**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.

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,

(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

- 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 \(\frac{x}{n}\)

- 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