Here you can download CMAT 2022 – important MAH – CET Average Questions PDF by Cracku. Very Important MAH – CET 2022 and These questions will help your MAH – CET preparation. So kindly download the PDF for reference and do more practice.<\/p>\n
Download Average Questions for MAH – CET<\/a><\/p>\n Enroll to MAH-CET Crash Course<\/a><\/p>\n Question 1:\u00a0<\/b>Consider the set S = {2, 3, 4, …., 2n+1}, where n is a positive integer larger than 2007. Define X as the average of the odd integers in S and Y as the average of the even integers in S. What is the value of X – Y ?<\/p>\n a)\u00a00<\/p>\n b)\u00a01<\/p>\n c)\u00a0(1\/2)*n<\/p>\n d)\u00a0(n+1)\/2n<\/p>\n e)\u00a02008<\/p>\n Question 2:\u00a0<\/b>Ten years ago, the ages of the members of a joint family of eight people added up to 231 years. Three years later, one member died at the age of 60 years and a child was born during the same year. After another three years, one more member died, again at 60, and a child was born during the same year. The current average age of this eight-member joint family is nearest to a)\u00a023 years<\/p>\n b)\u00a022 years<\/p>\n c)\u00a021 years<\/p>\n d)\u00a025 years<\/p>\n e)\u00a024 years<\/p>\n Question 3:\u00a0<\/b>A college has raised 75% of the amount it needs for a new building by receiving an average donation of Rs. 600 from the people already solicited. The people already solicited represent 60% of the people the college will ask for donations. If the college is to raise exactly the amount needed for the new building, what should be the average donation from the remaining people to be solicited?<\/p>\n a)\u00a0Rs. 300<\/p>\n b)\u00a0Rs. 250<\/p>\n c)\u00a0Rs. 400<\/p>\n d)\u00a0500<\/p>\n Question 4:\u00a0<\/b>Three classes X, Y and Z take an algebra test. a)\u00a081<\/p>\n b)\u00a081.5<\/p>\n c)\u00a082<\/p>\n d)\u00a084.5<\/p>\n Question 5:\u00a0<\/b>Consider a sequence of seven consecutive integers. The average of the first five integers is n. The average of all the seven integers is: a)\u00a0n<\/p>\n b)\u00a0n+1<\/p>\n c)\u00a0kn, where k is a function of n<\/p>\n d)\u00a0n+(2\/7)<\/p>\n Take Free MAH-CET mock tests here<\/a><\/p>\n Instructions<\/b>There are 60 students in a class. These students are divided into three groups A, B and C of 15, 20 and 25 students each. The groups A and C are combined to form group D.<\/p>\n Question 6:\u00a0<\/b>What is the average weight of the students in group D?<\/p>\n a)\u00a0More than the average weight of A<\/p>\n b)\u00a0More than the average weight of C<\/p>\n c)\u00a0Less than the average weight of C<\/p>\n d)\u00a0Cannot be determined<\/p>\n Question 7:\u00a0<\/b>The average marks of a student in 10 papers are 80. If the highest and the lowest scores are not considered, the average is 81. If his highest score is 92, find the lowest.<\/p>\n a)\u00a055<\/p>\n b)\u00a060<\/p>\n c)\u00a062<\/p>\n d)\u00a0Cannot be determined<\/p>\n Question 8:\u00a0<\/b>Total expenses of a boarding house are partly fixed and partly varying linearly with the number of\u00a0boarders. The average expense per boarder is Rs. 700 when there are 25 boarders and Rs. 600\u00a0when there are 50 boarders. What is the average expense per boarder when there are 100 boarders?<\/p>\n a)\u00a0550<\/p>\n b)\u00a0580<\/p>\n c)\u00a0540<\/p>\n d)\u00a0560<\/p>\n Question 9:\u00a0<\/b>Prof. Suman takes a number of quizzes for a course. All the quizzes are out of 100. A student a)\u00a06<\/p>\n b)\u00a07<\/p>\n c)\u00a08<\/p>\n d)\u00a09<\/p>\n e)\u00a0None of these<\/p>\n Question 10:\u00a0<\/b>Prof. Bee noticed something peculiar while entering the quiz marks of his five students into a spreadsheet. The spreadsheet was programmed to calculate the average after each score was entered. Prof. Bee entered the marks in a random order and noticed that after each mark was entered, the average was always an integer. In ascending order, the marks of the students were 71, 76, 80, 82 and 91. What were the fourth and fifth marks that Prof. Bee entered?<\/p>\n a)\u00a071 and 82<\/p>\n b)\u00a071 and 76<\/p>\n c)\u00a071 and 80<\/p>\n d)\u00a076 and 80<\/p>\n e)\u00a091 and 80<\/p>\n Get 5 MAH-CET mocks at just Rs.299<\/a><\/p>\n Answers & Solutions:<\/strong><\/span><\/p>\n 1)\u00a0Answer\u00a0(B)<\/strong><\/p>\n The odd numbers in the set are 3, 5, 7, …2n+1<\/p>\n Sum of the odd numbers = 3+5+7+…+(2n+1) = $n^2 + 2n$<\/p>\n Average of odd numbers =\u00a0$n^2 + 2n$\/n = n+2<\/p>\n Sum of even numbers = 2 + 4 + 6 + … + 2n = 2(1+2+3+…+n) = 2*n*(n+1)\/2 = n(n+1)<\/p>\n Average of even numbers = n(n+1)\/n = n+1<\/p>\n So, difference between the averages of even and odd numbers = 1<\/p>\n 2)\u00a0Answer\u00a0(E)<\/strong><\/p>\n Ten years ago, the total age of the family is 231 years.<\/p>\n Seven years ago, (Just before the death of the first person), the total age of the family would have been 231+8*3 = 231+24 = 255. After the death of one member, the total age is 255-60 =\u00a0195 years.<\/p>\n Since a child takes birth in the same year, the number of members remain the same i.e. (7+1) = 8<\/p>\n Four years ago, (i.e. 6 years after start date)\u00a0one of the member of age 60 dies, Since a child takes birth in the same year, the number of members remain the same i.e. (7+1) = 8<\/p>\n After 4 more years, the current total age of the family is = 8×4 + 159 = 191 years<\/p>\n The average age is 191\/8 = 23.875 years = 24 years (approx)<\/p>\n Alternatively, 3)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Let there be total 100 people whom the college will ask for donation. Out of these 60 people have already given average donation of 600 Rs. Thus total amount generated by 60 people is 36000. This is 75% of total amount required . so the amount remaining is 12000 which should be generated from remaining 40 people. So average amount needed is 12000\/40 = 300<\/p>\n 4)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Let x , y and z be no. of students in class X, Y ,Z respectively.<\/p>\n From 1st condition we have<\/p>\n 83*x+76*y = 79*x+79*y which give 4x = 3y.<\/p>\n Next we have 76*y + 85*z = 81(y+z) which give 4z = 5y .<\/p>\n Now overall average of all the classes can be given as $\\frac{83x+76y+85z}{x+y+z}$<\/p>\n Substitute the relations in above equation we get,<\/p>\n $\\frac{83x+76y+85z}{x+y+z}$ \u00a0= (83*3\/4 + 76 + 85*5\/4)\/(3\/4 + 1 + 5\/4) = 978\/12 = 81.5<\/p>\n 5)\u00a0Answer\u00a0(B)<\/strong><\/p>\n The first five numbers could be n-2, n-1, n, n+1, n+2. The next two number would then be, n+3 and n+4, in which case, the average of all the 7 numbers would be $\\frac{(5n+2n+7)}{7}$ = n+1<\/p>\n 6)\u00a0Answer\u00a0(D)<\/strong><\/p>\n As data regarding weights of people is not given, hence we can’t determine the avg. weight of people in group D<\/p>\n 7)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Total marks = 80 x 10 = 800 8)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Let the fixed income be x and the number of boarders be y.<\/p>\n x + 25y = 17500<\/p>\n x + 50y = 30000<\/p>\n => y = 500 and x = 5000<\/p>\n x + 100y = 5000 + 50000 = 55000<\/p>\n Average expense = $\\frac{55000}{100}$ = Rs.550.<\/p>\n 9)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Grade\u00a0A $\\geq$ 90 and Grade B = 87 to 89<\/p>\n If Ramesh scores 70 instead of 97, => Change of marks = 97 – 70 = 27<\/p>\n It creates a change from grade\u00a0A to B, this means an overall change in average by<\/p>\n = Minimum marks for grade A – Minimum marks for Grade B = 90 – 87 = 3<\/p>\n $\\therefore$ Number of subjects = $\\frac{27}{3} = 9$<\/p>\n 10)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Marks = 71, 76, 80, 82 and 91<\/p>\n For average to be an integer each time, the sum of numbers entered after 2nd entry should be divisible by 2, after third entry should be divisible by 3 and so on.<\/p>\n The first two numbers have to be both odd or both even, so that their sum is even and can be divisible by 2.<\/p>\n Also, the sum of first four numbers should also be even.<\/p>\n => Numbers entered are either $OOEEE$ or $EEOOE$ \u00a0 (where O -> odd and E -> even)<\/p>\n Case 1\u00a0: First two numbers are 71 and 91 in any order.<\/p>\n Average = $\\frac{71 + 91}{2} = 81$<\/p>\n Now, sum of 71 and 91 is multiple of 3, so the third number has to be a multiple of 3, which is not possible.<\/p>\n Case 2\u00a0: First two numbers can be = $(76,80) , (76,82) , (80,82)$<\/p>\n Now, following above criteria, only 2nd option is possible<\/p>\n So, third number has to be 91, average = $\\frac{76 + 82 + 91}{3} = 83$<\/p>\n $\\therefore$ The fourth and fifth marks that Prof. Bee entered = 71 and 80<\/p>\n
\n[CAT 2007]<\/p>\n
\nThe average score in class X is 83.
\nThe average score in class Y is 76.
\nThe average score in class Z is 85.
\nThe average score of all students in classes X and Y together is 79.
\nThe average score of all students in classes Y and Z together is 81.
\nWhat is the average for all the three classes?<\/p>\n
\n[CAT 2000]<\/p>\n
\ncan get an A grade in the course if the average of her scores is more than or equal to 90.Grade B is awarded to a student if the average of her scores is between 87 and 89 (both included). If the average is below 87, the student gets a C grade. Ramesh is preparing for the last quiz and he realizes that he will score a minimum of 97 to get an A grade. After the quiz, he realizes that he will score 70, and he will just manage a B. How many quizzes did Prof. Suman take?<\/p>\n
\nThis is because, in 3 years, every person in the family would have aged by 3 years,
\nTotal change in age = 231+24 = 255<\/p>\n
\ntherefore,\u00a0total age of the family is 195+24-60 = 159 years.<\/p>\n
\nSince the number of members is always the same throughout
\nThe 2 older members dropped their age by 60
\nSo, after 10yrs, total age = 231 + 8*10 – 2*60\u00a0= 191
\nAverage age = 191\/8 = 23.875 $\\simeq$ 24<\/p>\n
\nTotal marks except highest and lowest marks = 81 x 8 = 648
\nSo Summation of highest marks and lowest marks will be = 800 – 648 = 152
\nWhen highest marks is 92, lowest marks will be = 152-92 = 60<\/p>\n