Question

six years ago the ratio of ages of A&B was 1:3 and after six years the ratio becomes 2:3.find the sum of present age of A&B.​

Answer 1

▪Assumptions :-

Let

Present age of A be = x years

and

Present age of B be = y years

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▪Given :-

1》Six years ago the ratio of ages of A & B was 1:3

2》After six years the ratio will be 2:3.

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▪To Calculate :-

Sum of present ages of A and B.

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▪Solution :-

《 Six years ago 》

Age of A = (x - 6) years

and

Age of B = (y - 6) years

According to the Given Condition:

$\bf \frac{x - 6}{y - 6} = \frac{1}{3} \\ \\ \implies \sf3(x - 6) = y - 6 \\ \\ \implies \sf 3x - 18 = y - 6 \\ \\ \implies \bf 3x - y = 12 \: \: \: \: .\: . \: .\: \{i \}$

《 After Six years 》

Age of A = (x + 6) years

and

Age of B = (y + 6) years

According to the Given Condition:

$\bf\frac{x + 6}{y + 6} = \frac{2}{3} \\ \\ \implies \sf 3(x + 6) = 2(y + 6) \\ \\ \implies \sf 3x + 18 = 2y + 12 \\ \\ \implies \bf3x - 2y = - 6 \: \: \: \:. \: . \: . \: \{ii \}$

Subtracting {ii} from {i} we get,

$\Large\purple{ \underline{\boxed{ \bf y = 18 }}}$

Putting Value of y in {i} We get,

$\sf3x = 12 + 18 \\ \\ \implies \sf3x = 30 \\ \\ \implies\Large \purple{ \underline{\boxed{ \bf x = 10 }}}$

Hence ,

Present age of A = 10 years

and

Present age of B = 18 years

So,

$\bf Sum \: of \: Present \: Ages \\ \bf of \: A \: and \: B = 10 + 18 \\ \\ = \Large \purple{ \underline{\boxed{ \bf 28 \: years}}}$

$\Large \red{\mathfrak{  \text{W}hich \:   \: is  \:  \: the  \:  \: required} }\\ \huge \red{\mathfrak{ \text{ A}nswer.}}$

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Answer 2

Given :-

Six years ago the ratio of ages of A&B was 1:3 and after six years the ratio becomes 2:3

To Find :-

Sum of present ages

Solution :-

Six years ago

A - 6/B - 6 = 1/3

3(A - 6) = 1(B - 6)

3A - 18 = B - 6

3A - B = 18 - 6

3A - B = 12 (1)

Now

A + 6/B + 6 = 2/3

3(A + 6) = 2(B + 6)

3A + 18 = 2B + 12

3A - 2B = 12 - 18

3A - 2B = -6

Subtract both

B = 18

Using 1

3a - 18 = 12

3a = 30

a = 30/3

a = 10

Sum = 18 + 10 = 28 years