This question requires us to determine the present age of an individual named Q, given specific relationships between the present ages of P and Q, and their ages after 9 years. We can solve this using algebra by setting up and solving equations.
First, let's define variables for the unknown ages:
The problem states: "Two times the present age of P exceeds the present age of Q by 15 years". This translates to the algebraic equation:
$2P = Q + 15 \quad (Equation \, 1)$
We need to consider their ages after 9 years:
The problem states: "After 9 years, twice Q's age will be 14 years less than twice P's age at that time". This gives us the equation:
$2(Q + 9) = 2(P + 9) - 14 \quad (Equation \, 2)$
Our goal is to find the value of $Q$. We have a system of two linear equations with two variables:
Let's simplify the second equation:
$2Q + 18 = 2P + 18 - 14$
Combine the constants on the right side:
$2Q + 18 = 2P + 4$
Rearrange the terms to make it easier to substitute or solve:
$2Q = 2P + 4 - 18$
$2Q = 2P - 14 \quad (Simplified \, Equation \, 2)$
From Equation 1, we know that $2P = Q + 15$. We can substitute this expression for $2P$ into the simplified Equation 2:
$2Q = (Q + 15) - 14$
Now, we have a single equation with only the variable $Q$. Let's solve it:
$2Q = Q + 15 - 14$
$2Q = Q + 1$
Subtract $Q$ from both sides of the equation:
$2Q - Q = 1$
$Q = 1$
The value of $Q$ represents the present age of Q. Therefore, Q's present age is 1 year.
To ensure our answer is correct, let's find P's age and check the conditions.
Using Equation 1 with $Q=1$:
$2P = 1 + 15$
$2P = 16$
$P = 8$
So, P is currently 8 years old and Q is 1 year old.
Now, let's check the condition after 9 years:
Twice Q's age = $2 \times 10 = 20$.
Twice P's age = $2 \times 17 = 34$.
Is twice Q's age (20) equal to 14 less than twice P's age (34)?
$34 - 14 = 20$
Yes, $20 = 20$. The conditions are met, confirming our answer.
The present age of Q is calculated to be 1 year.