Quantitative Aptitude - SPLessons

Age Problems

Chapter 9

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Age Problems

shape Description

Problems on Ages are most frequently appearing questions in various competitive exams that include Quantitative Aptitude section. By analyzing the equations from the given data and assuming the unknown values, the age problems are solved.


Algebra is a very powerful branch of Mathematics which can be used to solve the Problems on Ages. Algebra helps in transforming word problems into mathematical expressions in the form of equations using variables to denote unknown quantities or parameters and thus, providing numerous techniques to solve these mathematical equations and hence, determining the answer to the problem. Identifying key information, organizing information, using mathematical expressions to assume unknown values and thus solving mathematical expressions for the unknown values will help us identify solutions.


shape Methods

Calculate Present age:

If the current age of a person be X, then


  • age after n years = X + n

  • age n years ago = X – n

  • n times the age = nX

  • If ages in the numerical are mentioned in ratio A : B, then A : B will be AX and BX


Example 1:
What is John’s present age, if after 10 years his age will be 5 times his age 5 years back.

Solution:

    Let John’s present age be x

    John’s age before 5 years = (x – 5)

    John’s age after 10 years = (x + 10)

    We are given that, John’s age after 10 years (x + 10) is 5 times his age 5 years back (x – 5)

    Therefore,

    (x + 10) = 5 (x – 5)

    Solving the equation, we get

    x + 10 = 5x – 25

    4x = 35

    x = 8.75 years


Example 2:
Rahul is 15 years elder than Rohan. If 5 years ago, Rahul was 3 times as old as Rohan, then find Rahul’s present age.

Solution:

    Let age of Rohan be y

    Rahul is 15 years elder than Rohan = (y + 15). So Rahul’s age 5 years ago = (y + 15 – 5)

    Rohan’s age before 5 years = (y – 5)

    5 years ago, Rahul is 3 times as old as Rohan

    (y + 15 – 5) = 3 (y – 5)

    (y + 10) = (3y – 15)

    2y = 25

    y = 12.5

    Rohan’s age = 12.5 years

    Rahul’s age = (y + 15) = (12.5 + 15) = 27.5 years.


Numerical to Determine Ages in ratio form:

If sum of ages of x and y is A and ratio of their ages is p : q respectively, then u can determine age of y by using the formula shown below:


    Age of y = \(\frac{Ratio \ of \ y}{Sum \ of \ ratios} \times sum \ of \ ages\)

    Age of y = \(\frac{q}{p + q} \times A\)


Examples 1:
One year ago, ratio of Harry and Peter age’s was 5 : 6 respectively. After 4 years, this ratio becomes 6 : 7. How old is Peter?

Solution:

    We are given that age ratio of Harry : Pitter = 5 : 6

    Harry’s age = 5x and Peter’s age = 6x

    One year ago, their age was 5x and 6x. Hence at present, Harry’s age = 5x +1 and Peter’s age = 6x +1

    After 4 years,

    Harry’s age = (5x +1) + 4 = (5x + 5)

    Peter’s age = (6x +1) + 4 = (6x + 5)

    After 4 years, this ratio becomes 6 : 7. Therefore,

    \(\frac{Harry’s Age}{6}\) = \(\frac{Peter’s Age}{7}\)

    \(\frac{(5X + 5)}{(6X + 5)}\) = \(\frac{6}{7}\)

    7 (5x + 5) = 6 (6x + 5)

    X = 5

    Peter’s present age = (6x + 1) = (6 x 5 + 1) = 31 years

    Harry’s present age = (5x + 1) = (5 x 5 + 1) = 26 years


Examples 2:

Age of mother 10 years ago was 3 times the age of her son. After 10 years, mother’s age will be twice that of his son. Find the ratio of their present ages.

Solution:

    We are given that, age of mother 10 years ago was 3 times the age of her son

    So, let age of son be x and as mother’s age is 3 times the age of her son, let it be 3x, three years ago.

    At present: Mother’s age will be (3x + 10) and son’s age will be (x + 10)

    After 10 years: Mother’s age will be (3x + 10) +10 and son’s age will be (x + 10) + 10

    Mother’s age is twice that of son

    (3x + 10) +10 = 2 [(x + 10) + 10]

    (3x + 20) = 2[x + 20]

    Solving the equation, we get x = 20

    We are asked to find the present ratio.

    (3x + 10) : (x + 10) = 70 : 30 = 7 : 3


Numerical to Determine Age of a Person before x Years:

Example 1:
Sharad is 60 years old and Santosh is 80 years old. How many years ago was the ratio of their ages 4 : 6?

Solution:

    Here, we have to calculate: How many years ago the ratio of their ages was 4 : 6

    Let us assume x years ago

    At present: Sharad is 60 years and Santosh is 80 years

    x years ago: Sharad’s age = (60 – x) and Santosh’s age = (80 – x)

    Ratio of their ages x years ago was 4 : 6

    \(\frac{(60 – x)}{(80 – x)}\) = \(\frac{4}{6}\)

    6(60 – x) = 4(80 – x)

    360 – 6x = 320 – 4x

    x = 20

    Therefore, 20 years ago, the ratio of their ages was 4 : 6


Example 2:
The ratio of Rohan’s age 4 years ago and Rahul’s age after 4 years is 1 : 1. If at present, the ratio of their ages is 5 : 3, then find the ratio between Rohan’s age 4 years hence and Rahul’s age 4 years ago.

Solution:

    1) At present: Ratio of their ages = 5 : 3. Therefore, 5 : 3 will be 5x and 3x.
    Rohan’s age 4 years ago = 5x – 4
    Rahul’s age after 4 years = 3x + 4


    2) Ratio of Rohan’s age 4 years ago and Rahul’s age after 4 years is 1 : 1
    Therefore, \(\frac{(5x – 4)}{(3x + 4)}\) = \(\frac{1}{1}\)
    Solving, we get x = 4


    3) We are asked to find the ratio between Rohan’s age 4 years hence and Rahul’s age 4 years ago.
    Rohan’s age : (5x + 4)
    Rahul’s age: (3x – 4)
    Ratio of Rahul’s age and Rohan’s age
    \(\frac{(5x + 4)}{(3x – 4)}\) = \(\frac{24}{8}\) = \(\frac{3}{1}\) = 3 : 1


Numericals to Determine Age of a Person after x Years:

Example 1:
Father is 3 times more aged than his daughter. If after 5 years, he would be 3 times of daughter’s age, then further after 5 years, how many times he would be of his daughter’s age?

Solution:

    Let daughter’s age be x and father’s age be 3x.

    Father’s age is 3 times more aged than his daughter, therefore father’s present age = x + 3x = 4x

    After 5 years, father’s age is 3 times more than his daughter age.

    (4x + 5) = 3 (x + 5)

    (4x+5)=3 (x+5)

    (4x + 5) = 3 (x + 5)

    x = 10

    After 5 years it was (4x + 5), then after further 5 years, father’s age = (4x +10) and daughter’s age = (x + 10)

    \(\frac{(4x + 10)}{(x + 10)}\) = ?

    Substitute the value of x, we get

    \(\frac{[(4 × 10) + 10]}{[10 + 10]}\) = \(\frac{50}{20}\) = 2.5

    After further 5 years, father will be 2.5 times of daughter’s age.


Example 2:
5 years ago, sister’s age was 5 times the age of her brother and the sum of present ages of sister and brother is 34 years. What will be the age of her brother after 6 years?

Solution:

    Let present age of brother be x and sister’s age be 34 – x.

    Past Age (5 Yrs Ago) Present Age Future Age (After 6 Yrs)
    Brother (x – 5) x (x + 6) = ?
    Sister (34 – x) – 5 (30 – x)


    We are given, 5 years ago sister’s age was 5 times the age of her brother.

    Therefore,

    (34 – x) – 5 = 5 (x – 5)

    34 – x – 5 = 5x – 25

    5x + x = 34 – 5 +25

    6x = 54

    x = 9

    Future age (after 6 yrs) = (x + 6) = (9 + 6) = 15 years

shape Formulae

  • If the present age is \(x\), then n times the age is n\(x\)

  • If the present age is \(x\), then age of n years later = \(x\) + n

  • If the present age is \(x\), then age of n years ago = \(x\) – n

  • If the ages are in the ratios of a : b, then a\(x\) and b\(x\)

  • If the current age is \(x\), then \(\frac{1}{n}\) of the age is \(\frac{x}{n}\)

shape Samples

1. Harry’s age after 25 years will be 5 times his age 3 years back. What is the present age of Harry?

Solution:

    Let the Harry‘s present age be \(x\)

    After 25 years is \(x\) + 25

    3 years back is \(x\) – 3

    Therefore the equation is \(x\) + 25 = 5(\(x\) – 3)

    ⇒\(x\) + 25 = 5\(x\) – 15

    ⇒25 + 15 = 5\(x\) – \(x\)

    ⇒40 = 4\(x\)

    ⇒\(x\) = \(\frac{40}{4}\)

    ⇒\(x\) = 10 years

    Therefore, present age of Harry is 10 years


2. The product of the ages of Charlie and James is 120. If twice the age of James is more than Charlie’s age by 8years, What is James’s age?

Solution:

    Let Charlie’s age be \(x\)

    Then James’s age = \(\frac{120}{x} \quad\)(since \(x\) +\(y\) = 120, in order to eliminate \(y\) variable replace \(y\) = \(\frac{120}{x}\))

    Given that 2(\(\frac{120}{x}\)) – \(x\) = 8

    ⇒\(\frac{120 – (x)^2}{x}\) = 8

    ⇒240 – \((x)^2\) = 8\(x\)

    ⇒\((x)^2\) + 8\(x\) – 240 = 0

    Now factorize the equation,

    ⇒\(x^2\) + 20\(x\) – 12\(x\) – 240

    ⇒\(x(x + 20)\) – 12\((x + 20)\)

    ⇒\((x + 20)\)\((x – 12)\)

    ⇒\(x = -20\) and \(x = 12\)

    Hence \(x\) = 12 years

    Therefore, James’s age is \(\frac{120}{12}\) = 10 years.


3. Henry is 30 years and George is 25 years old.How many years ago was the ratio of their ages 3: 5?

Solution:

    Given data

    Henry’s age is 30 years

    George’s age is 25 years

    \(\frac{(30 – x)}{(25 -x)}\) = \(\frac{3}{5}\)

    ⇒\(5(30 – x)\) = \(3(25 – x)\)

    ⇒\(150 – 5x\) = \(75 – 3x\)

    ⇒\(150 – 75\) = \(5x – 3x\)

    ⇒\(75\) = \(2x\)

    ⇒\(x\) = \(\frac{75}{2}\)

    ⇒\(x\) = \(37.5\)

    Therefore, 37.5 years ago was their ages 3 : 5.


4. A son’s present age is two fifth of the age of his father. After 8 years, he will be one -half of the age of his father. How old is the father at present?

Solution:

    Let the present age of son be \(x\)

    Present age of father be \(y\)

    Given that \(x\) = (\(\frac{2}{5}\))\(y\)

    ⇔\(y\) = (\(\frac{5}{2}\))\(x\)

    8 years later i.e. \(x + 8\) = \(\frac{1}{2}\) x \((y + 8)\)

    Substitute the value of \(y\), and simplify

    ⇒\(x + 8\) = \(\frac{1}{2}\)(\((\frac{5}{2})x + 8\))

    ⇒\(2x + 16\) = \(\frac{5x + 16}{2}\)

    ⇒\(4x + 32\) = \(5x + 16\)

    ⇒\(5x – 4x\) = \(32 – 16\)

    ⇒\(x\) = \(16\)

    Son’s age is 16 years

    Therefore, father’s age is \(\frac{(5 \ * \ 16)}{2}\) = 40 years


5. One year ago Alice was four times as old as her daughter Amelie. Six years hence, Alice age exceeds her daughter’s age 9 years. The ratios of the present ages of Alice and her daughter is?

Solution:

    Let Alice’s age be \(x\)

    Amelie’s age be \(y\)

    Given that \((x -1)\) = \(4(y -1)\)

    ⇒\((x -1)\) = \((4y -4)\)

    ⇒\((x -4y)\) = \(-4 + 1\)

    ⇒\((x -4y)\) = \(-3\) ——-(i)

    \(x + 6\) = \((y + 6) + 9\)

    ⇒\(x + 6\) = \(y + 15\)

    ⇒\(x – y\) = \(15 – 6\)

    ⇒\(x – y\) = \(9\) ——–(ii)

    By solving equations (i) and (ii)

    \(y\) = 4

    \(x\) = 13

    Therefore, the ratio of ages of Alice and Amelie is x : y = 13 years : 4 years