Here you can download SSC CHSL & MTS 2022 – important SSC CHSL & MTS Percentages Questions PDF by Cracku. Very Important SSC CHSL & MTS 2022 and These questions will help your SSC CHSL & MTS preparation. So kindly download the PDF for reference and do more practice.<\/p>\n
Download Percentages Questions for SSC CHSL and MTS <\/a><\/p>\n Enroll to 15 SSC CHSL 2022 Mocks At Just Rs. 149<\/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>Weight of A is 20% more than weight of B, whose weight is 30% more than weight of C. By how much percent weight of A is more than weight of C?<\/p>\n a)\u00a044<\/p>\n b)\u00a056<\/p>\n c)\u00a069<\/p>\n d)\u00a035.89<\/p>\n 3)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n Weight of B is 30% more than\u00a0weight of C.<\/p>\n B =\u00a0$\\frac{130}{100}\\times$C<\/p>\n Weight of A is 20% more than weight of B.<\/p>\n A =\u00a0$\\frac{120}{100}\\times$B =\u00a0$\\frac{120}{100}\\times\\frac{130}{100}\\times$C =\u00a0$\\frac{156}{100}$C<\/p>\n Required percentage =\u00a0$\\frac{\\frac{156}{100}C-C}{C}\\times100$<\/p>\n =\u00a0$\\frac{56}{100}\\times100$<\/p>\n = 56%<\/p>\n Hence, the correct answer is Option B<\/p>\n Question 4:\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 4)\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 5:\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 5)\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 Take a free SSC CHSL Tier-1 mock test<\/a><\/p>\n Download SSC CGL Tier-1 Previous Papers PDF<\/a><\/p>\n Question 6:\u00a0<\/b>The volume of the water in two tanks, A and B,is in the ratio of 6 : 5. The volume of water in tank A is increased by 30%. By what percentage should the volume of water in tank B be increased so that both the tanks have the same volume of water?<\/p>\n a)\u00a056%<\/p>\n b)\u00a030%<\/p>\n c)\u00a018%<\/p>\n d)\u00a015%<\/p>\n 6)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let the volume of water in tanks A and B are $6p$ and $5p$ respectively<\/p>\n Volume of water in tank A after increasing 30% = $\\frac{130}{100}\\times6p=7.8p$<\/p>\n Volume of water to be increased in tank B to have same volume as tank A = $7.8p-5p=2.8p$<\/p>\n $\\therefore\\ $Percentage increase in volume of water in tank B = $\\frac{2.8p}{5p}\\times100=56\\%$<\/p>\n Hence, the correct answer is Option A<\/p>\n Question 7:\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 7)\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 8:\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 8)\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 9:\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 9)\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 10:\u00a0<\/b>Sudha sold an article to Renu for \u20b9576 at a loss of 20%. Renu spent a sum of \u20b9224 on its transportation and sold it to Raghu at a price which would have given Sudha profit of 24%. The percentage of gain for Renu is:<\/p>\n a)\u00a013.2%<\/p>\n b)\u00a010.5%<\/p>\n c)\u00a012.9%<\/p>\n d)\u00a011.6%<\/p>\n 10)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/p>\n Cost price for Sudha = $\\frac{576}{80} \\times$ 100 = 720<\/p>\n Cost price for Renu = 576<\/p>\n Final cost\u00a0price for Renu = 576 + 224 = 800<\/p>\n Selling price for Renu = 24% profit of sudha<\/p>\n = 720 $ \\times\u00a0\\frac{124}{100}$ = 892.8<\/p>\n Profit for Renu =\u00a0892.8 -800 = 92.8<\/p>\n The percentage of gain for Renu = $\\frac{92.8}{800} \\times$ 100 = 11.6%<\/p>\n Question 11:\u00a0<\/b>In an examination in which the full marks were 500, A scored 25% more marks than B, B scored 60% more marks than C and C scored 20% less marks than D. If A scored 80% marks, then the percentage of marks obtained by D is:<\/p>\n a)\u00a065%<\/p>\n b)\u00a060%<\/p>\n c)\u00a050%<\/p>\n d)\u00a054%<\/p>\n 11)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/p>\n Given, Total marks = 500<\/p>\n A scored 80% marks in the examination<\/p>\n $\\Rightarrow$\u00a0 A =\u00a0$\\frac{80}{100}\\times500$ = 400<\/p>\n A scored 25% more marks than B<\/p>\n $\\Rightarrow$\u00a0 A =\u00a0$\\frac{125}{100}$B<\/p>\n $\\Rightarrow$\u00a0 400 =\u00a0$\\frac{125}{100}$B<\/p>\n $\\Rightarrow$\u00a0 400 = $\\frac{5}{4}$B<\/p>\n $\\Rightarrow$\u00a0 B = 320<\/p>\n B scored 60% more marks than C<\/p>\n $\\Rightarrow$\u00a0 B =\u00a0$\\frac{160}{100}$C<\/p>\n $\\Rightarrow$\u00a0 320 =\u00a0$\\frac{160}{100}$C<\/p>\n $\\Rightarrow$\u00a0 C = 200<\/p>\n C scored 20% less marks than D<\/p>\n $\\Rightarrow$\u00a0 C =\u00a0$\\frac{80}{100}$D<\/p>\n $\\Rightarrow$\u00a0 200 =\u00a0$\\frac{80}{100}$D<\/p>\n $\\Rightarrow$\u00a0 D = 250<\/p>\n $\\therefore\\ $Percentage of marks obtained by D =\u00a0$\\frac{250}{500}$ = 50%<\/p>\n Hence, the correct answer is Option C<\/p>\n Question 12:\u00a0<\/b>A, B and C donate 8%, 7% and 9%,of their salaries, respectively to a charitable trust. The salaries of A and B are same and the difference between their donations is \u20b9259. Thetotal donation of A and B is \u20b91,185 more than that of C. The total donation ofA and C is what percentage ofthe total salaries of A, B and C? (Correct to one decimalplace)<\/p>\n a)\u00a06.2%<\/p>\n b)\u00a05.8%<\/p>\n c)\u00a06.4%<\/p>\n d)\u00a07.1%<\/p>\n 12)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let the salaries of A, B and C be x, x, y.<\/p>\n Donation of A, B and C = 0.08x, 0.07x and 0.09z<\/p>\n Difference between donations of A and B\u00a0= 259<\/p>\n 0.08x – 0.07x = 259<\/p>\n x = 259\/0.01 = 25900<\/p>\n Total donation of A and B = donation of A + 1185<\/p>\n 0.08x + 0.07x =\u00a00.09z + 1185<\/p>\n 0.15x =\u00a00.09z + 1185<\/p>\n 0.09z =\u00a00.15 $\\times $ 25900 – 1185 = 2700<\/p>\n z = 2700\/0.09 = 30000<\/p>\n Total donation of A and C =\u00a00.08x +\u00a00.09z<\/p>\n =0.08 $\\times$ 25900 + 2700 = 2072 + 2700 = 4772<\/p>\n Total salaries of A, B and C =\u00a025900 +\u00a025900 +30000\u00a0= 81800<\/p>\n Required percentage = $\\frac{4772}{81800} \\times 100 = 5.8%<\/p>\n Question 13:\u00a0<\/b>Sonu saves 15% of her income. If her income increases by 20% and she still saves the same amount as before, then what is the percentage increase in her expenditure? (correct to one decimal place)<\/p>\n a)\u00a023.5<\/p>\n b)\u00a022.8<\/p>\n c)\u00a023.8<\/p>\n d)\u00a0242<\/p>\n 13)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let the initially income of Sonu be Rs.100.<\/p>\n saving = 15%<\/p>\n Expenditure = 100 – 15 = 85%<\/p>\n Saving = 100 -85 = Rs.15<\/p>\n Expenditure = 85% of 100 = Rs.85<\/p>\n Income after increment = 100 $\\times \\frac{120}{100}$ = Rs.120<\/p>\n Saving = 15<\/p>\n Expenditure = 120 – 15 = 105<\/p>\n Increment in expenditure = 105 – 85 = Rs.20<\/p>\n Percentage\u00a0Increment in expenditure = $\\frac{20}{85} \\times$ 100 = 23.52%<\/p>\n Question 14:\u00a0<\/b>The population of town B is 300%more than that of town A.For the next two years. the population ofA increases by x% per year and that of B decreases by the same percentage per year. After 2 years, if the population of A and B become equal, then the value of x is ………<\/p>\n a)\u00a0$30\\frac{2}{3}$<\/p>\n b)\u00a0$33\\frac{1}{3}$<\/p>\n c)\u00a040<\/p>\n d)\u00a025<\/p>\n 14)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let the population of town A be a.<\/p>\n Population of town B = $a + 300 \\times \\frac{300}{100} = 4a$<\/p>\n After the next 2 year –<\/p>\n a $\\times (1+\\frac{x}{100})^2 = 4a \\times (1-\\frac{x}{100})^2$<\/p>\n $(100+x)^2 = 4 \\times (100 – x)^2$<\/p>\n 100 + x = 2(100 – x)<\/p>\n 3x = 100<\/p>\n x = 33$\\frac{1}{3}$<\/p>\n Question 15:\u00a0<\/b>A person can save 25% of his income. If his income increases by 20% and still he saves the same amount as before, the percentage increase in his expenditure is ………..<\/p>\n a)\u00a0$26\\frac{2}{3}$<\/p>\n b)\u00a0$24$<\/p>\n c)\u00a0$25\\frac{1}{3}$<\/p>\n d)\u00a0$25$<\/p>\n 15)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let the initially income of person be Rs.100.<\/p>\n Saving\u00a0 = 100 $\\times \\frac{25}{100}$ = 25<\/p>\n expenditure = 100 -25 = 75<\/p>\n Final income = 100\u00a0$\\times \\frac{120}{100}$ = 120<\/p>\n Saving = 25<\/p>\n Expenditure = 120 – 25 = 95<\/p>\n Percentage increment in expenditure = $\\frac{95 – 75}{75} \\times 100 = 26\\frac{2}{3}$<\/p>\n Question 16:\u00a0<\/b>In a test consisting of 120 questions, Anuradha answered 65% of thefirst 60 questions correctly. What percentage of the remaining questions does she need to answer correctly to score 75% in the test?<\/p>\n a)\u00a080<\/p>\n b)\u00a085<\/p>\n c)\u00a084<\/p>\n d)\u00a090<\/p>\n 16)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n To score 75%, the total number of question need to be correct $=\\dfrac{120\\times 75}{100}=90$<\/p>\n In the first 60 questions, the corrected answer percentage $=\\dfrac{60\\times 65}{100}=39$<\/p>\n Hence, the remaining number of questions, need to be correct $=90-39=51$<\/p>\n Remaining questions, need to attempt $=60$<\/p>\n Hence, the corrected answer percentage should be $=\\dfrac{51\\times 100}{60}=85\\%$<\/p>\n Question 17:\u00a0<\/b>A is 75% less than B and C is 75% of the difference between A and B. C is what percentage more than A?<\/p>\n a)\u00a0125<\/p>\n b)\u00a0100<\/p>\n c)\u00a075<\/p>\n d)\u00a090<\/p>\n 17)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n As per the question,<\/p>\n Let B is = 100<\/p>\n So, as per the condition, A is 75% less than B, so A=25 Question 18:\u00a0<\/b>In an office, 70% of the total number of employees are females. 80% of the total number of employees, including 85 males, got promotion.If there are 105 female employees, then what percentage of female employees got promotion?<\/p>\n a)\u00a040%<\/p>\n b)\u00a0$33\\frac{1}{3}\\%$<\/p>\n c)\u00a035%<\/p>\n d)\u00a030%<\/p>\n 18)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n 105 female employee which is 70% of the total employee<\/p>\n so the total number of employee =$ 105 \\times \\dfrac{10}{7}$<\/p>\n $\\Rightarrow 150 $<\/p>\n 80% employee got a promotion so number of females who got a promotion = 120-85=35<\/p>\n so the %\u00a0 of female employee got is\u00a0 $\\dfrac{35}{105}\\times 100 $<\/p>\n $\\Rightarrow 33 \\dfrac{1}{3}$% Ans<\/p>\n Question 19:\u00a0<\/b>If A\u2019s income is 40% of B\u2019s income and B\u2019s income is 24% more than C\u2019s income, then by what percentage is C\u2019s\u00a0income more than A\u2019s income? (Your answer should be correct to one decimal place.)<\/p>\n a)\u00a075.6<\/p>\n b)\u00a0101.6<\/p>\n c)\u00a0104.2<\/p>\n d)\u00a050.4<\/p>\n 19)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let C income = 100<\/p>\n B’s income is 24% more than C<\/p>\n then B = 124<\/p>\n now A’s income is 40% of B income<\/p>\n so A income\u00a0 = $ 124 \\times \\dfrac{40}{100} $<\/p>\n $\\Rightarrow 49.6 $<\/p>\n so C’s income is 100- 49.6 = 50.4 (more than A)<\/p>\n then according to question $\\dfrac{50.4}{49.6}\\times 100 $<\/p>\n $\\Rightarrow 101.6 $Ans<\/p>\n Question 20:\u00a0<\/b>If A is 48% more than B and C is 60%less than the sum of A and B, then A is what percentage more than C? (Correct to one decimalplace.)<\/p>\n a)\u00a050.8<\/p>\n b)\u00a049.2<\/p>\n c)\u00a050.2<\/p>\n d)\u00a049.8<\/p>\n 20)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let B = 100 $x$<\/p>\n Here A =$ 148 x$<\/p>\n C = 60% less than A+B<\/p>\n = $\\dfrac{40}{100} \\times 248x $<\/p>\n $\\Rightarrow \\dfrac{992}{100}\u00a0\u00a0x $<\/p>\n $\\Rightarrow 99.2x $<\/p>\n Now A is more than C = $148x – 99.2x $<\/p>\n $\\Rightarrow 48.8x$<\/p>\n % more than C = $\\dfrac{48.2x}{99.2x}\\times 100$<\/p>\n $\\Rightarrow 49.19 $ %<\/p>\n $\\Rightarrow 49.2$ % Ans<\/p>\n Enroll to 15 SSC MTS 2022 Mocks At Just Rs. 149<\/a><\/p>\n
\nincrease?<\/p>\n
\nAs per the condition, c=$\\dfrac{(A-B)\\times 75}{100}=\\dfrac{(100-25)\\times 75}{100}=\\dfrac{225}{4}$
\nHence, the required percentage $=\\dfrac{\\dfrac{(225}{4}-25)\\times 100}{25}=\\dfrac{(225-100)\\times 100}{100}=125$<\/p>\n