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Practice Questions 2
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This lesson will help you answer the questions based on 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.

U
Unacademy user
SUCCESS MANTRA BY DEEPAK RAI
  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. 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?


  4. 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.


  5. 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?


  6. 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


  7. 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?


  8. 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


  9. Introduction Q6. A class has 20 boys and 10 girls. If the average weight of boys are interchanged with average weight of girls in the class, the new average weight of the class increases by 2. The average of average weight of boys and average weight of girls is how much more than the average weight of the class? Let the average weight of boys be x and that of girls be y. Sum of the weight of boys 20x Sum of the weight of girls 10y Average weight of the class 20x +10y/30 Now if their average weights are interchanged then average weight of the class - 20x+ 10y/30 + 2- 20y+10x - 6.


  10. Introduction xty/2 (20x+10y)/30 y-x/6 6/61


  11. 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


  12. 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 Solution: If to the set of 'n' consecutive natural number, next number is added then the average increases by 0.5. So, on including the next natural numbers the average becomes (xtn/2) Now, if from the set of 'n' consecutive natural numbers the smallest number is removed the average again increases by 0.5. So the average of the next 'n' natural numbers becomes [(x+n/2)+n/2] i.e. we have to remove the previous 'n' consecutive natural numbers.


  13. 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


  14. 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


  15. 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


  16. 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


  17. 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


  18. 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.


  19. 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)


  20. 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


  21. 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


  22. 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.


  23. 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


  24. THANK YOU