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This lesson will give you the Course overview of 32 must solve questions in Averages and allegations

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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Arju Sharma
10 months ago
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Arju Sharma
10 months ago
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  1. 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


  2. 32 must solve questions of AVERAGES AND ALLEGATIONS for CAT


  3. Course structure All three types of questions covered Easy, Medium & Difficult Useful for CAT, XAT, IIFT, NMAT, SNAP GRE, GMAT Bank PO, SSC & all other Government jobs


  4. Post your queries in the comment section, will be answered as soon as possible. If you like the lesson : Rate, revie share & recommend it to your friends


  5. Miscellaneous Question Q1. The average height of 22 toddlers increases by 2 inches when two of them leave this group. If the average height of these two toddlers is one third the average height of the original 22, then the average height, inches, of the remaining 20 toddlers is a.30 b.23 c.32 d.26


  6. Introduction Solution: Let the average height of 22 toddlers be x. Average height of two toddlers x/3 Now, Average increases by 2 inches when two leave the group So, sum of the heights of 20 toddlers 20(x + 2) So, 22x (x + 2) x 20+ 2x/3 30. So, the average height of the remaining 20 toddlers is 30 + 2 32 inches


  7. Introduction Q2. The average age of a family of 5 members is 20 years. If the age of the youngest member is 10 years, what was the approximate average age of the family just a day before the birth of the youngest member?


  8. Introduction Q2. The average age of a family of 5 members is 20 years. If the age of the youngest member is 10 years, what was the approximate average age of the family just a day before the birth of the youngest member? Solution: At present the total age of the family -5 x 20 - 100 The total age of the family at the time of the birth of the youngest member - [100 10 (10 x 4) 50 Therefore, average age of the family just a day before the birth of the youngest member -50-(4/365)/4 -12.5


  9. Introduction Q3. If the average of 36 numbers is 25, then how many minimum numbers from the set 147, 48, 49, ..., 77, 78) should be added so that the average moves to 30?


  10. Introduction Q3. If the average of 36 numbers is 25, then how many minimum numbers from the set 147, 48, 49, ..., 77, 78) should be added so that the average moves to 30? Solution Sum of 36 numbers-36 25-900 To make the average 30, I have to add x more numbers so (36 + X) x 30 1080 +30x So sum of rest should carry 180 extra value other than their average i.e. 30. To make it by minimum numbers we have to add values with maximum increase in average. So we add 78(48 +30), 77(47+30), 76(46+30) and 69(39+30) with 4 values.


  11. Introduction Q4. The average age of 6 members of a family is 37. A baby was born three years back and 2 old people expired 5 and 7 years back at the age of 58 and 63 respectively. What was the average age of the family 10 years back?


  12. Introduction Q4. The average age of 6 members of a family is 37. A baby was born three years back and 2 old people expired 5 and 7 years back at the age of 58 and 63 respectively. What was the average age of the family 10 years back? Solution: Sum of ages is 37 x 6 222. Sum of ages 3 yrs back 222-18 204. Sum of ages 5 years back 204 - 10+ 58 252 Sum of ages 7 yrs back 252 12+ 63 303 Sum of ages 10 yrs back 303-21 282 and now there are 7 people. New Average 282/7 40.28


  13. Introduction Q5. If a, b and c be non-zero numbers such that the average of 2a and 3b equals the average of b and 3c, then what is the average of a and b? Solution: 2a+3b/2-b+3c/2 ->2a+2b 3c a+b 3c/2 a+b/2-3c/4


  14. Introduction Q7. The average of 'n' consecutive natural numbers is x. Find the average of the next 'n' natural numbers. A.x+n B.x+2n C. x+n/2 D. None of these


  15. Introduction Q8. P and Q married each other when their ages were in the ratio of 4: 5 After 3 years of their marriage, they had a child. Two years after this, the average age of the family of the three was 19 years. What was the age of Q at the time of his/her marriage? Solution: Let the age of P and Q when they got maned be 4x and 5x respectively Now, according to the question, 5x+5+2+4x+5/3 19 =>9x + 12 = 57 ->9x 45 => x = 5 The age of Q at the time of his/her marriage was 5 x 5 25 years


  16. 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?


  17. 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.


  18. 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.


  19. 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?


  20. 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.


  21. 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?


  22. 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.


  23. 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?


  24. 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.


  25. 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


  26. Introduction Q19. There are 15 students in a class, who sat for maths semester exam. Average marks scored by 12 of them was 72. What will be the sum of marks of remaining 3 students, if addition of marks of each subsequent student increases the average by 1 mark.


  27. Introduction Q19. There are 15 students in a class, who sat for maths semester exam. Average marks scored by 12 of them was 72. What will be the sum of marks of remaining 3 students, if addition of marks of each subsequent student increases the average by 1 mark. Solution: 13th student if gets 72 marks then no change in average, if gets 72+ 13 85 average increases by 1. 14th student should get 73 + 14 and 15th student should get 74+ 15. So, Sum of marks of these 3 students will be 85 +87 + 89 261.


  28. 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?


  29. 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


  30. 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.


  31. 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


  32. 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


  33. 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:


  34. 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


  35. 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?


  36. 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


  37. 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


  38. 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.


  39. 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.


  40. 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