Question

4 years ago Seema's age was 3 times the age of
her son. The sum of their ages 4 years ago was 48 years. Find
their present ages.​

Answer 1

Let age of son = x

seemas age = 3x

4 years ago,

son = x-4

seema = 3x - 4

given

x-4 + 3x-4 = 48

4x - 8 = 48

4x = 56

x = 14

age of son = 14

age of seema = 42

Answer 2

❍ Let's say, that the age of Seema be x years and the age of her son be y years respectively.

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$\underline{\sf Case\;\it1\;:}$

• 4 years ago Seema's age was three times the age of her son. Therefore,

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$:\implies\sf{x-4=3(y-4)}\\\\\\:\implies\sf{x-4=3y-12}\\\\\\:\implies\sf{x-3y=-12+4}\\\\\\:\implies\sf{x-3y=-8}\qquad\qquad\bigg\lgroup\sf{eq^n\;1}\bigg\rgroup$

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$\underline{\sf Case\;\it2\;:}$

• The sum of their ages 4 years ago was 48 years. Therefore,

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$:\implies\sf{(x-4)+(y-4)=48}\\\\\\:\implies\sf{x+y-8=48}\\\\\\:\implies\sf{x+y=48+8}\\\\\\:\implies\sf{x+y=56}\qquad\qquad\bigg\lgroup\sf{eq^n\;2}\bigg\rgroup$

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$\;\;\;\;\;\;\;\;\small\underline{\bf{\dag}\frak{\;Subtracting\;eq^n\;1\;from\;eq^n\;2\;we\;get:}}$

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$:\implies\sf{-4y=-64}\\\\\\:\implies\sf{y=\cancel{\dfrac{-64}{-4}}}\\\\\\:\implies\underline{\boxed{\pink{\frak{y=16}}}}{\;\bigstar}$

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$\;\;\;\;\;\;\;\;\small\underline{\bf{\dag}\frak{\;Putting\;value\;of\;y\;in\;eq^n\;1\;:}}$

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$:\implies\sf{x-3y=-8}\\\\\\:\implies\sf{x-3(16)=-8}\\\\\\:\implies\sf{x-48=-8}\\\\\\:\implies\sf{x=-8+48}\\\\\\:\implies\underline{\boxed{\purple{\frak{x=40}}}}{\;\bigstar}$

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Hence,

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• Seema's present age = x = 40 years
• Her son's present age = y = 16 years

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$\therefore\;{\underline{\sf{Hence,\;their\;present\;ages\;are\;{\textsf{\textbf{40\;years}}}\;and\;{\textsf{\textbf{16\;years}}}}.}}$⠀⠀⠀⠀