When 15 is included in a list of natural numbers, their mean is increased by 2. When 1 is included in this new list, the mean of the numbers in the new list is decreased by 1. How many numbers were there in the original list?

### Concept & Strategy When a new number alters the mean, the difference between that incoming number and the resulting new mean equals the total change distributed across all previous items. We can build two linked equations from the two successive additions. ### Step-by-Step Solution Let the original number of items be $N$ and their original mean be $M$. * **Analyze the First Addition (15 is added):** The new list has $N+1$ items. The mean becomes $M+2$. The sum equation is: $$N \cdot M + 15 = (N + 1)(M + 2)$$ $$N \cdot M + 15 = N \cdot M + 2N + M + 2$$ $$15 = 2N + M + 2$$ $$2N + M = 13\ \ \text{--- (Equation 1)}$$ * **Analyze the Second Addition (1 is added):** The current list has $N+1$ items and a mean of $M+2$. We add a 1, so the new list has $N+2$ items. The mean decreases by 1, so it becomes $(M+2) - 1 = M+1$. The sum equation is: $$(N+1)(M+2) + 1 = (N+2)(M+1)$$ $$(N \cdot M + 2N + M + 2) + 1 = N \cdot M + N + 2M + 2$$ $$2N + M + 3 = N + 2M + 2$$ $$N + 1 = M\ \ \text{--- (Equation 2)}$$ * **Solve the System:** Substitute $M = N + 1$ into Equation 1: $$2N + (N + 1) = 13$$ $$3N + 1 = 13$$ $$3N = 12$$ $$N = 4$$ ### Exam Strategy & Shortcut Use the **Deviation Spread** mental framework to build the equations without expanding algebraic binomials. **First addition:** 15 comes in, making the new mean $M+2$. The excess value 15 brings is $15 - (M+2)$. This excess is what gave $+2$ to the original $N$ numbers. $N \times 2 = 15 - (M+2) \implies 2N + M = 13$. **Second addition:** 1 comes into a list that currently has $N+1$ members and a mean of $M+2$. The new mean drops by 1, becoming $M+1$. The deficit that 1 brings is $1 - (M+1) = -M$. This deficit is what caused the $-1$ drop for the current $N+1$ members. $(N+1) \times (-1) = -M \implies -N - 1 = -M \implies M = N+1$. Substitute $M = N+1$ directly into $2N + M = 13 \implies 3N + 1 = 13 \implies N = 4$. ### Common Pitfall Misreading "decreased by 1" in the second condition. Students often subtract 1 from the *original* mean $M$ (writing $M-1$) instead of subtracting 1 from the *current* mean of the new list (which is $M+2$, making it $M+1$). Pay close attention to relative baselines in successive problems. ### Final Answer **Therefore, the correct answer is 4.**