Enroll to XAT 2023 Crash Course<\/a><\/p>\n Question 1:\u00a0<\/b>What is the difference of mean and median of the given data : 4, 13, 8, 15, 9, 21, 18, 23, 35, 1?<\/p>\n a)\u00a00.7<\/p>\n b)\u00a01.7<\/p>\n c)\u00a01.2<\/p>\n d)\u00a02.1<\/p>\n 1)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n Mean:<\/p>\n No. of samples (n) = 10<\/p>\n Mean = $\\frac{\\sum x}{n} = \\frac{4+13+8+15+9+21+18+23+35+1}{10}= \\frac{147}{10} = 14.7$<\/p>\n Median:<\/p>\n Arranging the data in ascending order, we get:<\/p>\n 1, 4, 8, 9, 13, 15, 18, 21, 23, 35<\/p>\n n = 10 (even)<\/p>\n Therefore, median is the average of 5th and 6th term.<\/p>\n Median = $\u00a0\u00a0\\frac{13+15}{2} \u00a0= 14$<\/p>\n Mean – Median = 14.7 – 14 = 0.7<\/p>\n Therefore, Option A is correct.<\/p>\n Question 2:\u00a0<\/b>60% of a number is 168, then what is the number?<\/p>\n a)\u00a0280<\/p>\n b)\u00a0320<\/p>\n c)\u00a0240<\/p>\n d)\u00a0200<\/p>\n 2)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n 60% of the number is 168.<\/p>\n Let’s assume the number is ‘y’.<\/p>\n 60% of y =\u00a0168<\/p>\n 0.6y = 168<\/p>\n y =\u00a0280<\/p>\n Question 3:\u00a0<\/b>What is the value of: $5\u00a0\u00a0of\u00a0 5\u00a0 of\u00a0 5 \\div 5 + 5 – 6 \\div 3 \\times 4 + 2 + (3 \\div 6 \\times 2)?$<\/p>\n a)\u00a021<\/p>\n b)\u00a025<\/p>\n c)\u00a028<\/p>\n d)\u00a019<\/p>\n 3)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n $5\\times5\\times\\frac{5}{5}+5-\\frac{6}{3}\\times4+2+\\frac{3}{6}\\times2$<\/p>\n $25+5-8+2+1$<\/p>\n 25<\/p>\n Question 4:\u00a0<\/b>What is the median of the given data? a)\u00a061<\/p>\n b)\u00a058<\/p>\n c)\u00a057<\/p>\n d)\u00a055<\/p>\n 4)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n Arrange the given data in ascending order.<\/p>\n 41, 43, 46, 50, 55, 61, 68, 76, 85, 95<\/p>\n n = number of given data<\/p>\n n = 10<\/p>\n median =\u00a0$\\frac{\\left(\\frac{n}{2}\\right)^{th\\ term}\\ +\\ \\left(\\frac{n}{2}+1\\right)^{th\\ term}}{2}$<\/p>\n =\u00a0$\\frac{\\left(\\frac{10}{2}\\right)^{th\\ term}\\ +\\ \\left(\\frac{10}{2}+1\\right)^{th\\ term}}{2}$<\/p>\n = $\\frac{5^{th\\ term}\\ +\\ \\left(5+1\\right)^{th\\ term}}{2}$<\/p>\n = $\\frac{5^{th\\ term}\\ +\\ 6^{th\\ term}}{2}$<\/p>\n = $\\frac{55+61}{2}$<\/p>\n =\u00a0$\\frac{116}{2}$<\/p>\n = 58<\/p>\n Question 5:\u00a0<\/b>If A is the smallest three-digit number divisible by both 6 and 7 and B is the largest four-digit number divisible by both 6 and 7, then what is the value of (B – A)?<\/p>\n a)\u00a09912<\/p>\n b)\u00a09870<\/p>\n c)\u00a09996<\/p>\n d)\u00a09954<\/p>\n 5)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n LCM of (6, 7) = 42<\/p>\n 6 = $2\\times3$<\/p>\n 7 = 7<\/p>\n As per the information given in the question, the value of both A and B will be\u00a0divisible by both 6 and 7. The LCM of these is 42. It means that\u00a0the value of both A and B will be the multiple of 42.<\/p>\n If A is the smallest three-digit number divisible by both 6 and 7.<\/p>\n The\u00a0smallest three-digit number is 100. When we divide 100 by 42, then the remainder will be\u00a016. So we need to add (42-16) in 100.<\/p>\n So A = 100+(42-16)<\/p>\n A = 100+26<\/p>\n A =\u00a0126<\/strong><\/p>\n B is the largest four-digit number divisible by both 6 and 7.<\/p>\n The\u00a0largest four-digit number is 9999. When we 9999 by 42, then the remainder will be 3. So we need to subtract 3 from 9999.<\/p>\n So B = 9999-3<\/p>\n B = 9996<\/strong><\/p>\n Value of (B – A) =\u00a09996-126<\/p>\n =\u00a09870<\/p>\n Take XAT 2023 Mock Tests<\/a><\/p>\n Question 6:\u00a0<\/b>A set of data is as under: 4, 2, 3, 2, 7, 4, 8, 5, 2, 4, 5, 6, 2, 5, 6, 6, 5, 4, 6, 5, 3, 5, 4, 3 What is the mode of the set?<\/p>\n a)\u00a02<\/p>\n b)\u00a05<\/p>\n c)\u00a06<\/p>\n d)\u00a04<\/p>\n 6)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n 4, 2, 3, 2, 7, 4, 8, 5, 2, 4, 5, 6, 2, 5, 6, 6, 5, 4, 6, 5, 3, 5, 4, 3<\/p>\n Arrange the given data in ascending order.<\/p>\n 2, 2, 2, 2, 3, 3, 3,\u00a04, 4,\u00a04,\u00a04, 4, 5, 5,\u00a05, 5, 5, 5,\u00a06, 6, 6, 6, 7, 8<\/p>\n Mode = maximum number of times a number is available in the given data set<\/p>\n = 5 (It is available six times in the given data set.)<\/p>\n Question 7:\u00a0<\/b>The median of the first seven prime numbers is:<\/p>\n a)\u00a07<\/p>\n b)\u00a05<\/p>\n c)\u00a013<\/p>\n d)\u00a011<\/p>\n 7)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n first, seven prime numbers are 2, 3, 5, 7, 11, 13, 17.<\/p>\n Here the number of terms are 7. So n = 7.<\/p>\n median =\u00a0$\\left(\\frac{n+1}{2}\\right)^{th}\\ term$<\/p>\n =\u00a0$\\left(\\frac{7+1}{2}\\right)^{th}\\ term$<\/p>\n =\u00a0$\\left(\\frac{8}{2}\\right)^{th}\\ term$<\/p>\n = $4^{th}\\ term$<\/p>\n = 7<\/p>\n Question 8:\u00a0<\/b>What is the sum of the digits of the largest six-digit number divisible by 3, 4, 5 and 6?<\/p>\n a)\u00a045<\/p>\n b)\u00a042<\/p>\n c)\u00a039<\/p>\n d)\u00a048<\/p>\n 8)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n As we know that the largest\u00a0six-digit number is 999999.<\/p>\n The required number should be divisible by 3, 4, 5 and 6.<\/p>\n So the LCM of\u00a03, 4, 5 and 6 is 60.<\/p>\n When\u00a0999999 is divided by 60, then the remainder will be 39.<\/p>\n $999999=60\\times16666+39$<\/p>\n Required number =\u00a0999999-39 =\u00a0999960<\/p>\n the sum of the digits of the required\u00a0number =\u00a09+9+9+9+6+0 =\u00a042<\/strong><\/p>\n Question 9:\u00a0<\/b>What is the difference between the mean and median of the given data: a)\u00a05<\/p>\n b)\u00a01.5<\/p>\n c)\u00a03<\/p>\n d)\u00a04.5<\/p>\n 9)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n $Mean=\\frac{sum\\ of\\ data}{number\\ of\\ data}$<\/p>\n $=\\frac{4+6+3+7+10+13+16+5}{8}$<\/p>\n $=\\frac{64}{8}$<\/p>\n = 8<\/p>\n For\u00a0median, first we need to arrange the given data in ascending order from left to right.<\/p>\n 3,\u00a04,\u00a05,\u00a06,\u00a07,\u00a010, 13, 16<\/p>\n n = number of data = 8<\/p>\n median =\u00a0$\\frac{\\left(\\frac{n}{2}\\right)^{th}\\ term\\ +\\ \\left(\\frac{n}{2}+1\\right)^{th}\\ term}{2}$<\/p>\n = $\\frac{\\left(\\frac{8}{2}\\right)^{th}\\ term\\ +\\ \\left(\\frac{8}{2}+1\\right)^{th}\\ term}{2}$<\/p>\n = $\\frac{4^{th}\\ term\\ +\\ \\left(4+1\\right)^{th}\\ term}{2}$<\/p>\n = $\\frac{4^{th}\\ term\\ +\\ 5^{th}\\ term}{2}$<\/p>\n = $\\frac{6+7}{2}$<\/p>\n = $\\frac{13}{2}$<\/p>\n = 6.5<\/p>\n difference between the mean and median = 8-6.5<\/p>\n = 1.5<\/p>\n Question 10:\u00a0<\/b>What is the difference between the mean and the median of the following data: a)\u00a02.3<\/p>\n b)\u00a00.4<\/p>\n c)\u00a01.8<\/p>\n d)\u00a01.1<\/p>\n 10)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/p>\n First, arrange the given data in ascending order from left to right.<\/p>\n 2,\u00a05, 7, 8, 9, 10, 12,\u00a013, 14, 26<\/p>\n There are ten numbers. So n = 10.<\/p>\n median =\u00a0$\\frac{\\left(\\frac{n}{2}\\right)^{th}\\ term\\ +\\ \\left(\\frac{n}{2}+1\\right)^{th}\\ term}{2}$<\/p>\n =\u00a0$\\frac{\\left(\\frac{10}{2}\\right)^{th}\\ term\\ +\\ \\left(\\frac{10}{2}+1\\right)^{th}\\ term}{2}$<\/p>\n = $\\frac{5^{th}\\ term\\ +\\ \\left(5+1\\right)^{th}\\ term}{2}$<\/p>\n = $\\frac{5^{th}\\ term\\ +\\ 6^{th}\\ term}{2}$<\/p>\n =\u00a0$\\frac{9+10}{2}$<\/p>\n = $\\frac{19}{2}$<\/p>\n = 9.5<\/p>\n mean =\u00a0$\\frac{sum\\ of\\ data}{number\\ of\\ data}$<\/p>\n =\u00a0$\\frac{2+5+7+8+9+10+12+13+14+26}{10}$<\/p>\n =\u00a0$\\frac{106}{10}$<\/p>\n =\u00a010.6<\/p>\n difference between the mean and the median =\u00a010.6-9.5<\/p>\n = 1.1<\/p>\n Question 11:\u00a0<\/b>Let x be the median of data: 33, 42, 28, 49, 32, 37, 52, 57, 35, 41. a)\u00a078<\/p>\n b)\u00a078.5<\/p>\n c)\u00a079.5<\/p>\n d)\u00a079<\/p>\n 11)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n 33, 42, 28, 49, 32, 37, 52, 57, 35, 41<\/p>\n Arrange the given data in ascending order from left to right.<\/p>\n 28, 32, 33, 35,\u00a037, 41,\u00a042, 49, 52, 57<\/p>\n Median =\u00a0$\\frac{\\left(\\frac{n}{2}\\right)^{th}\\ term\\ +\\left(\\frac{n}{2}+1\\right)^{th}\\ term}{2}$<\/p>\n Here n = the number of data<\/p>\n Median = x =\u00a0$\\frac{\\left(\\frac{10}{2}\\right)^{th}\\ term\\ +\\left(\\frac{10}{2}+1\\right)^{th}\\ term}{2}$<\/p>\n =\u00a0$\\frac{5^{th}\\ term\\ +\\left(5+1\\right)^{th}\\ term}{2}$<\/p>\n = $\\frac{5^{th}\\ term\\ +6^{th}\\ term}{2}$<\/p>\n =\u00a0$\\frac{37+41}{2}$<\/p>\n = $\\frac{78}{2}$<\/p>\n x = 39<\/strong><\/p>\n If 32 is replaced by 36 and 41 by 63, then the median of the data, so obtained, is y.<\/p>\n 28, 36, 33, 35, 37, 63, 42, 49, 52, 57<\/p>\n Arrange the given data in ascending order from left to right.<\/p>\n 28, 33, 35, 36,\u00a037,\u00a042, 49, 52, 57,\u00a063<\/p>\n Median = y = $\\frac{\\left(\\frac{10}{2}\\right)^{th}\\ term\\ +\\left(\\frac{10}{2}+1\\right)^{th}\\ term}{2}$<\/p>\n = $\\frac{5^{th}\\ term\\ +\\left(5+1\\right)^{th}\\ term}{2}$<\/p>\n = $\\frac{5^{th}\\ term\\ +6^{th}\\ term}{2}$<\/p>\n = $\\frac{37 + 42}{2}$<\/p>\n =\u00a0$\\frac{79}{2}$<\/p>\n y =\u00a039.5<\/strong><\/p>\n value of (x + y)<\/strong> =\u00a0(39+39.5)<\/p>\n =\u00a078.5<\/p>\n Question 12:\u00a0<\/b>The following table shows the weight of 20 students: a)\u00a051 and 48<\/p>\n b)\u00a053 and 56<\/p>\n c)\u00a060 and 53<\/p>\n d)\u00a048 and 53<\/p>\n 12)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/p>\n Mode is the maximum number of students have the same weight. So here 6 students have 48 kg weight. Hence 48 will be the mode.<\/strong><\/p>\n For\u00a0Median, first, we need to arrange the given weight in the ascending order from left to right.<\/p>\n Total number of students = n =\u00a020<\/p>\n Median =\u00a0$\\frac{\\left(\\frac{n}{2}\\right)th+\\left(\\frac{n}{2}+1\\right)th}{2}$<\/p>\n =\u00a0$\\frac{\\left(\\frac{20}{2}\\right)th+\\left(\\frac{20}{2}+1\\right)th}{2}$<\/p>\n = $\\frac{\\left(10\\right)th+\\left(10+1\\right)th}{2}$<\/p>\n =\u00a0$\\frac{\\left(10\\right)th+\\left(11\\right)th}{2}$<\/p>\n =\u00a0$\\frac{53+53}{2}$<\/p>\n = 53<\/p>\n Question 13:\u00a0<\/b>The mode of the given data 2, 5, 5, 7, 2, 6, 8, 6, 9, 6 is:<\/p>\n a)\u00a07<\/p>\n b)\u00a06<\/p>\n c)\u00a05<\/p>\n d)\u00a02<\/p>\n 13)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n Arrange the given data in ascending order which is given below.<\/p>\n 2, 2,\u00a05, 5,\u00a06, 6, 6, 7,\u00a08, 9<\/strong><\/p>\n The maximum number of times a number is occurring is called mode. Here ‘6’ is occurring three times. So ‘6’ will be the mode.<\/p>\n Question 14:\u00a0<\/b>x is the greatest number by which, when 2460, 2633 and 2806 are divided, the remainder in each case is the same. What is the sum of digits of x?<\/p>\n a)\u00a011<\/p>\n b)\u00a010<\/p>\n c)\u00a013<\/p>\n d)\u00a09<\/p>\n 14)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n As given in the question, the\u00a0remainder is the same in each. So (2633-2460),\u00a0(2806-2460) and\u00a0(2806-2633) should be completely divisible by ‘x’.<\/p>\n (2633-2460) = 173<\/p>\n (2806-2460) =\u00a0346<\/p>\n (2806-2633) = 173<\/p>\n As we know that\u00a0173 is a prime number. So the HCF of (173, 346 and\u00a0173) will be\u00a0173.<\/p>\n Hence x = 173<\/p>\n Sum of digits of x =\u00a01+7+3<\/p>\n = 11<\/p>\n Question 15:\u00a0<\/b>What is the sum of the median and mode of the data a)\u00a013<\/p>\n b)\u00a011<\/p>\n c)\u00a012<\/p>\n d)\u00a014<\/p>\n 15)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n Arrange the given data in ascending order which is given below.<\/p>\n 1,\u00a01, 1, 2, 3, 3, 4,\u00a05, 5,\u00a06, 6, 6, 6,\u00a07,\u00a08,\u00a09, 9<\/p>\n Median = middle term of the given data after arranging them in\u00a0ascending order<\/p>\n Here 17 numbers are given. So the median will be the 9th number from the left end. It’s\u00a05.<\/p>\n Mode = maximum number of times a number is occurring in the given data<\/p>\n = 6 (It is coming four times in the given data)<\/p>\n Sum of the median and mode of the data = 5+6<\/p>\n = 11<\/p>\n Question 16:\u00a0<\/b>The mode of 2, 2, 3, 3, 5, 5, 5, 7, 8, 8, 9, 10 is:<\/p>\n a)\u00a05<\/p>\n b)\u00a02<\/p>\n c)\u00a03<\/p>\n d)\u00a06<\/p>\n 16)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n Mode : The value that appears most often in a set of given data values.<\/p>\n Given Data :\u00a02, 2, 3, 3, 5, 5, 5, 7, 8, 8, 9, 10<\/p>\n So, Mode of the given data is 5.<\/p>\n Hence,\u00a0Option\u00a0<\/strong>A\u00a0<\/strong>is correct.<\/p>\n Question 17:\u00a0<\/b>The mode of the following data is 36. What is the value of x ? a)\u00a011<\/p>\n b)\u00a015<\/p>\n c)\u00a013<\/p>\n d)\u00a012<\/p>\n 17)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/p>\n As per given data,<\/p>\n Class interval of 30-40 has highest frequency, thatswhy it is modal class<\/p>\n As we know,<\/p>\n M =\u00a0$l+\\left\\{\\frac{\\left(f_1-f_0\\right)}{2f_1-f_0-f_2}\\right\\}\\times\\ h$<\/p>\n where, h= size of the class interval,<\/p>\n l = lower limit of the modal class,<\/p>\n $f_{1=}$ frequency of the modal class,<\/p>\n $f_{_0=}$ frequency of the class preceding the modal class<\/p>\n $f_{_2=}$ frequency of the class succeeding the modal class<\/p>\n putting the values from the given data\u00a0:<\/p>\n $36=30+\\frac{\\left(16-10\\right)}{2\\times\\ 16-10-x}\\times\\ 10$<\/p>\n $36-30=\\frac{6}{22-x}\\times\\ \\ 10$<\/p>\n $22-x=10$<\/p>\n $x=12$<\/p>\n Hence, Option D<\/strong> is correct.<\/p>\n Question 18:\u00a0<\/b>When 6892, 7105 and 7531 are divided by the greatest number x, then the remainder in each case is y. What is the value of (x – y)?<\/p>\n a)\u00a0123<\/p>\n b)\u00a0137<\/p>\n c)\u00a0147<\/p>\n d)\u00a0113<\/p>\n 18)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n We have to find HCF of given numbers : 6892, 7105, 7531<\/p>\n 7105 – 6892 = 213<\/p>\n 7531 – 7105 = 426<\/p>\n 426 – 213 = 213<\/p>\n So, Either the difference or the factor of difference is the\u00a0HCF of those given number.<\/p>\n Here , 213 is the HCF.<\/p>\n When 6892, 7105, 7531 is divided by 213 we get 76 as an remainder<\/p>\n So, x = 213 and y = 76<\/p>\n According to Question :<\/p>\n x – y = 213 – 76 = 137<\/p>\n Hence,\u00a0Option<\/strong> B<\/strong> is correct.\u00a0\u00a0<\/strong><\/p>\n Question 19:\u00a0<\/b>The sum of the perfect square between 120 and 300 is:<\/p>\n a)\u00a01400<\/p>\n b)\u00a01024<\/p>\n c)\u00a01296<\/p>\n d)\u00a01204<\/p>\n 19)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n Sum of the squares of n consecutive numbers =<\/p>\n The sum of the perfect square between 120 and 300 =\u00a0$11^2+12^2+13^2+14^2+15^2+16^2+17^2$<\/p>\n $=\\frac{17\\left(17+1\\right)\\left(2\\left(17+1\\right)\\right)}{6}-\\frac{10\\left(10+1\\right)\\left(2\\left(10\\right)+1\\right)}{6}$<\/p>\n $=\\frac{17\\left(18\\right)\\left(35\\right)}{6}-\\frac{10\\left(11\\right)\\left(21\\right)}{6}$<\/p>\n $=51\\times35-11\\times35$<\/p>\n $=35\\left(51-11\\right)$<\/p>\n $=35\\left(40\\right)$<\/p>\n $=1400$<\/p>\n Hence, the correct answer is Option A<\/p>\n Question 20:\u00a0<\/b>The difference between the greatest and the least four digit numbers that begins with 3 and ends with 5 is:<\/p>\n a)\u00a0990<\/p>\n b)\u00a0900<\/p>\n c)\u00a0909<\/p>\n d)\u00a0999<\/p>\n 20)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n The greatest four digit number that begins with 3 and ends with 5 = 3995<\/p>\n The least four digit number that begins with 3 and ends with 5 = 3005<\/p>\n $\\therefore\\ $The difference between the greatest and the least four digit numbers that begins with 3 and ends with 5 = 3995 – 3005 = 990<\/p>\n Question 21:\u00a0<\/b>If a nine-digit number 389 x 6378 y is divisible by 72, then the value of $\\sqrt{6x + 7y}$ will be:<\/p>\n a)\u00a06<\/p>\n b)\u00a0$\\sqrt{13}$<\/p>\n c)\u00a0$\\sqrt{46}$<\/p>\n d)\u00a08<\/p>\n 21)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/p>\n 389x6378y is divisible by 72,<\/p>\n Factor of\u00a072 = 8 $\\times$ 9<\/p>\n So, number is divisible by 8 and 9 both.<\/p>\n Divisibility rule for 8,<\/p>\n 78y (last three digits should be divisible by 8)<\/p>\n 784 is divisible by 8 so, Divisibility rule of 9,<\/p>\n 3 + 8 + 9 + x + 6 + 3 + 7+ 8 +\u00a04<\/p>\n = 48 + x<\/p>\n 54 is divisible by 9<\/p>\n So, x = 54 – 48 = 6 Question 22:\u00a0<\/b>When 7897, 8110 and 8536 are divided by the greatest number x, then the remainder in each case is the same. The sum of the digits of x is:<\/p>\n a)\u00a014<\/p>\n b)\u00a05<\/p>\n c)\u00a09<\/p>\n d)\u00a06<\/p>\n 22)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let the remainder be k. Question 23:\u00a0<\/b>One of the factors of $(8^{2k} + 5^{2k})$, where k is an odd number, is:<\/p>\n a)\u00a086<\/p>\n b)\u00a088<\/p>\n c)\u00a084<\/p>\n d)\u00a089<\/p>\n 23)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/p>\n $(8^{2k} + 5^{2k})$,k is odd nuber so, Question 24:\u00a0<\/b>The sum of the digits of a two-digit number is $\\frac{1}{7}$ of the number. The units digit is 4 less than the tens digit. If the number obtained on reversing its digits is divided by 7, the remainder will be:<\/p>\n a)\u00a04<\/p>\n b)\u00a05<\/p>\n c)\u00a01<\/p>\n d)\u00a06<\/p>\n 24)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let the number be (10a + b). Question 25:\u00a0<\/b>If the 11-digit number 5678x43267y is divisible by 72, then the value of $\\sqrt{5x + 8y}$ is:<\/p>\n a)\u00a06<\/p>\n b)\u00a04<\/p>\n c)\u00a07<\/p>\n d)\u00a08<\/p>\n 25)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n 11-digit number 5678x43267y is divisible by 72.
\n41, 43, 46, 50, 85, 61, 76, 55, 68, 95<\/p>\n
\n4, 6, 3, 7, 10, 13, 16 and 5?<\/p>\n
\n5, 7, 8, 13, 12, 14, 9, 2, 26, 10 ?<\/p>\n
\nIf 32 is replaced by 36 and 41 by 63, then the median of the data, so obtained, is y. What is the value of (x + y)?<\/p>\n
\n![](\"https:\/\/cracku.in\/media\/uploads\/Screenshot_58_m0YRzhK.png\")
\nThe mode and median of the above data are, respectively:<\/p>\n![](\"https:\/\/cracku.in\/media\/uploads\/DeepinScreenshot_select-area_20220506102810.png\")
<\/p>\n
\n8, 1, 5, 4, 9, 6, 3, 6, 1, 3, 6, 9, 1, 7, 2, 6, 5 ?<\/p>\n\n
\n![](\"https:\/\/cracku.in\/media\/uploads\/Screenshot_5_rSrqvwb.png\")
<\/p>\n
\nValue of y = 4<\/p>\n
\nValue of $\\sqrt{6x + 7y}$
\n= $\\sqrt{6 \\times 6 + 7 \\times 4}$
\n= $\\sqrt{36 + 28}$ =$\\sqrt{64}$
\n= 8<\/p>\n
\n7897 – k = ax
\n8110 – k = bx
\n8536 – k = cx
\nCommon factor is x.
\nSo difference between the numbers,
\n8110 – 7897 = 213
\n8536 – 8110 = 426
\n8536 – 7897 = 639
\nHCF of 213, 426 and 639 is 213.
\nx = 213
\nSum of the digits of x = 2 +\u00a01 +\u00a03 = 6<\/p>\n
\nLet the k be 1.
\n=\u00a0$(8^{2} + 5^{2})$
\n= 64 +\u00a025 = 89<\/p>\n
\nATQ,
\na + b = $\\frac{10a + b}{7}$
\n7a + 7b = 10a +\u00a0b
\n6b = 3a
\n2b = a —(1)
\na – b = 4 —(2)
\nFrom eq (1) and (2),
\n2b – b = 4
\nb = 4
\na = 4 $\\times 2$ = 8
\nNumber =\u00a010a + b = 10 $\\times 8 + 4 = 84
\nreverse of the number = 48
\nRemainder after divide by 7 = 48\/7 = 6<\/p>\n
\nIt will be divisible by 9 and 8.
\nFor the divisiblity by 8,
\n67y divisible by 8.
\nSo value of y =\u00a0 2
\nFor the divisiblity by 9,
\n(5+6+7+8+x+4+3+2+6+7+y = 50 + x) divisible by 8.
\nSo value of x = 54 – 50 = 4
\n$\\sqrt{5x + 8y}$
\n=\u00a0$\\sqrt{5\\times 4 + 8\\times 2}$
\n= $\\sqrt{36}$ = 6<\/p>\n