PE 4 - Average | Arithmetic - Average

Answer: Option B

Explanation :

Let ‘n’ be the number of numbers Anjali wanted to write on the board (including the missed number).

Let ‘m’ be the missed number.

Sum of all number, excluding the missed number = n(n + 1)/2 – m.

Average of these numbers (excluding the missed number) = $38\frac{1}{11}$ = $\frac{419}{11}$ = $\frac{{\displaystyle \frac{n(n+1)}{2}}-m}{n}$

The total no. of natural numbers she wanted to write on the black board (i.e., n) should be a multiple of 11. 

∴ n = 11k   …(1)

Now we know that the average of first n natural numbers is $\frac{(n+1)}{2}$. [Since the sum of first n natural nos. is $\frac{n(n+1)}{2}$].

Hence, no. of natural number = 2 × average – 1.

Even if she misses out one of the numbers the final average would be slightly less than $\frac{(n+1)}{2}$.

∴ The number of natural number written on the board would be around 2 × $\frac{419}{11}$ – 1 ≈ 75   …(2)

From (1) and (2)

The nearest possible number of 11k form around 75 is 77.

Hence, she wanted to write 77 numbers on the board.

∴ Sum of first 77 natural number = $\frac{77\times 78}{2}$ = 3003.

Sum of the number which she added up = 77 × $\frac{419}{11}$ = 2933.

Hence the number erased = 3003 - 2933 = 70.

Hence, option (b).