x = 30$.
\nTherefore, option A is the right answer.<\/p>\n
\nAccording to question, => $P \\propto (6x)^3$.
\n$\\Rightarrow$ $P=216kx^3$ ————–(i)
\nLet weight of each broken piece be $x,2x$ and $3x$ and price $P_{1}, P_{2} and P_{3}$ respectively.
\nSo $P_{1}=k(x)^3=kx^3$.
\n$P_{2}=k(2x)^3=8kx^3$.
\n$P_{3}=k(3x)^3=27kx^3$.
\nSo, the total price of all 3 diamonds combined = $ kx^3+ 8kx^3+ 27kx^3= 36kx^3$.
\nIt is given that $36kx^3 = 21600.$
\n$\\Rightarrow kx^3 = \\frac{21600}{36} = 600$
\nSubstituting the value in equation (i)
\n$\\Rightarrow P=216 * 600$ = Rs. 129600.
\nTherefore, option\u00a0D is the right answer.<\/p>\n