Download Percentage Questions for XAT PDF \u2013 XAT Percentage questions pdf by Cracku. Practice XAT solved Percentage Questions paper tests and These sample questions are to analyse the XAT exam pattern. Top 20 very Important Percentage Questions for XAT based on asked questions in previous exam papers. \u00a0The XAT question papers contain actual questions asked with answers, and solutions.<\/p>\n
Download Percentage Questions for XAT<\/a><\/p>\n Enroll to XAT 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>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 2)\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 3:\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 3)\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 4:\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 4)\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 5:\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 5)\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 6:\u00a0<\/b>The average of some numbers is 54.6. If 75% of the numbers are increased by 5.6 each, and the rest are decreased by 8.4 each, then what is the average of the numbers so obtained?<\/p>\n a)\u00a055.6<\/p>\n b)\u00a055.8<\/p>\n c)\u00a056.7<\/p>\n d)\u00a056.3<\/p>\n 6)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/p>\n New average of the numbers = (average + increment\/decrements)<\/p>\n The average of some numbers =\u00a054.6<\/p>\n 75% of the numbers are increased by 5.6 each, and the rest are decreased by 8.4 each so,<\/p>\n New average of the numbers =\u00a054.6 + (5.6 \\times 75\/100 – 8.4 \\times 25\/100) = 54.6 +\u00a04.2 – 2.1 = 56.7<\/p>\n Take XAT 2023 Mock Tests<\/a><\/p>\n Question 7:\u00a0<\/b>The population of a city increased by 30%in the first year and decreased by 15%\u00a0in the next year. If the present population is 11.050 then the population 2 years ago was:<\/p>\n a)\u00a010,000<\/p>\n b)\u00a099,500<\/p>\n c)\u00a099,000<\/p>\n d)\u00a010,050<\/p>\n 7)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let 2 years ago\u00a0the population of a city be x.<\/p>\n Population after 1 year = x $\\times$ \\frac{130}{100}$ = 1.3x<\/p>\n Population after 2 year =\u00a01.3x\u00a0$\\times$ \\frac{85}{100}$ = 1.105x<\/p>\n 11050 = 1.105x<\/p>\n x = 10000<\/p>\n The population 2 years ago was 10000.<\/p>\n Question 8:\u00a0<\/b>The average of the marks of 30 boys is 88, and when the top two scores were\u00a0excluded, the average marks reduced to 87.5. If the top two scores differ by 2, then the highest\u00a0mark is:<\/p>\n a)\u00a092<\/p>\n b)\u00a096<\/p>\n c)\u00a094<\/p>\n d)\u00a090<\/p>\n 8)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n Total marks of 30 boys = 88 $\\times 30 = 2640<\/p>\n ($\\because$ Average = sum of all term\/number of terms)<\/p>\n when the top two scores were excluded, the average marks reduced to 87.5<\/p>\n Total marks of 28 boys = 87.5 $\\times 28 = 2450<\/p>\n Sum of the\u00a0top two score = 2640 – 2450 = 190 —(1)<\/p>\n difference of two score = 2<\/p>\n Let the top score be x so 2nd top score = x – 2<\/p>\n From eq(1),<\/p>\n x + x – 2 = 190<\/p>\n x = 192\/2 = 96<\/p>\n Question 9:\u00a0<\/b>The average of five consecutive odd numbers is m. If the next three odd numbers are also included, then what is the increase in the average?<\/p>\n a)\u00a03<\/p>\n b)\u00a00<\/p>\n c)\u00a017<\/p>\n d)\u00a08<\/p>\n 9)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let the first odd number is x. Then the five consecutive odd numbers will be<\/p>\n x, x+2, x+4, x+6, x+8 whose sum is<\/p>\n 5x + 20 and the average will be (5x + 20 )\/5 = x+ 4 which is given to be M. So x= M-4<\/p>\n the next 3 odd numbers would be<\/p>\n x+10, x+12, x+14 whose sum is 3x+36<\/p>\n The sum of 10 numbers is (5x + 20)+(3x+36)<\/p>\n = 8x + 56. So the new average is 8(x+7)\/8 = x+7<\/p>\n on substituting x = M-4 in new average, we get<\/p>\n M-4+7 = M+3<\/p>\n The new average is M+3 Question 10:\u00a0<\/b>The average weight of some students in a class was 58.4 kg. When 5 students having the average weight 62.8 kg joined the class, the average weight of all students in the class increased by 0.55 kg. The number of students initially in the class, were:<\/p>\n a)\u00a030<\/p>\n b)\u00a035<\/p>\n c)\u00a025<\/p>\n d)\u00a040<\/p>\n 10)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n The average weight of some students in a class = 58.4 kg<\/p>\n Let the total number of students be n.<\/p>\n Total weight of n students = 58.4n kg<\/p>\n The average weight of 5 students =\u00a062.8 kg<\/p>\n Total weight of 5 students =\u00a062.8 $\\times 5$ = 314 kg<\/p>\n The average weight of (n +\u00a05) students = 58.4 + 0.55 = 58.95 kg<\/p>\n Total weight of\u00a0 (n + 5) students =\u00a058.95(n + 5)\u00a0kg<\/p>\n Total weight of n students +\u00a0Total weight of 5 students =\u00a058.95(n + 5)<\/p>\n 58.4n +\u00a0314 =\u00a058.95(n + 5)<\/p>\n 58.4n + 314 =\u00a058.95n + 294.75<\/p>\n 0.55n = 19.25<\/p>\n n = 1925\/55 = 35<\/p>\n $\\therefore$ 35\u00a0students were\u00a0initially in the class.<\/p>\n Question 11:\u00a0<\/b>The income of A is 60% less than that of B, and the expenditure of A is equal to 60%of B\u2019s expenditure. If A\u2019s income is equal to 70% of B\u2019s expenditure, then whatis the ratio of the savings of A and B?<\/p>\n a)\u00a03 : 8<\/p>\n b)\u00a05 : 9<\/p>\n c)\u00a04 : 7<\/p>\n d)\u00a02 : 15<\/p>\n 11)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let the income of B be 100.<\/p>\n Income of A = 100 $\\times$ 40\/100 = 40<\/p>\n A\u2019s income is equal to 70% of B\u2019s expenditure so,<\/p>\n 40 =\u00a0 70% of B\u2019s expenditure<\/p>\n Expenditure of B = \\frac{40}{70} \\times 100 = 57.14 = 400\/7<\/p>\n Saving of B = 100 – 400\/7 = 300\/7<\/p>\n The expenditure of A is equal to 60%of B\u2019s expenditure so,<\/p>\n Expenditure of\u00a0A =\u00a060% of B\u2019s expenditure<\/p>\n Expenditure of A = 60 $\\times \\frac{400}{7 \\times100}$ = 240\/7<\/p>\n Saving of A = 40 – 240\/7 = 40\/7<\/p>\n The ratio of the savings of A and B = 40\/7 :\u00a0300\/7 = 2 :\u00a015<\/p>\n Question 12:\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 12)\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 13:\u00a0<\/b>Ina class, the average score of thirty students on a test is 69. Later on it was found that the score of one student was wrongly read as 88 instead of 58. The actual average score is:<\/p>\n a)\u00a058<\/p>\n b)\u00a088<\/p>\n c)\u00a069<\/p>\n d)\u00a068<\/p>\n 13)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/p>\n When data read wrongly\u00a0then\u00a0total\u00a0score of thirty students on a test = 69 $\\times$ 30 = 2070<\/p>\n (Average = sum of all terms\/number of terms)<\/p>\n Difference in data = 88 – 58 = 30<\/p>\n Actual score = 2070 – 30 = 2040<\/p>\n Actual average score = 2040\/30 = 68<\/p>\n Question 14:\u00a0<\/b>If the difference between 62% and 80% of a number is 198, then the difference between 92% and 56% of the number will be:<\/p>\n a)\u00a01100<\/p>\n b)\u00a03564<\/p>\n c)\u00a0396<\/p>\n d)\u00a0360<\/p>\n 14)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/p>\n Difference between 62% and 80% = 198<\/p>\n 18% = 198<\/p>\n Difference between 92% and 56% = 36%<\/p>\n 18% $\\times 2 = 198 \\times$ 2<\/p>\n 36% = 396<\/p>\n Difference between 92% and 56% = 396<\/p>\n Question 15:\u00a0<\/b>The average of five consecutive even numbers is M.If the next five even numbers are also included, the average of ten numbers will be:<\/p>\n a)\u00a0M + 5<\/p>\n b)\u00a011<\/p>\n c)\u00a010<\/p>\n d)\u00a0M + 10<\/p>\n 15)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n The average of five consecutive even numbers = M<\/p>\n Sum of the 5 number = 5M<\/p>\n Let the first even number be x.<\/p>\n $\\frac{x + x + 2 + x + 4 +\u00a0x +\u00a06 +\u00a0x +\u00a08}{5} = M$<\/p>\n 5x = 5M – 20<\/p>\n x = M – 4<\/p>\n Next 5 numbers = x + 10,\u00a0x + 12,\u00a0x + 14,\u00a0x + 16,\u00a0x + 18,<\/p>\n On put the value of x,<\/p>\n Numbers = M + 6, M + 8, M + 10, M + 12, M + 12<\/p>\n average of ten numbers = $\\frac{5M +\u00a0M + 6 + M + 8 +\u00a0M + 10 +\u00a0M + 12 + M + 14}{10}$<\/p>\n = (10M +\u00a050)\/10 = M + 5<\/p>\n Question 16:\u00a0<\/b>The average score in Mathematics of 90 students of sections A and B together is 49. The number of students in A was 25% more than that of B, and the average score of the students in B was 20% higherthan that of the students in A. What is the average score of the students in A?<\/p>\n a)\u00a044.5<\/p>\n b)\u00a045<\/p>\n c)\u00a044<\/p>\n d)\u00a045.5<\/p>\n 16)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let the number of students in section B be x.<\/p>\n The number of students in A was 25% more than that of B. So,<\/p>\n The number of students in A = x $\\times \\frac{125}{100}$ = 1.25x<\/p>\n Total student = 90<\/p>\n x + 1.25x = 90<\/p>\n x = 40<\/p>\n The number of students in B = 40<\/p>\n The number of students in A = 90 – 40 = 50<\/p>\n Let the average score of the section A be p. Average score of the section B = 120% of p = 1.2p<\/p>\n Total score of the section B = 40 $\\times 1.2p = 48p<\/p>\n Total score of 90 students = 49 $\\times 90 = 4410<\/p>\n Total score of the section A +\u00a0Total score of\u00a0the section B = 4410<\/p>\n 50p + 48p = 4410<\/p>\n p = 4410\/98 = 45<\/p>\n $\\therefore$Average score of the section A is 45.<\/p>\n Question 17:\u00a0<\/b>By what number must the given number be multiplied to increase the number by 25%<\/p>\n a)\u00a0$\\frac{3}{4}$<\/p>\n b)\u00a0$\\frac{5}{4}$<\/p>\n c)\u00a0$\\frac{2}{5}$<\/p>\n d)\u00a03<\/p>\n 17)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n Number to increase 25% = 125% = 125\/100 = 5\/4<\/p>\n Question 18:\u00a0<\/b>The average of 60 student\u2019s results is 38. If the average of the first 22 students is 36, and that of the last 32 students is 32, then the average result of the remaining students is:<\/p>\n a)\u00a081<\/p>\n b)\u00a077<\/p>\n c)\u00a065<\/p>\n d)\u00a052<\/p>\n 18)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n Total sum of result of 60 students = average $\\times number of students = 38 \\times 60 = 2280$<\/p>\n Sum of result of first 22\u00a0students = $ 36 \\times 22$ =\u00a0792<\/p>\n Sum of result of last 32 students = $ 32 \\times 32$ = 1024<\/p>\n Sum of result of remaining students =\u00a02280 – 792 – 1024 = 462<\/p>\n Number of remaining students = 60 – 22 – 32 = 6<\/p>\n Average result of the remaining students = 462\/6 = 77<\/p>\n Question 19:\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 19)\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 20:\u00a0<\/b>24 students collected money for donation. The average contribution was \u20b950. Later on, their teacher also contributed some money. Now the average contribution is \u20b956. The teacher\u2019s contribution is:<\/p>\n a)\u00a0\u20b956<\/p>\n b)\u00a0\u20b9200<\/p>\n c)\u00a0\u20b9106<\/p>\n d)\u00a0\u20b9194<\/p>\n 20)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n Total money contribution by 24 students = average $\\times no. of students = 50 \\times 24$ = 1200<\/p>\n Total money contribution by 24 students and their teacher =\u00a056 $\\times 25$ = 1400<\/p>\n Teacher\u2019s contribution = 1400 – 1200 = Rs.200<\/p>\n Question 21:\u00a0<\/b>In a school, 4 % of the students did not appear for the annual exams. 10% of the students who appeared for the exams could not pass the exam. Out of the remaining students, 50% got distinction marks and 432 students passed the exam but could not get distinction marks. The total number of students in the school is:<\/p>\n a)\u00a01000<\/p>\n b)\u00a01200<\/p>\n c)\u00a0960<\/p>\n d)\u00a0878<\/p>\n 21)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let the total students in the school be 100%.<\/p>\n Appeared student in the exam = 100 – 4 = 96%<\/p>\n Number of student which are not pass in the exam = 96% $\\times \\frac{10}{100}$ = 9.6%<\/p>\n Number of student which are pass in the exam = 96 – 9.6 = 86.4%<\/p>\n Number of student which are not get distinction marks in the exam = 432<\/p>\n 50% of the remaining students = 432<\/p>\n 86.4% $\\times \\frac{50}{100}$ = 432<\/p>\n 43.2% = 432<\/p>\n Total students(100%) = $\\frac{432}{43.2} \\times$ 100 = 1000<\/p>\n Question 22:\u00a0<\/b>A manufacturer sells cooking gas stoves to shopkeepers at 10% profit, and in turn they sell the cooking gas stoves to customer to earn 15% profit. If a customer gets a cooking gas stove for \u20b97,590, then what is its manufacturing cost?<\/p>\n a)\u00a0\u20b9 6,500<\/p>\n b)\u00a0\u20b95,000<\/p>\n c)\u00a0\u20b95,090<\/p>\n d)\u00a0\u20b96,000<\/p>\n 22)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/p>\n Selling price for shopkeeper = Rs.7590<\/p>\n Profit = 15%<\/p>\n Cost price\u00a0for shopkeeper = $\\frac{7590}{115} \\times 100$ = Rs.6600<\/p>\n Selling price for\u00a0manufacturer =\u00a0Rs.6600<\/p>\n Profit = 10%<\/p>\n Cost price for\u00a0manufacturer = $\\frac{6600}{110} \\times 100$ = Rs.6000<\/p>\n Question 23:\u00a0<\/b>The average of 4 terms is 30 and the Ist term is $\\frac{1}{3}$ of the sum of the remaining terms. What is the first term?<\/p>\n a)\u00a060<\/p>\n b)\u00a030<\/p>\n c)\u00a040<\/p>\n d)\u00a020<\/p>\n 23)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n Sum of the 4 terms = average $\\times$ no. of terms = 30 $\\times$ 4 = 120<\/p>\n 1st term =\u00a0 $\\frac{1}{3}$ of the sum of the remaining terms<\/p>\n 1st term = $\\frac{1}{3}$ of the sum of the (all terms – )<\/p>\n $\\frac{first term}{3} +\u00a0first term = 120 \\times\u00a0\\frac{1}{3}$<\/p>\n 1st term\u00a0$\\times$ $\\frac{4}{3}$ = 40<\/p>\n 1st term = 30<\/p>\n Question 24:\u00a0<\/b>When 2 is subtracted from each of the given n numbers, then the sum of the numbers so obtained is 102. When 5 is subtracted from each of them, then the sum of the numbers so obtained is 12. What is the average of the given n numbers?<\/p>\n a)\u00a06.2<\/p>\n b)\u00a06.6<\/p>\n c)\u00a05.8<\/p>\n d)\u00a05.4<\/p>\n 24)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let for ‘n’ numbers the average be ‘x’.<\/p>\n So, the total sum of ‘n’ numbers would be ‘nx’.<\/p>\n If 2 is subtracted from each ‘n’ numbers, then the resulted value to be subtracted becomes = 2n<\/p>\n Thus, value of the total sum now = (nx – 2n)<\/p>\n Given that, this value equals to 102.<\/p>\n So, nx – 2n = 102 …(1)<\/p>\n Again when 5 is subtracted from each ‘n’ numbers, then the resulted value to be subtracted becomes = 5n<\/p>\n Thus, value of the total sum now = (nx – 5n)<\/p>\n Given that, this value equals to 12.<\/p>\n So, nx – 5n = 12 …(2)<\/p>\n Subtracting (2) from (1), we get:<\/p>\n nx – 2n – (nx – 5n) = 102 – 12<\/p>\n \u21d2 -2n + 5n = 90<\/p>\n \u21d2 3n = 90<\/p>\n \u21d2 n = 90\/3 = 30<\/p>\n There are 30 numbers.<\/p>\n Putting n = 30, in eqn.(1), we get:<\/p>\n (30)x – 2(30) = 102<\/p>\n \u21d2 30x – 60 = 102<\/p>\n \u21d2 30x = 162<\/p>\n \u21d2 x = 162\/30 = 5.4<\/p>\n $\\therefore$ The average of 30 numbers is 5.4.<\/p>\n Question 25:\u00a0<\/b>The average height of 5 boys is 175 cm. A sixth boy joined the group and the average height of all the boys in the group now increased by 1 cm. The height of the sixth boy is:<\/p>\n a)\u00a0180 cm<\/p>\n b)\u00a0181 cm<\/p>\n c)\u00a0175 cm<\/p>\n d)\u00a0179 cm<\/p>\n 25)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n The average height of 5 boys = 175 cm<\/p>\n Average = sum of all terms\/number of terms<\/p>\n Sum of the height of 5 boys = 175 $\\times$ 5 = 875<\/p>\n Average height after joining of 6th boys = 175 + 1 = 176<\/p>\n Sum of the height of 6 boys = 176 $\\times$ 6 = 1056<\/p>\n Height of 6th boy = 1056 – 875 = 181 cm<\/p>\n
\nincrease?<\/p>\n
\nIncrement in the average = m + 3 – m = 3<\/p>\n
\nTotal score of the section A = number of students $times$ average = 50p<\/p>\n