Percentages is one of the most important topics in the CMAT, and also it is an important section. One can utilize this article which consists of the most important questions regarding Percentages. Cracku provides you with the Top 20 very Important Percentages Questions for CMAT based on the questions asked in previous exam papers. Click on the link below to download the Percentages Questions for CMAT PDF with detailed answers.<\/p>\n
Download Percentages Questions for CMAT<\/a><\/p>\n Enroll to CMAT 2023 Crash Course<\/a><\/p>\n Question 1:\u00a0<\/b>A girl spends 76% of her income. If her income increases by 18% and her expenditure increases by 25%,then what is the percentage increase or decrease in her savings (correct to one decimal place)?<\/p>\n a)\u00a06.9%, decrease<\/p>\n b)\u00a04.2%, decrease<\/p>\n c)\u00a05.7%, increase<\/p>\n d)\u00a08.4%, increase<\/p>\n 1)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let the income of girl is 100<\/p>\n Expenditure is 76% of income<\/p>\n i.e,\u00a0$\\frac{76}{100}\\times\\ 100$<\/p>\n = 76<\/p>\n Saving = Income – expenditure<\/p>\n 100 – 76 = 24<\/p>\n According to question,<\/p>\n income is increased by 18%<\/p>\n Increased income =\u00a0$\\frac{18}{100}\\times\\ 100\\ +\\ 100=118$<\/p>\n Expenditure is increased by 25%<\/p>\n increased expenditure =\u00a0$\\frac{25}{100}\\times\\ 76\\ +\\ 76=95$<\/p>\n New saving = 118 – 95 = 23<\/p>\n % decrease in saving =\u00a0$\\frac{\\left(24-23\\right)}{24}\\times\\ 100$<\/p>\n i.e;\u00a0$4.16\\ \\simeq\\ 4.2\\ \\%$<\/p>\n Hence, Option B is correct.<\/p>\n Question 2:\u00a0<\/b>The total number of students in a school is 1400, out of which 35% of the students are girls and the rest are boys. If 80% of the boys and 90% of the girls passed in an annual examination, then the percentage of the students who failed is:<\/p>\n a)\u00a016.5<\/p>\n b)\u00a021.5<\/p>\n c)\u00a017.4<\/p>\n d)\u00a015.8<\/p>\n 2)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n Total number of students in a school is 1400.<\/p>\n 35% of the students are girls and the rest are boys.<\/p>\n Number of girls =\u00a0$\\frac{35}{100}\\times1400$ = 490<\/p>\n Number of boys = 1400 – 490 = 910<\/p>\n 80% of the boys and 90% of the girls passed in an annual examination.<\/p>\n Number of students passed in the examination =\u00a0$\\frac{80}{100}\\times910+\\frac{90}{100}\\times490$ =\u00a0$728+441$ = 1169<\/p>\n Number of students failed in the examination = 1400 – 1169 = 231<\/p>\n Percentage of the students failed in the examination =\u00a0$\\frac{231}{1400}\\times100$<\/p>\n = 16.5%<\/p>\n Hence, the correct answer is Option A<\/p>\n Question 3:\u00a0<\/b>A woman earns \u20b9 1,000\/day. After some weeks, she earns \u20b91,160\/day. By how much percentage did her earnings a)\u00a018%<\/p>\n b)\u00a016%<\/p>\n c)\u00a017%<\/p>\n d)\u00a015%<\/p>\n 3)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n Increase in earnings of the woman = 1160 – 1000 = \u20b9 160\/day<\/p>\n $\\therefore\\ $Percentage increase in earnings of the woman =\u00a0$\\frac{160}{1000}\\times100=$ 16%<\/p>\n Hence, the correct answer is Option B<\/p>\n Question 4:\u00a0<\/b>Sachin\u2019s income is 25% more than Dileep\u2019s income. By how much percentage is Dileep\u2019s income less than Sachin\u2019s income ?<\/p>\n a)\u00a015%<\/p>\n b)\u00a020%<\/p>\n c)\u00a018%<\/p>\n d)\u00a022%<\/p>\n 4)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let the income of Sachin = S<\/p>\n Income of Dileep = D<\/p>\n Given,\u00a0 Sachin’s income is 25% more than Dileep’s income<\/p>\n $=$> \u00a0$\\text{S}=\\frac{125}{100}\\text{D}$<\/p>\n $=$> \u00a0$\\text{S}=\\frac{5}{4}\\text{D}$<\/p>\n $\\therefore\\ $Required Percentage = $\\frac{S-D}{S}\\times100$<\/p>\n =\u00a0$\\frac{\\frac{5}{4}D-D}{\\frac{5}{4}D}\\times100$<\/p>\n = $\\frac{\\frac{D}{4}}{\\frac{5D}{4}}\\times100$<\/p>\n =\u00a0$\\frac{1}{5}\\times100$<\/p>\n = 20%<\/p>\n $\\therefore\\ $Dileep’s income is 20% less than Sachin’s income<\/p>\n Hence, the correct answer is Option B<\/p>\n Question 5:\u00a0<\/b>Rita’s income is 15% less than Richa’s income. By what percent is Richa’s income more than Rita’s income?<\/p>\n a)\u00a0$14 \\frac {11}{17}%$<\/p>\n b)\u00a0$15 \\frac {11}{17}%$<\/p>\n c)\u00a0$16 \\frac {11}{17}%$<\/p>\n d)\u00a0$17 \\frac {11}{17}%$<\/p>\n 5)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let the income of Rita = $T$<\/p>\n Let the income of Richa = $C$<\/p>\n Given, Rita’s income is 15% less than Richa’s income<\/p>\n $=$>\u00a0\u00a0 $T=\\frac{85}{100}C$<\/p>\n $\\therefore\\ $Required Percentage = $\\frac{C-T}{T}\\times100$<\/p>\n $=\\frac{C-\\frac{85}{100}C}{\\frac{85}{100}C}\\times100$<\/p>\n $=\\frac{\\frac{15}{100}C}{\\frac{85}{100}C}\\times100$<\/p>\n $=\\frac{15}{85}\\times100$<\/p>\n $=\\frac{300}{17}$<\/p>\n $=17\\frac{11}{17}$<\/p>\n Hence, the correct answer is Option D<\/p>\n Question 6:\u00a0<\/b>Sachin scored 120 runs, which included 6 boundaries and 4 sixes. What percentage of his total score did he make by running between the wickets?<\/p>\n a)\u00a0$45 %$<\/p>\n b)\u00a0$46 \\frac{4}{9} %$<\/p>\n c)\u00a0$60 %$<\/p>\n d)\u00a0$33 \\frac{1}{3} %$<\/p>\n 6)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/p>\n Total runs scored by Sachin =120<\/p>\n Runs scored in boundaries =\u00a0$\\left(6\\times4\\right)+\\left(4\\times6\\right)=24+24=48$<\/p>\n Runs scored by running between wickets =\u00a0$120-48=72$<\/p>\n $\\therefore\\ $Required Percentage = $\\frac{72}{120}\\times100=60\\%$<\/p>\n Hence, the correct answer is Option C<\/p>\n Question 7:\u00a0<\/b>A and B spend 60% and 75% of their incomes, respectively. If the savings of A are 20% more than that of B. then by what percentage is the income of A less than the income of B?<\/p>\n a)\u00a015<\/p>\n b)\u00a020<\/p>\n c)\u00a010<\/p>\n d)\u00a025<\/p>\n 7)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let the income of A and B be\u00a0X and Y respectively.<\/p>\n Saving of A = X – $\\frac{60}{100} \\times$ X = 0.4X<\/p>\n Saving of B = Y – $\\frac{75}{100} \\times$ Y = 0.25Y<\/p>\n If the savings of A are 20% more than that of B then,<\/p>\n Saving of A = 0.25Y $\\times \\frac{120}{100} = 3Y\/10<\/p>\n 0.4X = 3Y\/10<\/p>\n 4X = 3Y<\/p>\n Ratio of income of A and B = 3 : 4<\/p>\n Required percentage = $\\frac{(4 – 3)}{4} \\times$ 100 = 25%<\/p>\n Question 8:\u00a0<\/b>The price of cooking oil increased by 25%. Find by how much percentage a family must reduce its consumption in order to maintain the same budget?<\/p>\n a)\u00a070%<\/p>\n b)\u00a080%<\/p>\n c)\u00a030%<\/p>\n d)\u00a020%<\/p>\n 8)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/p>\n By the formula,<\/p>\n Decrements in the consumption in order to maintain the same budget\u00a0= $\\frac{increment in the rate}{100 + increment in the rate} \\times 100$<\/p>\n = $\\frac{25}{100 + 25} \\times 100 = \\frac{25}{125} \\times 100 = 20%<\/p>\n Question 9:\u00a0<\/b>Anu spends 68% of her monthly income. If her monthly income increases by 20% and her monthly savings increase by $9\\frac{3}{8}\\%$, then the percentage increase in her monthly expenditure is:<\/p>\n a)\u00a020%<\/p>\n b)\u00a025%<\/p>\n c)\u00a022%<\/p>\n d)\u00a032%<\/p>\n 9)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let the initial salary of Anu be 100%.<\/p>\n Initially Expenditure of Anu = 68%<\/p>\n Saving = 100 -68 = 32%<\/p>\n Salary after 20% increment =\u00a0\u00a0100 $\\times 120\/100 = 120%$<\/p>\n Saving after increment = 32% $\\times \\frac{100 +\u00a09\\frac{3}{8}}{100}$ = 32% $\\times \\frac{109.375}{100}$ = 35%<\/p>\n Expenditure = 120 – 35 = 85%<\/p>\n Percentage increase in her monthly expenditure = $\\frac{85 – 68}{68} \\times 100 =\u00a0\\frac{17}{68} \\times 100$ = 25%<\/p>\n Question 10:\u00a0<\/b>The price of sugaris increased by 20%. By what percentage must one cut down on the consumption of sugar, so that no extra amount has to be incurred on sugar?<\/p>\n a)\u00a0$16\\frac{2}{3}\\%$<\/p>\n b)\u00a0$20\\%$<\/p>\n c)\u00a0$80\\%$<\/p>\n d)\u00a0$83\\frac{1}{3}\\%$<\/p>\n 10)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n Percentage decrements in the\u00a0consumption of sugar = $\\frac{rate}{100 + rate} \\times 100$<\/p>\n =\u00a0$\\frac{20}{100 + 20} \\times 100$ =\u00a0$\\frac{20}{120} \\times 100$ = $16 \\frac{2}{3}$%<\/p>\n
\nincrease?<\/p>\n