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.

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.

Example 1: What will be the sum of first 40 natural numbers?

Solution:From the above formula, we can calculate, 1 + 2 + 3+..........+ 40 (Here, n=40)

40*41/2 = 820

Example 2: A book has 60 pages numbered 1, 2, 3, 4.......60. What is the sum of all the page numbers of this book?

Solution: Here, we have to find the sum of the first 60 natural numbers. Using the formula given above,

60*61/2 = 1830

Example 3: The sum of the pages of a book is 630. Find the number of pages in this book.

Solution: Here, let us consider that there are 'n' pages in the book whose sum is 630.
⇒ n(n+1)/2 = 630
⇒ n*(n + 1) = 1260 ⇒ n²+ n - 1260 = 0 which is a quadratic equation and its roots are 35 and - 36. Since, the number of pages of the book cannot be negative, so the answer is 35.
We can also solve this question by hit and trial.
We have n (n+1) = 1260
As n² ≈ 1260 (To simplify the calculations we have taken n and n+1 as equal)
⇒ 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.

Example 4: The page numbers of a book were added and the sum obtained was 1053, but one page was added twice. Find the page number which was added twice?

Solution: Let the total pages were 'n'
Hence 1 + 2 + 3 + 4+ ........+ n = 1053
⇒ n(n+1)/2 = 1053
⇒ n(n+1) = 2106.

As n² ≈ 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

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

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

Example 5: A teacher asked his students to find the sum of all pages of their book. One student gave the answer as 640.The teacher told him that he has not added one page.
Find the page number which the student forgot to add?

Solution: Let the total pages are 'n'. We have
1 + 2 + 3 + .........+n = 640
⇒ n(n+1)/2 = 640
⇒ n(n + 1) = 1280 ⇒ ≈ n² = 1280. Now, the perfect square closest 1280 is 1296, which is the square of 36. Since the student did not add one page, so 640 is lesser than the actual sum and 1280 is also less than the actual value. Hence, we will take n = 36. Hence, initially, 36 pages were added. So their sum will be -
36*37/2 = 666
As the given sum is 640, so the student forgot to add page number 666 - 640 = 26.

Example 6: 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?

Solution: Let the total pages are 'n'. We have
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.