Question

the age of a man 15 years ago was 5 times his son's age of that time . His age 10 years ago was thrice his son's age of that time. After how many years will their combined age becomes 80 years

Answer 1

Answer:

10 years.

Step-by-step explanation:

Let the son's present age be "X" and father's age

be 'y'.

Age of son before 15 years = x - 15

Age of the father = 5(x - 15) = y - 15

y = 5x -75 + 15

y = 5x - 60 ___ equation (1)

Age of son before 10 years = X -10

Age of father before 10 years = 3(x - 10)

= y - 10

y = 3x - 30+10

y = 3x - 20 ___ equation (2)

Equate both the situations.

5x - 60 = 3x - 20

5x - 3x = -20 + 60

2x = 40

X= 20 years.

Substitute X value in equation (2).

y = 3(20) - 20

y = 60 - 20

y = 40 years.

Let the number of years be "M".

20+M +40+M = 80 because the ages of both

increase .

60 + 2M = 80

2M = 80 - 60

2M = 20 years

M = 20/2

M = 10 years.

Hence , after 10 years they combined ages become 80 years.

Answer 2

Answer:

After ten years, the combined ages becomes $80$ years.

Step-by-step explanation:

Let son's present age be $x$ years and father's age is $y$ years.

According to question, before $15$ years

Son's age $=x-15$

Father's age $5(x-15)=y-15$

Hence, sum is $y=5x-75+15$

$y=5x-60$ ---(1)

Similarly before ten years,

Sum of son's and father's age

$y=3x-20$  ---(2)

Comparing both, $5x-3x=60-20$

$2x=40$

$x=20$

Using it in second equation

$y=3(20)-20$

$y=40$

Assuming number of years as $A$

Hence, $20+A+40+A=80$

$60+2A=80$

$2A=20$

$A=10$

So, their combine ages will be $80$ after $10$ years.

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