A father is 7 times as old as his son. Three years ago, the father was 13 times as old as his son. What are their presen

Question

A father is 7 times as old as his son. Three
years
ago, the father was 13 times as old as his son.
What are their
present
ages?​

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Adalynn 1 month 2021-08-13T13:57:10+00:00 2 Answers 0 views 0

Answers ( )

    0
    2021-08-13T13:58:10+00:00

    Solution :

    Let the present age of son’s be r years & the present age of father’s be 7r years respectively;

    A/q

    \underbrace{\bf{3\:years\:ago\::}}}

    The son age was (r-3) years.

    The father age was (7r-3) years.

    \longrightarrow\sf{(7r - 3) = 13(r-3)}\\\\\longrightarrow\sf{7r-3 = 13r - 39}\\\\\longrightarrow\sf{7r-13r = -39 + 3}\\\\\longrightarrow\sf{-6r = -36}\\\\\longrightarrow\sf{r = \cancel{-36/-6}}\\\\\longrightarrow\bf{r = 6\:years}

    Thus;

    The present age of son’s will be 6 years & the present age of father’s will be 7r = 7 × 6 = 42 years.

    0
    2021-08-13T13:58:48+00:00

    Answer:

    present age of son x, father is 7x

    3years ago

    7x-3=13x-39

    36=4x

    x=9,son is 9,father is 45

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