Download CMAT Ratio and Proportion questions with solutions PDF by Cracku. Practice CMAT solved Ratio and Proportion Questions paper tests, which are the practice question to have a firm grasp on the Ration and Proportion topic in the XAT exam. Top 20 very Important Ration and Proportion Questions for CMAT based on the questions asked in previous exam papers. Click on the link below to download the Ration and Proportion Questions for CMAT PDF with detailed solutions.<\/p>\n
Download Ratio and Proportion Questions for CMAT<\/a><\/p>\n Enroll to CMAT 2023 Crash Course<\/a><\/p>\n Question 1:\u00a0<\/b>If $a : b : c = 1 : 3 : 5$, what is the value of $\\frac{4a – b + 2c}{3(a + b + c)}$ ?<\/p>\n a)\u00a0$\\frac{8}{27}$<\/p>\n b)\u00a0$\\frac{10}{27}$<\/p>\n c)\u00a0$\\frac{11}{27}$<\/p>\n d)\u00a0$\\frac{1}{3}$<\/p>\n 1)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let a = x; b= 3x, c=5x<\/p>\n $\\frac{4a – b + 2c}{3(a + b + c)}=\\frac{\\left(4x-3x+10x\\right)}{3\\left(x+3x+5x\\right)}=\\frac{11}{27}$<\/p>\n Question 2:\u00a0<\/b>A earns \u20b9640 per day and works for 8 hours. B earns \u20b9360 per day and works for 6 hours. The ratio of per day wages of A to that of B is:<\/p>\n a)\u00a016 : 9<\/p>\n b)\u00a04 : 5<\/p>\n c)\u00a05 : 4<\/p>\n d)\u00a09 : 16<\/p>\n 2)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n Wage of A per day =\u00a0\u20b9640<\/p>\n Wage of B per day = \u20b9360<\/p>\n $\\therefore$ Ratio of per day wages of A to that of B = 640 : 360 = 16 : 9<\/p>\n Hence, the correct answer is Option A<\/p>\n Question 3:\u00a0<\/b>Two numbers are in the ratio 3 : 4. On increasing each of them by 30, the ratio becomes 9 : 10. The numbers are:<\/p>\n a)\u00a030, 40<\/p>\n b)\u00a018, 24<\/p>\n c)\u00a015, 20<\/p>\n d)\u00a012, 16<\/p>\n 3)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let the two numbers be $3x$, $4x$<\/p>\n According to the problem<\/p>\n $\\ \\frac{3x+30}{4x+30}=\\frac{9}{10}$<\/p>\n $=$> $30x+300=36x+270$<\/p>\n $=$> $6x=30$<\/p>\n $=$> $x=5$<\/p>\n $\\therefore\\ $The numbers are $3x,4x=3(5),4(5)=15,20$<\/p>\n Hence, the correct answer is Option C<\/p>\n Question 4:\u00a0<\/b>The ratio between the speeds of two trains is 2 : 5. If the first train runs 350 km in 5 hours, then the sum of the speed (in km\/h) of both the trains is:<\/p>\n a)\u00a0265<\/p>\n b)\u00a0350<\/p>\n c)\u00a0245<\/p>\n d)\u00a0180<\/p>\n 4)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let the speed of two trains be 2p and 5p respectively<\/p>\n Speed of first train = $\\frac{\\ 350}{5}$ = 70 km\/hr<\/p>\n $=$> 2p = 70<\/p>\n $=$> p = 35<\/p>\n $\\therefore\\ $Sum of the speed of both trains = 2p+5p = 7p = 7(35) = 245 km\/hr<\/p>\n Hence, the correct answer is Option C<\/p>\n Question 5:\u00a0<\/b>Two numbers are in the ratio 3 : 4. On increasing each of them by 30, the ratio becomes 9 : 10. The sum of the numbers is:<\/p>\n a)\u00a035<\/p>\n b)\u00a030<\/p>\n c)\u00a032<\/p>\n d)\u00a025<\/p>\n 5)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let the two numbers be $3x$, $4x$<\/p>\n According to the problem<\/p>\n $\\ \\frac{3x+30}{4x+30}=\\frac{9}{10}$<\/p>\n $=$> \u00a0$30x+300=36x+270$<\/p>\n $=$>\u00a0\u00a0$6x=30$<\/p>\n $=$> \u00a0$x=5$<\/p>\n $\\therefore\\ $Sum of the numbers$=3x+4x=7x=7\\left(5\\right)=35$<\/p>\n Hence, the correct answer is Option A<\/p>\n Question 6:\u00a0<\/b>The ratio between the speeds of two trains is 2 : 5. If the first train covers 350 km in 5 hours, then the speed (in km\/h) of the second train is:<\/p>\n a)\u00a0180<\/p>\n b)\u00a0175<\/p>\n c)\u00a0165<\/p>\n d)\u00a0150<\/p>\n 6)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let the speed of two trains be 2p and 5p respectively<\/p>\n Speed of first train = $\\frac{\\ 350}{5}$ = 70 km\/hr<\/p>\n $=$>\u00a0 2p = 70<\/p>\n $=$> \u00a0 p = 35<\/p>\n $\\therefore\\ $The speed of second train = 5p = 5(35) = 175 km\/hr<\/p>\n Hence, the correct answer is Option B<\/p>\n Question 7:\u00a0<\/b>Two numbers are in the ratio of 7 : 5. On diminishing each of them by 40, the ratio becomes 27 : 17. The sum of the numbers is:<\/p>\n a)\u00a0300<\/p>\n b)\u00a0240<\/p>\n c)\u00a0325<\/p>\n d)\u00a0275<\/p>\n 7)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n let two numbers be X1 and X2<\/p>\n Acc to question X1 : X2 = 7 : 5 i,e X1\\X2 = 7\/5 => X1= (7\/5)*X2 =1.4*X2……….Eq1<\/p>\n (X1-40)\/(X2-40) = 27\/17 …..Eq2<\/p>\n Putting Eq1 in Eq2<\/p>\n (1.4*X2 – 40) \/ (X2-40) = 27\/17<\/p>\n =>\u00a0(1.4*X2\u00a0– 40)\u00a0*17 = 27*\u00a0(X2-40)<\/p>\n => 23.8*X2 – 680 = 27 X2-1080<\/p>\n => 3.2*X2 = 400<\/p>\n => X2=125 …..Eq3<\/p>\n Put value of X2 in Eq1<\/p>\n X1= 1.4*X2<\/p>\n X1= 1.4*125<\/p>\n X1=175<\/p>\n ie sum of two numbers X1+X2=125+175=300<\/p>\n Question 8:\u00a0<\/b>Two numbers are in the ratio 7 : 5. On diminishing each of them by 40, the ratio becomes 27 : 17. The difference between the numbers is:<\/p>\n a)\u00a025<\/p>\n b)\u00a050<\/p>\n c)\u00a075<\/p>\n d)\u00a040<\/p>\n 8)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let the numbers are $x$ and $y$<\/p>\n According to questions,\u00a0 Ratio of $x$ and $y$ are $7:5$<\/p>\n $\\therefore \\frac{x}{y}=\\frac{7}{5}$<\/p>\n $\\Rightarrow y=\\frac{5x}{7}$<\/p>\n After diminishing both of them by 40, Ratio becomes $27:17$<\/p>\n $\\therefore \\frac{(x-40)}{(y-40)} = \\frac{27}{17}$<\/p>\n $\\Rightarrow 17(x-40)=27(y-40)$<\/p>\n $\\Rightarrow 17x – 17\\times40 = 27y – 27\\times40$<\/p>\n $\\Rightarrow 27y – 17x=27\\times40 – 17\\times40$<\/p>\n $\\Rightarrow 27y – 17x=40\\times(27-17)$<\/p>\n $\\Rightarrow 27y – 17x=400$<\/p>\n Putting value of $y$ in above equation,<\/p>\n $\\Rightarrow 27(\\frac{5x}{7}) – 17x=400$<\/p>\n $\\Rightarrow \\frac{135x}{7} – 17x=400$<\/p>\n $\\Rightarrow \\frac{135x-119x}{7}=400$<\/p>\n $\\Rightarrow 16x=2800$<\/p>\n $\\Rightarrow x=\\frac{2800}{16}=175$<\/p>\n $\\Rightarrow y=\\frac{5x}{7}=\\frac{5\\times175}{7}=5\\times25=125$<\/p>\n $\\therefore$ Difference b\/w both numbers =$ 175-125=50$<\/p>\n Question 9:\u00a0<\/b>Three numbers are in the ratio $\\frac{1}{2} : \\frac{2}{3} : \\frac{3}{4}$. The difference between the greatest and the smallest number is 27. The smallest number is:<\/p>\n a)\u00a072<\/p>\n b)\u00a054<\/p>\n c)\u00a081<\/p>\n d)\u00a040<\/p>\n 9)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n $\\frac{1}{2} : \\frac{2}{3} : \\frac{3}{4}$<\/p>\n $\\frac{1}{2}*12 : \\frac{2}{3} *12: \\frac{3}{4}$*12<\/p>\n $6 : 8 : 9$<\/p>\n Let the numbers are 6x, 8x and 9x<\/p>\n By the problem 9x – 6x = 27<\/p>\n i.e x =9<\/p>\n hence the smallest number is 6x i.e 54<\/p>\n Question 10:\u00a0<\/b>Incomes of A and B are in the ratio 5 : 3 and their expenditures are in the ratio 9 : 5. If income of A is twice the expenditure of B, then what is the ratio of savings of A and B?<\/p>\n a)\u00a03:2<\/p>\n b)\u00a03:4<\/p>\n c)\u00a02:3<\/p>\n d)\u00a01:1<\/p>\n 10)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let income of A be 5x Question 11:\u00a0<\/b>If $x + y + z = 19, x^2 + y^2 + z^2 = 133$, then the value of $x^3 + y^3 + z^3 – 3xyz$ is:<\/p>\n a)\u00a0361<\/p>\n b)\u00a0352<\/p>\n c)\u00a0380<\/p>\n d)\u00a0342<\/p>\n 11)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n $\\left(x+y+z\\right)=19$<\/p>\n Squaring both side,<\/p>\n $\\left(x+y+z^{ }\\right)^2=19^2$<\/p>\n $x^2+y^2+z^2+2\\left(xy+yz+xz\\right)=361$<\/p>\n $133+2\\left(xy+yz+xz\\right)=361$<\/p>\n $\\left(xy+yz+xz\\right)=114$<\/p>\n $x^3 + y^3 + z^3 – 3xyz=(x+y+z)(x^2+y^2+z^2-xy-yz-xz$<\/p>\n $x^3 + y^3 + z^3 – 3xyz=19 (133-114) = 361.<\/p>\n Question 12:\u00a0<\/b>Two numbers A and arein the ratio 5 : 2, If 4 is added to each number then this ratio becomes 9 : 4. If 5 is subtracted from each ofthe original numbers,then the ratio of A and B will be:<\/p>\n a)\u00a08 : 3<\/p>\n b)\u00a07 : 2<\/p>\n c)\u00a03 : 1<\/p>\n d)\u00a04 : 1<\/p>\n 12)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let numbers be 5x and 2x Question 13:\u00a0<\/b>In an examination, the success to failure ratio was 5 : 2. Had the number of failures been 14 more, then the success to failure ratio would have been 9 : 5. The total number of candidates who appeared for the examination was:<\/p>\n a)\u00a0126<\/p>\n b)\u00a0210<\/p>\n c)\u00a0196<\/p>\n d)\u00a0203<\/p>\n 13)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/p>\n As per the question $\\dfrac{\\text{number of success student}}{\\text{number of failed student}}=\\dfrac{5}{2}$ $\\dfrac{5k-14}{2k+14}=\\dfrac{9}{5}$<\/p>\n Hence, $25k-70=18k+126$ Question 14:\u00a0<\/b>\u20b92100 is divided among P, Q and R in such a way that P’s share is half of the combined share of Q and R, and Q’s share is one-fourth of the combined share of P and R. By what amount is R’s share more than that of P?<\/p>\n a)\u00a0\u20b9200<\/p>\n b)\u00a0\u20b9250<\/p>\n c)\u00a0\u20b9500<\/p>\n d)\u00a0\u20b9280<\/p>\n 14)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/p>\n According to the question,<\/p>\n $P+Q+R=2100$………………….(1)<\/p>\n P’s share is half of the combined share of Q and R<\/p>\n $=$> \u00a0$P=\\frac{\\ Q+R}{2}$<\/p>\n $=$> \u00a0$2P=\\ Q+R$……………………..(2)<\/p>\n From (1) and (2), \u00a0$P+2P=2100$<\/p>\n $=$>\u00a0\u00a0$P=700$<\/p>\n Q’s share is one-fourth of the combined share of P and R<\/p>\n $=$> \u00a0$Q=\\frac{\\ P+R}{4}$<\/p>\n $=$> \u00a0$4Q=\\ P+R$……………………..(3)<\/p>\n From (1) and (3), \u00a0$Q+4Q=2100$<\/p>\n $=$> \u00a0$Q=420$<\/p>\n Substituting $P=700$ and $Q=420$ in equation(1)<\/p>\n $700+420+R=2100$<\/p>\n $=$> \u00a0$R=980$<\/p>\n Now, \u00a0$R-P=980-700=280$<\/p>\n So, R’s share is 280 more than P’s share<\/p>\n Hence, the correct answer is Option D<\/p>\n Question 15:\u00a0<\/b>The sum of the squares of three numbers is 336 and the ratio of the first and the second as also of the second and the third is 1 : 2. Then find the difference of third and first number?<\/p>\n a)\u00a010<\/p>\n b)\u00a012<\/p>\n c)\u00a014<\/p>\n d)\u00a016<\/p>\n 15)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let the number be x, 2x, 4x<\/p>\n According to the question,<\/p>\n $x^2+4x^2+16x^2=336$<\/p>\n $21x^2=336$<\/p>\n $x^2=16$<\/p>\n $x=4$<\/p>\n Required difference = 16 – 4= 12<\/p>\n Question 16:\u00a0<\/b>The ratio of the incomes of A and B is 2 : 3 and that of their expenditures is 1 : 2. If 90% of B\u2019s expenditure is equal to the income of A, then what is the ratio of the savings of A and B?<\/p>\n a)\u00a08:7<\/p>\n b)\u00a01:1<\/p>\n c)\u00a09:8<\/p>\n d)\u00a03:2<\/p>\n 16)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n consider incomes of A=2x<\/p>\n consider incomes of B=3x<\/p>\n consider Expenditure of A=y<\/p>\n consider Expenditure of\u00a0B=2y<\/p>\n In the question given 90% of B’s expenditure is equal to the income of A<\/p>\n $\\frac{90}{100}$\u00d72y=2x<\/p>\n $\\frac{x}{y}$=$\\frac{9}{10}$<\/p>\n x=$\\frac{9}{10}$y<\/p>\n Formula to find savings is<\/p>\n Savings= Income-Expenditure<\/p>\n Given Income=$\\frac{2x}{3x}$<\/p>\n Expenditure=$\\frac{y}{2y}$<\/p>\n Savings=$\\frac{2x-y}{3x-2y}$<\/p>\n Substitute x value in the above savings formula then we get<\/p>\n Savings=(2\u00d7$\\frac{9}{10}$y-y)\u00f7(3\u00d7$\\frac{9}{10}$y-2y)<\/p>\n Savings=$\\frac{8\u00f710}{7\u00f710}$<\/p>\n Hence savings=8:7<\/p>\n Question 17:\u00a0<\/b>A sum of \u20b94,360 was to be divided among A,B, C and D in the ratio of 3 : 4: 5 : 8, but it was divided in the ratio of $\\frac{1}{3} : \\frac{1}{4} : \\frac{1}{5} :\u00a0\\frac{1}{8}$ by mistake. As a result:<\/p>\n a)\u00a0D received \u20b91,144 less<\/p>\n b)\u00a0B received \u20b9318 more<\/p>\n c)\u00a0A received \u20b9956 more<\/p>\n d)\u00a0C received \u20b9132 less<\/p>\n 17)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n Sum was to be divided in the ratio\u00a03:4:5:8<\/p>\n Let the sum of A,B,C and D is 3x,4x,5x and 8x respectively<\/p>\n According to question,<\/p>\n we get 20x=4360 By Equating , the ratio will be\u00a0\u00a0 ;\u00a0 40 : 34 : 24 : 15<\/p>\n New sum of A,\u00a0B, C and D will be 40x, 30x, 24x and 15x respectively<\/p>\n Solving in the same way, we get<\/p>\n A = 1600 , B = 1200, C = 960 and D = 600<\/p>\n So we can say D got 1144 less<\/p>\n Hence\u00a0Option A\u00a0<\/strong>is correct.<\/p>\n Question 18:\u00a0<\/b>If x + y + z = 19, xyz = 216 and xy + yz + zx = 114, then the value of $\\sqrt{x^3 + y^3 + z^3 + xyz}$ is:<\/p>\n a)\u00a028<\/p>\n b)\u00a030<\/p>\n c)\u00a035<\/p>\n d)\u00a032<\/p>\n 18)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/p>\n $we\\ have,\\ x+y+z=19$ $19\\left(133-114\\right)+3\\times\\ 216$<\/p>\n $361+648\\ =\\ 1009$<\/p>\n $we\\ get\\ ,\\ x^3+y^3+z^3=1009$ Question 19:\u00a0<\/b>The prices of two articles are in the ratio 4 : 5. If the price of the first article is increased by x% and that of the other is decreased by 30%, then the new prices of A and B will be in the ratio 10 : 7. The value of x is:<\/p>\n a)\u00a024.5<\/p>\n b)\u00a022.5<\/p>\n c)\u00a025<\/p>\n d)\u00a020<\/p>\n 19)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/p>\n Ratio of price of two articles = 4 : 5<\/p>\n Let the price of two articles are 4p and 5p respectively<\/p>\n The price of the first article is increased by x%<\/p>\n New price of the first article =\u00a0$\\frac{100+x}{100}\\times$4p<\/p>\n The price of the second article is decreased by 30%<\/p>\n New price of the second article =\u00a0$\\frac{70}{100}\\times$5p<\/p>\n Ratio of the new prices of articles = 10 : 7<\/p>\n $\\Rightarrow$ \u00a0$\\frac{\\frac{100+x}{100}\\times4p}{\\frac{70}{100}\\times5p}=\\frac{10}{7}$<\/p>\n $\\Rightarrow$ \u00a0$\\frac{\\left(100+x\\right)\\times4}{70\\times5}=\\frac{10}{7}$<\/p>\n $\\Rightarrow$ \u00a0$\\frac{\\left(100+x\\right)\\times2}{10\\times5}=\\frac{5}{1}$<\/p>\n $\\Rightarrow$\u00a0 200 + 2x = 250<\/p>\n $\\Rightarrow$\u00a0 2x = 50<\/p>\n $\\Rightarrow$\u00a0 x = 25<\/p>\n Hence, the correct answer is Option C<\/p>\n
\nso income of B will be 3x
\nNow let expenditure of A be 9y
\nso expenditure of B will be 5y
\nNow as per given condition
\nwe get 5x=10y
\nor x = 2y
\nso 3x will be 6y
\nNow saving of A = 5x-9y =y\u00a0\u00a0\u00a0 (1)
\nsaving of B will be 3x-5y = y (2)
\nTherefore ratio of savings = 1:1<\/p>\n
\nNow when 4 is added to each number , the ratio becomes 9:4
\nso we get 5x+4:2x+4 =9:4
\ncross multiplying we get
\n20x+16=18x+36
\nwe get x=10
\nso numbers are 50 and 20
\nNow when 5 is subtracted from the numebers
\nThe numbers will become :45 ,15
\nand we get ratio as 3:1<\/p>\n
\nHence, number of success student $5k$ and number of failed student $=2k$
\nso, total number of appeared student $=5k+2k=7k$
\nAccording to the question’s condition,<\/p>\n
\n$25k-18k=126+70$
\n$7k=196$
\n$k=\\dfrac{196}{7}=28$
\nHence, total number of candidates appeared in the exam $=7\\times k =7\\times28=196$<\/p>\n
\nx=218
\nso we get A=654 ; B =872 ; 1090;D=1744
\nBut by mistake it was divided in the ratio ;\u00a0$\\frac{1}{3}:\\frac{1}{4}:\\frac{1}{5}:\\frac{1}{8}$<\/p>\n
\n$xy+yz+zx=114$
\n$xyz=216$
\n$(x+y+z)^2=x^2+y^2+z^2+2(xy+yz+zx)$
\n$361=x^2+y^2+z^2+228$
\n$we\\ get\\ ,\\ x^2+y^2+z^2=133$
\n$Now,\\ x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-yz-zx)$<\/p>\n
\n$Now\\sqrt{\\ x^3+y^3+z^3+xyz}=\\sqrt{\\ 1009+216}\\ =\\ 35$<\/p>\n