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Quantitative Methods Session 1 Chapter 1 - AVERAGE

Quantitative Methods Session 1 Chapter 1 - AVERAGE. Pranjoy Arup Das. Average summarises a set of unequal numerical data. The numerical data values may be termed as observations.

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Quantitative Methods Session 1 Chapter 1 - AVERAGE

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  1. Quantitative MethodsSession 1 Chapter 1 - AVERAGE Pranjoy Arup Das

  2. Average summarises a set of unequal numerical data. • The numerical data values may be termed as observations. • Eg. The weights (in kgs) of each student of a class of 10 students are 60, 45, 75, 55, 63, 45, 80, 62, 58, 46. • The Average weight of these 10 students can be found out by the following : Average value of ‘n’ observations (Avgn)= = > OBSERVATIONS

  3. POINTS TO REMEMBER: • All observations have to be in the same unit. • Average value will have the same unit as that of the observations. Average value of ‘n’ observations (An) = = > An * Nn = Sn Eg. The average marks scored by 5 students is 40. What is the sum of the actual marks / Total marks scored by all the 5 students? Here A5 = 40, N5= 5 Since S5 = A5 * N5 => = 40 * 5 => S5 = 200

  4. EG. The average of the numbers 10, 10, m, 15, 4 is 7.8. Find the value of m. Avg5 = (10 + 10 + m + 15+ 4) / 5 = 7.8 => (39 + m) = 7.8 * 5 => m = 39 – 39 = 0 RSA, Ex 9, Pg 124 . The average of 11 nos. is 70. If the average of the first 6 numbers is 68 and that of the last 6 numbers is 73, find the 6th number. Let the 11 nos be : a b c d e f g h i j k It is given that Avg11 = (a+b+c+d+e+f+g+h+i+j +k) / 11 = 70 Therefore (a+b+c+d+e+f+g+h+i+j+k) = 70 * 11 = 770 1st SIX SIXTH No. LAST SIX

  5. Therefore a+b+c+d+e+f+g+h+i+j + k = 770……..(i) It is given that Avg1st 6 = 68 => (a+b+c+d+e+f) /6= 68 => (a+b+c+d+e+f )= 68 * 6 = 408……..(ii) Also it is given that AvgLast 6 = 73 => (k+j+i+h+g+ f) /6= 73 => (k+j+i+h+g+ f) = 73 * 6 = 438………(iii) Adding (ii) & (iii) we get , (a+b+c+d+e+f)+ (k+j+i+h+g+ f) = 846 => (a+b+c+d+e+f+ g+h+i+j + k) +f = 846 From (i) we get, 770 + f = 846 Therefore, f = 846 – 770 = 76

  6. RSA, Ex. 6, Page 124: Three tears ago, the average age of a family of 5 members was 17 years. A baby having been born, the average age of the family is the same today, i.e., 17 years. What is the present age of the baby? Soln: Let us assume the ages of the family members 3 years ago were F1, F2, F3, F4 & F5 Three years ago, Avg5 = 17 => (F1+F2+F3+F4+F5) / 5 = 17 => (F1 + F2 + F3 + F4 + F5) = 85 years ……..(i) After 3 years the ages of the 5 member have become (F1 +3) , (F2+3), (F3+3), (F4+3), (F5+3) Also a baby was born during. Let us assume the present age of the baby is F6. It is given that the present average of the 6 members of the family is 17 years. Therefore, Avg6 = 17 • {(F1+3) + (F2+3) + (F3+3) + (F4+3) + (F5+3) + F6} /6 = 17 • (F1+F2+F3+F4+F5)+F6+15 = 17 * 6 = 102 • From (i) we can get, 85 + F6 + 15 = 102 => F6 = 102 – 100 = 2 years

  7. RSA, Ex. 11, Page 124: A ship sails out to a mark at a speed of 20km/hr and sails back to the starting point at a speed of 12 km/hr. What is the average speed of the whole journey? Soln: PLEASE NOTE : Speed = Distance / Time To find out average speed we have to do Let us assume that the distance covered by the ship one way is x Km. Therefore the total distance covered by the ship = x + x = 2x Kms For the onward journey , the ship travels at 20 km per hour. That means at that speed a distance of 20 km can be covered in 1 hour So a distance of x Km will be covered in hours

  8. For the return journey , the ship travels at 12 km per hour. That means at that speed a distance of 12 km can be covered in 1 hour So a distance of ‘x’ Km will be covered in hours Therefore total time taken = = Therefore Average speed = = ___________ Km / hour

  9. RSA, Exercise 6A. Problem no 33, Page 127: A man travels 160 kms by car @ 64 Km/hr and another 160 kms @ 80 Km / hr. What is the average speed of the whole journey? Soln: PLEASE NOTE : Speed = Distance / Time To find out average speed we have to do The total distance covered by the man = 160+ 160 = 320 Kms For the onward journey , the man travels at 64 km per hour. That means at that speed a distance of 64 km can be covered in 1 hour So a distance of 160 Km will be covered in

  10. For the return journey , the man travels at 80 km per hour. That means at that speed a distance of 80 km can be covered in 1 hour So a distance of ‘x’ Km will be covered in Therefore total time taken = 2.5 + 2 = 4.5 Hours Therefore Average speed =

  11. Class Assignment • The average of 25, 32, 16, 40, x and 5 is 21. What is the value of x? • The average of 11 nos is 10.9. The avg of 1st six is 10.5 and avg of last six is 11.4. Find the sixth number? • A man covers half of his journey at 6km/hr and remaining half at 3km/hr. What is his average speed? • The avg weight of 120 students of a class is 56 kg. If the avg weight of the boys is 60 kgs and the avg weight of the girls is 50 kgs, how many boys and how many girls are there in the class? • The average of 8 persons in a committee gets increased by 2 years when two men aged 35 years and 45 years are substituted by two women. What is the average age of these two women? • In an examination, a pupils average marks were 63 per paper. Had he scored 20 more marks in Geography and 2 more marks in History, his average marks per paper would have been 65. How many papers were there in total in the examination? • The average monthly salary of workers in a factory is Rs. 8,500/-. If the average monthly salary of 7 technicians is Rs. 10000/- and the avg monthly salary of the rest is Rs. 7800/-, what is the total number of workers? • The avg weight of 3 men A,B & C is 84 kg. When D joins the group, the avg becomes 80 kgs. E, whose weight is 3 kgs more than D replaces A and the avg weight of B,C,D & E now becomes 79 Kgs. What is A’s weight?

  12. Practice problems • RSA, Exercise 6A, Page 125 • Problem nos. 1, 2 18, 19, 22, 29, 30, 33, 34, 39, 43, 44, 46, 48, 57, 59, 64, 73, 74, 75, 77, 78, 80, 82.

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