Question 51

Logical Reasoning Aptitude Easy

A set of consecutive positive integers beginning with 1 is written on the blackboard. A student came along and erased one number. The average of the remaining numbers is&nbsp;<span class="math-tex">\(35 \dfrac{7}{17}\)</span>. What was the number erased?

(A) 7
(B) 8
(C) 9
(D) None of the above
View Dynamic Solution & Explanation
Correct Solution: Option A

Step-by-step Solution:

Step-by-Step Algebraic Solution

To solve this problem, we can set up an equation for the average of the remaining numbers and use it to find the original number of integers (n) and the erased number (x).

1. Formulating the Equations

  • Let 'n' be the largest integer in the original set {1, 2, ..., n}.
  • The sum of these 'n' integers is given by the formula: $S_n = \frac{n(n+1)}{2}$.
  • Let 'x' be the integer that was erased. The number of integers remaining is (n-1).
  • The sum of the remaining integers is $S_{new} = S_n - x = \frac{n(n+1)}{2} - x$.
  • The average of the remaining integers is given as $35\frac{7}{17}$, which is $\frac{35 \times 17 + 7}{17} = \frac{602}{17}$.

The equation for the average is: $\frac{S_{new}}{n-1} = \frac{\frac{n(n+1)}{2} - x}{n-1} = \frac{602}{17}$.

2. Estimating the Value of 'n'

The average of the remaining numbers ($\approx 35.4$) must be very close to the average of the original set of numbers, which is approximately $\frac{n}{2}$.

So, $\frac{n}{2} \approx 35.4$, which means $n \approx 70.8$. Thus, 'n' is likely to be 69, 70 or 71.

3. Finding the Exact Value of 'n'

From our average equation, the sum of the remaining numbers is $S_{new} = (n-1) \times \frac{602}{17}$.

Since the sum of integers must be an integer, (n-1) must be a multiple of 17.

Let's look for a multiple of 17 near our estimate for n ($ \approx 70 $):

  • $17 \times 3 = 51 \rightarrow n-1 = 51 \rightarrow n = 52$ (too low)
  • $17 \times 4 = 68 \rightarrow n-1 = 68 \rightarrow n = 69$ (This is a strong candidate)
  • $17 \times 5 = 85 \rightarrow n-1 = 85 \rightarrow n = 86$ (too high)

So, the original number of integers, n, must be 69.

4. Calculating the Erased Number 'x'

Now that we know n=69, we can find the original sum and the new sum.

  • Original Sum ($S_{69}$): $\frac{69(69+1)}{2} = \frac{69 \times 70}{2} = 69 \times 35 = 2415$.
  • New Sum ($S_{new}$): $(69-1) \times \frac{602}{17} = 68 \times \frac{602}{17} = 4 \times 602 = 2408$.

The erased number 'x' is the difference between the original sum and the new sum:

x = Original Sum - New Sum = 2415 - 2408 = 7.

Conclusion

The number that was erased is 7.