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

Introduction Q23. Seven years ago at the time of their marriage, the average age of a man and his wife was 28 years. At present, they have two children. Their daughter is 2 years older than their son. One year after the birth of the daughter, the average age of the man, wife and their daughter was 21 years, then the present age of the son is Solution: Let the daughter be born 'x' years after their marriage. Then, average age of the man, wife and their daughter one year after the birth of the daughter (28+(x+1))x 2+1/3- 59+2x/3 -21 Present age of the daughter 7-2 5 years Hence, the present age of the son 5-2 3 years

Introduction Q24. Salary of a person on 01-01-2001 is Rs. 6,400 per month with an increment Rs. 600 per month due on 01- 09-2001, 01-09-2002, 01-09-2003, 01-09-2004 and 01-09-2005. If his monthly salary on 01-01-2006 increases by 40% of average monthly salary during last five years, what monthly salary did he draw in February 2006?

Introduction Q24. Salary of a person on 01-01-2001 is Rs. 6,400 per month with an increment Rs. 600 per month due on 01- 09-2001, 01-09-2002, 01-09-2003, 01-09-2004 and 01-09-2005. If his monthly salary on 01-01-2006 increases by 40% of average monthly salary during last five years, what monthly salary did he draw in February 2006? Solution: His total salary in 2001 6400 x 8+ (6400+ 600) x 4 Rs. 79,200 His total salary in 2002 7000x 8+ (7000 + 600) x 4 Rs. 86,400 His total salary in 2003 7600 x 8+ (7600+ 600) x 4 Rs. 93,600 His total salary in 2004 8200 x 8+ (8200+ 600) x 4 Rs. 1,00,800 His total salary in 2005 8800 x 8+ (8800+ 600) x 4 Rs. 108,000 His average monthly salary during last 5 years 79200 + 86400+93600+100800 +108000/5x12 - Rs. 7,800 His monthly salary in December 2005 8800+ 600 Rs. 9,400 His monthly salary in February 2006 9400+7800*40/100 Rs. 12,520.

Introduction Q25. A milk vendor has 2 cans of milk. The first contains 25% water and the rest milk. The second contains 50% water. How much milk should he mix from each of the containers so as to get 12 litres of milk such that the ratio water to milk is 3:52 A.4 litre,8 litres B.6 litres,6 litres C.5 litres,7 litres D.7 litre,5 litres

Introduction Q25. A milk vendor has 2 cans of milk. The first contains 25% water and the rest milk. The second contains 50% water. How much milk should he mix fro each of the containers so as to get 12 litres of milk such that the ratio water to milk is 3:52 A.4 litre,8 litres B.6 litres,6 litres C.5 litres,7 litres D.7 litre,5 litres Solution: Let the cost of 1 litre milk be Re. 1 Milk in 1 litre mix. in 1st can = litre, CP. of 1 litre mix. in 1st can Re 3/4 Milk in 1 litre mix. in 2nd can 1/2 litre. C.P. of 1 litre mix. in 2nd can Re. 1/2 Milk in 1 litre of final mix,-5/8 litre. Mean price = Re 5/8 By the rule of alligation, we have:

Introduction CP 011 litre mixture in 1st can . of 1 litre mixture in 21 d can Mean Price 8 8 .: Ratio of two mixtures=-:-m 1 : 1 So, quantity of mixture taken from each can-1/2*126

Introduction Q26. In what ratio must a person mix three kinds of wheat costing him Rs 1.20,Rs 1.44 and Rs 1.74 per Kg so that the mixture may be worth Rs 1.41 per Kg? A. 11:77:7 B. 25:45:8 C. 27:23:6 D. 11:45:7

Introduction Q26. In what ratio must a person mix three kinds of wheat costing him Rs 1.20,Rs 1.44 and Rs 1.74 per Kg so that the mixture may be worth Rs 1.41 per Kg? A. 11:77:7 B. 25:45:8 C. 27:23:6 D. 11:45:7 Solution: Mix wheat of first and third kind to get a mixture worth Rs 1.41 per Kg. C.P of 1 Kg wheat of 1st kind 120p C.P of 1 Kg wheat of 3rd kind 174p Mean Price 141p

Introduction C.P of 1 Kg Wheat of 1st kind (120p) C.P of 1 Kg wheat of 3rd kind (174p) Mean Price (141p) 21 By Alligation rule: Quantity of 1st kind of wheat/ Quantity of 3rd kind of wheat-33/21-11/7 So,they must be mixed in the ratio of 11:7

Introduction By allegation Rule: Quantity of 1st kind of wheat/ Quantity of 3rd kind of wheat-3/21-1/7 So, they must be mixed in the ratio of 1:7. Thus, Quantity of 2nd kind of wheat/Quantity of 3rd kind of wheat-(Quantity of 1st kind of wheat/Quantity of 3rd kind of wheat)x (Quantity of 2nd kind o wheat/Quantity of 1st kind of wheat) Quantity of 2nd kind of wheat/Quantity of 3rd kind of wheat-(11/7x7/1)-(11/1) Thus,Quantities of wheat of 1st.2nd:3rd-11:77:7

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 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? Solution: 13*25 -X-13.125*24 x-10

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 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 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.. Solution: 10 students have scored 600 marks amongst them, and no one is allowed to sco lesser than 40 or higher than 100. The idea now is to maximize what the highest scorer gets.

Introduction The 5 least scores have an average of 55, which means that they have scored 55 x 5 - 275 marks amongst them. This leaves 325 marks to be shared amongst the top 5 students. Lets call them a, b, c, d and e. Now, in order to maximize what the t scorer e" gets, all the others have to get the least possible scores (and at the same time, they should also get distinct integers.) The least possible score of the top 5 should be at least equal to the highest of the bottom 5. Now we want to make sure that the highest of the bottom 5 is the least possible. This can be done by making all scores equal to 55. If some scores are less than 55, some other scores have to be higher than 55 to compensate and make the average 55. Thus the highest score is the least only when the range is 0.

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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Vishukant Goel

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