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Aditya8624
If we take 153 the earlier page, do we have to consider the page that is being read as it is mentioned not to include that, it a bit confusing there, do we need to subtract that one page?
No, there is no subtraction of one page, that is overcomplicating things. Just stick to exactly what the passage says:

"At exactly two of those pages (the earlier page and the later page), the total number of pages remaining to complete the assignment, not including the page currently being read, is equal to the value expressed by the last two digits of the number on that page."

As a simple example, if we were on page 3 of a 5-page book, there would be 5 - 3 = 2 pages remaining, not including the page currently being read. Just pages 4 and 5 would be remaining.

Also, as a general thing, we can re-evaluate our approach if we see that none of the answer choices work. If you did decide to subtract one, you would see that none of the answers would work, and that would be a clue that this extra step is not necessary.­
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The students in a literature class have been assigned to read pages 85 through 206 of a novel. At exactly two of those pages (the earlier page and the later page), the total number of pages remaining to complete the assignment, not including the page currently being read, is equal to the value expressed by the last two digits of the number on that page. For example, the value expressed by the last two digits of 107 is 7.

Select for Earlier page the number on the earlier page and for Later page the number on the later page. Make only two selections, one in each column.

­
An apt question to use the options.

Thu option plus the last two digits should give you 206.
153+53 = 206 = 203+03

But say, you got in a PS and it asked how many such numbers exist­

It cannot be a 2-digit number, becuas ethen you are looking at double the number and the maximum possible sum would be 99+99 = 198
Any 3-digit number, abc, can be written as 100a+10b+c, so 206 = 100a+10b+c+(10b+c) = 100a+20b+2c ...... 103 = 50a+10b+c..(c would always be 3)
If a=1, then 103=50+10b+c ..... 53 = 10b+c....As b and c are digits, b=5 and c=3.....NUmber = 153
If a=2, then 103=100+10b+c ..... 03 = 10b+c....As b and c are digits, b=0 and c=3.....Number = 203
Thus two possibilities as a = 3 will give us sum>300.
Earlier = 153 and Later = 203
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Once we know what the question wants, it's very easy to find the answer from choice.
During my mock exam, it's so confusing and I end up by guessing the number. So sad.
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Solving the question was quite straightforward but reading and understanding the question and picturizing it took majority of time.
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This is rather easy question. I tend to complicate all my solutions. Any suggestions for me??
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This is rather easy question. I tend to complicate all my solutions. Any suggestions for me??
Just stick to the wording that has been given, and do not get any outside knowledge to it.
Do not assume a question is going to be easy or difficult.­
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Thought process: As last 2 digits would be the pages remaining, it simply means pages remaining <100. Becomes easy by elimination after this as we narrow down the range from pg nos. 85-206 to pg nos. 107-206. Then looking at Options- we get 153 and 203. Hope I have explained. Thanks.­
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For me this question got me bamboozled in 3 ways:

1. "pages 85" confused me with "85 pages" and number 85 doesn't need to be in the question. The choice starts with 91 which is already higher than 85
2. "not including the page currently being read" is also misguiding . It doesn't need to be in the question­ and the question will still work perfectly fine.­
(It's like saying 1+2 = 3 not inculding 5....)
3. "At exactly two of those pages...." also somehows leads me to think that I need to sum up last 2 digits of those 2 pages.... because I thought the page that is being read is fixed and not moving/varied.­
2.+3. well "not including the page currently being read" is what saying that there can be more than 1 page being read. The choices (2PA) also give this clue already, I guess. But the paragraph structure of this question is what killing me.­
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