Question

Ages of two people different by 16 years if six year ago the elder one was three times as old as the younger one find their present ages.

With proper explanation or will be reported ​

Answer 1

Given:

1. The age of 2 people differ by 16 years.
2. 6 years ago, the elder one was thrice as old as the younger one.

Find:

Their present ages.

Solution:

Let the present age of the Elder person be x years and the present age of the Younger person be y years.

Then, the difference of their ages is:

x-y = 16.

x - y - 16 = 0. → Equation 1.

6 years ago, their ages are x-6 and y-6 respectively.

Also, it is given that:

x-6 = 3(y-6).

x-6 = 3y - 18.

x - 3y + 12 = 0. → Equation 2.

Equation 1 - Equation 2:

2y - 28 =0.

2y = 28.

y = 14.

Substituting the value of y in Equation 1:

x-(14) = 16.

x = 16+14.

x = 30.

∴ The present ages of the Elder person and Younger person are 30 and 14 years respectively.

Answer 2

Let present age of one people be "M" years and present age of another people be "N" years.

Ages of two people differ by 16.

Let M be the elder person and N be the younger person.

According to question,

→ Elder person - younger person = 16

→ M - N = 16

→ M = 16 + N ____ (eq 1)

Six years ago, the elder one was three times as old as the younger one

According to question,

• Age of elder person = (M - 6) years
• Age of younger person = (N - 6) years

→ (M - 6) = 3(N - 6)

→ M - 6 = 3N - 18

→ M - 3N = - 12

→ 16 + N - 3N = - 12 [From (eq 1)]

→ - 2N = - 12 - 16

→ - 2N = - 28

→ N = 14

Put value of N in (eq 1)

→ M = 16 + 14

→ M = 30

∴ Present age of elder person is 30 years and present age of younger person is 14 years.