Question

The number of pairs of integers \((x , y)\) satisfying \(x≥y≥-20\) and \(2x+5y=99\) is
[This Question was asked as TITA]

• A. 18
• B. 11
• C. 15
• D. 17

Answer 1

The Correct Option is D

Given the equation: \(2x + 5y = 99\), find the number of integer solutions \((x, y)\) where \(x \ge y\).

Step 1: Express x in terms of y

Rearranging the equation gives: \(2x = 99 - 5y \Rightarrow x = \frac{99 - 5y}{2}\).

For \(x\) to be an integer, \(99 - 5y\) must be an even number.

Since 99 is odd, \(5y\) must also be odd for their difference to be even. Therefore, \(y\) must be an odd integer.

Step 2: Determine the range of integer values for y

We need to find odd integer values of \(y\) that yield integer values for \(x\) and satisfy \(x \ge y\).

• Let's test some odd values for \(y\):
  • If \(y = -19\), then \(x = \frac{99 - 5(-19)}{2} = \frac{99 + 95}{2} = 97\). Here, \(x = 97 \ge y = -19\).
  • If \(y = 13\), then \(x = \frac{99 - 5(13)}{2} = \frac{99 - 65}{2} = 17\). Here, \(x = 17 \ge y = 13\).

These examples suggest that \(y\) can range from \(-19\) to \(13\), with \(y\) being odd integers.

Step 3: Count the number of valid y values

The odd integers from \(-19\) to \(13\) form an arithmetic progression (AP) with:

• First term: \(a = -19\)
• Last term: \(t_n = 13\)
• Common difference: \(d = 2\)

Using the AP nth term formula, \(t_n = a + (n - 1)d\):

\(13 = -19 + (n - 1) \cdot 2\)

\((n - 1) \cdot 2 = 13 + 19 = 32\)

\(n - 1 = \frac{32}{2} = 16\)

\(n = 17\)

There are 17 possible odd integer values for \(y\).

Step 4: Verify the condition \(x \ge y\) for all solutions

We confirmed that \(x \ge y\) holds for the minimum \(y\) value (\(y = -19, x = 97\)) and the maximum \(y\) value (\(y = 13, x = 17\)).

Since \(x = \frac{99 - 5y}{2}\) is a decreasing function of \(y\), and the condition \(x \ge y\) is met at the boundary values, it will hold for all 17 integer solutions.

Final Answer: (D) 17