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

We are given the equation: 
\(2x + 5y = 99\)

We need to find the number of integer solutions \((x, y)\) such that \(x \ge y\).

Step 1: Express x in terms of y

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

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

Since 99 is odd, \(5y\) must also be odd so that the subtraction becomes even. So, \(y\) must be odd.

Step 2: Find integer values of y such that x is also an integer

Try different odd values of y to see for which values x is also an integer.

• When \(y = -19\), \(x = \frac{99 - 5(-19)}{2} = \frac{99 + 95}{2} = 97\)
• When \(y = 13\), \(x = \frac{99 - 5(13)}{2} = \frac{99 - 65}{2} = 17\)

So the range of y is from \(-19\) to \(13\), and all values are odd integers.

Step 3: Count the number of such y values (odd numbers from -19 to 13)

This forms an arithmetic progression (AP) with: 
First term: \(a = -19\),
Last term: \(t_n = 13\),
Common difference: \(d = 2\)

Use the nth term formula of an AP: 
\(t_n = a + (n - 1)d\)

Plug in values: 
\(13 = -19 + (n - 1) \cdot 2\)
\(\Rightarrow (n - 1) \cdot 2 = 32\)\)
\(\Rightarrow n - 1 = 16 \Rightarrow n = 17\)\)

Step 4: Check if all such (x, y) pairs satisfy \(x \ge y\)

We already checked the extremes: 
- For \(y = -19, x = 97 \Rightarrow x > y\)
- For \(y = 13, x = 17 \Rightarrow x > y\)
Since \(x = \frac{99 - 5y}{2}\) decreases as y increases, and \(x \ge y\) holds at both ends, it will hold for all 17 values.

Final Answer: (D) 17