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__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:**

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__Memory Based Example Problems base on various types__

__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−0.2⇒59A+45=60A−12⇒45+12=60A−59A⇒A=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