Biology, asked by aaryashet09, 3 months ago

Usage of water use in litres​

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Answered by itzpikachu76
1

Answer:

GiveN:-

A father is 7 times as old as his son, two years ago the Father was 13 times as old as his son.

To FinD:-

The present age of son.

SolutioN:-

Let the present age of the son be x years.

Let the present age of the father be 7x years.

2 years ago their ages were:-

Son = (x - 2) years.

Father = (7x - 2) years.

It is said that two years ago the Father was 13 times as old as his son.

According to the question,

\large\implies{\sf{7x-2=13(x-2)}}⟹7x−2=13(x−2)

\large\implies{\sf{7x-2=13x-26}}⟹7x−2=13x−26

\large\implies{\sf{-2+26=13x-7x}}⟹−2+26=13x−7x

\large\implies{\sf{24=6x}}⟹24=6x

\large\implies{\sf{\dfrac{24}{6}=x}}⟹

6

24

=x

\large\implies{\sf{\dfrac{\cancel{24}}{\cancel{6}}=x}}⟹

6

24

=x

\large\implies{\sf{4=x}}⟹4=x

\large\therefore\boxed{\bf{x=4.}}∴

x=4.

Their present ages:-

Son's age = x = 4 years.

Father's age = 7x = 7 × 4 = 28 years.

The present age of the son is 4 years.

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