Introduction Q9. Average of 5 consecutive 2 digit numbers is 36 less than the average of the 5 numbers formed by reversing the digits of those 5 numbers. If the difference between the largest number in both the cases is 54, then which of the following can be the sum of the original 5 numbers?

Introduction Q9. Average of 5 consecutive 2 digit numbers is 36 less than the average of the 5 numbers formed by reversing the digits of those 5 numbers. If the difference between the largest number in both the cases is 54, then which of the following can be the sum of the original 5 numbers? Solution: Difference between a two digit number and a number with reversed digit has to be 54. So, there are 3 cases possible 71-17-54,82-28-54,93-39-54 So, the largest of five original two-digit numbers must be among: 17, 28 or 39 Since, it is given that the numbers ar consecutive, so when we take 28 as largest among the 5 the sum comes out to be 130, which is one of the options.

Introduction Q10. The average age of husband, wife and their child 3 years ago from now was 27 years and that of wife and child 5 years ago from now was 20 years. The age (in years) of the husband 5 years ago from now wa:s

Introduction Q10. The average age of husband, wife and their child 3 years ago from now was 27 years and that of wife and child 5 years ago from now was 20 years. The age (in years) of the husband 5 years ago from now wa:s Solution: Let the Age of husband wife and child is h.w.c. Then htw+c-9/3-27 and w+c-10/2-20 By solving these 2 equations we will get h-40 years 5 year ago age of husband-35 years

Introduction Q11. he average of n numbers is 41. If 2/3rd of these numbers are increased by 9 and the remaining 1/3rd are decreased by 6, then find the new average.

Introduction Q11. he average of n numbers is 41. If 2/3rd of these numbers are increased by 9 and the remaining 1/3rd are decreased by 6, then find the new average. Solution: Sum of n numbers 41n Now 2/3 rd of n numbers are increased by 9. So, sum should get increased by 2/3 x n x 9. Also 1/3rd of n numbers are decreased by 6. So, sum should get decreased by 1/3 xnx 6. New total 41n +(2n/3)9 -(n/3) x 6 45n And hence new average 45n/n45

Introduction Q12. The average weight of students in a class is 60 kg. When two new students having weight 90 kg and 98 kg join the class and one student of weight 42 kg leaves the class, the new average of the class becomes 62 kg. How many students were there in the class initially?

Introduction Q12. The average weight of students in a class is 60 kg. When two new students having weight 90 kg and 98 kg join the class and one student of weight 42 kg leaves the class, the new average of the class becomes 62 kg. How many students were there in the class initially? Solution: Let N be the number of students present in class initially. Total weight = 60 N New weight -62 60 N + 90 + 98-42 and hence average 60N+90+ 56/(N-1)+2 Solving it, we gef N 42

Introduction Q13. From the set of 1st n natural numbers,one of the number is erased and the average of the remaining number comes out to be 16 1/10.Find the erased number.

Introduction Q13. From the set of 1st n natural numbers,one of the number is erased and the average of the remaining number comes out to be 16 1/10.Find the erased number. Solution: The average of first 'n' natural numbers deviate by maximum of 0.5 on erasing one of the numbers, so it can be concluded that the average would have been around 16 before erasing the number also. Further the average of first 'n' natural numbers is n +1/2 which is around 16 when n is close to 31 (i.e.n+1/2-16> n-31) So, taking n as 31, the sum of first 31 natural numbers is 496 (i.e. 31*32/2) and the given sum is 483 (i.e. 161/10 *30) so the erased number 496-483 13.

Introduction Q14. Due to a man leaving the group of 6 people, the average weight of the group drops from 63 to 60. Find the weight of the man who left the group? Solution: Let every person has Rs. 1 coins equivalent to their weight. While leaving, this person took away 63 coins of his own plus 3 coins from each of the remaining 5 individuals. So, weight number of coins he had 63+ 5 x 3 78.

Introduction Q16. A boat has a maximum capacity to carry 1200 kg. There are 20 children and 20 adults waiting on the river bank to cross the river in the boat Each child weighs 30kg and each adult weighs 50kg. Ten of the children are also carrying a school bag weighing 5kg. If every child on the boat must be accompanied by an adult, what is the maximum number of people who can board the boat?

Introduction Q17. The average salary of marketing department, having 22 employees working, of ABC company is 3.7 lac more than the average salary of operations department, having 15 employees working, of same company. If the average salary of all 37 employees is 5.6 lacs, find the average salary (in lac) of marketing department.

Introduction Q17. The average salary of marketing department, having 22 employees working, of ABC company is 3.7 lac more than the average salary of operations department, having 15 employees working, of same company. If the average salary of all 37 employees is 5.6 lacs, find the average salary (in lac) of marketing department. Solution: Let average salary of operations department be x lac. Then, average salary of marketing department becomes (x + 3.7) lac. So, 22(x + 3.7) + 15x 37 x 5.6 x- 3.4 Hence, average salary of marketing department- 7.1 lac

Introduction Q18. The average monthly salary of 12 workers and 3 managers in a factory was Rs. 600. When one of the managers, whose salary was Rs. 720, and c worker, whose salary was Rs 300, were replaced with a new manager and a new worker, where the salary of the new worker was Rs 200, then the average salary of the team dropped down to Rs.580. What is the salary (in Rs.) of the new manager?

Introduction Q18. The average monthly salary of 12 workers and 3 managers in a factory was Rs. 600. When one of the managers, whose salary was Rs. 720, and c worker, whose salary was Rs 300, were replaced with a new manager and a new worker, where the salary of the new worker was Rs 200, then the average salary of the team dropped down to Rs.580. What is the salary (in Rs.) of the new manager? Solution: The total salary of the 15 employees 15 x 600 9000 Total salary reduced after the change of two employees by 15 x 20 out of which 100 dropped due to the new worker. So, 200 must have dropped due to the new manager As the salary of the exiting manager is 720, the salary of the new manager 720 - 200 520

Introduction Q20. On any given day, the bank balance of a person A is the sum of his bank balance on the previous day and his bank balance on the next day. If the bank balance of A on 18th November 2007 and 19th November 2007 is Rs.4000 and Rs.2000 respectively, then what will be his bank balance (in Rs.) on 16thNovember 2008? (Assume that the bank balance of A can be negative.)

Introduction Q21. In an examination, 40% of the candidates wrote their answers in Hindi and the remaining candidates in English. The average marks of the candidates who wrote the exam in Hindi is 74 and the average marks of the candidates who wrote the exam in English is 77. What is the average marks of all the candidates?

Introduction Q22. Ryan international B-school has the following student profile. The average age of students at the school is 27 years. There are 215 students in the school. If another student David is added to the group the average age increases by 0.1, while if Raul is added the average age decreases by 0.1. If Anton is removed from the group and David and Raul are added the average age decreases by 0.1. Find Anton's approximate age? Solution: The students Raul and David nullify the change in average age. So the average of the group of 217 is also 27. Now if we remove Anton, the average age reduces by 0.1. Hence the change in sum of ages 0.1 x 217 21.7 Since the average age reduces Anton must be above the average, Anton's age 27 21.7 48.7 wars

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

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 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 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 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 Mix wheat of 1st kind and 2nd kind to obtain a mixture worth of Rs. 1.41per Kg C.P of 1 Kg Wheat of 1st kind (120p) C.P of 1 Kg wheat of 3rd kind (144p) Mean Price (141p) 21 By Alligation rule:

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 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 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 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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Abhishek Bhati

a year ago

thank you for the lesson madam.

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