Download important IIFT Averages Questions PDF based on previously asked questions in IIFT and other MBA exams. Practice Averages questions and answers for IIFT exam.<\/p>\n
Download Averages Questions for IIFT PDF<\/a><\/p>\n IIFT Previous year question\u00a0 answer PDF<\/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>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 3:\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 4:\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 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 5:\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 Enroll To CAT 3 Months Crash Course<\/a><\/p>\n Question 6:\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 7:\u00a0<\/b>Total expenses of a boarding house are partly fixed and partly varying linearly with tile number of boarders. The average expense per boarder is Rs. 700 when there are 25 boarders and Rs. 600 when 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 8:\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 9:\u00a0<\/b>2 years ago, one-fifth of Amita\u2019s age was equal to one-fourth of the age of Sumita, and the average of their age was 27 years. If the age of Paramita is also considered, the average age of three of them declines to 24. What will be the average age of Sumita and Paramita 3 years from now?<\/p>\n a)\u00a025 years<\/p>\n b)\u00a026 years<\/p>\n c)\u00a027 years<\/p>\n d)\u00a0cannot be determined<\/p>\n Question 10:\u00a0<\/b>The average of 7 consecutive numbers is P. If the next three numbers are also added, the average shall<\/p>\n a)\u00a0remain unchanged<\/p>\n b)\u00a0increase by 1<\/p>\n c)\u00a0increase by 1.5<\/p>\n d)\u00a0increase by 2<\/p>\n IIFT Free Mock Test<\/a><\/p>\n Enroll for CAT\/MBA Courses<\/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(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 3)\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 4)\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 5)\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 6)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Total marks = 80 x 10 = 800 Top 500+ Free Questions for IIFT<\/a><\/p>\n IIFT Previous year question\u00a0 answer PDF<\/a><\/p>\n 7)\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 8)\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 Quant Formula For IIFT PDF<\/a><\/p>\n IIFT Free Mock Test<\/a><\/p>\n 9)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Let ‘A’, ‘S’ and ‘P’ be Amita’s, Sumita’s and Paramita’s present age.<\/p>\n It is given that 2 years ago, one-fifth of Amita\u2019s age was equal to one-fourth of the age of Sumita, and the average of their age was 27 years.<\/p>\n $\\dfrac{(A-2)+(S-2)}{2} = 27$<\/p>\n $A+S = 58$\u00a0 … (1)<\/p>\n Also,\u00a0$\\dfrac{A-2}{5} = \\dfrac{S-2}{4}$<\/p>\n $4A-8 = 5S-10$<\/p>\n $5S – 4A = 2$ … (2)<\/p>\n From equation (1) and (2) we can say that S = 26, A = 32.<\/p>\n Average age of Amita, Sumita and Paramita before 2 years = 24.<\/p>\n $\\dfrac{(A-2)+(S-2)+(P-2)}{3} = 24$<\/p>\n $A+S+P = 78$. Hence, P = 20.<\/p>\n Therefore, the average age of Sumita and Paramita 3 years from now? = $\\dfrac{(S+3)+(P+3)}{2}$ = $\\dfrac{(26+3)+(20+3)}{2}$ = 26 years.<\/p>\n Hence, option B is the correct answer.<\/p>\n <\/p>\n 10)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Let the 7 consecutive numbers be a-3, a-2, a-1, a, a+1, a+2 and a+3.
\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
\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
\nSum of the numbers = 7a and the average of these numbers = a
\nIf next 3 numbers a+4, 4+5 and a+6 are also added then the average of these 10\u00a0 numbers = $\\dfrac{7a+a+4+a+5+a+6}{10} = a+1.5$
\nThus, the average increases by 1.5
\nHence, option C is the correct answer.<\/p>\n