Every page of a book is numbered from 1 without any omission. One pag

Question

Every page of a book is numbered from 1 without any omission. One page of the book is torn out. The summation of the numbers of the remaining pages is 1256. Which page is torn out?

A. 9~10 
B. 21~22 
C. 31~32 
D. 43~44 
E. 51-52

yashikaaggarwal · Answer

Sum of n consecutive integers = n(n+1)/2 
Let n be 50.
Sum of 50 consecutive integers is 50*51/2 = 1275>1256

Let n = 49
Sum of 49 consecutive integers is 49*50/2 = 1225<1256

So the book is of 50 pages and torn page has to be a no. Between 1 to 50
And 2 no. Will be k,l 
K+L+1256= 1275
K+L= 1275-1256
K+L=19 ----(1)

Since a page consist of 2 consecutive no. On front and back page therefore. K and L are two consecutive no. N,N+1 ----(2)
N+N+1 = 19
2N = 18
N= 9
K and L are 9&10 respectively
Answer is A

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nick1816 · Answer

Sum of all the pages of the book  = \frac{n*(n+1)}{2}

\frac{n*(n+1)}{2} - (x+x+1) = 1256

n^2+n -4x -2514 = 0

Since n is real and an integer,  b^2-4ac  is a perfect square

b^2-4ac = 1- 4*( -4x-2514) = 16x+10057

Nearest square of 10057 is  (101)^2

16x+10057 = 10201 
 16x = 144 
 x=9

A

EgmatQuantExpert · Answer

Solution

Given  
In this question, we are given that
 • Every page of a book is numbered from 1 without any omission
• One page of the book is torn out
• The summation of the numbers of the remaining pages is 1256

To find  
We need to determine
 • The page numbers of the page which is torn out

Approach and Working out  
We can use the options here to ease the calculation
 • If 9~10 are the page numbers of the torn page, then sum of the all the page numbers = 1256 + 9 + 10 = 1275 = 25 x 51 = 50 x 51 x ½ = sum of first 50 consecutive numbers

Thus, option A is the correct answer.

Correct Answer: Option A

Piyush1993 · Answer

An easy way out, works for similar questions as well.
assume the correct sum is 1256 with total of n pages.
n(n+1)/2 =1256 
n(n+1) =2512

Find an "n" which gets a close value to the number 2512. Easily, n=50
If n was 50, sum of pages would be n(n+1)/2
n(n+1) =2550

If, n was 50, the difference in the actual sum and reduced sum is 2550-2512 =38
But note that is the not the actual difference, this is double of the actual difference, because 2550 (or 2512) is double the sum.
Sum of torn pages =38/2 =19

Hence A. This looks lengthy but when worked out, it becomes a lot easier.

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MathRevolution · Answer

=>
Assume the original book has  n  pages and the page numbers of the torn page are  k  and  k+1.

Then we have  1 + 2 + 3 + … + n = \frac{n(n+1)}{2} = 1256 + k + ( k + 1 ).

Since  k ≥ 1 , we have  \frac{n(n+1)}{2} = 1256 + k + k + 1 ≥ 1256 + 1 + 2 = 1259, \frac{n(n+1)}{2} ≥ 1259,  or  n(n+1) ≥ 2518.

Then we have  n ≥ 50.

Since  n ≥ k , we have  \frac{n(n+1)}{2} ≤ 1256 + k + k + 1 ≤ 1256 + n + n + 1, \frac{n(n+1)}{2} ≤ 1257 + 2n, n(n+1) ≤ 2514+ 4n, n^2 + n - 4n ≤ 2514 , or  n^2 – 3n ≤ 2514.  Then  n ≤ 51.

Case 1)  n = 50

We have  \frac{(50·51)}{2} = 1256 + k + (k + 1)  or  1275 = 1256 + 2k + 1. 
 
Then  1275 = 1257 + 2k, 2k = 18 , and  k = 9.

Then, we have  k = 9.

Case 2)  n = 51

\frac{(51·52)}{2} = 1256 + k + (k + 1)  or  1326 = 1256 + 2k + 1.  Then  1326 = 1257 + 2k, 2k = 69,  and  k = \frac{69}{2} .

Then, we have  k = \frac{69}{2} , which is not an integer.

Thus, the page numbers of the torn page are 9 and 10.

Therefore, A is the answer.
Answer: A

bumpbot · Answer

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Every page of a book is numbered from 1 without any omission. One pag

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