By analyzing the equations from the given data and assuming the unknown values, theÂ age problems are solved. From that data, the unknown terms are also solved.

If the current age of a person be X, then

- age after n years = X + n
- age n years ago = X â€“ 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 Johnâ€™s present age, if after 10 years his age will be 5 times his age 5 years back.

**Solution**:

- Let Johnâ€™s present age be x

Johnâ€™s age before 5 years = (x – 5)

Johnâ€™s age after 10 years = (x + 10)

We are given that, Johnâ€™s age after 10 years (x + 10) is 5 times his age 5 years back (x â€“ 5)

Therefore,

(x + 10) = 5 (x â€“ 5)

Solving the equation, we get

x + 10 = 5x â€“ 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 Rahulâ€™s age 5 years ago = (y + 15 â€“ 5)

Rohanâ€™s age before 5 years = (y â€“ 5)

5 years ago,

(y + 15 â€“ 5) = 3 (y â€“ 5)

(y + 10) = (3y â€“ 15)

2y = 25

y = 12.5

Rohanâ€™s age = 12.5 years

Rahulâ€™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 ageâ€™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

Harryâ€™s age = 5x and Peterâ€™s age = 6x

One year ago, their age was 5x and 6x. Hence at present, Harryâ€™s age = 5x +1 and Peterâ€™s age = 6x +1

After 4 years,

Harryâ€™s age = (5x +1) + 4 = (5x + 5)

Peterâ€™s age = (6x +1) + 4 = (6x + 5)

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

\(\frac{Harryâ€™s Age}{6}\) = \(\frac{Peterâ€™s Age}{7}\)

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

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

X = 5

Peterâ€™s present age = (6x + 1) = (6 x 5 + 1) = 31 years

Harryâ€™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, motherâ€™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 motherâ€™s age is 3 times the age of her son, let it be 3x, three years ago.

At present: Motherâ€™s age will be (3x + 10) and sonâ€™s age will be (x + 10)

After 10 years: Motherâ€™s age will be (3x + 10) +10 and sonâ€™s age will be (x + 10) + 10

Motherâ€™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: Sharadâ€™s age = (60 â€“ x) and Santoshâ€™s age = (80 â€“ x)

Ratio of their ages x years ago was 4 : 6

\(\frac{(60 â€“ x)}{(80 â€“ x)}\) = \(\frac{4}{6}\)

6(60 â€“ x) = 4(80 â€“ x)

360 â€“ 6x = 320 â€“ 4x

x = 20

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

**Example 2**:

The ratio of Rohanâ€™s age 4 years ago and Rahulâ€™s age after 4 years is 1 : 1. If at present, the ratio of their ages is 5 : 3, then find the ratio between Rohanâ€™s age 4 years hence and Rahulâ€™s age 4 years ago.

**Solution**:

- 1) At present: Ratio of their ages = 5 : 3. Therefore, 5 : 3 will be 5x and 3x.

Rohanâ€™s age 4 years ago = 5x â€“ 4

Rahulâ€™s age after 4 years = 3x + 4

2) Ratio of Rohanâ€™s age 4 years ago and Rahulâ€™s age after 4 years is 1 : 1

Therefore, \(\frac{(5x â€“ 4)}{(3x + 4)}\) = \(\frac{1}{1}\)

Solving, we get x = 4

3) We are asked to find the ratio between Rohanâ€™s age 4 years hence and Rahulâ€™s age 4 years ago.

Rohanâ€™s age : (5x + 4)

Rahulâ€™s age: (3x â€“ 4)

Ratio of Rahulâ€™s age and Rohanâ€™s age

\(\frac{(5x + 4)}{(3x â€“ 4)}\) = \(\frac{24}{8}\) = \(\frac{3}{1}\) = 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 daughterâ€™s age, then further after 5 years, how many times he would be of his daughterâ€™s age?

**Solution**:

- Let daughterâ€™s age be x and fatherâ€™s age be 3x.

Fatherâ€™s age is 3 times more aged than his daughter, therefore fatherâ€™s present age = x + 3x = 4x

After 5 years, fatherâ€™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, fatherâ€™s age = (4x +10) and daughterâ€™s age = (x + 10)

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

Substitute the value of x, we get

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

After further 5 years, father will be 2.5 times of daughterâ€™s age.

**Example 2**:

5 years ago, sisterâ€™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 sisterâ€™s age be 34 â€“ x.

Past Age (5 Yrs Ago) | Present Age | Future Age (After 6 Yrs) | |
---|---|---|---|

Brother | (x â€“ 5) | x | (x + 6) = ? |

Sister | (34 â€“ x) – 5 | (30 â€“ x) |

We are given, 5 years ago sisterâ€™s age was 5 times the age of her brother.

Therefore,

(34 â€“ x) â€“ 5 = 5 (x â€“ 5)

34 â€“ x â€“ 5 = 5x â€“ 25

5x + x = 34 â€“ 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)

â‡’\(x\) + 25 = 5\(x\) – 15

â‡’25 + 15 = 5\(x\) – \(x\)

â‡’40 = 4\(x\)

â‡’\(x\) = \(\frac{40}{4}\)

â‡’\(x\) = 10 years

Therefore, present age of Harry is 10 years

**2. The product of the ages of CharlieÂ and JamesÂ 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

â‡’\(\frac{120 – (x)^2}{x}\) = 8

â‡’240 – \((x)^2\) = 8\(x\)

â‡’\((x)^2\) + 8\(x\) – 240 = 0

Now factorize the equation,

â‡’\(x^2\) + 20\(x\) – 12\(x\) – 240

â‡’\(x(x + 20)\) – 12\((x + 20)\)

â‡’\((x + 20)\)\((x – 12)\)

â‡’\(x = -20\) and \(x = 12\)

Hence \(x\) = 12 years

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

**3. HenryÂ is 30 years and GeorgeÂ 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

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

â‡’\(5(30 – x)\) = \(3(25 – x)\)

â‡’\(150 – 5x\) = \(75 – 3x\)

â‡’\(150 – 75\) = \(5x – 3x\)

â‡’\(75\) = \(2x\)

â‡’\(x\) = \(\frac{75}{2}\)

â‡’\(x\) = \(37.5\)

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

â‡”\(y\) = (\(\frac{5}{2}\))\(x\)

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

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

â‡’\(x + 8\) = \(\frac{1}{2}\)(\((\frac{5}{2})x + 8\))

â‡’\(2x + 16\) = \(\frac{5x + 16}{2}\)

â‡’\(4x + 32\) = \(5x + 16\)

â‡’\(5x – 4x\) = \(32 – 16\)

â‡’\(x\) = \(16\)

Son’s age is 16 years

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

**5. One year ago AliceÂ 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 \((x -1)\) = \(4(y -1)\)

â‡’\((x -1)\) = \((4y -4)\)

â‡’\((x -4y)\) = \(-4 + 1\)

â‡’\((x -4y)\) = \(-3\) ——-(i)

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

â‡’\(x + 6\) = \(y + 15\)

â‡’\(x – y\) = \(15 – 6\)

â‡’\(x – y\) = \(9\) ——–(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