Question

Five years ago a man was seven times as old as his son. Five years hence, the father will be three times as old as his son. Find their present ages.

Answer 1

Define their age 5 years ago:

Son = x

Father = 7x

Express their current age in term of x:

Son = x + 5

Father = 7x + 5

Express their age 5 years later in term of x:

Son = x + 5 + 5 = x + 10

Father = 7x + 5 + 5 = 7x + 10

Form equation and solve for x:

Father will be 3 times as old as his son

7x + 10 = 3( x + 10)

7x + 10 = 3x + 30

4x = 20

x = 5

Find their present age:

Son = x + 5 = 5 + 5 = 10 years old

Father = 7x + 5 = 7(5) + 5 = 40 years old

Answer: The son is 10 years old and the father is 40 years old

Answer 2

Heya!
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▪Let the present age of man be x years

▪Let the present age of Son be y years


=> Five Years ago ,


=> ( x - 5 ) =7 ( y - 5 )


=> x - 5 = 7y - 35


=> x - 7y = -35 + 5


=> x - 7y = -30.........................(1)




=> Five Years later ,


=> ( x + 5 ) = 3( y + 5 )


=> x + 5 = 3y + 15


=> x - 3y = 15 - 5


=> x - 3y = 10.........................(2)



♠Solving (1) and (2) by Elimination Method,



=> ( x - x ) + ( -7y -3y ) = (-30 - 10)


=> 4y = 40


=>y = 10✔
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▪Plug value of y in Eq. (1)


=> x - 7 ( 10 ) = -30

=> x - 70 = -30

=> x = 40✔
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♠Present Age of Man is 40 Years !

♠Present Age of Son is 10 Years !


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