Exam - 20 Math: Topic: Average, Mean, Problems on Ages

Question: The average of 5 consecutive numbers is 34. What is the largest number?

Solution:
For 5 consecutive numbers, the average is the middle number.
So, middle number = 34

The numbers are = 32, 33, 34, 35, 36

∴ Largest number = 34 + 2 = 36

Question: The average of 10 numbers is 15. The average of 8 of these numbers is also 15. What is the average of the remaining two numbers?

Solution:
Average of 10 numbers = 15
∴ Total sum of 10 numbers = 10 × 15 = 150

And,
Average of 8 numbers = 15
∴ Sum of 8 numbers = 8 × 15 = 120

∴ Sum of remaining 2 numbers = 150 - 120 = 30

∴ Average of remaining 2 numbers = 30/2 = 15

Question: Rifat's age after 15 years will be 5 times his age 5 years back. What is the present age of Rifat?

Solution:
Let Rifat's present age be x years.
Then, Rifat's age after 15 years = (x + 15) years.
Rifat's age 5 years back = (x - 5) years.

Therefore, 
x + 15 = 5(x - 5)
⇒ x + 15 = 5x - 25 
⇒ 5x - x = 15 + 25
⇒ 4x = 40 
⇒ x = 40/4 = 10 
∴ x = 10

Hence, Rifat's present age = 10 years.

Question: A batsman scored an average of 45 runs in 6 innings. How many runs should he score in the 7th innings to increase his average by 3?

Solution:
Given that,
Average runs in 6 innings = 45

∴ Total runs in 6 innings = Average × Number of innings
= 45 × 6
= 270

He wants to increase his average by 3.
New average after 7 innings = 45 + 3
= 48

Therefore, total runs needed in 7 innings = 48 × 7
= 336

∴  Runs to be scored in the 7th innings = 336 - 270
= 66

So, the batsman must score 66 runs in his 7th innings to increase his average by 3.

Question: The average weight (in kg) of a family of five members, whose weights are 40 kg, 49 kg, 57 kg, 66 kg and 38 kg, is:

Solution:
Given weights,
40, 49, 57, 66, 38

∴ Total weight = 40 + 49 + 57 + 66 + 38 = 250
And number of members = 5

∴ Average weight = 250/5 = 50

So, the average weight of the five family members is 50 kg.

Question: The product of the ages of Ankit and Nikita is 240. If twice the age of Nikita is more than Ankit's age by 4 years, What is Nikita's age?
 
Solution:
Given that,
Product of the ages of Ankit and Nikita = 240
And two times age of Nikita is more than Ankit's age by 4 years

Now,
Let Ankit's age be x years, then, Nikita's age = 240/x years
Two times age of Nikita is more than Ankit's age by 4 years

ATQ,
2 × (240/x) - x = 4
⇒ 480 - x2 = 4x
⇒ x² + 4x - 480 = 0
⇒ x² + 24x - 20x - 480 = 0
⇒ x(x + 24) - 20(x + 24) = 0
⇒ (x + 24)(x - 20) = 0
∴ x + 24 = 0
⇒ x = - 24 [Neglecting the negative value]
Or, x - 20 = 0
∴ x = 20

∴ Nikita's age = 240/20 = 12 Years

 So, the Nikita's age is 12 years

Question: The average age of 15 students is 18 years. If the age of the teacher is also included, the average age increases by 1 year. What is the age of the teacher?

Solution:
Given that,
Average age of 15 students = 18 years

∴ Total age of 15 students = 15 × 18
= 270 years

When the teacher is included,
New average = 18 + 1 = 19 years
And total persons = 16

∴ New total age = 16 × 19
= 304 years

∴ Teacher’s age = 304 - 270
= 34 years

So, the age of the teacher is 34 years

Question: If mean of 14, 13, 18, 16, k, (k + 3) is 13, then what will be the mean of k, 8, 9, 11, 5, 10, 6?

Solution:
The mean of 14, 13, 18, 16, k, (k + 3) = 13 which means
{14 + 13 + 18 + 16 + k + (k + 3)}/6 = 13
⇒ 2k + 64 = 78
⇒ 2k = 78 - 64
⇒ 2k = 14
⇒ k = 14/2
∴ k = 7

Then, the mean of k, 8, 9, 11, 5, 10, 6 is
= (k + 8 + 9 + 11 + 5 + 10 + 6)/7
= (7 + 8 + 9 + 11 + 5 + 10 + 6)/7
= 56/7
= 8

Question: The present age of a father is 3 years more than three times the age of his son. Three years hence, father's age will be 10 years more than twice the age of the son. Find the present age of the father.

Solution: 
Let the son's present age be x years.
Then, father's present age = (3x + 3) years

ATQ,
(3x + 3 + 3) = 2(x + 3) + 10
⇒ 3x + 6 = 2x + 16 
⇒ 3x - 2x = 16 - 6
∴ x = 10. 

Hence, father's present age = (3x + 3) = {(3 × 10) + 3} years = 33 years.

Question: The weighted mean of marks in Mathematics, Science, and English is 82. The marks in Mathematics and Science are 76 and 88, with weights 3 and 4, respectively. If the weight of English is 3, find the marks in English. 

Solution:
Given that,
Marks in Mathematics, x₁ = 76, Weight, w₁ = 3
Marks in Science, x₂ = 88, Weight, w₂ = 4
Weight of English, w₃ = 3
And weighted Mean = 82

Let the marks in English = x₃

We know, 
Weighted Mean = [(w₁ × x₁) + (w₂ × x₂) + (w₃ × x₃)] ÷ (w₁ + w₂ + w₃)
82 = [(3 × 76) + (4 × 88) + (3 × x₃)] ÷ (3 + 4 + 3)
⇒ 82 = [228 + 352 + 3x₃] ÷ 10
⇒ 580 + 3x₃ = 82 × 10
⇒ 3x₃ = 820 - 580
⇒ 3x₃ = 240
⇒ x₃ = 240 ÷ 3
∴ x₃ = 80

So, the marks in English are 80. 

Question: One year ago, the ratio of Rahul's and Amir's age was 6 :  7 respectively. Four years hence, this ratio would become 7 : 8. How old is Amir?

Solution: 
Let Rahul's and Amir's ages one year ago be 6x and 7x years respectively.
Then,
Rahul's age 4 years hence = (6x + 1) + 4 = (6x + 5) years.
And Amir's age 4 years hence = (7x + 1) + 4 = (7x + 5) years. 

Therefore,
(6x + 5) : (7x + 5) = 7 : 8
⇒ (6x + 5)/(7x + 5) = 7/8
⇒ 8(6x + 5) = 7(7x + 5)
⇒ 48x + 40 = 49x + 35
∴ x = 5

Hence, Amir's present age = (7x + 1) = 35 + 1 = 36 years

Question: The average weight of P and his three friends is 55 kg. If P is 4 kg more than the average weight of his three friends, what is P's weight (in kg)?

Solution: 
Average weight of P and his 3 friends = 55 kg
P's weight is 4 kg more than the average weight of his 3 friends.

Let the average weight of the 3 friends be x kg.
∴ Total weight of the 3 friends = 3x kg
And weight of P = (x + 4) kg

∴ Total weight of all 4 persons = 4 × 55 = 220 kg

ATQ,
3x + (x + 4) = 220
⇒ 4x + 4 = 220
⇒ 4x = 220 - 4 = 216
⇒ x = 216/4
∴ x = 54 kg

∴ P's weight = x + 4 = 54 + 4 = 58 kg

Shortcut Trick
Total weight of P and his 3 friends = 4 × 55 = 220 kg.
Let P's weight be P. The average of the 3 friends is (P − 4) kg.

Total weight = P + 3 × (P - 4) = 220
⇒ 4P - 12 = 220
⇒ 4P = 232
⇒ P = 232/4 = 58 kg.

Question: What is the mean of the first 99 natural numbers?

Solution:
The first 99 natural numbers are: 1, 2, 3, ..., 99.

We know,
Mean of the first n natural numbers = (n + 1)/2
∴ Mean = (99 + 1)/2 ; [Here, n = 99]
= 100/2
= 50
So, the mean of the first 99 natural numbers is 50.

Alternatively:
Sum = (99 × 100)/2 = 4950
Mean = Sum/99 = 4950/99 = 50
So, the average is exactly 50.

Question: Father’s present age is four times his son’s present age. After 8 years, the father’s age will be two and a half times his son’s age. After another 8 years, what will be the ratio of the father’s age to his son’s age?

Solution:
Let the son’s present age = x years
So, father’s present age = 4x years

After 8 years,
Son’s age = x + 8 years
Father’s age = 4x + 8 years

According to the question,
4x + 8 = (5/2)(x + 8)
⇒ 2(4x + 8) = 5(x + 8)
⇒ 8x + 16 = 5x + 40
⇒ 8x - 5x = 40 - 16
⇒ 3x = 24
⇒ x = 8

So,
Son’s present age = 8 years
Father’s present age = 4 × 8 = 32 years

After another 8 years means after total 16 years from present,
Son’s age = 8 + 16 = 24 years
Father’s age = 32 + 16 = 48 years

Therefore,
Father’s age : Son’s age = 48 : 24
= 2 : 1

Question: 20 students of a college went to a hotel. 19 of them spent Tk. 175 each on their meal and the 20th student spent Tk. 19 more than the average of all the 20. Find the total money spent by them. 

Solution:
Let the average spending of all 20 students = x
So, total spending = 20x 

Given that, 
19 students spent 175 each
∴ Total = 19 × 175 = 3325
20th student spent = x + 19

ATQ, 
3325 + (x + 19) = 20x
⇒ 20x - x = 3325 + 19
⇒ 19x = 3344
⇒ x = 3344/19
∴ x = 176

∴ Total money spent = 20x = 20 × 176 = Tk. 3520 

Question: The average age of A, B, C, D is 24 years. Average of A, B, C is 22 years and average of B, C, D is 26 years. Find average of B and C

Solution:
Let the ages of A, B, C, D be a, b, c, d respectively.

Total age of A, B, C, D,
a + b + c + d = 24 × 4 = 96 ...…(1)

Total age of A, B, C,
a + b + c = 22 × 3 = 66 ...…(2)

Total age of B, C, D,
b + c + d = 26 × 3 = 78 ....…(3) 

Add equations (2) and (3),
⇒ (a + b + c) + (b + c + d) = 66 + 78 = 144
∴ a + 2(b + c) + d = 144 ...…(4)

Subtract equation (1) from equation (4),
⇒ a + 2(b + c) + d - (a + b + c + d) = 144 - 96
∴ b + c = 48

∴ Average age of B and C = (b + c)/2 = 48/2 = 24 years.

Question: The average of 10 numbers is 60. The average of the first 5 numbers is 55 and the average of the next 3 numbers is 65. The 9th number is 10 less than the 10th number. Find the 10th number.

Solution: 
Total sum of 10 numbers = 10 × 60 = 600
Sum of first 5 numbers = 5 × 55 = 275
Sum of next 3 numbers (6th, 7th, and 8th) = 3 × 65 = 195

∴ Sum of first 8 numbers = 275 + 195 = 470
∴ Sum of 9th + 10th number = 600 - 470 = 130

Let the 10th number = x
Then, 9th number = x - 10
So,
(x - 10) + x = 130
⇒ 2x - 10 = 130
⇒ 2x = 140
⇒ x = 140/2
∴ x = 70

So, the 10th number is 70.

Question: The sum of ages of 5 children born at the intervals of 3 years each is 50 years. What is the age of the youngest child?

Solution: 
Let the ages of children are,
x, (x + 3), (x + 6), (x + 9) and (x + 12) years.

Then,
⇒ x + (x + 3) + (x + 6) + (x + 9) + (x + 12) = 50
⇒ 5x + 30 = 50
⇒ 5x = 20
∴ x = 4. 

∴ Age of the youngest child = x = 4 

Question: The ages of 5 people are in the ratio 2 : 3 : 4 : 10 : 11. If their average age is 18 years, find the age of the eldest person.

Solution:
Given that,
The ages of 5 people are in the ratio 2 : 3 : 4 : 10 : 11
And average age = 18 years

Let the common ratio multiplier = x
So their ages are: 2x, 3x, 4x, 10x, 11x

∴ Total age = 2x + 3x + 4x + 10x + 11x = 30x

Now,
30x/5 = 18
⇒ 30x = 90
⇒ x = 90/30
∴ x = 3

Age of the eldest person = 11x = 11 × 3 = 33 years

So, the eldest person is 33 years old.

Question: A man is 24 years older than his son. In two years, his age will be twice the age of his son. The present age of his son is:

Solution: 
Let the son's present age be x years.
Then, man's present age = (x + 24) years.

ATQ,
(x + 24) + 2 = 2(x + 2)
⇒ x + 26 = 2x + 4
⇒ 2x - x = 26 - 4
∴ x = 22

So, the present age of his son is 22 years.

Question: The average of x and y is 45, and the average of y and z is 50. If y = 42, then find the value of (x + z).

Solution:
Given that, 
y = 42

Average of x and y = 45
∴ x + y = 45 × 2
⇒ x + y = 90
⇒ x + 42 = 90
⇒ x = 90 - 42
⇒ x = 48

Average of y and z = 50
∴ y + z = 50 × 2
⇒ y + z = 100
⇒ 42 + z = 100
⇒ z = 100 - 42
⇒ z = 58

Now, x + z = 48 + 58 = 106

Question: The sum of the present ages of a father and his son is 60 years. Six years ago, father's age was five times the age of the son. After 6 years, son's age will be-

Solution: 
Let the present ages of son and father be x and (60 - x) years respectively.
Then,
⇒ (60 - x) - 6 = 5(x - 6) 
⇒ 54 - x = 5x - 30
⇒ 6x = 84
∴ x = 14. 
∴ Present age of son = 14 years

So, After 6 years, son's age = 14 + 6 = 20 years

Question: The average of six numbers is x and the average of three of these is y. If the average of the remaining three is z, then which one is correct?

Solution:
Total sum of six numbers = 6x
Total sum of three numbers = 3y
Total sum of the other numbers = 3z

Now,
6x = 3y + 3z
⇒ x = 3(y + z)/6
⇒ x = (y + z)/2
∴ 2x = y + z

Question: The present ages of three persons are in the ratio 4 : 7 : 9. Eight years ago, the sum of their ages was 56. Find their present ages (in years).

Solution:
Let the present ages of the three persons be 4x, 7x, and 9x years.
Eight years ago, their ages were,
(4x - 8), (7x - 8) and (9x - 8)

According to the problem, 
⇒ (4x - 8) + (7x - 8) + (9x - 8) = 56
⇒ 20x - 24 = 56
⇒ 20x = 80
⇒ x = 80/20
∴ x = 4
∴ First person = 4x = 4 × 4 = 16 years
∴ Second person = 7x = 7 × 4 = 28 years
∴ Third person = 9x = 9 × 4 = 36 years

So, the present ages of the three persons are 16 years, 28 years, and 36 years respectively

Question: If the mean exceeds the mode by 48 and the median is 12, find the mean.

Solution:
Given,
Median = 12
Mean - Mode = 48

We know,
Mode = 3Median - 2Mean

So,
Mean - Mode = Mean - (3Median - 2Mean)
= Mean - 3Median + 2Mean
= 3Mean - 3Median
= 3(Mean - Median)

Therefore, Mean - Mode = 3(Mean - Median)
⇒ 48 = 3(Mean - 12)
⇒ Mean - 12 = 48/3
⇒ Mean - 12 = 16
⇒ Mean = 16 + 12
∴ Mean = 28

Answer: গ) 28