## Quantitative Aptitude Notes - Averages

### Quantitative Aptitude Notes - Averages

What is Average?
The result obtained by adding several quantities together and then dividing this total by the number of quantities is called Average. The main term of average is equal sharing of a value among all, where it may share persons or things. We obtain the average of a number using formula that is sum of observations divide by Number of observations.
Average = Sum of Quantities/Number of quantities =>S = A × N
Important Points to Remember:
a) If all the numbers are increased by 'a’, then their average will also be increased by 'a’.
b) If all the numbers are decreased by ‘a’, then their average will also be decreased by 'a’.
c) If all the numbers are multiplied by 'a’, then their average will also be multiplied by 'a'.
d) If all the numbers are divided by 'a', then their average will also be divided by 'a’.
Important Formulae on Averages:

### Memory Based Example Problems base on various types

Type 1: Relationship between Averages and Numbers
1. Find the average of the following set of scores 216, 463, 154, 605,446, 336
Solution:
S = 216 + 463 + 154 + 605 + 446 + 336 = 2220; N = 6
A = S/N = 2220/6 = 370
2. The average of four consecutive even numbers A, B, C and D respectively is 55. What is the product of A and C?
Solution:
The average of A, B, C and D = Average of B and C
But B and C are consecutive even numbers. Their average will be equal to the odd number in between them (Which is 55)
=> B = 55 – 1 = 54; C = 55 + 1 = 56
=> A = B – 2 = 52
=> A × C = 52 × 56 = 2912
3. Average of four consecutive odd numbers is 106. What is the third number in ascending order?
Solution:
A, B, C and D be the four consecutive odd numbers in ascending order
Their average = Average of B and C = the even number between B and C = 106
=> B = 106 – 1 = 105; C = 106 + 1 = 107
Therefore, the third number in ascending order = C = 107
4. The average of five positive integers is 55.8 if average of first two integers is 4 and the average of fourth and fifth integers is 69.5. Then find the third integer.
Solution:
55.8 × 5 = 279; 49 × 2 = 98; 69.5 × 2 = 139
Therefore third number = 279 – 98 – 139 = 42
Type 2: Questions based on Partial Average
1. In a college, 16 girls has the average age is 18 years and 14 boys has the average age 17 years. What would be the average age of entire college?
Solution:
16 girls has the average age is 18 years (16 × 18) = 288
4 boys has the average age 17 years (14 × 17) = 238
Average age = 288 + 238/30 = 526/30 = 17.54
2. The average salary of 25 employee in a company per month is Rs.6000. If the manager’s salary also added then the average is increases by Rs.500. What would be the salary of Manager?
Solution:
Average salary of employee in company = 6000
When added one member salary 25 + 1 = 26 = 6500
So, (26 x 6500) – (25 x 6000) = 169000 – 150000 = 19000
3. The average wages of a worker during a fortnight comprising 15 consecutive working days was Rs.90 per day. During the first 7 days, his average wages was Rs.87/day and the average wages during the last 7 days was Rs.92 /day. What was his wage on the 8th day?
Solution:
The total wages earned during the 15 days that the worker worked =15×90= Rs.1350
The total wages earned during the first 7 days = 7×87 = Rs. 609
The total wages earned during the last 7 days = 7×92 = Rs. 644
Total wages earned during the 15 days = wages during first 7 days + wage on 8thday + wages during the last 7 days.
1350=609+1350=609+ wage on 8th day +644
Wage on 8thday = 1350 – 609 – 644 = Rs. 97
4. 40% of the employees in a factory are workers. All the remaining employees are executives. The annual income of each worker is Rs. 390. The annual income of each executive is Rs. 420. What is the average annual income of all the employees in the factory together?
Solution:
Let the number of employees be x.
40% of employees are workers = 2x/5; Number of executives = 3x/5
The annual income of each worker is Rs. 390.
Hence, the total annual income of all the workers together = 2x/5 × 390 = 156x
Also, the annual income of each executive is Rs. 420.
Hence, the total income of all the executives together = 3x/5 × 420 = 252x
Hence, the total income of the employees = 156x + 252x = 408x
The average income of all the employees together equals = 408x/x = 408
5. The average annual income of Ramesh and Suresh is Rs. 3800. The average annual income of Suresh and Pratap was Rs. 4800, and the average annual incomes of Pratap and Ramesh was Rs. 5800. What is the average of the incomes of the three?
Solution:
Average of Ramesh and Suresh (R + S)/2 = 3800 => Total income (R + S) = 3800 x 2 = 7600
Average of Surech and Pratap (S +P)/2= 4800 => Total (S + P) = 4800 x 2 = 9600
Average of Pratap and Ramesh (P + R)/2 = 5800 => Total (P + R) = 5800 x 2 = 11600
Total of three (2R + 2P + 2S) = 7600 + 9600 + 11600 = 28800 => R + P + S = 14400
Therefore average = 14400/3 = 4800
6. How many students are there in the class who has an average age of 15?
I. The ratio between the boys and girls is 4:3, total ages of whole class is 420.
II. The ratio between the average age of girls and boys is 13:15.
Solution:
From statement I, it is not cleared about the total strength of students.
From statement II, only the ratio between the ages of boys and girls is given, which is inadequate.
On combining we get that total number of student is multiple of 7 but we do not get the total number, so both the statement I and II are not sufficient to answer the question.
7. On a school’s Annual day sweets were to be distributed amongst 112 children. But on that particular day, 32 children were absent. Thus, the remaining children got extra 6 sweets. How many sweets did each child originally supposed to get?
Solution:
Children attended the function = 112 – 32 = 80
Total extra sweets = 80 * 6 =480 due to absence of children.
Original share of each child = 480 / 32 = 15.
Alternative Method:
112x = ( 112-32) * ( x+6) => 32x = 6*80 => x= 6 * 80/ 32 = 15
8. Arithmetic mean of the scores of a group of students in a test was 52. The brightest 20% of them secured a mean score of 80 and the dullest 25% a mean score of 31. The mean score of the remaining 55% is:
Solution:
Remaining students = 100% - 20% - 25% = 55%
Let remaining students got a mean score of x marks.
Then, 20% of 80 + 25% of 31+ 55% of x = 52 => 16+ 7.75 + 55% of x = 52
Therefore, 55% of x = 52 – 16 – 7.75 = 28.25 => x = 2825/100 * 100/55 = 565/11 = 51.4% (approx.)
Type 3: Questions based on Replacement/ without Replacement
1. When a student weighing 45 kg left a class, the average weight of the remaining 59 students increased by 200 g. What is the average weight of the remaining 59 students?
Solution:
Let the average weight of the 59 students be A.
Therefore, the total weight of the 59 of them will be 59 A.
The questions states that when the weight of this student who left is added, the total weight of the class= 59A+ 45
When this student is also included, the average weight decreases by 0.2 kg.
59A+45/60=A−0.259A+45=60A−1245+12=60A−59AA=57
2. There were 35students in a hostel. Due to the admission of 7 new students the expenses of the mess were increased by Rs.42 per day while the average expenditure per head diminished by Re.1. What was the original expenditure of the mess?
Solution:
Let the original average expenditure be Rs.x.
Then 42(x – 1) – 35x = 42 => 7x = 84 => x = 12
Therefore original expenditure = Rs.(35 x 12) = Rs.420
3. The average age of 40 students of a class is 18 years. When 20 new students are admitted to the same class, the average age of the students of the class is increased by six months. The average age of newly admitted students is:
Solution:
Total age of 40 students = 40 * 18 = 720
Total age of 60 students = 60 * 18.5 = 1110
Total age of 20 new students = 1110-720=390
Average age of 20 new students = 390/20 = 19 years 6 months.
Alternative Method: Average age of 60 students = 18 years 6 months
Total increase in age of 40 students = 40 * 6 = 20 years
Increase in age of new students = 20/20= 1year
Therefore, Avg. of new 20 students = 18 years 6 months + 1 year = 19 years 6 months.
Type 4: Questions based on Mistaken Average
1. The average of 8 observations was 25.5. It was noticed later that two of those observations were wrongly taken. One observation was 14 more than the original value and the other observation was wrongly taken as 31 instead of 13. What will be the correct average of those 8 observations?
Solution:
Let correct average =x
Then, correct total =8x
Obtained total =8×25.5=204
204−14 − (31−13) = 8x => x=21.5
2. The arithmetic mean of 100 numbers was computed as 89.05. It was later found that two numbers 92 and 83 have been misreads 192 and 33 respectively. What is the correct arithmetic mean of the numbers?
Solution:
Sum of marks were wrongly increased by = (192 + 33) * (92+83) = 50
Average was wrongly increased by 50 / 100 = 0.5
Therefore, correct mean = 89.05 – 0.5 = 88.55
3. In an examination, the average marks of all the students calculated to be 58 marks. It was later found that marks of 60 students were wrongly written as 70 instead of 50. If the corrected average is 55, find the total no. Of students who took the exam.
Solution:
Total decrease in marks = 60 * (70-50) = 1200
Decrease in average = 58 – 55 = 3.
Therefore, No. of students = 1200 / 3 = 400
Type 5: Questions based on Cricket
1. A cricketer has completed 10 innings and his average is 21.5 runs. How many runs must be make in his next innings so as to rise is average to 24?
Solution:
Total of 10 innings = 21.5 x 10 = 215
Suppose he needs a score of x in 11th innings then average in 11 innings = (215 + x)/11
(215 + x)/11 = 24 => x = (24 x 11) – 215 = 264 – 215 = 49
2. In a cricket team, the average age of eleven players in 28 years. What is the age of the captain?
I. The captain is eleven years older than the youngest player.
II. The average age of 10 players, other than the captain is 27.3 years.
III. Leaving aside the captain and the youngest player, the average ages of three groups of three players each are 25 years, 28 years and 30 years respectively.
Solution:
Total age of 11 players = (28 x 11) years = 308 years.
I. C = Y + 11 C - Y = 11 .... (i)
II. Total age of 10 players (excluding captain) = (27.3 x 10) years = 273 years.
Age of captain = (308 - 273) years = 35 years.
Thus, C = 35. .... (ii)
From (i) and (ii), we get Y = 24
III. Total age of 9 players = [ (25 x 3) + (28 x 3) + (30 x 3)] years = 249 years.
C + Y = (308 - 249) = 59 .... (iii)
From (i) and (iii), we get C = 35.
Thus, II alone gives the answer.
Also, I and III together give the answer.
3. A cricketer had a certain average of runs for his 64th innings. In his 65th inning, he is bowled out for no score on his part. This brings down his average by 2 runs. His new average of runs is:
Solution:
Let original average be x. Then 64x = 65(x-2) => x = 130
Therefore, New average = 130-2 = 128 runs.
Alternative Method: Decrease in average = 2 runs
Total decrease in 64 innings = 64*2 = 128 runs => New average = 0+ 128 = 128 runs.
4. The batting average of a cricket player for 64 innings in 62 runs. His highest score exceeds his lowest score by 180 runs. Excluding these two innings, the average of the remaining innings becomes 60 runs. His highest score is
Solution:
Total runs of the two innings = 2* 62 + 62 * (64-62)= 124+ 124 = 248
Highest score – Lowest score = 180 runs => Highest Score = (240 + 180) / 2 = 214 runs
5. Vinod's bowling average till yesterday was 19.2 . Today he took 7 more wickets and conceded 84 runs, there by his average decreased by 0.2 . How many wickets had he taken till yesterday?
Solution:
Bowling average = (Total runs conceded) / (Number of wickets taken)
bowling average till yesterday = 19.2
Let total runs conceded till yesterday =r
total wickets taken till yesterday =w
r / w = 19.2 => r = 19.2w ------(1)
bowling average till today = 19.2 - 0.2 = 19
Total wickets taken till today = (w+7)
Total runs conceded till today = (r+84)
(r+84)/(w+7) = 19 --------(2)
Using (1) and (2)
(19.2w + 84)/(w + 7) = 19 => 19.2w + 84 = 19w + 133 => 0.2w = 49 => w = 245
Type 6: Miscellaneous
1. How many candidates were interviewed everyday by the panel A out of the three panels A, B and C?
I. The three panels on average interview 15 candidates every day.
II. Out of a total of 45 candidates interviewed everyday by the three panels, the number of candidates interviewed by panel A is more by 2 than the candidates interviewed by panel c and is more by 1 than the candidates interviewed by panel B.
Solution:
I. Total candidates interviewed by 3 panels = (15 x 3) = 45.
II. Let x candidates be interviewed by C.
Number of candidates interviewed by A = (x + 2).
Number of candidates interviewed by B = (x + 1).
Therefore x + (x + 2) + (x + 1) = 45 =>3x = 42 =>x = 14
Hence II alone sufficient while I alone not sufficient to answer
2. The average weight of a class of 25 students is 30kg. The average weight of girls is 5 kg more than that of boys. If the class teacher's weight, which is between 64kg and 106 kg is included, the average weight of the male members of the class equals that of the female members. What is the number of girls in the class, if the average weight of the boys(in kg) is an integer?
Solution:
Let average weight of boys =x kg
Then, average weight of girls =(x+5)kg
Let number of boys = n
Then, number of girls = (25- n)
nx+(25- n)(x+5)=25×30 => 5x - n=125 (1)
Let class teacher's weight =y
(nx+y)/(n+1)=(x+5) => y = 5n+x+5
From (1), x = (125+n) / 5
From (2), x=y - 5n - 5
Therefore,
( 125+n)/ 5 = y - 5n - 5 => n = (5y - 150) / 26 (3)
Given, 64 ≤ y ≤ 106
y=64 n ≥ 7 y=106 => n ≤ 14
Therefore, 7 ≤ n ≤ 14
From (3), y = (26n + 150)/5 = 26n/5 + 30
Only for n=10, y is an integer ( Since xx is an integer, from (2), it is clear that y is also an integer)
Thus we get n=10
Therefore, number of girls in the class =25 - n = 15
3. A family consists of two grandparents, two parents and three grandchildren. The average age of the grandparents is 67 years, that of the parents is 35 years and that of the grandchildren is 6 years. The average age of the family is
Solution:
Total age of the grandparents = 67 × 2
Total age of the parents = 35 × 2
Total age of the grandchildren = 6 × 3
Average age of the family = ((67×2) + (35×2) + (6×3))/7 = (134+70+18)/7 = 222/7 = 31 5/7
4. Average temperature from 9th to 16th of a month is 30° C and that from 10th to 17th is 31°C. What is the temperature on 17th , if temperature on 9th is 35° C ?
Solution:
Difference between temperature on 9th and 17th = 8 * (31° C - 30° C) = 8° C
Temperature for 8 days including 17th is more than that of 8 days including 9th.
Therefore, Temperature on 17th is more than the temperature on 9th.
Therefore, Temperature on 17th = Temperature on 9th + Difference = 35° C + 8° C
5. Some students planned a trip and estimated their total expenses to be Rs. 500. However, 5 of them could not go for the trip and as a result average expenditure of the remaining students is increased by Rs. 5. How many students have gone for trip?
Solution:
Let the total no. of students be x.
Then, 500/x = 500/(x – 5) – 5
500/x + 5 = 500/(x – 5) => (500 + 5x)/x = 500/(x – 5)
(500 + 5x) (x – 5) = 500x => 500x + 5x2 – 2500 – 25x = 500x
5x2 – 25x – 2500 = 0 => x = 25, - 20. We cannot take negative value. So, x = 25.
6. A ship 40 km from shore springs a leak which admits 3 ¾ quintals of water in 12 minutes. 60quintals would suffice to sink the ship, but its pump can throw out 12 quintals of water in one hour. Find the average rate of sailing so that it may reach the shore just it begins to sink.
Solution:
In 12 min. leak admits = 15/4 quintals
In one hour leak admits = 15/4 * 60/12 =75/4 quintals
In one hour pumps throw out = 12 quintals
Water left in the ship in one hour = 75/4 – 12 =27/4 quintals
27/4 quintals of water is left in the ship in 1 hour
60 quintals of water is left in the ship in = 1*60*4 / 27 = 80/9
Now in 80/9hrs ship runs = 40 km
1 hr the ship runs = 40 * 9 / 80 =4.5 km/hr