Question

five year ago a man was 7 time as old as his son . Five year hence the father will be three times as old as his son . find their present ages. please solve it in easy method for class 7​

Answer 1

Step-by-step explanation:

Let present age age of the son be x years

Therefore, five year ago son's age =(x - 5) years.

Five year hence son's age = (x + 5) years.

According to the problem,

Five year ago father was 7 times as old as his son

Therefore,

Five year ago father's age = 7*(x - 5) years.

So, Present age of father = (7*(x - 5) + 5) years

= (7x - 35 +5 ) years

= (7x - 30) years

Therefore,

Five years hence father's age = (7x - 30 + 5) years

= (7x - 25) years.

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Now,

Given that,

Five year hence,the father will be three times as old as his son,

Therefore,

7x - 25 = 3*(x + 5)

→7x - 25 = 3x + 15

→7x - 3x = 15 + 25

→4x = 40

→x = 40/4

→x = 10

So, Son's present age is 10 years.

_____________________

Five year hence father's age = 7x - 25

Putting the value of x, We get,

Five year hence father's age = 7*10 - 25

= 70 - 25

= 45 years

Therefore,

Present age of father = (45 - 5) years = 40 years.

____________________

So, required Father's & Son's present age 40 years and 10 years respectively.

______________________________

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Answer 2

Answer:

hey mate:-

Step-by-step explanation:

Dear friend, Let the past age of son = x,

Let the past age of father = 7x ,

So, Present ages are x+5 and 7x+5 ,

Future age of son = x+10 ,

Future age of father = 7x+10,

Given 7x+10 = 3*(x+10) 7x+10 ,

= 3x+30 7x-3x ,

= 30-10 4x ,

= 20 x ,

= 20/4 ,

x = 5 ,

7x = 35 ,

x+5 = 10,

7x+5 = 40

age of son= 10 years,

age of father=40 years,

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