A whole number is added to 100 and the same number is subtracted from 100. the sum of the two resulting numbers so obtained is :
200
The question asks us to find the sum of resulting numbers when a specific operation is performed on the number 100 using a whole number.
We are given a whole number. Let's call this whole number 'x'. A whole number is any non-negative integer (0, 1, 2, 3, ...).
According to the problem, two steps are performed:
We need to find the sum of the two numbers obtained from these two steps. This is the sum of resulting numbers.
Let the chosen whole number be represented by the variable \(x\).
Now, we need to find the sum of these two resulting numbers:
Sum = (First resulting number) + (Second resulting number)
Sum = \((100 + x) + (100 - x)\)
Let's simplify the expression for the sum:
Sum = \(100 + x + 100 - x\)
We can rearrange the terms:
Sum = \( (100 + 100) + (x - x) \)
Sum = \( 200 + 0 \)
Sum = \( 200 \)
As we can see from the calculation, the sum of the two resulting numbers is always 200, regardless of which whole number \(x\) was chosen. The \(+x\) and \(-x\) terms cancel each other out.
Let's take an example:
In both examples, the sum of resulting numbers is 200. This confirms our algebraic result.
Therefore, the sum of the two resulting numbers so obtained is 200.
Which of the following statements is not true?
Suppose a2 + b2 = 4(a + 3b -10), where a and b are two real numbers. Then which of the following is true?
If m and n are two positive real numbers such that 9m2 + n2 = 40 and mn = 4, then the value of 3m + n is:
Consider the following statements :
1. If n is a natural number, then the number \(\frac{n\left(n^2+2\right)}{3}\) is also a natural number.
2. If m is an odd integer, then the number \(\frac{\mathrm{m}^4+4 \mathrm{~m}^2+11}{16}\) is an integer.
Which of the statements given above is/are correct ?
Which composite number can divide the sum of the first 12 natural numbers?