Question

The age of A and B are in ratio 4:3 . After 6 years their ages will be in the ratio 11:9. What is A's present age?

Answer 1

Answer:

Step-by-step explanation:

let the age of A be : x

let the age of B be: y

now x/y=4/3

3x=4y

now in condition 2

x+6/y+6 = 11/9

9x+54 = 11y + 66

3(3x) = 11y+ 66 -54

3(4y) = 12 +11y

12y -11y = 12

y =12

now

sub y in 1

x=4 x 12/3

x = 4x4

x =16

please amrk brainlist

Answer 2

$\bold{Given,}$

$\mathsf{Age \: of \: \bold{A} \: and \bold{B} \: are \: in \: ratio \: 4 : 3}$

$\mathsf{Let \: the \: common \: ratio \: be }$$\bold{x}$

So, the ratio becomes

$\mathsf{4x \: and \: 3x}$

$\large{ACCORDING \: TO \: QUESTION }$

$\mathsf{After \: six \: years \: age \: will \: becomes \: (4x + 6) \: and \: (3x + 6 )\: in \: ratio \: 11 :9}$

$\huge \mathsf{\frac{4x + 6}{3x + 6 } = \frac{11}{9}}$

$\mathsf{On \: cross \: multiplication, }$

$\mathsf{9(4x + 6) =11 (3x + 6)}$

$\mathsf{36x + 54 = 33x + 66}$

$\mathsf{On \: taking \: like \: terms \: aside}$

$\mathsf{36x - 33x = 66 - 54}$

$\mathsf{3x = 12}$

$\mathsf{x = \frac{12}{3}}$

$\huge \boxed{ \mathsf{x = 4}}$

$\mathsf{So, \: real \: age \: will \: be}$

$\boxed{ \mathsf{A = 4 \times 4 = 16\: years}}$

$\boxed{ \mathsf{B= 3 \times 4 = 12 \: years}}$