Abhishek Khurana B.Com (Honours) from Panjab University MBA in Marketing from NMIMS 2009-2011 Plus Educator on Unacademy 8 years teaching experience 6 times 99 percentiler in CAT

32 must solve questions of AVERAGES AND ALLEGATIONS for CAT

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Introduction Q27. Natural No. from 1 to 25 are written on a board. One of the number is erased and the average of the remaining numbers is 13.125.Find the number Erased?

Introduction Q28. Natural No. from 1 to n are written on a board. One of the number is erased and the average of the remaining numbers is 35 5/17 .Find the number Erased? Solution: n(n+1)/2-x 35 5/17 *(n+1) sum/(n-1)-35*17+5/17 n-1 has to be a multiple of 17. as average is around 35 the total number of number will be in the vicinity of70. So n-l is 17*4-68(nearer to 70) so n-69 post erasing average is nearly equal to 35. Before erasing average -35

Introduction 35*69- 600/17*68 solving this equation we will get x-15.

Introduction Q29. Average age of a class of 30 students and a teacher reduces by 0.5if we exclude the teacher.lf initial average is 14 years, find the age of class teacher.

Introduction Q29. Average age of a class of 30 students and a teacher reduces by 0.5if we exclude the teacher. If initial average is 14 years, find the age of class Solution: Age of teacher-Total age of students and teacher - Total age of students-31*14-30*13.5-434-405-29 years

Introduction Q30. Q.1: Consider a class of 40 students whose average weight is 40 kgs. m new students join this class whose average weight is n kgs. If it is known that m + n 50, what is the maximum possible average weight of the class now? 40.18 kgs 40.56 kgs 40.67 kgs 40.49 kgs

Introduction Q30. Consider a class of 40 students whose average weight is 40 kgs. m new students join this class whose average weight is n kgs. If it is known that m + n 50, what is the maximum possible average weight of the class now? 40.18 kgs 40.56 kgs 40.67 kgs 40.49 kgs Solution: If the overall average weight has to increase after the new people are added, the average weight of the new entrants has to be higher than 40. So, n> 40

Introduction Consequently, m has to be < 10 (as n + m 50) Working with the "differences" approach, we know that the total additional weight added by "m" students would be (n - 40) each, above the already existing average of 40. m(n - 40) is the total extra additional weight added, which is shared amongst 40 + m students. So, m* (n-40)/(m+40) has to be maximum for the overall average to be maximum. At this point, use the trial and error approach (or else, go with the answer options) to arrive at the answer. The maximum average occurs when m 5,and n 45 And the dverage i5 40- (45-0)*5/45 - 40 + 5/9 - 40.56 kgs

Introduction Q31.The average score in an examination of 10 students of a class is 60. If the scores of the top five students are not considered, the average score of the remaining students falls by 5. The pass mark was 40 and the maximum mark was 100. It is also known that none of the students failed. If each of the top five scorers had distinct integral scores, the maximum possible score of the topper is.. 1.99 2.100 3.87 4.95

Introduction So now, we have the lowest value that the top 5 can score, which is 55. The others have to get distinct integer scores, and as few marks as possible, so that "e" gets th maximum So, 55 + 56 + 57 + 58 + e = 325 e = 99 marks. Answer choice (A)

Introduction Q32. 5 Scores in a classroom are broken into 5 different ranges, 51-60, 61-70, 71-80, 81-90 and 91-100. The number of students who have scored in each range is given below. 51 to 60 3 students, 61 to 70 8 students, 71 to 80 7 students, 81 to 90-4 students 91 to 100 3 students Furthermore, we know that the number of students who scored 76 or more is atleast one more than those who scored below 75. What is the minimum possible average overall of this class? 1.72 2.71.2 3.70.6 4.69.2

Introduction Q32. 5 Scores in a classroom are broken into 5 different ranges, 51-60, 61-70, 71-80, 81-90 and 91-100. The number of students who have scored in each range is given below. 51 to 60 3 students, 61 to 70 8 students, 71 to 80 7 students, 81 to 90-4 students 91 to 100 3 students Furthermore, we know that the number of students who scored 76 or more is atleast one more than those who scored below 75. What is the minimum possible average overall of this class? 1.72 2.71.2 3.70.6 4.69.2

Introduction Solution Let's employ the idea of a total of 25 students (all of the same weight) sitting see-saw, which has numbers from 51 to 100 marked on it. At least as many students are sitting on 76 (or to its right), as there are sitting to the left of 75. Now this means that you can have only one person sitting to the left of 75 and all the rest sitting beyond 76. But you can't do that, as you have other constraints as well First of all, you have to seat 3 students from 51 to 60, and 8 students from 61 to 70. Secondly, you also have to make sure that the average is the least. This means that the see-saw should be tilting as much to the left as possible, which in turn means that the number of people sitting to the left of 75 should be the highest possible. This makes it 12 students to the left of 75, and the remaining 13 students on 76 or to its right. Next, how do you ensure that the average is least, i.e. how do you ensure that the balance tilts as much as possible to the left? Make each student score as little as possible given the constraints.

Introduction Solution: So, the first 3 students only score 51 each. The next 8 students score only 61 e Students are now fixed. The 12th student has to be below 75, so seat him on 71. The remaining 6 students (who are in the 71 to 80 range) have to score 76. The next 4 score 81 and the next 3 score 91. his would give you the least average. The lowest possible average would be: (3 51)+ (8 *61) + 71 + (6 76)+ (4* 81) + (3 *91)/25 => 70.6

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Abhishek Khurana

I am Abhishek Khurana. I completed my B.Com(H) from Panjab University in 2008 & my MBA from NMIMS in 2011.

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