Question

the ages of x and y are in the ratio 8:3 . six years later hence, their ages will be in the ratio of 9:4 . find their present ages

Answer 1

Let the common multiple be a

So, let the present ages of x be 8a years and y be 3a

After 6 years,

Age of x = 8a + 6

Age of y = 3a + 6

According to the given condition

8a+6/3a+6 = 9/4

☛ 4(8a + 6) = 9(3a + 6)

☛ 32a + 24 = 27a + 54

☛ 32a - 27a = 54 - 24

☛ 5a = 30

☛ a = 6

∴ Present age of x ☛ 8a

☛ 8 × 6

☛ $\large{\boxed{\boxed{\rm{48\:years}}}}$

∴ Present age of y ☛ 3a

☛ 3 × 6

☛ $\large{\boxed{\boxed{\rm{18\:years}}}}$

Answer 2

$\sf\underline{Step-by-step \: explanation}$

Let the common variables be x

There ages before 6 years =

$\tt\dfrac{8x}{3x}$

[m : n = m/n]

According to the question...

● $\tt\dfrac{8x \: + \: 6}{3x \: + \: 6}$ = $\tt\dfrac{9}{4}$

● 4(8x + 6) = 9(3x + 6) [Using cross multiplication to find out x]

● 32x + 24 = 27x + 54

● 27x - 32x = 24 - 54
[Bringing like terms together]

● -5x = -30

● -x = -6

● x = 6 [Minuses cancel each other]

So the age of x and y will be;

x = $\sf{\boxed{\boxed{48 \: years}}}$

[x = 8x = 8 × 6 = 48]

y =$\sf{\boxed{\boxed{18\:years}}}$

[y = 3x = 3 × 6 = 18]

$\bold{Verification,}$

After 6 years...

● 48 + 6 : 18 + 6 = 9 : 4

● 54 : 24 = 9 : 4

● 18 : 8 = 9 : 4

● 9 : 4 = 9 : 4

$\sf\underline{Hence \: Verified}$

Present ages...

x = 48 years

y = 18 years.