08 Average Notes

08 AVERAGE AVERAGE Average Average Speed The average is nothing but the sum of all observations divided by the number of observations. This is also known as arithmetic mean of the given observations or average value or mean value. i.e. If a person cover a certain distance at a speed of A km/h and again cover the same distance at a speed of B km/h, then the average speed during the whole journey will be Sum of given observations/Quantities Average(A) = Number of observations/Quantities If distance ‘A’ is covered with speed a , distance ‘B’ is covered with speed b and distance ‘c’ is covered with speed c, then for the whole journey: Some key terms:• • Average of a given term is always lies in range at given data. i.e. Lowest Quantity  Average  Greatest quantity If the quantities of given data are equal then the average will also be the same as quantities. i.e. Average = Greatest / Lowest quantity Value of all quantities = Average 2AB AB A + B + C + ......... A B C + + + ....... a b c Sum of first n natural no. = n n  1 2 Average of first n natural no. = * Average speed = n  1 2 Sum of squares of first n natural no. = n n 1 2n 1 6 Avg. of squares of first n natural no.= * Helping point:- n  1 2n  1 6 Sum of cubes of first n natu- Sum of arithmetic progression whose first term is “a” last term is  n n  1  ral no. =   2   [a + (n - 1)d] . Average of cubes of first n Sn If ‘0’ is one of the quantities of a given data, then that ‘0’ will also be included while calculating average. 2 n n  1 4 2 n  2a  (n 1)d  2 Sum of geometric progression whose first term is [a], last term is [ar n-1] and common ratio is (r) (No greatest or lowest exist) • *  a[r n - 1] r -1 if r  1  a[1 - r n ] 1- r if r  1 natural = * * Sum of first n natural odd no. = n² Avg. of first n natural odd no. =n Sum of first n natural Even no. = n(n+1) Average of first n natural Even no. = (n + 1) Examples 1. = 2701 Find the average of first 73 numbers: (a) 37 (c) 73 (d) 72 Sol. (a) 1 + 2 + 3 ------- + 72 + 73  n (n  1) 2 Average Average  (b) 36 2701  37 73 n 1  37 2 The average of 47 numbers is Alteranate:- Average  2. 106 459. If each of the number is divided by 17, find the new average: (a) 27 (b) 28 (c) 21 (d) 26 Sol. (a) 459  27 , 17 Note : When all the no. are divided or multiplied by an arbitrary no. then the average also get divided or multiplied accordingly. 3. The average of 107 number is 33. If each of the number is multiplied by 13, find the new average: (a) 439 (b) 429 (c) 419 (d) 423 Sol. (b) 33 × 13 = 429 Hence, Answer will be divided 3 and 11 4. The average of 11 result is 50. If the average of the first six result is 49 and that of the last six is 52, the sixth result is : (a) 48 (b) 50 (c) 52 (d) 56 Sol. (d) A+B+C+D+E+F+G+H+I+J+K 11 The average of 1st 6 result = 49 The average of the last 6 result = 52 The Average of total result is= 50 6th result = (52 × 6 + 49 × 6) – (11 × 50) = 56 5. The average of 50 numbers is 45. The average of 50 number and 3 new numbers is 51. The average of the three new number will be : (a) 153 (b) 151 (c) 157 (d) 351 Sol. (b) Total of 3 Numbers = (50 + 3) × 51– 50 × 45 = 453 The average of 3 new result will be  6. 453 151 3 The average of salary of 20 workers in an office is Rs.2100 per month. If the manager’s salary is added, the average becomes Rs.2200 per month. The manager’s annual salary is (in Rs.) : (a) 4200 (b) 4000 (c) 48,000 (d) 50,400 Sol. (d) manager's salary = 2200 + 20(2200 – 2100) = 4200 7. Then manager's annual salary = 4200 ×12 = Rs. 50,400 In a class, there are 10 students at the age of 15 years, 15 at the age of 16 years and 5 at the age of 14 years. What is the average age of a student: (a) 15 2 3 (b) 15 (c) 14 3 4 (d) 15 1 3 Sol. (b)Average age of a student = 10 ´ 15 + 15 ´ 16 + 5 ´ 14 10 + 15 + 5 460 1 = = 15 3 30 8. The average age of 54 boys in a class is 21 years. If the ‘lecture-spell check’s age is included the average age of the boys and lecturer becomes 21 year 6 month. What is the lecturer’s age : (a) 48 years 3 month (b) 48 year (c) 47 year 6 month (d) 48 year 6 month Sol. (d) Lecturer's age is = 21.5 + 54(21.5 – 21) = 48.5 years 9. A family consist of grandparents, parent and three children. The average age of the grandparents is 67 years, that of the parents is 35 years and that of the children is 6 years. What is the average age of the family? (a) 28 4 7 (b) 31 5 7 (c) 32 1 7 (d) 32 5 7 Sol. (b) Required average = 67 ´ 2 + 35 ´ 2 + 6 ´ 3 2+ 2+ 3 107 = 31 5 years 7 10. A library has an average of 510 visitors on friday and 240 on other days. The average number of visitors per day in a month at 30 days begining with a friday is : (a) 250 (b) 276 (c) 280 (d) 285 Sol. (d) If the month starts with a Friday, then there will be 5 Friday in the month Required average æ 510 ´ 5 + 240 ´ 25 ö 8550 ÷= ø 30 30 = çè = 285 11. A student was asked to find the arithmetic mean of the following 12 numbers: 3, 11, 7, 9, 15, 13, 8, 19, 17, 21, 14 and x He found the mean to be 12. The value of x will be. (a) 3 (c) 7 (b) 7 (d) 31 Sol. (b) mean = 3 +11+ 9 + 7 +15 +13 + 8 +19 +17 + 21+14 + x 12 According to question, 137 + x = 12 12 \ 137 + x = 144 \ x = 144 – 137 = 7 12. The average height of 16 boys in a class is 50.25 inches and 8 boys is 45.15 inches. Find the average height of all boys in the class : (a) 47.55 inches (b) 48 inches (c) 48.55 inches (d) 49.25 inches Sol. (c) The required av erage height æ 50.25´ 16+ 45.15´ 8 ö ÷ = çè ø 16+ 8 Average = 48.55 13. The average score of a cricketer for 5 matches is 38.9 runs. If the average of first three mathes is 42, find the average for the last two matches : (a) 33.25 (b) 33.5 (c) 34.25 (d) 35 Sol. (c) The average of last 2 games = 5  38.9  42  3 68.5  2 2 = 34.25 14. The average runs of a cricket player of 5 innings was 62. How many runs must he make in his next innings so as to increase his average of runs by 4? (a) 88 (b) 87 (c) 86 (d) 84 Sol. (c) The runs to be scored by him in 6th innings = 62 + 6  4 = 86 15. In the first 20 overs of a cricket game, the run rate was only 3.2. What should be the run rate in remaining 30 overs to reach the target 262? (a) 5.6 (b) 7.2 (c) 6.6 (d) 8.8 Sol. (c) Score of 20 overs = 3.2  20 = 64 Score of 30 overs = 262 – 64 = 198 Average of remaining overs  198  6.6 30 16. If the average of first 75 innings is 35. How much should he scored in his 76 innings to increase his average by 2 runs? (a) 186 (b) 189 (c) 187 (d) 188 Sol. (c) Score of 76th innings = 35 + 2  76 = 187 17. A car travels from Delhi to Agra at the rate of 20km/hour and from Agra to Delhi at the rate of 30km/hour. What is average speed whole journey ? Average (a) 18km/hr (b) 20km/hr (c) 25km/hr (d) 24km/hr/. Sol. (d) Average speed = 2 (20 ´ 30) = 24 km/hr 20 + 30 18. The average weight of five persons sitting in a boat is 38 kg. The average weight of the boat and the persons sitting in the boat is 52 kg. What is the weight of the boat ? (a) 228 kg (b) 122 kg (c) 232 kg (d) 242 kg Sol. (b) Weight of the boat = 52  (5+Boat) – 5  38 = 312 – 190 = 122 kg 19. The mean of 50 observations was 36. It was found later that an observation 48 was wrongly taken as 23. The corrected (new) mean is : (a) 35.2 (b) 36.1 (c) 36.5 (d) 39.1 Sol. (c) Total sum of 50 observations = 50  36 = 1800 The correct mean 1800 23  48 1825   36.5  50 50 20. The average of eight successive numbers is 6.5. The average of the smallest and the greatest numbers among them will be : (a) 4 (b) 6.5 (c) 7.5 (d) 9 Sol. (b) x + x + 1 + x + 2 + x + 3 + x + 4 + x + 5 + x + 6 + x + 7 = 6.5  8 = 52  8x + 28 = 52  8x = 52 - 28 = 24 x=3  Required average = 3  10 = 6.5 2 21. The average temperature of the first 4 days of a week was 37°C and that of the last 4 days of the week was 41°C. If the average 108 temperature of the whole week was 39°C, the temperature of the fourth days was : (a) 38°C (b) 38.5°C (c) 39°C (d) 40°C Sol. (c) M + T + W + TH = 4  37 = 148°C........(i) TH + F + S + S = 4  41 = 164°C........(ii) M + T +.......S + S = 7  39 = 273°C......(iii)  Temperature of 4th days = 148 + 164 – 273 = 39°C 22. The average of five numbers is 7. When three new numbers are included, the average of the eight numbers becomes 8.5. The average of the three new numbers is : (a) 9 (b) 10.5 (c) 11 (d) 11.5 Sol. (c) Total sum of new three no. = 8  8.5 – 5  7 = 68 – 35 = 33  Required average  33  11 3 23. The average of x numbers is y and Avg. of y numbers is x. Then the average of all the numbers taken together is : x y (a) 2xy (c) x 2 y 2 x y 2xy (b) x  y x y (d) x  y Sol. (b) Sum of x numbers = xy Sum of y no. = xy Required average = xy + xy 2x y = x+y x+y 24. 5 members of a team are weighed Respectively and calculation of their Avg. weight is done after each member is weighed. If the average weight increase by one kg each time, how much heavier is the last player than the first one ? (a) 5 kg. (b) 8 kg. (c) 4 kg. (d) 20 kg. Sol. (b) Average1 = x, weight1 = x Average2 = x + 1, Weight2 = x + 2 Average3 = x + 2, Weight3 = x + 4 Average4 = x + 3, Weight4 = x + 6 Average5 = x + 4, Weight5 = x + 8 Hence, Difference between 5th and 1st = 8 kg. 25. The average score of a group of 20 students in a test was 52. The brightest 20% of them secured a average score of 80 and the dullest 25% a average score of 31. The mean score of remaining : (a) 45% (b) 50% (c) 51.4% (d) 54.6% 109 1 5 Sol. (c) 20%  , 25%  1 L.C.M  20 4 remaining average 1 1 (80) - 20 ´ (31) 5 4 (20 – 5 – 4 =11) 20 ´ 52 - 20 ´ = = 51.4 Average