Question

The toral ages of A,B and C at present is 90 years. Ten years ago the ratio of this ages 1:2:3. The present age of B?​

Answer 1

Given :

• Sum of ages of A , B and C is 90 years.
• Ten years ago the ratio of their ages were in the ratio 1:2:3.

To Find :

• Present Age of B.

Solution :

$\longmapsto\tt{Let\:Present\:Age\:of\:A=1x+10}$

$\longmapsto\tt{Let\:Present\:Age\:of\:B=2x+10}$

$\longmapsto\tt{Let\:Present\:Age\:of\:C=3x+10}$

A.T.Q :

$\longmapsto\tt{1x+10+2x+10+3x+10=90}$

$\longmapsto\tt{6x+30=90}$

$\longmapsto\tt{6x=90-30}$

$\longmapsto\tt{6x=60}$

$\longmapsto\tt{x=\cancel\dfrac{60}{6}}$

$\longmapsto\tt\bf{x=10}$

Value of x is 10 ...

Therefore :

$\longmapsto\tt{Present\:Age\:of\:B=2(10)+10}$

$\longmapsto\tt{20+10}$

$\longmapsto\tt\bf{30\:yrs.}$

Answer 2

Given :-

• Sum of Ages of A, B, C = 90 years.

• Ten year ago the ratio of their ages = 1:2:3.

To Find :-

• Present age of B.

Solution :-

$\implies\sf{Let, \: the \: present \: age \: of \: a = \: }{\textsf{\textbf{1x + 10}}} \\ \\ \implies\sf{Let, \: the \: present \: age \: of \: b = \: }{\textsf{\textbf{2x + 10}}} \\ \\ \implies\sf{Let, \: the \: present \: age \: of \: c = \: }{\textsf{\textbf{3x + 10}}} \\ \\ \implies\sf{1x + 10 + 2x + 10 + 3x + 10 = \: }{\textsf{\textbf{90}}} \\ \\ \implies\sf{6x + 30 = \: }{\textsf{\textbf{90}}} \\ \\ \implies\sf{6x = \: }{\textsf{\textbf{90 - 30 = 60}}} \\ \\ \implies\sf{x = \: }{\textsf{\textbf{60/6}}} \\ \\ \implies\sf{x = \: }{\textsf{\textbf{10}}}$

Value of x = 10.

$\therefore\sf{The \: present \: age \: of \: b = \: }{\textsf{\textbf{2(10) + 10}}} \\ \\ \implies\sf{20 + 10 = \: }{\underline{\boxed{\textsf{\textbf{\red{30 year}}}}}}$