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Five years ago, A's age was four times the age of B. Five years hence A's age will be twice of the age of B. Find their present ages.
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Let us consider B’s age = X

Then A’s age = 4X

Five years ago

⇒ A’s age = 4X + 5

⇒ B’s age = X + 5

Five years hence

⇒ A’s age = 4X + 5 + 5

⇒ A’s age = 4X + 10

⇒ B’s age = X + 5 + 5

⇒ B’s age = X + 10

From the given question

4X + 10 = 2(X + 10)

4X + 10 = 2X + 20

4X – 2X = 10

2X = 10

X = 5

A’s age = 4X + 5

A’s age = 25 years

B’s age = X + 5

B’s age = 10 years

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Five years ago,A's age was four times the age of B. 5 years hence, A's age will be twice of the age of B. Find their present ages.

Five years ago:

As it's Given 5 years ago A's age was 4 times the age of B

.°. Equation formed as follows:

Now Let's solve this equation:

Five years hence:

As it's given 5 years hence A's age will be twice of the age of B.

.°. Equation formed as follows:

Now put value of y in Equation 2

multiply each digit by 4

we got value of x i.e.25

Put value of x in equation 1:

As we have supposed A's age as x

.°. A's age = 25

As we have supposed B's age as y

.°. B's age = 10

Now Let's Verify their ages:

Put age of A and B in Equation 1:

☆Hence Verified ☆

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And all we are done! ✔

:D

step-by-step explanation:

b

step-by-step explanation: