Math, asked by aryac2904, 2 months ago

The sum of the ages of a father and his two sons is 52 years. One son is older than the other by 3 years. When the age of the older son will be same as the present age of the father, the sum of the ages of the three will be 136 years. Find each of their present ages.​

Answers

Answered by dreamrob
1

Answer:

The sum of the ages of a father and his two sons is 52 years.

One son is older than the other by 3 years.

When the age of the older son will be the same as the present age of the father, the sum of the ages of the three will be 136 years.

To find:

Find each of their present ages.​

Solution:

Let, father's age = x

Let, younger son's age = y

So, the age of elder son = y + 3

Sum of their ages = 52

x + y + (y + 3) = 52

x + 2y = 52 - 3

x + 2y = 49

x = 49 - 2y

After T years, the age of the older son will be the same as the present age of the father.

So, After T years, the age of the elder son = y + 3 + T

So, y + 3 + T = x = 49 - 2y

y + 3 + T = 49 - 2y

T = 49 - 2y - y - 3

T = 46 - 3y

So, after 46 - 3y years the age of the older son will be the same as the present age of the father.

So,

Age of father after 46 - 3y years = (49 - 2y) + (46 - 3y) = 95 - 5y

Age of younger son after 46 - 3y years = y + (46 - 3y) = 46 - 2y

Age of older son after 46 - 3y years = (y + 3) + (46 - 3y) = 49 - 2y

So, (95 - 5y) + (46 - 2y) + (49 - 2y) = 136

190 - 9y = 136

9y = 190 - 136

9y = 54

y = 6

The present age of younger son = y = 6 years

The present age of older son = y + 3 = 6 + 3 = 9 years

The present age of father = x = 49 - 2y = 49 - 2(6) = 49 - 12 = 37 years

Answered by RvChaudharY50
0

Given :- The sum of the ages of a father and his two sons is 52 years. One son is older than the other by 3 years. When the age of the older son will be same as the present age of the father, the sum of the ages of the three will be 136 years.

To Find :-

  • Find each of their present ages. ?

Answer :-

→ Let present age of father is F years , Elder son is E years and younger son is Y years .

so,

→ F + E + Y = 52 ------ Eqn.(1)

→ E - Y = 3 ------ Eqn.(2)

→ (F + F - E) + (E + F - E) + (Y + F - E) = 136

→ 2F - E + F + F - E + Y = 136

→ 4F - 2E + Y = 136 ------ Eqn.(3)

multiply Eqn.(1) by 4 and subtracting Eqn.(3) from that,

→ 4(F + E + Y) - (4F - 2E + Y) = 52*4 - 136

→ 4F - 4F + 4E + 2E + 4Y - Y = 72

→ 6E + 3Y = 72

→ 2E + Y = 24 ------- Eqn.(4)

putting value of E from Eqn.(2) in Eqn.(4) now,

→ 2(3 + Y) + Y = 24

→ 6 + 3Y = 24

→ 3Y = 18

→ Y = 6 years .

Putting value of Y in Eqn.(2)

→ E - 6 = 3

→ E = 9 years .

putting values of E and Y in Eqn.(1),

→ F + 9 + 6 = 52

→ F = 52 - 15

→ F = 37 years .

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