Ten years ago, A was half of B in age. If the ratio of their present ages is 3:4. What will be the total of their present age

Question

Ten years ago, A was half of B in age.
If the ratio of their present ages is
3:4. What will be the total of their
present ages?
(a) 20 years
(C) 30 years
(b) 25 years
(d) 35 years​

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Hadley 2 weeks 2021-09-05T06:30:46+00:00 1 Answer 0 views 0

Answers ( )

    0
    2021-09-05T06:32:02+00:00

    \Large{\underline{\underline{\red{GiVen:-}}}}

    The Age of Father = 2 year more than five times of the age of the son.

    The Sum of their age = 38 years.

    \Large{\underline{\underline{\green{To Find:-}}}}

    The Age of each of them.

    \Large{\underline{\underline{\red{SoLuTioN:-}}}}

    \implies\sf{Let, \: the \: age \: of \: son \:  = \:  }{\textsf{\textbf{x}}} \\  \\ \implies\sf{As, \: the \: age \: of \: Father \: is \: 2 \: year \: more \: than \: five \: times \: of \: son \: age} \\  \\ \implies\sf{So, \: the \: age \: of \: father \:  =  \: }{\textsf{\textbf{5x + 2}}} \\  \\ \implies\sf{x + 5x + 2 =  \: }{\textsf{\textbf{36}}} \\  \\ \implies\sf{6x + 2 = \: }{\textsf{\textbf{38}}} \\  \\ \implies\sf{6x = 38 - 2 =  \: }{\textsf{\textbf{36}}} \\  \\ \implies\sf{x = \cancel{36/6}} \\  \\ \implies\sf{x =  \: }{\textsf{\textbf{6}}}

    Value of x = 6.

    Therefore,

    The age of son = 6.

    The age of father = 5(6) + 2 = 32.

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