Question

Ten years later, A will be twice as old as B and five years ago, A was three times as old as B. What are the present ages of A and B ?

A’s age = 50 years,and B’s age = 20 years

A’s age = 50 years,and B’s age = 25 years

A’s age = 30 years,and B’s age = 20 years

A’s age = 50 years,and B’s age = 30


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Answer 1

Answer:

Let son's age = x

Two years ago son's age= x-2.

His Father's age at that time = 3(x-2).

Present age of Father = 3x-6+2=3x-4

Two years hence father's age=3x-4+2=3x-2.

Two years hence son's age = x+2.

Given :-5*(x+2)=2*(3x-2)

5x + 10= 6x - 4

10+4=6x-5x

14=x

SON'S PRESENT AGE : 14 years.

FATHER'S PTESENT AGE: 38 years.

Answer 2

Answer:

Present age of A is 50 years and present age of B is 20 years m

Step-by-step explanation:

Given :-

• 10 years later, A will be twice as old as B.
• 5 years ago, A was three times as old as B.

To find :-

• Present age of A & B.

Solution :-

Let the present age of A be x and the present age of B be y.

10 years later,

• Age of A = (x+10) years
• Age of B = (y+10) years

According to the 1st condition,

$\to\sf{x+10=2(y+10)}$

$\to\sf{x+10=2y+20}$

$\to\sf{x=2y+20-10}$

$\to\sf{x=2y+10...............(i)}$

5 years ago,

• Age of A = (x-5) years
• Age of B = (y-5) years

According to the 2nd condition ,

$\to \sf \: x - 5 = 3(y - 5) \\ \\ \to \sf \: 2y + 10 - 5 = 3y - 15 \: \{put \: x = 2y + 10 \} \\ \\ \to \sf \: 2y + 5 = 3y - 15 \\ \\ \to \sf \: 2y - 3y = - 15 - 5 \\ \\ \to \sf \: - y = - 20 \\ \\ \to \sf \: y = 20$

• Present age of B = 20 years.

Now put x = 2y+10 in eq(i).

x = 2×20+10

→ x = 40+10

→ x = 50

• Present age of A = 50 years