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Solving Equations with Linear Expressions on One Side and Numbers on the Other Side

Five years ag...

Question

Five years ago, mother was seven times as old as her daughter. Five years hence (from present), mothers age will be three times the age of her daughter. Find the present age of mother.

30 y e a r s

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35 y e a r s

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40 y e a r s

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45 y e a r s

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Solution

The correct option is C $40 y e a r s$
Let present age of daughter be x.
So $5$ years ago her age $= x - 5$
Her mother's age is $7 (x - 5)$
Including $5$ years from past to present and then to future $= 10$
So $7 x - 35 + 10 = 3 (x - 5 + 10)$
$7 x - 25 = 3 x + 15$
$4 x = 40$
$x = 10$
so daughters age $= 10$
mother's age $= 4 \times 10 = 40$ .

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Five years ago, mother was seven times as old as her daughter. Five years hence (from present), mothers age will be three times the age of her daughter. Find the present age of mother.

The correct option is C 40 yearsLet present age of daughter be x.So 5 years ago her age =x−5Her mother's age is 7(x−5)Including 5 years from past to present and then to future =10So 7x−35+10=3(x−5+10)7x−25=3x+154x=40x=10so daughters age=10mother's age=4×10=40.