The pages of a report are numbered consecutively from 1 to 1

The same end result as Bunnuel, (and more time consuming, but at least more intuitive)
X*(1 + X)/2 -(10-X)*((X+1 + 10)/2) = 1 ==> X^2 + X - 56 = 0 ==> (X-7)*(X+8) = 0 , X = 7 or X = -8
Reponse X = 7
Brother Karamazov

1, 2, 3, 4, 5, 6, 7,............ 8, 9, 10

Huge numbers are at the right; started addition from there

28 ................................... 27

x = 7

Answer = D

There are 10 consecutive pages. Sum of first few = sum of last few. Note that first few are small numbers and last few are greater ones. So obviously, the split cannot be half-half. At least first 6 numbers need to be together. The question is whether 7 is in the first group or second group.
Sum of first 7 positive integers = 7*8/2 = 28
Sum of 8+9+10 = 9*3 = 27

This is the correct answer so x = 7

Answer (D)

\(Sum(1 to 10) = \frac{(10*11)}{2} = 55\)
n+(n+1) = 55
n=27
Since sum of first X digits greater by 1, Sum (1 to X) = 28
Now, we can either use :
\(Sum(1 to X) = \frac{[X*(X+1)]}{2}\)
\(28 = \frac{[X*(X+1)]}{2}\)
\(X^2+X-56=0\)
X=7
OR
Plug in ans choices in equation: \(28 = \frac{[X*(X+1)]}{2}\)

FYI....
Sum of sequence = [Half no of terms]*[Sum of first term & last term]

I've gone through the question by the choice approach as these are small numbers.

as our peers said the first few numbers are small and the last are big.

so I started with 6. sum upto 6 is 21
from 7 to 10 is 34

then gone to 7. sum upto 7 is 28
sum from 8 to 10 is 27. bingo

Careful while selecting the choice. question asked abt x
answer choice D

Solution:

The sum of all the page numbers is 1 + 2 + … + 10 = 55. The sum of the page numbers from 1 to x, inclusive, is (1 + x)/2 * x = x(1 + x)/2. Therefore, the sum of the page numbers from x + 1 to 10, inclusive, is 55 - x(1 + x)/2. From the information given in the problem, we can create the equation:

x(1 + x)/2 = 55 - x(1 + x)/2 + 1

x(1 + x) = 56

x^2 + x - 56 = 0

(x + 8)(x - 7) = 0

x = -8 or x = 7

Since x can’t be negative, x must be 7.


Alternate solution:

If you have difficulty solving this problem algebraically, you can just check the given choices numerically. Let’s start with 6 in choice C; if it works, then that is the answer and if it doesn’t, we can then decide to move forward to 7 or move backward to 5.

If x = 6, we have the sum of the numbers 1 to 6, inclusive, as 1 + 2 + … + 6 = 21 and the sum of the numbers 7 to 10, inclusive, as 7 + 8 + 9 + 10 = 34. We see that x is not 6 and since 21 is less than 34, we should increase the value of x.

So let x = 7. We have the sum of the numbers 1 to 7, inclusive, as 1 + 2 + … + 7 = 28 and the sum of the numbers 8 to 10, inclusive, as 8 + 9 + 10 = 27. We see that 28 is exactly 1 more than 27, therefore, x must be 7.

Answer: D

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