One commonly asked question-type is where you are asked to determine the number of pages added twice/missed when the sum of pages is given.

Take a look at the following question.

A page is torn from a novel. The sum of the remaining digits is 312. What are the page numbers on page torn from this novel?

If you don't know the concept of average and the approach to solve these type of questions, you'll generally approach this problem by making complex equations. That will lead to cumbersome calculations and will end up taking a lot of time. Through this article, we'll see how to tackle such questions using the concept of averages.

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Before we start the topic, let us first recall a basic relation related to natural numbers, i.e. the sum of first 'n' natural numbers.

Sum of first 'n' natural numbers= 1+ 2 + 3 + 4 +.......+n = n(n+1)/2

Let us see how we can use this concept in finding the missing page of a book.

40*41/2 = 820

60*61/2 = 1830

⇒ n(n+1)/2 = 630

⇒ n*(n + 1) = 1260 ⇒ n

We can also solve this question by hit and trial.

We have n (n+1) = 1260

As n

⇒ n = 35 as the square of 35 is 1225 and we have reduced n + 1 to n.

The above illustrations were to familiarize with the concept. Now, let us see some advanced level questions from this topic that is more relevant for CAT and other competitive examinations.

Hence 1 + 2 + 3 + 4+ ........+ n = 1053

⇒ n(n+1)/2 = 1053

⇒ n(n+1) = 2106.

As n^{2} ≈ 2106 (To simplify calculations, we have taken n and n+1 as equal)

Now, perfect square closest to 2106 is 2116, which is the square of 46.

Here, one page was added twice. It means that 1053 is greater than the actual sum; therefore, 2106 is also greater than the actual value.

So, we will not take n = 46, as it will further increase the sum.

So, it has to be a number, which appears immediately before 46 i.e. 45.

Hence, we can conclude that 45 pages were added. Now, the sum of 45 pages will be

Now, perfect square closest to 2106 is 2116, which is the square of 46.

Here, one page was added twice. It means that 1053 is greater than the actual sum; therefore, 2106 is also greater than the actual value.

So, we will not take n = 46, as it will further increase the sum.

So, it has to be a number, which appears immediately before 46 i.e. 45.

Hence, we can conclude that 45 pages were added. Now, the sum of 45 pages will be

1 + 2 + 3 +.........+ 45 = 45*46/2 = 1035

Since the given sum is 1053, so the page number added twice was 1053 - 1035 = 18.

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Find the page number which the student forgot to add?

1 + 2 + 3 + .........+n = 640

⇒ n(n+1)/2 = 640

⇒ n(n + 1) = 1280 ⇒ ≈ n

36*37/2 = 666

As the given sum is 640, so the student forgot to add page number 666 - 640 = 26.

1 + 2 + 3 + .........+n = 312

⇒ n(n+1)/2 = 312

n (n+1) = 624

The square nearest to 624 is 625.This is the square of 25. Hence, we will take n= 25.

Therefore, sum of the pages= n(n+1)/2 = 25*26/2=25*13=325

Missing number = 325-312=13.

Now, we have to think of 2 consecutive numbers whose sum is 13.

x+x+1=13 ⇒2×=12

x=6.

Now, we have to think of 2 consecutive numbers whose sum is 13.

x+x+1=13 ⇒2×=12

x=6.