Question

A father is 7 times as old as his son. Two yrs ago,the father was 13 times as old as his son. What are their present ages.(explain with step by step)

Answer 1

Given :

A father is 7 times as old as his son. Two yrs ago,the father was 13 times as old as his son.

To find :

Present ages of father and his son

Solution :

★ Let the present age of son be x and his father be 7x

• Two years ago
• Son's age = (x - 2)
• Father's age = (7x - 2)

According to the given condition

$\implies\sf 7x-2=13(x - 2) \\ \\ \\ \implies\sf 7x - 2=13x - 26 \\ \\ \\ \implies\sf 26-2=13x-7x \\ \\ \\ \implies\sf 24=6x \\ \\ \\ \implies\sf x=\cancel\dfrac{24}{6}=4$

★${\boxed{\sf{\red{Present\:age\:of\:son=x=4years}}}}$

★${\boxed{\sf{\red{Present\:age\:of\:father=7x=28years}}}}$

Answer 2

Answer:

Father's age=28years

son's age =4years

Step-by-step explanation:

let son's age be x

let Father's age be 7x

2 years ago their ages =

father's age =7x-2

son's age =x-2

13(x-2)=7x-2

13x-26=7x-2

13x-7x= -2+26

6x=24

x=24/6

x=4

so son's age=4years

Father's age =7×4

=28years

I hope it helped you

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