Enroll to 15 SSC MTS 2022 Mocks At Just Rs. 149<\/a><\/p>\n Question 1:\u00a0<\/b>A fruit seller has a sale of \u20b910,435, \u20b99,927,\u20b910,855 \u20b910,230 and \u20b99,562 for five consecutive months.How much sale (in \u20b9) must he have in the sixth month so that he gets an average sale of \u20b910,500?<\/p>\n a)\u00a09,231<\/p>\n b)\u00a08,231<\/p>\n c)\u00a08,991<\/p>\n d)\u00a011,991<\/p>\n 1)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/p>\n As we know,<\/p>\n Average = $\\frac{Sumofobservation}{total\\ number\\ of\\ observation}$<\/p>\n Let the sale of 6th month be x<\/p>\n Average =\u00a0$\\frac{\\left(10435+9927+10855+10230+9562+x\\right)}{6}=10500$<\/p>\n =\u00a0$51009\\ +\\ x=63000$<\/p>\n x = 11991<\/p>\n Hence, option D is correct.<\/p>\n Question 2:\u00a0<\/b>The mean of 100 items is 47.It was discovered that three items which should have been 60, 70, 80 were wrongly read as 40, 20, 50, respectively. The correct mean is:<\/p>\n a)\u00a048<\/p>\n b)\u00a047<\/p>\n c)\u00a050<\/p>\n d)\u00a051<\/p>\n 2)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n Given,<\/p>\n Mean of 100 items = 47<\/p>\n Sum of observation =\u00a0$mean\\times\\ number\\ of\\ observation$<\/p>\n i.e; $47\\times\\ 100\\ =\\ 4700$<\/p>\n Correct sum = 4700 + (60 + 70 + 80) – (40 + 20 +\u00a050)<\/p>\n i.e; 4800<\/p>\n Correct mean =\u00a0$\\frac{Correct\\ sum}{number\\ of\\ observation}$<\/p>\n i.e\u00a0$\\frac{4800}{100}=\\ 48$<\/p>\n Hence, Option A is correct<\/p>\n Question 3:\u00a0<\/b>The average weight of 6 people increased by 2.5 kg when a new person came in place of other person weighing 55 kg. what can be the weight of new person (in kg)?<\/p>\n a)\u00a070<\/p>\n b)\u00a062.5<\/p>\n c)\u00a060<\/p>\n d)\u00a075.5<\/p>\n 3)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n Total weight increased = number of persons\u00a0$\\times\\ $ increased weight<\/p>\n i.e;$6\\times\\ 2.5=15$<\/p>\n Weight of new person = weight of replaced person + Total increased weight<\/p>\n i.e; 55 + 15 = 70 kg<\/p>\n Hence, Option A is correct.<\/p>\n Question 4:\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 4)\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 5:\u00a0<\/b>The average of eleven numbers is 56. The average of first three numbers is 52 and that of next five numbers is 60. The 9th and 10th number are 3 and 1 more than the 11th number respectively. What is the average of 9th and 11th numbers?<\/p>\n a)\u00a053.5<\/p>\n b)\u00a052<\/p>\n c)\u00a052.5<\/p>\n d)\u00a054<\/p>\n 5)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n The average of first three numbers is 52.<\/p>\n Sum of the first three numbers = 52 x 3 = 156<\/p>\n The average of next five numbers is 60.<\/p>\n Sum of the next five numbers = 60 x 5 = 300<\/p>\n Let the 11th number be ‘n’.<\/p>\n The average of eleven numbers is 56.<\/p>\n Sum of the eleven numbers = 56 x 11 = 616<\/p>\n 156 + 300 + (n + 3) + (n + 1) + n = 616<\/p>\n 3n + 460 = 616<\/p>\n 3n = 156<\/p>\n n = 52<\/p>\n Average of 9th and 11th numbers = $\\frac{(n+3)+n}{2}$<\/p>\n = $\\frac{107}{2}$<\/p>\n = 53.5<\/p>\n Hence, the correct answer is Option 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>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 6)\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 7:\u00a0<\/b>The average of 23 numbers is 51. The average of first 12 numbers is 49 and the average of last 12 numbers is 54. If the twelfth number is removed, then the average of the remaining numbers (correct to two decimal places) is:<\/p>\n a)\u00a053:25<\/p>\n b)\u00a050.45<\/p>\n c)\u00a051.75<\/p>\n d)\u00a052.65<\/p>\n 7)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n The average of 23 numbers is 51.<\/p>\n Sum of the 23 numbers = 51 x 23 = 1173…….(1)<\/p>\n The average of first 12 numbers is 49.<\/p>\n Sum of the first 12 numbers = 49 x 12 = 588<\/p>\n Sum of the first 11 numbers + 12th number = 588…..(2)<\/p>\n The average of last 12 numbers is 54.<\/p>\n Sum of the last 12 numbers = 54 x 12 = 648<\/p>\n 12th number +\u00a0Sum of the last 11 numbers = 648…..(3)<\/p>\n Adding (2) and (3),<\/p>\n Sum of the first 11 numbers + 12th number +\u00a012th number + Sum of the last 11 numbers = 588 + 648<\/p>\n (Sum of the first 11 numbers + 12th number +\u00a0Sum of the last 11 numbers) +\u00a012th number = 1236<\/p>\n Sum of the 23 numbers +\u00a012th number = 1236<\/p>\n 1173 +\u00a012th number = 1236<\/p>\n 12th number = 63<\/p>\n Required average =\u00a0$\\frac{1173-63}{22}$<\/p>\n = 50.4545<\/p>\n Hence, the correct answer is Option B<\/p>\n Question 8:\u00a0<\/b>The average weight of a certain number of students in a class is 55.5 kg. If 4 students with average weight 60 kg join the class, then the average weight of all students in the class increases by 360 g. The number of students in the class, initially, is:<\/p>\n a)\u00a046<\/p>\n b)\u00a031<\/p>\n c)\u00a041<\/p>\n d)\u00a036<\/p>\n 8)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let the initial number of students = n<\/p>\n Average of the weight of ‘n’ students = 55.55<\/p>\n Sum of the weight of ‘n’ students = 55.55n<\/p>\n Sum of the weight of 4 students = 60 x 4 = 240 kg<\/p>\n According to the problem,<\/p>\n $\\frac{55.55n+240}{n+4}=55.55+\\frac{360}{1000}$<\/p>\n $\\frac{55.55n+240}{n+4}=55.55+0.36$<\/p>\n $\\frac{55.55n+240}{n+4}=55.86$<\/p>\n 55.55n + 240 = 55.86n + 223.44<\/p>\n 0.36n = 16.56<\/p>\n n = 46<\/p>\n The initial number of students = n = 46<\/p>\n Hence, the correct answer is Option A<\/p>\n Question 9:\u00a0<\/b>The average monthly salary of 60 employees of a factory is \u20b929900. If two officers are getting \u20b990000 each and the average salary of 8 supervisors is \u20b965000, then what is the average salary (in \u20b9) of the remaining employees?<\/p>\n a)\u00a022680<\/p>\n b)\u00a029080<\/p>\n c)\u00a021080<\/p>\n d)\u00a021880<\/p>\n 9)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/p>\n The average monthly salary of 60 employees of the factory is \u20b929900.<\/p>\n Total monthly salary of 60 employees of the factory = 29900 x 60 = \u20b91794000<\/p>\n Two officers are getting \u20b990000 each.<\/p>\n Sum of the salary of two officers = 2 x 90000 = \u20b9180000<\/p>\n The average salary of 8 supervisors is \u20b965000.<\/p>\n Total salary of 8 supervisors = 65000 x 8 =\u00a0\u20b9520000<\/p>\n Total salary of remaining 50 employees of the factory =\u00a01794000 –\u00a0180000 –\u00a0520000 =\u00a0\u20b91094000<\/p>\n Average of remaining 50 employees of the factory =\u00a0$\\frac{1094000}{50}$ =\u00a0\u20b921880<\/p>\n Hence, the correct answer is Option D<\/p>\n Question 10:\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 10)\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 11:\u00a0<\/b>The average daily production of toys in a factory in the month of December is 512. If the average production during first 20 days is 515 and that of the last 13 days is 510, then what is the average of production on 19 and 20 December?<\/p>\n a)\u00a01058<\/p>\n b)\u00a0513<\/p>\n c)\u00a0529<\/p>\n d)\u00a0512<\/p>\n 11)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/p>\n The average daily production of toys in a factory in the month of December is 512.<\/p>\n Total production of toys in the month of December = 512 x 31 = 15872……..(1)<\/p>\n The average production during first 20 days is 515.<\/p>\n Total production during first 20 days = 515 x 20 = 10300<\/p>\n Production during first 18 days + 19 December + 20 December = 10300……..(2)<\/p>\n The average production during last 13 days is 510.<\/p>\n Total production during last 13 days = 510 x 13 = 6630<\/p>\n 19 December + 20 December +\u00a0Production during last 11 days = 6630……..(3)<\/p>\n Solving (2) + (3) – (1), we get<\/p>\n 19 December + 20 December =\u00a010300 +\u00a0 6630 – 15872<\/p>\n 19 December + 20 December = 1058<\/p>\n Total production on 19 and 20 December = 1058<\/p>\n Average of production on 19 and 20 December =\u00a0$\\frac{1058}{2}$ = 529<\/p>\n Hence, the correct answer is Option C<\/p>\n Question 12:\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 12)\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 13:\u00a0<\/b>The average of 40 numbers is 36. The average of the first 25 numbers is 31 and the average of last 16 numbers is 43. Find the 25th number.<\/p>\n a)\u00a021<\/p>\n b)\u00a023<\/p>\n c)\u00a024<\/p>\n d)\u00a022<\/p>\n 13)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n Given, average of 40 numbers = 36<\/p>\n $\\Rightarrow$ Sum of 40 numbers = 36 x 40 = 1440………………(1)<\/p>\n The average of the first 25 numbers is 31<\/p>\n $\\Rightarrow$ Sum of first 25 numbers = 31 x 25 = 775<\/p>\n $\\Rightarrow$ Sum of first 24 numbers + 25th number = 775 ……(2)<\/p>\n The average of last 16 numbers is 43<\/p>\n $\\Rightarrow$ Sum of last 16 numbers = 43 x 16 = 688 ………….(3)<\/p>\n Adding equations (2) and (3),<\/p>\n Sum of first 24 numbers + 25th number +\u00a0Sum of last 16 numbers = 775 + 688<\/p>\n $\\Rightarrow$ \u00a0Sum of first 24 numbers +\u00a0Sum of last 16 numbers +\u00a025th number = 1463<\/p>\n $\\Rightarrow$\u00a0 Sum of 40 numbers + 25th number = 1463<\/p>\n $\\Rightarrow$ \u00a01440 +\u00a025th number = 1463<\/p>\n $\\Rightarrow$ \u00a025th number = 23<\/p>\n Hence, the correct answer is Option B<\/p>\n Question 14:\u00a0<\/b>Malti bought a TV for \u20b9 8,800 including GST at 10%. What is the original cost of the TV?<\/p>\n a)\u00a0\u20b98,800<\/p>\n b)\u00a0\u20b98,000<\/p>\n c)\u00a0\u20b97,920<\/p>\n d)\u00a0\u20b99,600<\/p>\n 14)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let the original cost of the TV = C<\/p>\n Cost of TV with 10% GST = $\\frac{110}{100}$C<\/p>\n Given, Cost of TV with 10% GST = \u20b9 8,800<\/p>\n $\\Rightarrow$ \u00a0$\\frac{110}{100}$C =\u00a0 8800<\/p>\n $\\Rightarrow$ \u00a0C = 8000<\/p>\n $\\therefore\\ $The original cost of the TV =\u00a0\u20b9 8,000<\/p>\n Hence, the correct answer is Option B<\/p>\n Question 15:\u00a0<\/b>The average of 24 numbers is 26. The average of the first 15 numbers is 23 and that of the last 8 number is 33. Find the 16$^{th}$ number.<\/p>\n a)\u00a016<\/p>\n b)\u00a017<\/p>\n c)\u00a015<\/p>\n d)\u00a018<\/p>\n 15)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/p>\n Given,\u00a0the average of the first 15 numbers is 23<\/p>\n $\\Rightarrow$\u00a0 Sum of first 15 numbers = 23 x 15 = 345<\/p>\n Average of last 8 numbers is 33<\/p>\n $\\Rightarrow$\u00a0 Sum of last 8 numbers = 33 x 8 = 264<\/p>\n Average of 24 numbers is 26<\/p>\n $\\Rightarrow$\u00a0 Sum of 24 numbers = 26 x 24 = 624<\/p>\n $\\Rightarrow$\u00a0 Sum of first 15 numbers + 16th number +\u00a0Sum of last 8 numbers = 624<\/p>\n $\\Rightarrow$\u00a0 345 + 16th number + 264 = 624<\/p>\n $\\Rightarrow$ \u00a016th number = 15<\/p>\n Hence, the correct answer is Option C<\/p>\n Question 16:\u00a0<\/b>The average of four consecutive odd natural numbers is eight less than the average of three consecutive even natural numbers. If the sum of these three even numbers is equal to the sum of above four odd numbers, then the average of four original odd numbers is:<\/p>\n a)\u00a032<\/p>\n b)\u00a024<\/p>\n c)\u00a018<\/p>\n d)\u00a036<\/p>\n 16)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let the four consecutive odd natural numbers are a, a+2, a+4, a+6<\/p>\n Let the three consecutive even natural numbers are b, b+2, b+6<\/p>\n Average of four consecutive odd natural numbers = $\\frac{a+a+2+a+4+a+6}{4}=\\frac{4a+12}{4}=a+3$<\/p>\n Average of three consecutive even natural numbers = $\\frac{b+b+2+b+4}{3}=\\frac{3b+6}{3}=b+2$<\/p>\n The average of four consecutive odd natural numbers is eight less than the average of three consecutive even natural numbers.<\/p>\n $\\Rightarrow$ \u00a0$a+3=b+2-8$<\/p>\n $\\Rightarrow$\u00a0\u00a0$b-a=9$ ………(1)<\/p>\n The sum of the three even numbers is equal to the sum of four odd numbers.<\/p>\n $\\Rightarrow$ \u00a0$b+b+2+b+4=a+a+2+a+4+a+6$<\/p>\n $\\Rightarrow$\u00a0 $3b=4a+6$<\/p>\n $\\Rightarrow$ \u00a0$3b-4a=6$ ……(2)<\/p>\n Solving 3(1)-(2) we get,\u00a0 $a = 21$<\/p>\n $\\therefore\\ $Average of four consecutive odd natural numbers =\u00a0$a+3=21+3=24$<\/p>\n Hence, the correct answer is Option B<\/p>\n Question 17:\u00a0<\/b>Ravinder invests \u20b9 3,750 which is equal to 15% of his monthly salary in a medical insurance policy. Later he invests 25% and 8% of his monthly salary on a child education policy, and mutual funds, respectively. The total amount left with him is:<\/p>\n a)\u00a0\u20b9 15,000<\/p>\n b)\u00a0\u20b9 8,000<\/p>\n c)\u00a0\u20b9 12,000<\/p>\n d)\u00a0\u20b9 13,000<\/p>\n 17)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let the monthly salary of Ravinder = M<\/p>\n Amount invested in medical insurance policy =\u00a0\u20b9 3,750<\/p>\n Percentage of amount invested in\u00a0medical insurance policy = 15%<\/p>\n $\\Rightarrow$ \u00a0$\\frac{15}{100}$M = 3750<\/p>\n $\\Rightarrow$\u00a0 M = 25000<\/p>\n Monthly salary of Ravinder = \u20b9 25,000<\/p>\n Percentage of salary left with him after investing in\u00a0medical insurance policy,\u00a0child education policy and mutual funds = 100 – 15 – 25 – 8 = 52%<\/p>\n $\\therefore\\ $Total amount left with him = $\\frac{52}{100}\\times25000$ = \u20b9 13,000<\/p>\n Hence, the correct answer is Option D<\/p>\n Question 18:\u00a0<\/b>The average of four consecutive even numbers is 27. By adding which number does the average become 28?<\/p>\n a)\u00a032<\/p>\n b)\u00a030<\/p>\n c)\u00a033<\/p>\n d)\u00a029<\/p>\n 18)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let the four consecutive even numbers are a, a+2, a+4, a+6<\/p>\n Average of four consecutive even numbers is 27<\/p>\n $\\Rightarrow$ \u00a0$\\frac{a+a+2+a+4+a+6}{4}=27$<\/p>\n $\\Rightarrow$ \u00a0$\\frac{4a+12}{4}=27$<\/p>\n $\\Rightarrow$ \u00a0$a+3=27$<\/p>\n $\\Rightarrow$ \u00a0$a=24$<\/p>\n The four consecutive even numbers are 24, 26, 28, 30<\/p>\n Let the number which is to be added = p<\/p>\n After adding p, the new average is 28<\/p>\n $\\Rightarrow$ \u00a0$\\frac{24+26+28+30+p}{5}=28$<\/p>\n $\\Rightarrow$ \u00a0$108+p=140$<\/p>\n $\\Rightarrow$ \u00a0$p=32$<\/p>\n $\\therefore\\ $The required number which is to be added is 32<\/p>\n Hence, the correct answer is Option A<\/p>\n Question 19:\u00a0<\/b>Several students have taken an exam. There was an error in the answer key which affected the marks of 48 students, and their average marks reduced from 78 to 66. The average of remaining students increased by 3.5 marks. This resulted the reduction of the average of all students by 4.5 marks. The number of students that attended the exam is:<\/p>\n a)\u00a096<\/p>\n b)\u00a084<\/p>\n c)\u00a0100<\/p>\n d)\u00a093<\/p>\n 19)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let the total number of students = T<\/p>\n Number of students whose average is reduced = 48<\/p>\n Number of students whose average is increased = T-48<\/p>\n Using Alligation and Mixture rule,<\/p>\n Ratio of 48 and T-48 = 16 : 15<\/p>\n $=$> \u00a0$\\frac{48}{T-48}=\\frac{16}{15}$<\/p>\n $=$>\u00a0 T-48 = 45<\/p>\n $=$>\u00a0 T = 93<\/p>\n $\\therefore\\ $Total number of students = 93<\/p>\n Hence, the correct answer is Option D<\/p>\n \u00a0<\/u><\/p>\n Another method<\/u><\/strong><\/p>\n Let the total number of students = T<\/p>\n Without error<\/u><\/p>\n Average of 48 students = 78<\/p>\n Sum of 48 students = 78 x 48<\/p>\n Let average of remaining T-48 students = p<\/p>\n Sum of T-48 students = p(T-48)<\/p>\n Total marks = (78 x 48) + p(T-48)<\/p>\n Let the average of total students = q<\/p>\n $\\Rightarrow$ (78 x 48) + p(T-48) = Tq………(1)<\/p>\n With error<\/u><\/p>\n Average of 48 students = 66<\/p>\n Sum of 48 students = 66 x 48<\/p>\n Average of T-48 students = (p+3.5)<\/p>\n Sum of T-48 students = (p+3.5)(T-48)<\/p>\n Total marks = (66 x 48)+(p+3.5)(T-48)<\/p>\n Average of total students = (q-4.5)<\/p>\n $\\Rightarrow$ (66 x 48)+(p+3.5)(T-48) = T(q-4.5)<\/p>\n $\\Rightarrow$ (66 x 48)+(p+3.5)(T-48) = Tq – 4.5T<\/p>\n $\\Rightarrow$ (66 x 48)+(p+3.5)(T-48) = (78 x 48) + p(T-48) – 4.5T [From (1)]<\/p>\n $\\Rightarrow$ (p+3.5)(T-48) – p(T-48) + 4.5T = (78 x 48) – (66 x 48)<\/p>\n $\\Rightarrow$ (T-48)(p+3.5-p) + 4.5T = 48(78 – 66)<\/p>\n $\\Rightarrow$ (T-48)3.5 + 4.5T = 48 x 12<\/p>\n $\\Rightarrow$ 3.5T – 48 x 3.5 + 4.5T = 48 x 12<\/p>\n $\\Rightarrow$ 8T = 48 x 15.5<\/p>\n $\\Rightarrow$ T = 93<\/p>\n Total number of students = 93<\/p>\n Hence, the correct answer is Option D Question 20:\u00a0<\/b>A shopkeeper pays 12% of the cost price as tax while purchasing an item whose cost is \u20b9 500. He wants to earn a profit of 20% after giving a discount of 16% on the marked price. So, the marked price should be:<\/p>\n a)\u00a0\u20b9800<\/p>\n b)\u00a0\u20b9780<\/p>\n c)\u00a0\u20b9960<\/p>\n d)\u00a0\u20b9840<\/p>\n 20)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n Given,<\/p>\n Cost of the item =\u00a0\u20b9 500<\/p>\n Tax% = 12%<\/p>\n Cost price of the shopkeeper =\u00a0$500+\\frac{12}{100}\\times500$ = \u20b9 560<\/p>\n Profit% = 20%<\/p>\n Selling price of the item = $\\frac{120}{100}\\times560$ = \u20b9 672<\/p>\n Let the Marked price = M<\/p>\n Discount% = 16%<\/p>\n $=$> \u00a0$\\frac{84}{100}\\times \\text{M}=672$<\/p>\n $=$> \u00a0$\\text{M}=800$<\/p>\n $\\therefore\\ $Marked price =\u00a0\u20b9 800<\/p>\n Hence, the correct answer is Option A<\/p>\n
\nincrease?<\/p>\n![](\"https:\/\/cracku.in\/media\/uploads\/349315.png\")
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\n<\/u><\/strong><\/p>\n