Question

If \(x\) and \(y\) are non-negative integers such that \(x + 9 = z, \ y + 1 = z\) and \(x + y < z + 5\), then the maximum possible value of \(2x + y\) equals
[This Question was asked as TITA]

• A. 18
• B. 21
• C. 23
• D. 28

Answer 1

The Correct Option is C

Given that:

$x + 9 = z = y + 1$  
and 
$x + y < z + 5$

Step 1: Express $x$ and $y$ in terms of $z$

From $x + 9 = z$, we get $x = z - 9$ 
From $z = y + 1$, we get $y = z - 1$

Step 2: Substitute into the inequality

$x + y < z + 5$ 
$\Rightarrow (z - 9) + (z - 1) < z + 5$ 
$\Rightarrow 2z - 10 < z + 5$ 
$\Rightarrow z - 10 < 5$ 
$\Rightarrow z < 15$

Step 3: Maximum possible value of $z$

Since $z < 15$, the maximum integer value of $z$ is $14$

Step 4: Find corresponding values of $x$ and $y$

$x = z - 9 = 14 - 9 = 5$ 
$y = z - 1 = 14 - 1 = 13$

Step 5: Compute the required expression

$2x + y = 2 \times 5 + 13 = 10 + 13 = \mathbf{23}$

Answer: (C): $23$

Answer 2

Given: 

• Equation 1: \( x + 9 = z \)
• Equation 2: \( y + 1 = z \)
• Equation 3: \( x + y < z + 5 \)

Step 1: Express \(x\) and \(y\) in terms of \(z\)

From Equation 1: \( x = z - 9 \)
From Equation 2: \( y = z - 1 \)

Step 2: Substitute into inequality Equation 3

\( x + y < z + 5 \)
\( \Rightarrow (z - 9) + (z - 1) < z + 5 \)
\( \Rightarrow 2z - 10 < z + 5 \)
\( \Rightarrow z < 15 \)

Step 3: Maximum possible integer value of \(z\)

Since \(z < 15\), the maximum possible integer value of \(z\) is 14.

Step 4: Calculate \(2x + y\)

We need to maximize \( 2x + y \)
Substitute \(x = z - 9\) and \(y = z - 1\):
\( 2x + y = 2(z - 9) + (z - 1) = 2z - 18 + z - 1 = 3z - 19 \)
Now, put \( z = 14 \):
\( 3 \times 14 - 19 = 42 - 19 = \mathbf{23} \)

Final Answer: Option (C): 23