Question

Radha was asked to find the average of N consecutive natural numbers starting from 1. By mistake, he added a number twice but he did not notice it. As a result, he obtained a 11 wrong average of 45 Find the number she added twice. ​

Answer 1

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The average of n consecutive natural numbers is given by n+12 (why?)

If you add x twice, the average increases by xn .(why?)

Putting this all together, we get n+12+xn=315

Pouring some simplification powders, we arrive at (57–5n)n=10x

First we rule out the case that he added the last number twice, i.e. x=n .

Putting x=n gives us n=75 which is impossible, so that handles it. Now we have x is strictly less than n .

Now, we notice that n divides the LHS, which is 10x, since n does not divide x (because x≤n ), n must divide 10. So that n is either 1,2,5 or 10 . We can rule out 1 , since that leaves no space for x. Now it’s just a game of trial and error. Putting n=2 leaves us with a fractional x, and putting n=5 gives integral x , but x>n .

$Putting \: \: \: n=10 , gives \: x=7$

Answer 2

The average of n consecutive natural numbers is given by n+12 (why?)

If you add x twice, the average increases by xn .(why?)

Putting this all together, we get n+12+xn=315

Pouring some simplification powders, we arrive at (57–5n)n=10x

First we rule out the case that he added the last number twice, i.e. x=n .

Putting x=n gives us n=75 which is impossible, so that handles it. Now we have x is strictly less than n .

Now, we notice that n divides the LHS, which is 10x, since n does not divide x (because x≤n ), n must divide 10. So that n is either 1,2,5 or 10 . We can rule out 1 , since that leaves no space for x. Now it’s just a game of trial and error. Putting n=2 leaves us with a fractional x, and putting n=5 gives integral x , but x>n .

Putting \: \: \: n=10 , gives \: x=7Puttingn=10,givesx=7