POSTAL ASSISTANT EXAM MATERIAL - QUANTITATIVE APTITUDE

POSTAL ASSISTANT EXAM MATERIAL - QUANTITATIVE APTITUDE

Written By Admin on April 16, 2014 | Wednesday, April 16, 2014

Disclaimer:- All the Information provided in this post are prepared & compiled by A. Praveen Kumar, SPM, Papannapet SO-502303, Telangana State for in good faith of Postal Assistant Exam Aspirants. Author of blog does not accepts any responsibility in relation to the accuracy, completeness, usefulness or otherwise, of contents.

AVERAGE

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Definition

Central Tendency of data is called an Average. Thus an average is a single value that indicates a group of values. Average can be calculated with the help of mean, median and mode. The most commonly used average is the Arithmetic Mean (AM). While some other average geometric mean and harmonic mean are also quite popular.

Arithmetic Mean is calculated by dividing the sum of all the number/quantities by the number of numbers/quantities. The mean of n numbers

x₁, x₂, x₃, … , x_n, is denoted by X and Calculated as

X =

= or

= ∑x_i / n

where ∑ (sigma) denotes the sum of the term of type I from 1 to n.

Average is a very simple but effective way of representing an entire group by a single value.

STIPES

(Standard Types In Problem Execution And Solving)

STIPE 01: To find the average when the number of quantities and their sum is given. We have the following formula, Average = Sum of the quantities / Number of quantities.

STIPE 02: To find the sum, when the number of quantities and their average is given. We have the following, Sum of quantities = Average × Number of quantities.

STIPE 03: To find the number of quantities, when the sum of quantities and average are given. We have the following formula, Number of quantities = Sum of quantities / Average.

AVERAGE SHORTCUT METHODS 1. Average = [Total of observations / No. of observations] 2. (i) When a person joins a group in case of increasing average Age weight of new comer = [ (Previous Age + No. of persons) * Increase in Average ](ii) In case of decreasing Average, Age (or) weight of new comer = [ (Previous Age - No. of persons) * Decrease in Average ] 3. When a person leaves a group and another person joins the group in the place of person left, then (i) In case of increasing average, Age (or) weight of new comer = [ (Age of person left + No. of persons) * Increase in Average ] (ii) In case of decreasing Average, Age (or) weight of new comer = [ (Age of person left - No. of persons) * Decrease in Average ] 4.When a person leaves the group but nobody joins this group, then (i) In the case of increasing Average, Age (or) weight of man left = [ (Previous Age - No. of present persons) * Increase in Average ] (ii) In case of decreasing Average, Age (or) weight of new comer = [ (Previous Age + No. of present persons) * Decrease in Average ]5. If a person travels a distance at a speed of x Km/hr returns to the original place of y Km/hr then average speed is [ 2.x.y / ( x + y ) ] 6.If half of the journey is travelled at speed of x km/hr and the next half at a speed of x km/hr. Then average speed during the whole journey is [ 2.x.y / (x + y) ] 7. If a person travels 3 equal distances at a speed of x Km/hr, y Km/hr,z km/hr. Then average speed during whole journey is [ 3.x.y / (x.y +y.x +z.x) ] EXCERCISE 1. Find the average of all prime numbers between 30 and 50. Sol. There are five prime numbers between 30 and 50. They are 31, 37, 41, 43 and 47. Required average = [(31 + 37 + 41 + 43 + 47) / 5] = 199/5 = 39.8. 2. Find the average of first 40 natural numbers. Sol. Sum of first n natural numbers = 40 * 41 / 2 = 820. Required average = 820 / 40 = 20.5 3. Find the average of first 20 multiples of 7. Sol. Required average = 7(1+2+3 ... + 20) / 20 = [7 * 20 * 21 / 20 * 2] = [147 / 2] = 73.5. 4. The average weight of 10 oarsmen in a boat is increased by 1.8 kg when one of the crew, who weighs 53kg is replaced by a new man. find the weight of the new man Sol. Total weight increased = (1.8 * 10)kg = 18kg. Weight of the new man = (53 + 18)kg = 71kg. 5. A batsman makes a score of 87 runs in the 17th inning and thus increases his average by 3. find his average after 17th inning. Sol. Let the average after 17th inning = x. Then, average after 16th inning = (x - 3) 16(x - 3) + 87 = 17x or x = (87 - 48) = 39. 6. The average age of the mother and her six children is 12 years which is reduced by 5 years if the age of the mother is excluded. How old is the mother? Sol: Therefore age of the mother = ( 12 × 7 – 7 × 6) = 42 years 7. The age of five numbers is 27. If one number is excluded, the average becomes 25. The excluded number is Sol: Therefore excluded number = (27 × 5) - ( 25 × 4) = 135 – 100 = 35. 8. Three years ago, the average age of A and B was 18 years. With C joining them, the average age becomes 22 years. How old is C now? Sol: Present age of (A + B) = (18 × 2 + 3 × 2) years = 42 years. Present age of (A + B + C) = (22 × 3) years = 66 years. Therefore C’s age = (66 – 42) years = 24 years. 9. The average age of 36 students in a group is 14 years. When teacher’s age is included to it, the average increases by one. What is the teacher’s age in years? SoL : Age of the teacher = ( 37 × 15 – 36 × 14 ) years = 51 years 10. The average runs of a cricket player of 10 innings was 32. How many runs must he makes in his next innings so as to increase his average of runs by 4? Sol: Average after 11 innings = 36. Therefore required number of runs = (36 × 11) – (32 × 10) = 396 - 320 = 76. 11. The average salary of all the workers in a workshop is Rs. 8000. The average salary of 7 technicians is Rs. 12000 and the average salary of the rest is Rs. 6000. The total number of workers in the workshop is Sol: Let the total number of workers be x. Then, 8000x = (12000 × 7) + 6000 ( x – 7) ‹=› 2000x = 42000 ‹=› x = 21. 12. Kamal obtained 76, 65, 82, 67 and 85 marks(out of 100) in English, Mathematics, Physics, Chemistry and Biology. What are his average marks? Sol: Average = 76 + 65 + 82 + 67 + 85 / 5) = (375 / 5) = 75. 13. In the first 10 over’s of a cricket game, the run rate was only 3.2. What should be the run rate in the remaining 40 0vers to reach the target of 282 runs? Sol: Required run rate = 282 – (3.2 × 10 / 40) = 240 / 40 = 6.25. 14. The average weight of a class of 24 students is 35 kg. If the weight of the teacher be included, the average rises by 400 g. The weight of the teacher is Sol: Weight of the teacher = (35.4 × 25 – 35 × 24)kg = 45 kg. 15. After replacing an old member by a new member, it was found that the average age of five numbers of a club is the same as it was 3 years ago. What is the difference between the ages of the replaced and the new member? Sol: Age decrease = (5 × 3) years = 15 years. So, the required difference = 15 years 16. The average weight of 8 person’s increases by 2.5 kg when a new person comes in place of one of them weighing 65 kg. What might be the weight of the new person? Sol: Total weight increased = (8 × 2.5) kg = 20 kg. Weight of new person = (65 + 20) kg = 85 kg. 17. In the first 10 overs of a cricket game, the run rate was only 3.2. What should be the run rate in the remaining 40 overs to reach the target of 282 runs? Sol: Runs scored in the first 10 overs = 10×3.2=32 Total runs = 282, remaining runs to be scored = 282 - 32 = 250 remaining over’s = 40, Run rate needed = 250 40 =6.25 18. A grocer has a sale of Rs. 6435, Rs. 6927, Rs. 6855, Rs. 7230 and Rs. 6562 for 5 consecutive months. How much sale must he have in the sixth month so that he gets an average sale of Rs. 6500? Sol: Let the sale in the sixth month = x Then 6435+6927+6855+7230+6562+x 6 =6500 => 6435 + 6927 + 6855+ 7230 + 6562 + x = 6 \times 6500 = 39000 => 34009 + x = 39000, x = 39000 - 34009 = 4991 19. The average of 20 numbers is zero. Of them, How many of them may be greater than zero, at the most? Sol: Average of 20 numbers = 0 =>Sum of 20 numbers 20 =0 => Sum of 20 numbers = 0 Hence at the most, there can be 19 positive numbers. (Such that if the sum of these 19 positive numbers is x, 20th number will be -x) 20. The average age of a class is 15.8 years. The average age of the boys in the class is 16.4 yrs while that of the girls is 15.4 years. What is the ratio of boys to girls in the class? Sol: Let the ratio is k: 1. Then, k*16.4 + 1 * 15.4= (k+l)*15.8 or(16.4 - 15.8)k=(15.8-15.4) or k=0.4/0.6= 2/3. Required ratio = 2/3 : 1 = 2 : 3 21. If a, b, c, d, e are five consecutive odd numbers, their average Is: Sol : Clearly=a+2, c=a+4,d=a+6 and e=a+8. Average (a + (a + 2) + (a + 4) + (a + 6) + (a + 8)) / 5 = (5a + 20) / 5 = (a + 4) 22. The average of first five multiples of 3 is Sol: average = (1 + 2 + 3 + 4 + 5) * 3/5 = 15 * 3 / 5 = 9 23. The average weight of 16 boys in a class is 50.25 kg and that of the remaining 8 boys is 45.15 kg. Find the average weights of all the boys in the class. Sol: Required average = 50.25 x 16 + 45.15 x 8 16 + 8 = 804 + 361.20 24 = 1165.20 24 = 48.55 24. A library has an average of 510 visitors on Sundays and 240 on other days. The average number of visitors per day in a month of 30 days beginning with a Sunday is: Sol: Since the month begins with a Sunday, to there will be five Sundays in the month. Required average = 510 x 5 + 240 x 25 30 = 8550 30 = 285 25. The average weight of 8 person's increases by 2.5 kg when a new person comes in place of one of them weighing 65 kg. What might be the weight of the new person? Sol: Total weight increased = (8 x 2.5) kg = 20 kg. Weight of new person = (65 + 20) kg = 85 kg. 26. The average (arithmetic mean) of 3 numbers is 60. If two of the numbers are 50 and 60, what is the third number? Sol : 3*60=180, which is the total number of points earned; the two numbers we do know are 50 and 60 which add up to 110 The third number is 180-110=70 since all three numbers must add up to 180 27. There are two sections A and B of a class, consisting of 36 and 44 students respectively. If the average weight of sections A is 40 kg and that of section b is 35 kg. Find the average weight of the whole class? Sol: Total weight of(36+44)Students (36x40+44x35) = 2980 kg. Average weight of the whole class = (2980 / 80) =37.25. 28. A batsman makes a score of 87 runs in the 17th inning and thus increases his averages by 3.Find his average after 17th inning? Sol: Let the average after 17th inning = x. Then, average after 16th inning = (x - 3) Average =16 (x-3)+87 = 17x or x=(87-48) = 39. 29. A students was asked to find the arithmetic mean of the numbers 3, 11, 7, 9, 15, 13, 8, 19, 17, 21, 14 and x. He found the mean to be 12. What should be the number in place of x? Sol: Clearly, we have (3+11+7+9+15+13+8+19+17+21+14+x/12) =12 Number in place of x is 137+x=144 x= 144-137 x= 7. . 30.David obtained 76, 65, 82, 67 and 85 marks (out in 100) in English, Mathematics, Physics, Chemistry and Biology.What are his average marks? Average = (76+65+82+67+85/5) = 375/5 = 75 31. The average of 20 numbers is zero. Of them, at the most, how many may be greater than zero? Average of 20 numbers = 0 Sum of 20 numbers =(0 x 20) =0. It is quite possible that 19 of these numbers may be positive and if there sum id a, then 20th number is (-a). 32. The average age of boys in a class is 16 years and that of the girls is 15 years. The average age for the whole class is Sol: Clearly, to find the average, we ought to know the numbers of boys, girls or students in the class, neither of which has been given. so the data provided is inadequate. 33. The average age of 36 students in a group is 14 years. When teacher's age is included to it, the average increases by one. What is the teacher's age in years? Sol: Age of the teacher is 37X15-34X14 years =51 years 34. The average of five numbers id 27. If one number is excluded,the average becomes 25. The excluded number is Sol: = 27X5- 25X4 Excluded Number is = 135-100 = 35 35. The average score of a cricketer for ten matches is 38.9 runs. If the average for the first six matches is 42. Then find the average for the last four matches? Required average =(38.9 x 10)-(42 x 6)/ 4 = 137 / 4. = 34.25 36. A motorist travel to a place 150 km away at an average speed of 50 km/hr and returns ar 30 km/hr. His average speed for the whole journey in km/hr is Average Speed = (2xy/x +y )km/hr =(2x50 x30/ 50+30) = 37.5 km/h