The difference between two numbers 2.when each number is rounded to the nearest hundred , the difference between them is 100. What Numbers Could these be?

The numbers could be (149, 151) or (249, 251), differing by 2, and rounding to 100 and 200,respectively.

Let's denote the two numbers as [tex]\( A \) and \( B \).[/tex]

1. The difference between the two numbers is 2:

 [tex]\[ |A - B| = 2 \][/tex]

2. When each number is rounded to the nearest hundred, the difference between them is 100.

Step-by-Step Solution:

**Step 1: Understanding the rounding impact**

When a number is rounded to the nearest hundred, it falls within the range of the midpoint between two hundreds. For example:

- Any number from 50 to 149 rounds to 100.

- Any number from 150 to 249 rounds to 200.

- And so on.

**Step 2: Establishing conditions based on rounding**

We know the rounded difference is 100, so the rounded values of ( A ) and ( B ) must differ by 100. Let's denote the rounded values as  [tex]\( A_r \)[/tex] and [tex]\( B_r \):[/tex]

[tex]\[|A_r - B_r| = 100\][/tex]

**Step 3: Analyzing the ranges for [tex]\( A \) and \( B \)[/tex] **

Given the possible ranges for rounding:

• If A rounds to Ar and B rounds to Br, the actual values A and B must be within +-50 of their

respective rounded values.

Potential Cases:

[tex]1. \( A_r = 100 \) and \( B_r = 200 \):\\ - For \( A \): \( 50 \leq A < 150 \)\\ - For \( B \): \( 150 \leq B < 250 \)[/tex]

[tex]2. \( A_r = 200 \) and \( B_r = 300 \):\\ - For \( A \): \( 150 \leq A < 250 \)\\ - For \( B \): \( 250 \leq B < 350 \)[/tex]

Step 4: Solving for exact numbers**

Since the difference between ( A ) and ( B ) is 2, we test within these ranges:

Case 1: ( A ) in [50, 150) and ( B ) in [150, 250)

  - Assume ( A ) is at the upper end of its range and ( B ) just 2 units more.

[tex]\[ B = A + 2 \][/tex]

  - Since ( B ) must round to 200 and \( A \) to 100:

[tex]\[ 50 \leq A < 150 \quad \text{and} \quad 150 \leq B < 250 \][/tex]

Check a specific example:

  - Let ( A = 149 ) (rounds to 100)

  - Then ( B = 149 + 2 = 151 ) (rounds to 200)

Thus, ( A = 149 ) and ( B = 151 ) is one solution.

Case 2: ( A ) in [150, 250) and ( B ) in [250, 350)

  - Assume ( A) is at the upper end of its range and ( B \) just 2 units more.

B = A + 2

  - Since ( B ) must round to 300 and ( A ) to 200:

    [tex]\[ 150 \leq A < 250 \quad \text{and} \quad 250 \leq B < 350 \][/tex]

Check a specific example:

• Let A 249 (rounds to 200)

• Then B 249 4- 2 251 (rounds to 300)

Thus, A = 249 and B=  251 is another solution.

Conclusion

The pairs (A, B) that satisfy the conditions are:

• (149, 151)

• (249,251)

These pairs of numbers differ by 2 and, when rounded to the nearest hundred, differ by 100.

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