Help needed

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Help needed

by Karthik12 » Mon Jul 28, 2008 1:00 am
Ricky took a novel to read from a bookseller and found that a page was missing. He went back to the bookseller and informed him about it, and also told him that the sum of the remaining pages is 10,000. Which of the following can be one of the page numbers missing from the novel?
1. 5
2. 11
3. 77
4. 153
5. option 1 and 3

Plz explain it
Source: — Problem Solving |

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by warlock » Mon Jul 28, 2008 2:21 am
this is by trail and error method that i (think) got it..

since all the numbers in the books are consecutive we can use the sum to n terms formula..which is

sigma(N) = N(N+1)
---------
2

therefore by trail and error when N = 141
we get the missing page as 11.

sigma(141) = (141*142)/2 = 10011.
subtract 11 and you get 10000.

not sure about the actual method..

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by parallel_chase » Mon Jul 28, 2008 2:50 am
I think the answer is 11. whats the OA?

The sum of the number of pages with the missing page is 10000.

Therefore if any of the options is true, then the sum would be

10005
10011
10077
10153

if we apply the sequence formula n/2[first term+last term] the total number of pages we could have is between 141-145 pages. This is because the sum of the all the integers between 141-145 is around 10000.

Now we take 142 as the number we get the sum = 10153, so we can assume that the answer is 153. Here 153 is a trap answer, when the total number of pages is 142 can 153 be the page number that could be missing?

Lets take 141, the total comes out to be 10011, hence 11 is the answer.

Let me know if the OA is 11, also kindly let me know if I have missed anything.

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by sudhir3127 » Mon Jul 28, 2008 2:54 am
Even i ended up with 11.. Let us know the OA. and source as well... i was wondering if such questions are Gmat Types.. as it involves trial and method based solution.

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by Karthik12 » Mon Jul 28, 2008 5:31 am
I dont know the correct option.

If you take the n is 141, you will get 10011. (Generally, both sides(front & back) are printed in novel.)

So the pages that were missing could be page numbers 5 and 6 where n=141.
if u take n=142, you will get 10153. When the missing page numbers could be 76 and 77.

Hence, the option is 5. Let me know if anyone have concerns abt this....

Then, i have the general doubt...what is the meaning of OA?

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by evansbd » Mon Jul 28, 2008 5:47 am
I was checking some of the approaches here and possibly wasn't calculating the values correctly. I don't see how the sum of 141...145 = 10000.

Also, I focused on answers 11 and 153 through POE because I figured that if option E was both answers 5 and 77 then its probably not either since both can't be missing (only one page missing). However, assuming I could do this POE quickly, I had little confidence as to which of the remaining answers were correct.

Chase:

Could you show a calculation on how you found your range 141-145, then I could see how you eliminated 153 as an option.

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by parallel_chase » Mon Jul 28, 2008 8:57 am
evansbd wrote:I was checking some of the approaches here and possibly wasn't calculating the values correctly. I don't see how the sum of 141...145 = 10000.

Also, I focused on answers 11 and 153 through POE because I figured that if option E was both answers 5 and 77 then its probably not either since both can't be missing (only one page missing). However, assuming I could do this POE quickly, I had little confidence as to which of the remaining answers were correct.

Chase:

Could you show a calculation on how you found your range 141-143, then I could see how you eliminated 153 as an option.
Your POE is absolutely right, you can easily eliminate 153. The reason why stressed on 153 is because it is a trap answer. This is a trial and error kind of question. Therefore some people could reach to 142 as the choice and select 153 as the answer.

I found the range 141-143 (sorry it was a typo) on the following basis.

if you want to find out the sum of integers from 1 to 141 you calculate it through following method:

n/2[first term + last term]

n = no. of integers = 141 (141-1+1)
first term = 1
last term = 141

(141*142)/2 = 10011

Let me know if you still have any doubts.