# PE 4 - Average | Arithmetic - Average

**PE 4 - Average | Arithmetic - Average**

Anjali wrote the first few natural numbers on black board but missed one of the numbers. She later calculated the sum of all the numbers that she wrote and divided it by what she thought was the number of numbers she had written. If the result she thus obtained was $38\frac{1}{11}$, find the number that she forgot to write.

- (a)
71

- (b)
70

- (c)
69

- (d)
None of these

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).

Workspace:

## Feedback

**Help us build a Free and Comprehensive Preparation portal for various competitive exams by providing
us your valuable feedback about Apti4All and how it can be improved.**