**QUESTION:-**When one number was erased from the set of natural numbers till N, average of the remaining numbers was 881/29. Which number was erased?

**OPTIONS****1)**6

**2)**7

**3)**8

**4)**9

**5)**10

Now we consider the average after deletion, 881/29

Since the denominator of the average after deleting one number is 29, value of N can be 30.

If the above fraction has been arrived at after removing the common multiple of 2 in both the denominator and numerator, value of N can be 29 × 2 + 1 = 59.

Similarly it can be 29 × 3 + 1 = 88 or 29 × 4 + 1 = 117 and so on.

Suppose N = 30.

Then sum of all the natural numbers till 30 is 30 × 31/2 = 465, which is less than 881.

So, N = 30 is not possible.

Suppose N = 59

Then sum of all the natural numbers till 59 is 59 × 60/2 = 1770, which is greater than 881 and hence possible.

Suppose the number deleted is x.

∴1770-x = 1762

∴ x = 8

Since we get an integer value of x, our assumption that N = 59 is true.

For higher values of N, x > N, which is not possible.

**Hence, option 3.**

.

## 0 comments:

## Post a Comment