Here you can download the Percentage Questions for SSC MTS PDF with solutions by Cracku. These are the most important Percentage questions PDF prepared by various sources also based on previous year’s papers. Utilize this PDF to Percentage for SSC MTS. You can find a list of Percentage in this PDF which help you to test yourself and practice. So you can click on the below link to download the PDF for reference and do more practice.<\/p>\n
Download Percentage Questions for SSC MTS<\/a><\/p>\n Enroll to 15 SSC MTS 2022 Mocks At Just Rs. 149<\/a><\/p>\n Question 1:\u00a0<\/b>The average weight of the boys in a class is 69.3 kg and that of the girls in the same class is 59.4 kg. If the average weight of all the boys and girls in the class is 63.8 kg, then the percentage of the number of boys in the class is:<\/p>\n a)\u00a0$55\\frac{5}{9}$<\/p>\n b)\u00a040<\/p>\n c)\u00a045<\/p>\n d)\u00a0$44\\frac{4}{9}$<\/p>\n 1)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let B be the number of Boys, G be number of girls<\/p>\n $ \\dfrac {total\u00a0 \u00a0weight\u00a0\u00a0 of\u00a0\u00a0 boys}\u00a0\u00a0{number\u00a0 of\u00a0 \u00a0boys} $= Any weight of boys<\/p>\n total weight of boys = (69.3)B<\/p>\n Similarly total weight of girls = (59.4) G<\/p>\n total weight of class = 69.3 B + 59.4 G<\/p>\n Any weight of class = $\\dfrac{total\u00a0 \u00a0weight\u00a0 \u00a0of\u00a0\u00a0 class}{total\u00a0 \u00a0Boys\u00a0 + Girls} $<\/p>\n $\\Rightarrow 63.8 = \\dfrac{ 63.3 B + 59.4 G} {B + G} $<\/p>\n $\\Rightarrow 63.8 B + 63.8 G = 69.3 B + 59.4 G $<\/p>\n $\\Rightarrow 4.4 G = 5.5 B $<\/p>\n $\\Rightarrow G = \\dfrac {5}{4} B $<\/p>\n % of Boys in class = $\\dfrac{B}{B+G} \\times 100 $<\/p>\n $\\Rightarrow \\dfrac{B}{B + \\dfrac{5}{4}B} \\times 100 $<\/p>\n $\\Rightarrow \\dfrac{B}{B+ \\dfrac{9}{4}B} \\times 100 $<\/p>\n $\\Rightarrow \\dfrac{4}{9} \\times 100 $<\/p>\n $\\Rightarrow \\dfrac {400}{9} $<\/p>\n $\\Rightarrow 44 \\dfrac{4}{9} $ Ans<\/p>\n Question 2:\u00a0<\/b>In a school, 60% of the numberof students are boys and the rest are girls. If 20% of the number of boys failed and 65% of the number of girls passed the examination, then the percentage of the total number of students who passed is:<\/p>\n a)\u00a068<\/p>\n b)\u00a072<\/p>\n c)\u00a074<\/p>\n d)\u00a078<\/p>\n 2)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let total number of students in school = 100<\/p>\n => Number of boys = 60 and number of girls = 40<\/p>\n Boys who passed = $\\frac{80}{100}\\times60=48$<\/p>\n Girls who passed = $\\frac{65}{100}\\times40=26$<\/p>\n $\\therefore$ Total percentage of students passed = $48+26=74\\%$<\/p>\n => Ans – (C)<\/p>\n Question 3:\u00a0<\/b>A man spends $\\frac{2}{3}rd$ of his income. If his income increases by 14% and the expenditure increases by 20%, then the\u00a0percentage increase in his savings will be<\/p>\n a)\u00a01%<\/p>\n b)\u00a02%<\/p>\n c)\u00a04%<\/p>\n d)\u00a06%<\/p>\n 3)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let initial income be Rs.\u00a0300<\/p>\n => Expenditure = Rs.\u00a0200 and savings = Rs. 100<\/p>\n Now, new income after 14% increase = $\\frac{114}{100}\\times300=Rs.$ $342$<\/p>\n Similarly, new expenditure = Rs. 240<\/p>\n => New savings = Rs. $(342-240)=Rs.$ $102$<\/p>\n $\\therefore$ Increase in savings = $\\frac{(102-100)}{100}\\times100=2\\%$<\/p>\n => Ans – (B)<\/p>\n Question 4:\u00a0<\/b>A is 40% less than B, and C is 40% of the sum of A and B. The difference between A and B is what percentage of C?<\/p>\n a)\u00a060.5%<\/p>\n b)\u00a064%<\/p>\n c)\u00a062.5%<\/p>\n d)\u00a060%<\/p>\n 4)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let $B = 10$, => $A = 6$<\/p>\n => $C = \\frac{40}{100}\\times(10+6)=6.4$<\/p>\n $\\therefore$ Required % = $\\frac{(B-A)}{C}\\times100$<\/p>\n = $\\frac{4}{6.4}\\times100=62.5\\%$<\/p>\n => Ans – (C)<\/p>\n Question 5:\u00a0<\/b>The monthly salary of a person was \u20b950,000. He used to spend on three heads- personal and family expenses (E), taxes (T), philanthropy (P), and rest were his savings. E was 50% of the income, T was 20% of E and P was 15% of T. When his salary got raised by 40%, he maintained the percentage level of E, but T became 30% of E and P became 20% of T. By whatpercentage is the new savings more or less than the earlier savings? (correct up to one decimal place)<\/p>\n a)\u00a016.4% more<\/p>\n b)\u00a08.2% more<\/p>\n c)\u00a016.4% less<\/p>\n d)\u00a08.2% less<\/p>\n 5)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n Initial salary of person = Rs. 50,000<\/p>\n Total % expenditure = $50+(\\frac{20}{100}\\times50)+(\\frac{15}{100}\\times\\frac{20}{100}\\times50)$<\/p>\n =\u00a0$50+10+1.5=61.5\\%$<\/p>\n => % savings initially = $100-61.5=38.5\\%$<\/p>\n Total savings = $\\frac{38.5}{100}\\times50,000=Rs.$ $19,250$<\/p>\n Now, new salary = Rs. 70,000<\/p>\n Similarly, total % expenditure =\u00a0$50+(\\frac{30}{100}\\times50)+(\\frac{20}{100}\\times\\frac{30}{100}\\times50)$<\/p>\n = $50+15+3=68\\%$<\/p>\n => % savings initially = $100-68=32\\%$<\/p>\n Total savings = $\\frac{32}{100}\\times70,000=Rs.$ $22,400$<\/p>\n $\\therefore$ New savings are\u00a0more<\/strong> by = $\\frac{22400-19250}{19250}\\times100\\approx16.4\\%$<\/p>\n => Ans – (A)<\/p>\n Take a free SSC MTS Tier-1 mock test<\/a><\/p>\n Download SSC CGL Tier-1 Previous Papers PDF<\/a><\/p>\n Question 6:\u00a0<\/b>If the word PHOTOGRAPH is spelt with \u2018F\u2019 in place of \u2018PH\u2019, then what would bethe percentage reduction in the number of letters?<\/p>\n a)\u00a025%<\/p>\n b)\u00a010%<\/p>\n c)\u00a020%<\/p>\n d)\u00a018%<\/p>\n 6)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/p>\n Number of letters in PHOTOGRAPH = 10<\/p>\n Number of letters in FOTOGRAF = 8<\/p>\n => % reduction = $\\frac{10-8}{10}\\times100=20\\%$<\/p>\n => Ans – (C)<\/p>\n Question 7:\u00a0<\/b>In a class, $83\\frac{1}{3}\\%$ of the number of studentsare girls and the rest are boys. If 60% of the number of boys and 80% of the\u00a0number of girls are present, then what percentage of the total number of students in the class is absent?<\/p>\n a)\u00a026$\\frac{2}{3}$<\/p>\n b)\u00a022$\\frac{2}{3}$<\/p>\n c)\u00a023$\\frac{1}{3}$<\/p>\n d)\u00a012$\\frac{1}{3}$<\/p>\n 7)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let the total students be x. Question 8:\u00a0<\/b>Raghav spends 80% of his income. If his income increases by 12% and the savings decrease by 10%, then what will be the percentage increase in his expenditure?<\/p>\n a)\u00a020.5<\/p>\n b)\u00a016<\/p>\n c)\u00a017.5<\/p>\n d)\u00a022<\/p>\n 8)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let the income of Raghav be Rs.100. Question 9:\u00a0<\/b>In an examination, 40% of the students who appeared were boys and the rest were girls. The pass percentage of the boys was 60% and the overall pass percentage was 56%. What was the pass percentage of the girls?<\/p>\n a)\u00a054$\\frac{2}{3}$<\/p>\n b)\u00a054$\\frac{1}{3}$<\/p>\n c)\u00a052$\\frac{2}{3}$<\/p>\n d)\u00a053$\\frac{1}{3}$<\/p>\n 9)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let’s assume the total number of students who appeared on the exam are 100.<\/p>\n boys appeared the exam = 40% of\u00a0100 = 40<\/p>\n girls appeared the exam = (100-40) = 60<\/p>\n boys who appeared and passed\u00a0the exam = 60% of\u00a040 = 24<\/p>\n overall students\u00a0who appeared and passed the exam = 56% of 100 = 56<\/p>\n girls who appeared and passed the exam =\u00a056-24 = 32<\/p>\n the pass percentage of the girls =\u00a0$\\frac{32\\ \\times\\ 100}{60}$ =\u00a053$\\frac{1}{3}$<\/p>\n therefore answer is option D<\/p>\n Question 10:\u00a0<\/b>Daily ticket of a metro train costs \u20b9120 and Monthly pass costs \u20b93276. If Sohan buys the Monthly pass and travels for 30 days in a month then what Percentage does he save?<\/p>\n a)\u00a018%<\/p>\n b)\u00a015%<\/p>\n c)\u00a012%<\/p>\n d)\u00a09%<\/p>\n 10)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/p>\n ticket cost per day is 120<\/p>\n if the man travels for 30 days,the cost would be 120$\\times\\ $30=3600<\/p>\n Given that monthly pass costs 3276<\/p>\n therefore the percentage of amount he saves is $\\frac{(3600-3276)}{3600}\\times100$<\/p>\n =\u00a0$\\frac{324}{3600}\\times100$<\/p>\n 9% is the correct answer.<\/p>\n Question 11:\u00a0<\/b>A box contains 780 bananas out of which 130 bananas are rotten and rest are of good quality. The percentage (correct to two decimal places) of good quality of bananas is:<\/p>\n a)\u00a071.12%<\/p>\n b)\u00a053.33%<\/p>\n c)\u00a065.35%<\/p>\n d)\u00a083.33%<\/p>\n 11)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/p>\n Out of 780 bananas, 130 are rotten means the remaining are 780-130=650<\/p>\n percentage of good quality bananas is $\\frac{650}{780}\\times100$=83.33%<\/p>\n hence answer is option D<\/p>\n Question 12:\u00a0<\/b>In spite of an increase in the price of a commodity by 20%, the overall expenditure on it increases by 12%. What is the percentage decrease in the quantity of commodities consumed?<\/p>\n a)\u00a0$7\\frac{1}{3}$<\/p>\n b)\u00a0$7\\frac{1}{2}$<\/p>\n c)\u00a08<\/p>\n d)\u00a0$6\\frac{2}{3}$<\/p>\n 12)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let’s assume the number of\u00a0commodities consumed initially is 10 and the price of each\u00a0commodity is 10.<\/p>\n Total\u00a0expenditure initially = $10\\times10$ = 100<\/p>\n In spite of an increase in the price of a commodity by 20%, the overall expenditure on it increases by 12%.<\/p>\n Price of each\u00a0commodity after increase = 10 of (100+20)%<\/p>\n = 10 of 120%<\/p>\n = 12<\/p>\n Total expenditure after the increase in the price = 100 of (100+12)%<\/p>\n = 100 of 112%<\/p>\n = 112<\/p>\n number of commodities consumed after the increase in the price =\u00a0$\\frac{112}{12}$<\/p>\n =\u00a0$\\frac{28}{3}$<\/p>\n Percentage decrease in the quantity of commodities consumed =\u00a0$\\frac{\\left(10-\\frac{28}{3}\\right)}{10}\\times\\ 100$<\/p>\n =\u00a0$\\frac{\\left(\\frac{30}{3}-\\frac{28}{3}\\right)}{10}\\times\\ 100$<\/p>\n = $\\frac{\\frac{2}{3}}{10}\\times\\ 100$<\/p>\n = $\\frac{2}{30}\\times\\ 100$<\/p>\n =\u00a0$\\frac{20}{3}$<\/p>\n = $6\\frac{2}{3}$ %<\/p>\n Question 13:\u00a0<\/b>In an examination, the average score of students in a class who passed is 69 and that of those who failed is 61. If the average score of all the students, who appeared in the examination is 66, then the percentage of students who passed is:<\/p>\n a)\u00a037.5<\/p>\n b)\u00a060<\/p>\n c)\u00a062.5<\/p>\n d)\u00a056<\/p>\n 13)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let’s assume the number of students who passed and failed the exam are P and F respectively.<\/p>\n 69P+61F = 66(P+F)<\/p>\n 69P+61F = 66P+66F<\/p>\n 69P-66P =\u00a066F-61F<\/p>\n 3P = 5F<\/p>\n P : F = 5 : 3<\/p>\n Percentage of students who passed =\u00a0$\\frac{5}{5+3}\\times\\ 100$<\/p>\n =\u00a0$\\frac{5}{8}\\times\\ 100$<\/p>\n = 62.5%<\/p>\n Question 14:\u00a0<\/b>In an examination, there are 800 boys and 600 girls. 40% boys and 60% girls passed the examination. The percentage (correct to two decimal places) of failed students from the total students is:<\/p>\n a)\u00a052.34%<\/p>\n b)\u00a050.36%<\/p>\n c)\u00a051.43%<\/p>\n d)\u00a053.57%<\/p>\n 14)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/p>\n given there are 800 boys<\/p>\n 40%of boys passed examination<\/p>\n i.e; $800\\times\\frac{40}{100}$ =320 boys passed exam<\/p>\n Given that there are 600 girls and 60% of girls passed exam<\/p>\n i.e; $600\\times\\frac{60}{100}$=360 girls<\/p>\n total number of passed students = 360+320=680<\/p>\n so no of failed students= 800+600-680<\/p>\n =720<\/p>\n so percentage of failed students = $\\frac{720\\ \\times\\ 100}{1400}$<\/p>\n 51.42%<\/p>\n hence answer is option C<\/p>\n Question 15:\u00a0<\/b>Manish\u2019s salary is half of Ravi\u2019s salary. Ravi\u2019s salary is how much percentage more than Manish\u2019s Salary?<\/p>\n a)\u00a0 b)\u00a025%<\/p>\n c)\u00a050%<\/p>\n d)\u00a075%<\/p>\n 15)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n Lets Ravi salary = 2x<\/p>\n And , Manish salary = x<\/p>\n Therefore,<\/p>\n percentage more\u00a0= $\\frac{\\left(2x-x\\right)}{x}\\times\\ 100$<\/p>\n =\u00a0$\\frac{x}{x}\\times\\ 100$<\/p>\n = 100%<\/p>\n Question 16:\u00a0<\/b>If the selling price of an article is 25% of its cost price, then what will be the loss percentage?<\/p>\n a)\u00a025%<\/p>\n b)\u00a060%<\/p>\n c)\u00a075%<\/p>\n d)\u00a050%<\/p>\n 16)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let the CP of the article be 100.<\/p>\n According to the question, SP of the article is 25 % of CP of the article , i.e. SP= 25<\/p>\n Hence, Loss = CP-SP = 100-25 = 75<\/p>\n Loss % = $\\frac{75\\ \\times\\ 100}{100}$ =\u00a075 %.<\/p>\n Question 17:\u00a0<\/b>A number is first increased by 40% and then it is increased by 30%. What is the net percentage increase?<\/p>\n a)\u00a082%<\/p>\n b)\u00a096%<\/p>\n c)\u00a072%<\/p>\n d)\u00a070%<\/p>\n 17)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let the number be 100<\/p>\n So when it is increased by 40 % then it becomes = 140.<\/p>\n Again it is increased by 30 % then it becomes = 30 % of 140 + 140 = 42 + 140 = 182<\/p>\n Hence, the net percentage change in the number is $\\frac{182-100}{100}$= 82 %.<\/p>\n Question 18:\u00a0<\/b>Side of a square is first increased by 10% and then decreased by 20%. What is the overall percentage change in area?<\/p>\n a)\u00a0Increase by 18.24%<\/p>\n b)\u00a0Decrease by 22.56%<\/p>\n c)\u00a0Decrease by 18.24%<\/p>\n d)\u00a0Increase by 22.56%<\/p>\n 18)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let the side of the square be 100<\/p>\n side increased by 10%<\/p>\n then S1 = $ 100 +( \\frac{10}{100} \\times 100 ) $<\/p>\n = 110<\/p>\n The resultant side then decreased by 20%<\/p>\n S2 = $ 110 – ( \\frac{20}{100} \\times 110 ) $<\/p>\n = 88<\/p>\n A1 = $ S1^2 $<\/p>\n = 12100<\/p>\n A2 = $ S2^2 $<\/p>\n = 7744<\/p>\n A2 < A1. That means the area is decreased from initially. Hence the change will also be a decrease.<\/p>\n overall percentage change in area = $ \\frac{10000 – 7744}{10000} \\times 100 $<\/p>\n = 22.56%<\/p>\n Question 19:\u00a0<\/b>If A is 200% more than B, then B is how much percentage less than A?<\/p>\n a)\u00a033.33%<\/p>\n b)\u00a050%<\/p>\n c)\u00a0100%<\/p>\n d)\u00a066.67%<\/p>\n 19)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let B = 100.<\/p>\n A is 200% more than B. Therefore $B\\times\\frac{100 + 200}{100} = A$<\/p>\n $100\\times\\frac{300}{100} = 300$<\/p>\n therefore A = 300.<\/p>\n To find how much percent B less than A,<\/p>\n $300\\times\\frac{100 – x}{100} = 100$<\/p>\n $100 – x = \\frac{100}{3}$<\/p>\n $x = 100 – \\frac{100}{3}$<\/p>\n $x = \\frac{200}{3}$<\/p>\n $x = 66.67$<\/p>\n hence B is 66.67% less than A.<\/p>\n Question 20:\u00a0<\/b>320 is how much percentage less than 400?<\/p>\n a)\u00a020%<\/p>\n b)\u00a018%<\/p>\n c)\u00a012%<\/p>\n d)\u00a015%<\/p>\n 20)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n = $\\frac{\\left(400-320\\right)}{400}\\times\\ 100$<\/p>\n =\u00a0$\\frac{\\left(80\\right)}{400}\\times\\ 100$<\/p>\n 20%<\/p>\n hence answer is option a<\/p>\n
\nGirls = $83\\frac{1}{3}\\%x = \\frac{250x}{300} = \\frac{5x}{6}$
\nBoys = x –\u00a0$\\frac{5x}{6} = \\frac{x}{6}$
\nAbsent boys = (100% – 60% = 40%)\u00a0of the number of boys
\n= $\\frac{x}{6} \\times \\frac{40}{100} = \\frac{x}{15}$
\nAbsent girls = (100% – 80% = 20%) of the number of girlss
\n= $\\frac{5x}{6} \\times \\frac{20}{100} = \\frac{x}{6}$
\nAbsent students =$\\frac{x}{15} +\u00a0\\frac{x}{6} = \\frac{7x}{30}$
\nPercentage\u00a0absent students = $\\frac{\\frac{7x}{30}}{x} \\times 100\u00a0 = 23 \\frac{1}{3}$<\/p>\n
\nExpenditure = 80
\nSaving = Rs.20
\nIncome increases by 12% and the savings decrease by 10%,
\nSo, income = 112,
\nSaving = 20 $\\times \\frac{90}{100} = 18$
\nExpenditure = 112 – 18 = 94
\nI<\/i>ncrement in his expenditure = 94 – 80 = 14
\nPercentage increment in expenditure = $\\frac{14}{80} \\times 100 = 20$<\/p>\n<\/del><\/u>100%<\/p>\n