Question

five years ago Mike was twice as old as but to the sum of their ages 6 years hence will be 40 years what are their present ages​

Answer 1

Question:

Five years ago Mike was twice as old as his brother but to the sum of their ages 6 years hence will be 40 years what are their present ages.

To find:

★ To find the present age of Mike and his brother.

Answer:

$\bigstar \: Mike \: age \: is \: \underline{ \: \bold{16 \: years} \: }$

$\bigstar \: Brother \: age \: is \: \underline{ \: \bold{13 \: years} \: }$

Given:

★ Five years ago Mike was twice as old as his brother.

$Mike \: age \: x= 2(y - 5)$

$\boxed{x= 2y - 10}--->eq(1)$

★ Sum of their ages 6 years hence will be 40 years.

$\implies x + 6 + y + 6 = 40$

$\implies x + y + 12 = 40$

$\implies \boxed{x + y = 28}--->eq(2)$

Step-by-step explanation:

Let Mike are be x .

Let his brother age be y .

$x + y = 28$

We know that, x value .

$\implies 2y - 10 + y = 28$

$\implies 3y = 28 + 10$

$\implies 3y = 38$

$\implies y = \frac{38}{3}$

$\implies \boxed{y = 12.66}$

Since age cannot be in decimal.

We can round off the value.

$\boxed{ \bold{y = 13}}$

∴ His brother age is 13 years.

Now substituting the value of y in eq(1)

$x= 2(13) - 10$

$\: = 26 - 10$

$\boxed{x = 16}$

∴ Mike’s age is 16 years .

Hint given in the question:

★ Five years ago Mike was twice as old as his brother.

★ Sum of their ages 6 years hence will be 40 years.

⚠️Note⚠️

We have to read the hint properly.

Because it forms the equation.

Verification:

$\boxed{x + y = 28}$

$\implies 16 + 13 \not= \:28$

$\implies \boxed{29 \not= \:28}$

Because we have round off the value of y to the nearest Whole Number (i.e) 13 .

So this can't be written as verified.