Question

2 years ago A’s age was 6 times of B’s age. 6 years after the ratio between the ages of A and B becomes 10 : 3. What is A’s present age?​

Answer 1

Answer:

44 years

Step-by-step explanation:

Let 2 years ago, the age of B = x years,

So, the age of A = 6x years,

At present,

Age of B = x + 2,

Age of A = 6x + 2

Again after 6 years,

Age of B = x + 8,

Age of A = 6x + 8,

According to the question,

$\frac{6x + 8}{x+8}=\frac{10}{3}$

$18x + 24 = 10x + 80$

$8x = 56$

$\implies x = 7$

Hence, the present age of A = 6(7) + 2 = 42 + 2 = 44 years

Answer 2

At present :

Age of A = x years ,

Age of B = y years

2 years ago :

Age of A = ( x - 2 ) years ,

Age of B = ( y - 2 ) years

/* According to the problem given

A's age was 6 times of B's age

=> x - 2 = 6( y - 2 )

=> x - 2 = 6y - 12

=> x = 6y - 12 + 2

=> x = 6y - 10 -----(1)

6 years after:

Age of A = ( x + 6 ) years

Age of B = ( y + 6 ) years

/* According to the problem given

$\frac{ age \: of \: A }{age \: of \: B } = \frac{10}{3}$

$\implies \frac{x+6}{y+6} = \frac{10}{3}$

$\implies \frac{6y-10+6}{y+6} = \frac{10}{3}\:[From \:(1) ]$

$\implies \frac{6y-4}{y+6} = \frac{10}{3}$

$\implies 3(6y-4) = 10(y+6)$

$\implies 18y - 12 = 10y + 60$

$\implies 18y - 10y = 60 + 12$

$\implies 8y = 72$

$\implies y = \frac{72}{8} = 9$

/* Put y = 9 in equation (1) , we get

$\implies x = 6 \times 9 - 10\\= 54 - 10 \\= 44$

Therefore.,

$\underline { \pink { At \: present }}$

$Age \: of \: A = 44 \: years$

$Age \: of \: B = 9 \: years$

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