**Quantitative Aptitude Ages Study Material**

**Quantitative Aptitude Ages ( Study Material)**

In “Problems based on age’s usually relationships between the ages of two or three persons at two different time are given and we are required to find the ages of these persons. Such problems can be solved by the knowledge of “Linear Equations”. However, in certain cases we can find short-cut methods of deriving direct formulae. This will save lot of precious time. Time-scale is divided into three parts-past, present and future. Problems usually state the relationship between ages of persons for two out of three parts of time-scale. These relationships are either in the form of ratio of their ages or alternatively as number of times one’s age is of the other. Sometimes, instead of stating the ratio of ages of two persons for two different parts of time-scale, only one such relationship is given. The other condition is stated in one of these forms–sum of their ages, difference of their ages or product of their ages. The main point that we have to see (e.g., in ca$e of two persons) that there are two unknowns (ages of two persons) to be found and hence we must have two unique relationships to solve these problems. Although types of problems on ages can be too many, we try to cover here all those types of problems which are commonly asked in competitive exams.

Ex. 1. The age of father 3 years ago was 8 times the age of his son. Presently the father’s age is 5 times that of the son. Find their present ages.

‘Sol. Let the present age of son be x years.

Then, the present age of father = 5x years.

3 years ago.

Son’s age = (x – 3) years

Father’s age = (5,x – 3) years

But 3 years ago the father’s age was 8 times that of his son. Therefore, we can write

8(x-3) = 5x-3 or, 8x – 24 = 5x – 3 or, 3x = 21 or, x’— 7 years and 5x = 35 years

We get the present age of son = 7 years and the present age of father = 35 years Ans.

Note: We could also state the above problem as given below:

3 years ago the ratio of ages of father to son was 8:1. At present the ratio of their ages is 5:1. Find their present ages.

In other words 8 times can be expressed as 8:1 or 8/1

Ex. 2. The age of A t1 years ago was n1 times the age of B. At present A’s age is n2 times that of B. Find their present ages.

Ex. 3. The present age of A is 9 times the present age of B. 3 years hence A’s age will be 5 times that of B. Find their present ages.

Sol. Let the present age of B be x years.

Then, the present age of A = 9x years.

3 years hence,

B’s age = (x + 3) years.

A’s age = (9x + 3) years.

But it is given that 3 years hence A’s age will be 5 times the age of B.

9x + 3 = 5(x+3) or,

9x +3 = 5x + 15 or, 4x =12

or, x = 3

B’s present age =3 years

A’s present age = 9 x 3 = 27 years Ans.

Ex. 4. The present age of A is n_{2} times the present age of B. t_{3} years hence A’s age will be n_{3} times that of B. Find their present ages.

Sol. Let B’s present age be x years.

Then, A’s present age= n_{2} x years.

t_{3} years hence,

B’s age = x + t_{3 }A’s age = (n_{2} x + t_{3})

According to the question, n_{2} x + t_{3} = n_{3} (x + t_{3})

Ex. 5. 4 years ago A’s age was 9 times the age of fl. 6 years hence A’s age will be 4 times fi’s age. Find their present ages. ”

Sol. Let the present age of fi be x years and the present age of A be y years.

4 years ago, y – 4 = 9 (x – 4)

or, 9x – y = 32 -(i)

6 years hence, y + 6 = 4 (x + 6)

or, 4x-y = -18 …(ii)

Here we have two unknowns (x and y) and a set of two simultaneous linear equations with x and y as variables. Since the number of equations is equal to the number of unknowns, we can find the Values of these unknowns by solving these two equations. In order to solve these equations, we multiply these equations by such factors that the coefficient of one of the variables become equal in both the equations.

Suppose we want to find the valuaof y. We need to make coefficients of x equal in both equations. We multiply equation (i) by 4 and equation (ii) by 9 and then subtract.

4 x Equation (i): 36x-4y= 128

9 x Equation (ii): 36x-9y=-162

5 y = 290

or,y = 58 years.

Now we can find the value of x by putting the value of y either in Equation (i) or in Equation (ii).

Equation (i) : 9x – y = 32

or, 9x=32 + 58

or, x =90/9 = 10 years.

Equation (ii): 4x-y =-18

or, 4x =-18 + 58

or, x =40/4 =10 years.

**Quantitative Aptitude Ages ( Study Material)**

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