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

Question

The pages of a report are numbered consecutively from 1 to 10. If the sum of the page numbers up to and including page number x of the report is equal to one more than the sum of the page numbers following page number x, then x =

A. 4
B. 5
C. 6
D. 7
E. 8

A. 4
B. 5
C. 6
D. 7
E. 8

Bunuel · Accepted Answer

The sum of the page numbers of a whole report is (1+10)/2*10=55;

So, the sum of the page numbers from 1 to x, inclusive is 28 and the sum of the page numbers from x to 10 is 27 (since we are told that the the difference is 1).

Now, the sum of the page numbers from 1 to x, inclusive is 28 means that (1+x)/2*x=28 --> (1+x)x=56 --> x=7.

Answer: D.

Or you can simply recognize that 27=8+9+10, so x=8-1=7.

sub3108 · Answer

Excellent Solution Bunuel, Thanks.

Ousmane · Answer

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

PareshGmat · Answer

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

KarishmaB · Answer

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)

colorblind · Answer

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{ }{2} 
 28 = \frac{ }{2} 
 X^2+X-56=0 
X=7
OR
Plug in ans choices in equation:  28 = \frac{ }{2}

FYI....
Sum of sequence =  *

venkivety · Answer

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

ScottTargetTestPrep · Answer

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

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