Age Problems - Key Formulas
Basic Age Relationships
If A is x years older than B:
If A is x times as old as B:
Age Ratios
If ages of A and B are in ratio m:n:
If sum of ages is S:
Future Ages
After t years:
New ratio after t years:
Past Ages
t years ago:
Past ratio t years ago:
Age Problems
Practice aptitude questions on age-related problems
Question 1:
B is twice as old as A. C is twice as old as B. If the difference between ages of A and C is 12 years, find the age of B.
Let the ages of A, B and C be x, y and z.
y = 2x, z = 2y
z = 4x
Difference between ages of x and z = 12 years
4x – x = 12 years
3x = 12 years, x = 4 years
y = 2x = 2 * 4 = 8 years
Question 2:
P is thrice as old as Q. The difference between the ages of P and Q is a multiple of 9. Find the minimum possible age of P, if it is known that Q's age is an odd prime number.
Let the age of P and Q be x and y.
The minimum possible age of Q can be 3 as the age is a prime number and an odd number.
The minimum difference between their ages can be 9 as the number should be a multiple of 9.
P's minimum possible age can be Q's minimum possible age + minimum possible difference.
3 + 9 = 12 years
Question 3:
The sum of ages of 2 brothers is 45 years. If their ages are in a ratio 4:5, find their ages after 10 years.
Let their ages be x and y, respectively.
x:y = 4:5
x / 4 = y / 5
5x = 4y
x + y = 45
4(x + y) = 45 * 4 (multiplying both sides by 4)
4x + 4y = 180
4x + 5x = 180
9x = 180, x = 20
y = 45 – x = 45 – 20 = 25
x = 20 years, y = 25 years
After 10 years:
x = 20 + 10 = 30 years
y = 25 + 10 = 35 years
Question 4:
Ram and Raju are cousins. If it is known that Ram is 30 years younger than Raju and the ratio of their ages is 28:7. Find their ages after 5 years.
Let their ages be x and y, respectively.
x:y = 28:7
x / 28 = y / 7
7x = 28y
x = 28y / 7 = 4y
Difference between their ages = 30 years
x – y = 4y – y = 30
3y = 30
y = 10
x = 4y = 10 * 4 = 40 years
After 5 years
x = 40 + 5 = 45 years
y = 10 + 5 = 15 years
Question 5:
X's age is 20% more than Y's age. If the sum of their ages is 20 more than the difference of their ages, find their respective ages.
Let the ages of X and Y be p and q.
p = q + 20% of q
(p + q) – (p – q) = 20
p + q – p + q = 20
2q = 20
q = 10
Y's age = 10 years
X's age is 20% more than Y's age.
10 + 20% of 10
10 + 10 * 20 / 100
10 + 2 = 12
Their ages are 12, 10.
Question 6:
Out of 3 friends one is 12 years old. The ratio of the ages of the other 2 friends a and b is 3:4. If the youngest among them is 12 years old, find the minimum possible age of the eldest friend.
In the question it is given that first friend is 12 years old also the youngest among them but nowhere it is given that two friends are not of the same age. Thus, in this question it will be assumed that two friends are 12 years old. Of which one is the first mentioned and the other is a.
a is 12 years old.
a:b = 3:4
b = a / 3 * 4, = 12 / 3 * 4 = 16 years
Total of their age = 12 + 12 + 16 = 40
Question 7:
A branch of a company have 2 employees m and n. If it is known that the ages of m and n are in a ratio 3:12, and the sum of their ages is 7 years and 6 months. Find the age of n.
The total of their ages is given in a combination of years and months so the first step will be to convert the total in one unit, here months.
7 years and 6 months = 7 * 12 + 6 = 84 + 6 = 90 months
Ratio of ages of m and n = 3:12
m / 3 = n / 12
12m = 3n
4m = n
m + 4m = 90
5m = 90
m = 18
n = 4m = 4 * 18 = 72
72 months = 72 / 12 = 6 years
Question 8:
A and B are sisters born after the year 2010. If the sum of their ages will be 19 in the year 2025, What will be the age of A in 2030. It is known that in the year 2024 the ratio of their age is 7:10.
The sum of their ages in 2025 = 19 years
The sum of their ages in 2024 = (sum in 2025) – 2
19 – 2 = 17
Ratio of ages of a and b in 2024 = 7:10
a / 7 = b / 10
10a = 7b
a + b = 17 years
10 (a + b) = 17 * 10 (multiplying both sides by 10
10a + 10b = 170
7b + 10b = 170
17b = 170
b = 170 / 17 = 10
a = 7b / 10 = 7 * 10 / 10 = 7 years
Age of a in 2024 = 7
Age of a in 2030 = 7 + 6 = 13
Question 9:
The sum of ages of 2 friends A and B is 42. If the ratio between the sum and difference of their ages is 2:1, find the age of A.
Let the ages of A and B be x and y.
(x + y):(x – y) = 2:1
x + y = 2(x – y)
x + y = 2x – 2y
3y = x
x + y = 42 (given)
3y + y = 42
4y = 42
y = 10.5
x = 3y = 3 * 10.5 = 31.5
Question 10:
If 2 members of a family are in a ratio 3:12, and the average of their ages is 35. Find the total of their ages after 11 years.
The average ages of 2 members = 35
The total of their ages = 35 * 2 = 70 years
The ratio of their ages = 3:12 = 1:4 (simplifying)
Let their ages be x and 4x
x + 4x = 70
5x = 70
x = 14, 4x = 56
The total of their ages after 11 years = (14 + 11) + (56 + 11) = 25 + 67 = 92 years.
Question 11:
When my sister was born, I was 1/10th of my father's age. This year my sister voted for the first time and I celebrated my 22nd birthday. Find my father's age when my sister was born.
My sister's age now: She voted for the first time that means she turned 18 years old.
My present age = 22 (given)
My age 18 years before = 22 – 18 = 4 years
When my sister was born, I was 1/10th of my father's age that means my father was 10 times of my age.
My father's age when my sister was born = my age * 10 = 4 * 10 = 40 years
Question 12:
When my parents married their age was in a ratio 11 : 10 (father's age : mother's age). This year they celebrated 15th anniversary. If the total of their ages at their silver jubilee anniversary will be 113, find my father's age now.
The total of my parent's age at their silver jubilee = 113 years (given)
The total of my parent's age when they got married = Total age after 25 years of marriage – 2 * 25
113 – 50 = 63
Total age 15 years before = 63 years
Individual age 15 years before:
Ratio of their ages then = 11 : 10
Let their ages then be in terms of x.
11x + 10x = 63
21x = 63
x = 63 / 21 = 3
Father's age then = 11x = 11 * 3 = 33
Father's age now = 33 + 15 = 48 years
Question 13:
My age in my sister's wedding was half my age now. My sister's age in her wedding was equal to my age now. If my sister got wedded 18 years back, find the ratio of my sister's age to my age 12 years later.
Let my present age be x.
My age at my sister's wedding = x / 2
My sister's age at her wedding = my current age = x
The wedding was 18 years back that means, x – x / 2 = 18
x – x / 2 = 18
2x – x = 18 * 2
x = 36
My current age = 36 years
My sister's age at her wedding = my current age = x = 36 years
My sister's current age = 36 + 18 = 54 years
My age after 12 years = 36 + 12 = 48 years
My sister's age after 12 years = 54 + 12 = 66 years
Ratio = 66 : 48 = 11 : 8 = 8 : 5.8 = 8 : 5
Question 14:
Ratio of ages of 2 people is 2 : 1. After 10 years the ratio of their ages is 3 : 2. Find their current ages.
Let their ages be in terms of x.
Their current ages = 2x, x
Their ages 10 years later = 2x + 10, x + 10
Ratio of their ages 10 years later = 3 : 2
(2x + 10) / 3 = (x + 10) / 2
2 * (2x + 10) = 3 * (x + 10)
4x + 20 = 3x + 30
4x – 3x = 30 – 20
x = 10
Their current ages = 2x, x = 10 * 2, 10 = 20, 10
Question 15:
T and E are sisters. The ratio of their ages when their brother was born was 6 : 5. The ratio of their ages when their brother was 12 years was 10 : 9. Find the age of T when her brother got married at the age of 28.
Ratio of ages of T and E at the time of their brother's birth = 6 : 5
Let these ages be in terms of x.
Their ages then = 6x, 5x
Ratio of their ages 12 years later = 10 : 9
Their ages after 12 years = 6x + 12, 5x + 12
(6x + 12) : (5x + 12) = 10 : 9
(6x + 12) / 10 = (5x + 12) / 9
9 * (6x + 12) = 10 * (5x + 12)
54x + 108 = 50x + 120
54x – 50x = 120 – 108
4x = 12
x = 12 / 4 = 3 years
Their age when brother was born = 6x, 5x = 18, 15 years
Their ages when their brother got married = Their ages at the time of brother's birth + 28
18 + 28, 15 + 28 = 46, 43 years
Question 16:
The ages of 3 brothers are in a ratio 3:5:11. If the difference between the ages of the youngest and the eldest brother is 24 years, find the total of their ages.
Let their ages be in relation to a constant term x.
The ages of the brother = 3x, 5x and 11x respectively.
The difference between the ages of the youngest and the eldest brother = 11x – 3x = 8x
The difference = 24 years (given)
8x = 24 years
x = 24 / 8 = 3 years
The sum of their ages = 3x + 5x + 11x = 19x
19x = 19 * 3 = 57 years
Question 17:
The ages of 3 employees of a company are in a ratio 1/2 : 2/8 : 3/4. If the total of their ages is 120, find their individual ages in ascending order.
The ratio of the ages of employees = 1/2 : 2/8 : 3/4.
Multiplying each ratio by 8:
1/2 * 8 : 2/8 * 8 : 3/4 * 8
4 : 2 : 6
Using a constant x in terms of the ratio = 4x, 2x and 6x
4x + 2x + 6x = 12x
12x = 120 given
x = 10
4x = 40, 2x = 20 and 6x = 60
Their ages are 40, 20, 60
Their ages in ascending order = 20, 40, 60
Question 18:
The ages of 3 people are x, y and z. It is known that 20% of x is equal to 40% of y and 50% of z. If the total of their ages is 190, find the age of the youngest person.
20% of x = 40% of y = 50% of z
20 * x / 100 = 40 * y / 100 = 50 * z / 100
x / 5 = 2y / 5 = z / 2
Multiplying all sides by 10:
2x = 4y = 5z
The ratio of their ages = 2x : 4y : 5z = 10 : 5 : 4
The sum of their ages = 190 (given)
Assuming a constant r.
10r + 5r + 4r = 190
19r = 190
r = 10
Minimum age of them = 4r = 4 * 10 = 40
Question 19:
There are 3 members in a family. What can be the age of the eldest person if the age of three members is in AP with a common difference of 18 months. It is known that the sum of their ages is 58 years 6 months.
Let the age of the youngest person be x.
The age of the 3 members will be x, x + 1.5 and x + 3 years
Total of their ages = x + (x + 1.5) + (x + 3) years = 3x + 4.5 years
3x + 4.5 years = 58.5 years
3x = 58.5 – 4.5 = 54 years
x = 54 / 3 = 18 years
The age of the eldest person = 18 + 3 = 21 years.
Question 20:
Ages of 3 individuals are in a ratio 3 : 1.5 : 4/3. If the sum of their ages is 70 years, find their individual ages.
The ratio of their ages = 3 : 1.5 : 4/3
Multiplying all sides by 6:
3 * 6 : 1.5 * 6 : 4/3 * 6
18 : 9 : 8
Taking a constant x, expressing all the terms in relation to x.
18x, 9x, 8x
The total of their ages = 70 (given)
18x + 9x + 8x = 70
35x = 70
x = 2
Their respective ages = 18x = 18 * 2, 9x = 9 * 2, 8x = 8 * 2
36, 18, 16 years.
Question 21:
There are 3 members in a family with ages in a ratio 3 : 4 : 5. If the difference between the ages of the eldest and the youngest is 14 years, find the age of the middle child after 25 years.
The ratio = 3 : 4 : 5
Using a constant x and representing their age in terms of x.
3x, 4x, 5x
The difference between the ages of the youngest and the eldest child = 14 years
5x – 3x = 2x = 14 years
2x = 14 years, x = 14 / 2 = 7 years
Present age of the middle child = 4x = 4 * 7 = 28 years
Age after 25 years = 28 + 25 = 53 years.
Question 22:
There are 5 members in a family. If their ages are in a series with equal gaps, find the age of the eldest person. It is known that the sum of their ages is 120 years, and no one is younger than 20 years and older than 30 years.
The only possible AP where minimum value is 20 and maximum value is 28 with 5 terms and a common difference 2 is, 20, 22, 24, 26 and 28. Thus, the age of the eldest person is 28 years.
Question 23:
There are 3 members in a family named a, b and c. A is twice as old as b, b is thrice as old as c. If the difference between the ages of a and c is 40 years, find the age of the youngest member after 11 years.
The youngest among them is c (as per the question)
Age of a, b and c are in a ratio 6 : 3 : 1 as a = 2b and b = 3c
Taking a constant x and expressing the ratio in terms of x, 6x, 3x, x.
The difference between the eldest and the youngest is 40 year.
6x – x = 40 years
5x = 40, x = 8 years
Age of the youngest member c is x = 8 years
Age of c after 11 years = 8 + 11 = 19 years
Question 24:
In an office there are 3 employees in a department named a, b and c. a is 120% of b and b is 50% of c. If they all started working at an age of 20 years, find the experience of a. It is known the total of their ages is 92 years.
Let the age of a, b and c be x, y and z.
B = 50% of c = 0.5c
A = 120% of b = 120 * 0.5c / 100 = 0.8c
X + y + z = 92 (given)
0.5c + 0.8c + c = 92
2.3c = 92
C = 40 years
A = 0.8 c = 32
It is known that all of them started working at an age of 20 years.
Experience of a = 32 – 20 = 12 years.
Question 25:
Out of 3 students a, b and c, a is 18 years old if the ages of the b and c are in a ratio 2 : 3 and the ratio of age of a and b is 6 : 5, find the average age of all the 3 students.
The ratio of ages of a and b is 6 : 5.
The ratio of ages of b and c is 2 : 3.
The age of a is 18 (given).
Age of b = 18 / 6 * 5 = 15 years
Age of c = 15 / 2 * 3 = 22.5 years
The total of their ages = 15 + 18 + 22.5 = 55.5 years
Average of their age = 55.5 / 3 = 18.5 years.
Question 26:
The ratio of ages of 3 people is 2 : 3 : 4 after 5 years the ratio of their ages will be 5 : 7 : 9. Find their current ages.
Let their ages be in terms of x.
Their age now = 2x, 3x and 4x, respectively.
Their ages after 5 years = 2x + 5, 3x + 5 and 4x + 5 years respectively.
The ratio of their ages after 5 years = 5 : 7 : 9
(2x + 5) / 5 = (3x + 5) / 7 = (4x + 5) / 9
Taking any 2 equations:
(2x + 5) / 5 = (3x + 5) / 7
7 * (2x + 5) = 5 * (3x + 5)
14x + 35 = 15x + 25
15x – 14x = 35 – 25
X = 10
Their current ages = 2x, 3x and 4x respectively = 20, 30 and 40 years.
Question 27:
A couple has 3 children. Each was born at a gap of 3 years and 4 months. If the age of the eldest brother is twice as old as the youngest brother, find the age of the middle brother.
Let the age of the youngest brother be x.
Age of the eldest brother = x + (3 years 4 months) * 2 = x + 6 years 8 months
It is given that age of the eldest brother is twice the age of the youngest brother.
2x = x + 6 years 8 months
2x - x = 6 years 8 months
x = 6 years 8 months
Age of the middle brother = age of the youngest brother + 3 years 4 months
= 6 years 8 months + 3 years 4 months = 10 years
Question 28:
13 years back my father's age was triple of my brother's age. Five years from now my father will be twice as old as my brother. Find my age now if I am twice of my brother's age.
Let my brother's age 13 years back be x.
My father's age 13 years back = 3x
My brother's age 5 years later = x + 13 + 5 = x + 18
My father's age 5 years later = 3x + 13 + 5 = 3x + 18
It is given that after 5 years my father will be twice as old as my brother.
3x + 18 = 2(x + 18)
3x + 18 = 2x + 36
3x - 2x = 36 - 18
x = 18
My brother's age now = x + 13 = 18 + 13 = 31 years
My age = twice of my brother's age now = 31 / 2 = 15.5 years
Question 29:
After 7 years my father will be thrice of my age. After 29 years he will be twice of my age, find my age now.
Let my age now be x.
My age after 7 years = 7 + x
My father's age after 7 years = thrice of my age = (7 + x) * 3 = 21 + 3x
My age after 29 years = x + 29
My father's age after 29 years = (x + 29) * 2 = 2x + 58
21 + 3x + (29 - 7) = 2x + 58
21 + 3x + 22 = 2x + 58
3x + 43 = 2x + 58
3x - 2x = 58 - 43
x = 15
Question 30:
5 years from now my age will be equal to my father's age when he got married. The ratio of my father's age to my mother's age at the time of their wedding was 7 : 6. If at the time of wedding my mother was 24 years old, find my age now.
Let my age now be x.
My father's age at the time of their wedding = my age 5 years later = x + 5
Ratio of age of my father to my mother at the time of their wedding = 7 : 6
My mother's age at their wedding = 24 years (given)
My father's age at his wedding = (24 / 6) * 7 = 4 * 7 = 28 years
My age after 5 years = my father's age at his wedding = 28 years
My current age = 28 - 5 = 23 years
Question 31:
Find the total age of 3 brothers, if, their ages are in the ratio 4 : 7 : 9 and the eldest of them is 20 years elder than the youngest brother.
Let the age of the brothers be in terms of x.
The age of the brothers is in a ratio = 4 : 7 : 9
The ages of the brothers are 4x, 7x and 9x.
The difference between the ages of the eldest and the youngest brother is 20 years.
9x - 4x = 20 years
5x = 20
x = 20 / 5 = 4 years
The total age of the brothers:
4x = 4 * 4 = 16 years
7x = 7 * 4 = 28 years
9x = 9 * 4 = 36 years
Total age = 16 + 28 + 36 = 80 years
Question 32:
The average age of 3 students is 21 years. If the ratio of their ages is x : y : z, find the value with which y is related. It is known that x : y = 1 : 2 and y : z = 1 : 3.
x : y = 1 : 2 and y : z = 1 : 3
x : y : z = 1 * 1 : 2 * 1 : 3 * 2
1 : 2 : 6
Let these ratios be in terms of a.
a + 2a + 6a = 21 * 3 = 63 years
9a = 63 years
a = 63 / 9 = 7 years
y = 2a = 7 * 2 = 14 years
Question 33:
3 friends a, b and c went for an internship in a company. They had different levels of experiences. If their age was in a ratio 4 : 5 : 6 and their experience was in a ratio 2 : 3 : 4 respectively, find the friend with the least age to start working. It is known that the ratio of their total age to their total experience is 10 : 1.
Let their age be in terms of x and their experience be in terms of y.
Their respective ages = 4 : 5 : 6 = 4x, 5x and 6x
Their respective experiences = 2 : 3 : 4 = 2y, 3y and 4y
Total age in terms of x = 4x + 5x + 6x = 15x
Total experience in terms of y = 2y + 3y + 4y = 9y
Ratio of age to experience = 10 : 1
15x : 9y = 10 : 1
15x / 10 = 9y / 1
1.5x = 9y
x = 9 / 1.5 y = 6y
x = 6y
Age and experience of a in terms of y = 4x, 2y = 6 * 4y, 2y = 24y, 2y
Age and experience of b in terms of y = 5x, 3y = 6 * 5y, 3y = 30y, 3y
Age and experience of c in terms of y = 6x, 4y = 6 * 6y, 4y = 36y, 4y
Age at which they started working:
a = 24y - 2y = 22y, b = 30y - 3y = 27y and c = 36y - 4y = 32y
Lowest of the above is 22y that is a.
Question 34:
3 friends are 1.5x, 2x and 2.5x years old. If the total of their ages is 45 years, find the age of second friend.
The total age of the three friends = 45 years
Total age of the friends in terms of x = 1.5x + 2x + 2.5x = 6x
6x = 45
x = 45 / 6 = 7.5 years = 7 years 6 months
Age of the second friend = 7 years 6 months * 2 = 15 years
Question 35:
A group of 3 friends went for a training of 6 months. After the training, the sum of their ages was 90 years and were in a ratio 5 : 6 : 7. Find their ages before the training.
Total age after the training = 90 years
Ratio of age after the training = 5 : 6 : 7
Let their ages be in terms of x.
Their ages after the training in terms of x = 5x, 6x and 7x
5x + 6x + 7x = 18x = 90
18x = 90
x = 90 / 18 = 5
x = 5
Their individual ages after the training = 5x = 25, 6x = 30, 7x = 35 years
Their ages before the training = 25 years - 6 months, 30 years - 6 months, 35 years - 6 months
24 years 6 months, 29 years 6 months, 34 years 6 months.
Question 36:
The age of 5 people are in a ratio 2 : 3 : 4 : 5 : 6. If the age of the eldest of them is 42 years, find the average of their ages.
Let the ages of the following people be in terms of a constant x. The age in terms of x will be 2x, 3x, 4x, 5x, and 6x. The age of the eldest person = 6x = 42 years 6x = 42 years X = 42 / 6 =7 years 2x = 2 * 7 = 14 years 3x = 3 * 7 = 21 years 4x = 4 * 7 = 28 years 5x = 5 * 7 = 35 years The respective ages of these 5 people are 14, 21, 28, 35 and 42 years. The sum of these ages = 14 + 21 + 28 + 35 + 42 = 140 Average age of the members = 140 / 5 = 28 years
Question 37:
There are 5 students in a batch with ages x, 1.1x, y, 1.1y and 1.2y. Find the average age of the batch if it is known that x = 1/2y and y – x = 10.
It is given that x = 1/2y and y – x = 10 X = 1/2y, y = 2x Y – x = 10, 2x – x = 10 X = 10 years Y = 2x = 2 * 10 = 20 years 1.1x = 1.1 * 10 = 11 years 1.1y = 1.1 * 20 = 22 years 1.2y = 1.2 * 20 = 24 years Age of the 5 students = 10, 20, 11, 22 and 24 years Sum ages of 5 students = 10 + 20 + 11 + 22 + 24 = 87 Average age of the 5 students = 87 / 5 = 17.4 years
Question 38:
The sum of ages of 10 individuals is 100 years. If their ages are in a ratio 1 : 2 : 3 : 4 : 5 : 6 : 8 : 6 : 10 : 5, find the average of the 5 youngest people.
Let the ages be in terms of a constant x. Their ages in terms of x = x, 2x, 3x, 4x, 5x, 6x, 8x, 6x, 10x and 5x Sum of the ages = 100 years (given) Sum of ages in terms of x = 50x 50x = 100, x = 2 The ages of 10 people = 2, 4, 6, 8, 10, 12, 16, 12, 20 and 10 years The 5 minimum ages = 2, 4, 6, 8 and 10 years Sum of their ages = 30 years Average age of 5 minimum ages = 30 / 5 = 6 years
Question 39:
There are 20 students in a class. If the average age of the students is 20 years including the age of the teacher and reduces by 10% when the age of the teacher is removed, find the age of the teacher.
The average age of the class including the age of the teacher = 20 * 21 (20 students and 1 Teacher makes 21) 20 * 21 = 420 The average without the age of teacher = 20 – 10% of 20 = 20 – 10 * 20 / 100 = 20 – 2 = 18 year Average age of 20 boys = 18 years Total age of 20 boys = 20 * 18 = 360 years Age of the teacher = 420 – 360 = 60 years
Question 40:
There are 7 members in a family. The eldest among them is 42 years old and the youngest among them is 20 years old. If the average age of the rest of the 5 members is 30 years find the average age of the family after 3 years.
The given ages = 42 and 20 years The average age of other 5 people = 30 years The total age of 5 people = 5 * 30 years = 150 years The total age of all the members = total age of the 5 people + age of the given two people 150 + 42 + 20 years = 212 years Average age of the entire family = 212 / 7 = 30.28 ≈ 30.3 years Average age of the family after 3 years = 30.3 + 3 = 33.3 years
Question 41:
There are 7 workers in a factory, out of which 4 share an average age of 40 years and the rest have an average age of 15% less than the average age of the 4 people. Find the total age of the union 5 years later.
The average age of 4 elder employees = 40 Total age of these 4 employees = 40 * 4 = 160 Average age of the rest 3 employees = 15% less than that of the senior employees Average age of the junior employees = 40 – 15% of 40 = 40 – 40 * 15 / 100 40 – 6 = 34 years Average age of 3 junior employees = 34 Total age of 3 junior employees = 34 * 3 = 102 years Total age of all the employees = 102 + 160 = 262 years Total age of the employees after 5 years = 262 + 5 * 7 = 262 + 35 = 297 years
Question 42:
There are 5 offices in a locality. Each office has 20 employees. If altogether the average age of the employees is 35.4 and the minimum age to join a company is 20 years, find the average experience of the employees.
Total age of the employees = average age * total number of employees Total number of employees = employees per office * number of offices = 20 * 5 = 100 Total age of the employees = 100 * 35.4 = 3540 Minimum age for getting a job = 100 Total age of the employees when they got the job = 100 * 20 = 2000 Total experience = 3540 – 2000 = 1540 years Average experience = 1540 / 100 = 15.4 years
Question 43:
Find the total age of a family of 7 people, if the age of eldest and youngest members is 50 and 21, respectively. And the members of the rest of the 5 members are in a ratio 2.5 : 3 : 3.5 : 4 : 4.5, the age of the 2nd youngest being 25 years.
The total age of the family = Age of the eldest + age of the youngest + age of the 5 rest members Age of the 5 members are in a ratio 2.5 : 3 : 3.5 : 4 : 4.5 Let these ages be in relation to a constant x. 2.5x = 25 (Given) X = 10 The age of the rest of the members of the family: 3x = 30, 3.5x = 35, 4x = 40 and 4.5x = 45 The total age of the family = 20 + 25 + 30 + 35 + 40 + 45 + 50 = 245 years
Question 44:
A is twice old as B, B is thrice old as C, C is 1.2 times as old as D. If the total of their ages is 114 years, find the age of A in terms of D.
Let the age of A, B, C and D be p, q, r and s. P = 2Q Q = 3R 2q = 3R * 2 = 6R P = 6R R = 1.2 S P = 2Q = 6 R = 6 * 1.2 S P = 7.2 S A = 7.2D
Question 45:
There are 5 members in a family. The eldest is 35 years older than the youngest. If the ratio of the family in descending order is 11 : 10 : 6 : 5 : 4, Find the average age of the family.
Let the ages of the family be in terms of x. 11 : 10 : 6 : 5 : 4 = 11x, 10x, 6x, 5x and 4x. The difference between the eldest and the youngest is 35 years(given). 11x – 4x = 35 years 7x = 35 years X = 35 / 7 = 5 years The ages of the family: 11x = 11 * 5 = 55 10x = 10 * 5 = 50 6x = 6 * 5 = 30 5x = 5 * 5 = 25 4x = 4 * 5 = 20 Total age of the family = 55 + 50 + 30 + 25 + 20 = 180 years Average age of the family = 180 / 5 = 36 years
Question 46:
Age of 7 people are in a ratio 9 : 12 : 13 : 14 : 5 : 9 : 10. If the total of their ages is 432, find the age of the eldest person.
Let the ages of the people be in terms of x. Their ages = 9x, 12x, 13x, 14x, 5x, 9x and 10x Total of their ages = 9x + 12x + 13x + 14x + 5x + 9x + 10x = 72x 72x = 432 years X = 432/72 = 6 years Age of the eldest person = 6 * 14 = 84 years
Question 47:
The age of 5 people are in a ratio 1 : 3 : 5 : 7 : 9. If the difference between the age of each individual when arranged in ascending order is 12 years, find their average age.
Let their ages be in terms of x.
Their ages in terms of x = x, 3x, 5x, 7x and 9x
Difference between their ages = 12 years
3x – x = 5x – 3x = 7x – 5x = 9x – 7x = 12 years
3x – x = 2x = 12 years
2x = 12 years
X = 12/2 = 6 years
X = 6 years
3x = 6 * 3 = 18 years
5x = 6 * 5 = 30 years
7x = 6 * 7 = 42 years
9x = 6 * 9 = 54 years
Total age of the group = 6 + 18 + 30 + 42 + 54 = 150 years
Average age = 150/5 = 30 years
Question 48:
If the average age of a group of 4 students is 20% more than the average age of a group of 7 students whose total age is 119 years, find the total age of the first group.
The total age of group 2 of students = 119 years
The average age of students of group 2 = total age/number of students
119/7 = 17 years
Average age of students of first group = 17 years + 20% of 17 years
17 years + 17 * 20/100 years = 17 + 3.4 years = 20.4 years
Total age of the students of the first group = 20.4 * 4
20.4 * 4 = 81.6 years
Question 49:
My sister is twice the age of my brother. My brother is 1/7th the age of my father. If the ratio of ages of my father to my mother is 17.5 : 18, find my sister's age. It Is known that my mother is 36 years old.
Age of my mother = 36 years (given)
Age of my father = (36/18) * 17.5 = 2 * 17.5
2 * 17.5 = 35
My father's age is 35 years old.
Brother's age = 1/7th of my father's age
Brother's age = 35/7 = 5 years
Sister's age is twice my brother's age.
Sister's age = 5 * 2 = 10 years
Question 50:
The ages of 6 friends are in a ratio 5 : 6 : 7 : 7 : 8 : 9. If the eldest of them is 8 years more than the mean of their ages, find the age of the 2nd youngest friend.
Let their ages be in terms of x.
Their ages are 5x, 6x, 7x, 7x, 8x and 9x.
The mean of their age in terms of x:
(5x + 6x + 7x + 7x + 8x + 9x)/6 = 42x/6 = 7x
Mean of the ages = 7x
Age of the eldest person = 9x
Difference between the ages of the eldest person and the mean of their ages:
Difference = 9x – 7x = 8 years
2x = 8 years
X = 8/2 = 4 years
X = 4 years
Age of the 2nd youngest = 6x
6x = 6 * 4 = 24 years
Question 51:
The median of age of 5 friends is 12 years. If their ages are in a ratio 6 : 5 : 8 : 3 : 4, find the age of the eldest friend.
Let their ages be in terms of x.
Their ages are 6x, 5x, 8x, 3x and 4x.
To find the median of their ages in terms we must arrange their ages in ascending order.
3x < 4x < 5x < 6x < 8x
Median of the series = 5x = 12 years given
X = 12/5 = 2.4 years
Age of the eldest friend = 8x
8x = 8 * 2.4 = 19.2 years
Question 52:
The total age of a group of 9 students is 171 years. The average age of another group half of the average age of first group. If there are 6 people in group, find the average age of both the groups put together.
The total age of group 1 = 171 years
The average age of first group:
171/9 = 19 years
The average age of the second group = half of the average age of the first group.
19/2 = 9.5 years
Total age of the second group:
9.5 * 6 = 57 years
Total age of the 2 groups = 57 + 171 = 228 years
Average age of both the groups = 228/15 = 15.2 years
Question 53:
There are 7 members in a clan. Out of them 3 shares an average age of 20 years. The average age of the rest 4 members is 18 years. Find the average age of the clan after a championship of 365 days.
Total age of the clan = age of the 1st group + age of the 2nd group
Total age of the first group = 3 * 20 years = 60 years
Total age of the second group = 4 * 18 years = 72 years
Total age of the clan = 60 + 72 = 132 years
Total age of the clan after 365 days = 132 + 7 * 1 = 132 + 7 = 139 years
Average age of the clan after a year = 139/7 = 19.85 years
Question 54:
A group has 4 friends named p, q, r and s. It is known that p = 2q, q = 2r and r = 2s. If the total of their ages is 105, find the total of the age of q.
Let the age of every individual be in terms of a constant x.
Now, assuming s = x, r = 2s = 2x, q = 2r = 4x, p = 2q = 8x.
Total of their ages = x + 2x + 4x + 8x = 15x
Given total of their age = 105 years
15x = 105 years
X = 105/15 = 7 years
Age of q = 4x = 7 * 4 = 28 years
Question 55:
Out of 7 members of a family a, b, c, d, e, f and g, a and b share an average age 20, c and d share an average age 35 and e, f and g share an average age of 50 years. Find the average age of the family nearest to 2nd place of decimal.
Total age of the family = sum of all the groups
Total age of group 1 = 20 * 2 = 40 years
Total age of group 2 = 35 * 2 = 70 years
Total age of the third group = 50 * 3 = 150 years
Total age of the entire family = 40 + 70 + 150 = 260 years
Average age of the entire family = 260/7 = 37.142 years ≈ 37.14 years