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A page is torn from a novel. The sum of the remaining page numbers is 14 Apr 2020, 09:08
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A page is torn from a novel. The sum of the remaining page numbers is 10,000. What could be the page-numbers on the torn page ?

I. 76 and 77
II. 33 and 34
III. 5 and 6.

A. I only
B. II only
C. III only
D. I and II only
E. I and III only


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A page is torn from a novel. The sum of the remaining page numbers is 14 Apr 2020, 09:23
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Bunuel
A page is torn from a novel. The sum of the remaining page numbers is 10,000. What could be the page-numbers on the torn page ?

I. 76 and 77
II. 33 and 34
III. 5 and 6.

A. I only
B. II only
C. III only
D. I and II only
E. I and III only


Are You Up For the Challenge: 700 Level Questions
Show more


CONCEPT: Sum of first n consecutive integers, \(1+2+3....+n = (\frac{1}{2})*n(n+1)\)

Here the sum should be close to 10,000 and greater than 10,000

Lets calculate (1/2)*n(n+1) = 10000
n*(n+1) = 20000
i.e.n ≈ 141

141*142 = 20022

i.e. (1/2)*141*142 = 10011

So ew are short of 11

OR

We are short of 142(next page)+11 = 153

I. 76 and 77 Sum of 76 and 77 = 153 POSSIBLE
II. 33 and 34
III. 5 and 6. Sum of 5 and 6 = 11 POSSIBLE

Answer: Option E
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A page is torn from a novel. The sum of the remaining page numbers is 14 Apr 2020, 13:13
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Bunuel
A page is torn from a novel. The sum of the remaining page numbers is 10,000. What could be the page-numbers on the torn page ?

I. 76 and 77
II. 33 and 34
III. 5 and 6.

A. I only
B. II only
C. III only
D. I and II only
E. I and III only


Are You Up For the Challenge: 700 Level Questions


CONCEPT: Sum of first n consecutive integers, \(1+2+3....+n = (\frac{1}{2})*n(n+1)\)

Here the sum should be close to 10,000 and greater than 10,000

Lets calculate (1/2)*n(n+1) = 10000
n*(n+1) = 20000
i.e.n ≈ 141

141*142 = 20022

i.e. (1/2)*141*142 = 10011

So ew are short of 11

OR

We are short of 142(next page)+11 = 153

I. 76 and 77 Sum of 76 and 77 = 153 POSSIBLE
II. 33 and 34
III. 5 and 6. Sum of 5 and 6 = 11 POSSIBLE

Answer: Option E
Show more

Hi Can you Pls explain how did you get 142 + 11= 153. Thanks in advance.
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Bunuel
A page is torn from a novel. The sum of the remaining page numbers is 10,000. What could be the page-numbers on the torn page ?

I. 76 and 77
II. 33 and 34
III. 5 and 6.

A. I only
B. II only
C. III only
D. I and II only
E. I and III only


Are You Up For the Challenge: 700 Level Questions


CONCEPT: Sum of first n consecutive integers, \(1+2+3....+n = (\frac{1}{2})*n(n+1)\)

Here the sum should be close to 10,000 and greater than 10,000

Lets calculate (1/2)*n(n+1) = 10000
n*(n+1) = 20000
i.e.n ≈ 141

141*142 = 20022

i.e. (1/2)*141*142 = 10011

So ew are short of 11

OR

We are short of 142(next page)+11 = 153

I. 76 and 77 Sum of 76 and 77 = 153 POSSIBLE
II. 33 and 34
III. 5 and 6. Sum of 5 and 6 = 11 POSSIBLE

Answer: Option E

Hi Can you Pls explain how did you get 142 + 11= 153. Thanks in advance.
Show more
Suppose instead of 141 pages in the book, if there were 142 pages then sum of the pages would have been 10153 then extra sum would have been 153
Which is 77 +76 =153
So this is also the possibility

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A page is torn from a novel. The sum of the remaining page numbers is 15 Apr 2020, 00:29
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Bunuel
A page is torn from a novel. The sum of the remaining page numbers is 10,000. What could be the page-numbers on the torn page ?

I. 76 and 77
II. 33 and 34
III. 5 and 6.

A. I only
B. II only
C. III only
D. I and II only
E. I and III only


Are You Up For the Challenge: 700 Level Questions


CONCEPT: Sum of first n consecutive integers, \(1+2+3....+n = (\frac{1}{2})*n(n+1)\)

Here the sum should be close to 10,000 and greater than 10,000

Lets calculate (1/2)*n(n+1) = 10000
n*(n+1) = 20000
i.e.n ≈ 141

141*142 = 20022

i.e. (1/2)*141*142 = 10011

So we are short of 11

OR

We are short of 142(next page)+11 = 153

I. 76 and 77 Sum of 76 and 77 = 153 POSSIBLE
II. 33 and 34
III. 5 and 6. Sum of 5 and 6 = 11 POSSIBLE

Answer: Option E

Hi Can you Pls explain how did you get 142 + 11= 153. Thanks in advance.
Show more

Hi AdiBatman

Sum of 1 to 141 is (1/2)*141*142 = 10011 so the sum of page number should be 11 more i.e. pages torn must have sum of page numbers 11

But

other possibilities is that there were 142 pages and in that case the missing sum would be 11+142(next page number) = 153

I hope this help!!! :)
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I think it's A because a book/novel can only have even pages
So for 141 pages is no chance
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I think it's A because a book/novel can only have even pages
So for 141 pages is no chance
Show more

rudi0setyawan

What makes you think that a book/novel can't have odd pages.

It's only the page number which can very well be even or odd

The numbering is done in many ways in books. A few pages in beginning in Roman count where the introduction is mentioned, the dedication letter is mentiones.

Towards the end there may be some advertise and all and the numbers mentioned on pages can well be Odd number too
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A page is torn from a novel. The sum of the remaining page numbers is 17 Apr 2020, 11:21
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Bunuel
A page is torn from a novel. The sum of the remaining page numbers is 10,000. What could be the page-numbers on the torn page ?

I. 76 and 77
II. 33 and 34
III. 5 and 6.

A. I only
B. II only
C. III only
D. I and II only
E. I and III only


Are You Up For the Challenge: 700 Level Questions
Show more

Recall that the sum of the consecutive integers from 1 to n is given by n(n + 1)/2. Let’s analyze each Roman numeral:

I: 76 and 77

If the page numbers on the torn page was 76 and 77, then the sum of the pages of the novel was 10,000 + 76 + 77 = 10,153. To determine whether 10,153 is the sum of consecutive integers, let’s solve n(n + 1)/2 = 10,153:

n(n + 1)/2 = 10153

n^2 + n = 20306

n^2 + n - 20306 = 0

(n + 143)(n - 142) = 0

n = -143 or n = 142

We see that 10,153 is the sum of the integers from 1 to 142; therefore it is possible that the page numbers on the torn page were 76 and 77.

II: 33 and 34

If the page numbers on the torn page was 33 and 34, then the sum of the pages of the novel was 10,000 + 33 + 34 = 10,067. To determine whether 10,067 is the sum of consecutive integers, let’s solve n(n + 1)/2 = 10,067:

n(n+1)/2 = 10067

n^2 + n = 20134

n^2 + n - 20134 = 0

This equation has no integer roots. Thus, the page numbers on the torn page cannot be 33 and 34.

III: 5 and 6

If the page numbers on the torn page was 5 and 6, then the sum of the pages of the novel was 10,000 + 5 + 6 = 10,011. To determine whether 10,011 is the sum of consecutive integers, let’s solve n(n + 1)/2 = 10,011:

n(n+1)/2 = 10011

n^2 + n = 20022

n^2 + n - 20022 = 0

(n + 142)(n - 141) = 0

We see that 10,011 is the sum of the integers from 1 to 141; therefore it is possible that the page numbers on the torn page were 5 and 6.

Answer: E
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Recall that the sum of the consecutive integers from 1 to n is given by n(n + 1)/2. Let’s analyze each Roman numeral:

I: 76 and 77

If the page numbers on the torn page was 76 and 77, then the sum of the pages of the novel was 10,000 + 76 + 77 = 10,153. To determine whether 10,153 is the sum of consecutive integers, let’s solve n(n + 1)/2 = 10,153:

n(n + 1)/2 = 10153

n^2 + n = 20306

n^2 + n - 20306 = 0

(n + 143)(n - 142) = 0

n = -143 or n = 142

We see that 10,153 is the sum of the integers from 1 to 142; therefore it is possible that the page numbers on the torn page were 76 and 77.

II: 33 and 34

If the page numbers on the torn page was 33 and 34, then the sum of the pages of the novel was 10,000 + 33 + 34 = 10,067. To determine whether 10,067 is the sum of consecutive integers, let’s solve n(n + 1)/2 = 10,067:

n(n+1)/2 = 10067

n^2 + n = 20134

n^2 + n - 20134 = 0

This equation has no integer roots. Thus, the page numbers on the torn page cannot be 33 and 34.

III: 5 and 6

If the page numbers on the torn page was 5 and 6, then the sum of the pages of the novel was 10,000 + 5 + 6 = 10,011. To determine whether 10,011 is the sum of consecutive integers, let’s solve n(n + 1)/2 = 10,011:

n(n+1)/2 = 10011

n^2 + n = 20022

n^2 + n - 20022 = 0

(n + 142)(n - 141) = 0

We see that 10,011 is the sum of the integers from 1 to 141; therefore it is possible that the page numbers on the torn page were 5 and 6.

Answer: E

[/quote]

Hi ScottTargetTestPrep,

Could you please tell how to split the middle term for such equations (n^2 + n - 20306 = 0
) in a quicker manner as I am finding difficulty in solving them.

Thanks!
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Recall that the sum of the consecutive integers from 1 to n is given by n(n + 1)/2. Let’s analyze each Roman numeral:

I: 76 and 77

If the page numbers on the torn page was 76 and 77, then the sum of the pages of the novel was 10,000 + 76 + 77 = 10,153. To determine whether 10,153 is the sum of consecutive integers, let’s solve n(n + 1)/2 = 10,153:

n(n + 1)/2 = 10153

n^2 + n = 20306

n^2 + n - 20306 = 0

(n + 143)(n - 142) = 0

n = -143 or n = 142

We see that 10,153 is the sum of the integers from 1 to 142; therefore it is possible that the page numbers on the torn page were 76 and 77.

II: 33 and 34

If the page numbers on the torn page was 33 and 34, then the sum of the pages of the novel was 10,000 + 33 + 34 = 10,067. To determine whether 10,067 is the sum of consecutive integers, let’s solve n(n + 1)/2 = 10,067:

n(n+1)/2 = 10067

n^2 + n = 20134

n^2 + n - 20134 = 0

This equation has no integer roots. Thus, the page numbers on the torn page cannot be 33 and 34.

III: 5 and 6

If the page numbers on the torn page was 5 and 6, then the sum of the pages of the novel was 10,000 + 5 + 6 = 10,011. To determine whether 10,011 is the sum of consecutive integers, let’s solve n(n + 1)/2 = 10,011:

n(n+1)/2 = 10011

n^2 + n = 20022

n^2 + n - 20022 = 0

(n + 142)(n - 141) = 0

We see that 10,011 is the sum of the integers from 1 to 141; therefore it is possible that the page numbers on the torn page were 5 and 6.

Answer: E

Show more

Hi ScottTargetTestPrep,

Could you please tell how to split the middle term for such equations (n^2 + n - 20306 = 0
) in a quicker manner as I am finding difficulty in solving them.

Thanks![/quote]

Solution:

To factor n^2 + n - 20306 = 0, we need to express 20306 as the product of two opposite sign factors where the sum of the factors is 1. We can begin by prime factorizing 20306:

20306 = 2 x 11 x 13 x 71

Using the prime factorization, we can determine the two factors with the required properties. Let’s not worry about the signs for the moment and concentrate on finding two factors with a difference of 1. For instance, if we take 2 as one of the factors and 11 x 13 x 71 as the other factor, the difference between the two factors will greatly exceed 1. If we move the factor of 11 from 11 x 13 x 71 to 2, we have 2 x 11 and 13 x 71 as the two factors. While the difference is less, compared to 2 and 11 x 13 x 71, it is still too much. It is the same case if we consider 2 x 13 and 11 x 71. However, if we consider 2 x 71 = 142 and 11 x 13 = 143, we see that we found two factors with a difference of 1. In order for the sum of these factors to be 1, then 143 must be positive and 142 must be negative; thus, we can express -20306 as 143 x -142. With these factors, we can write:

n^2 + n - 20306 = (n + 143)(n - 142)

We can open up the parentheses to verify that the product indeed equals n^2 + n - 20306.

Alternate Solution:

Recall that if a quadratic of the form n^2 + bn - c (where b and c are positive) is factorable, we are looking for two numbers such that their product is c and their difference is b. Now, if we suppose n^2 + n - 20306 is factorable, then we are looking for two numbers whose product is 20306 and whose difference is 1. That is the two numbers are very close to each other (since their difference is 1). In fact, if the two numbers are exactly the same, then the product of the two numbers becomes the square of either number. In other words, each number should be the square root of c or, in this case, 20306. Of course, the two numbers we are looking for here have a difference of 1 and even though they are not exactly the square root of 20306, they should be very close to the square root (one slightly larger and the other slightly smaller). Now, taking the square root of 20306, we have √20306 ≈ 142.5. We can guess one number is 142 and the other is 143. In fact, 142 x 143 = 20306. Therefore, n^2 + n - 20306 = (n + 143)(n - 142).

Hope these explanations help! Good luck!
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I found an arthematic progression series in this, 1 pg has 2 numbers So, 1st page = 1+2 = 3 , 2nd page = 3+4=7 and so on... So, lets assume 251 pages so, t251 = 3+(251-1)*4 (a + (n-1)*d) = 10,003. Now, we have 3 options which when added to 10,000 becomes 10011, 10153 and 10067. Check whether it comes in AP series which we identified. you will get 1& 3 option in AP series
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Hi, here's what I was trying to explain:

In a physical book, each sheet (or leaf) contains two pages — one on the front and one on the back.

So:

The 1st sheet has pages 1 and 2

The 2nd sheet has pages 3 and 4

The 3rd sheet has pages 5 and 6

...

The 38th sheet has pages 75 and 76

Now, if a single sheet is torn, then the two consecutive pages it contains must be an odd-even pair, like (1,2), (3,4), (5,6), ..., (75,76), etc.

Looking at the options:

Option I (76 and 77):
This is not valid — 76 is even and 77 is odd, meaning they would appear on different sheets. So this pair cannot be torn out together from a single sheet.

Option II (33 and 34):
This is a valid odd-even pair and would appear on the same sheet (Sheet 17).
However, when you do the math, removing 33 and 34 does not lead to a remaining sum of 10,000, so it fails mathematically.

Option III (5 and 6):
This pair is valid both physically (they're on the same sheet) and mathematically — removing them results in a total remaining sum of 10,000.

So, only Option III satisfies both the physical sheet structure and the math condition. That’s why I believe the question is flawed unless it clearly states that the torn pages are just "any two consecutive page numbers" — not necessarily on the same sheet.
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Hi, here's what I was trying to explain:

In a physical book, each sheet (or leaf) contains two pages — one on the front and one on the back.

So:

The 1st sheet has pages 1 and 2

The 2nd sheet has pages 3 and 4

The 3rd sheet has pages 5 and 6

...

The 38th sheet has pages 75 and 76

Now, if a single sheet is torn, then the two consecutive pages it contains must be an odd-even pair, like (1,2), (3,4), (5,6), ..., (75,76), etc.

Looking at the options:

Option I (76 and 77):
This is not valid — 76 is even and 77 is odd, meaning they would appear on different sheets. So this pair cannot be torn out together from a single sheet.

Option II (33 and 34):
This is a valid odd-even pair and would appear on the same sheet (Sheet 17).
However, when you do the math, removing 33 and 34 does not lead to a remaining sum of 10,000, so it fails mathematically.

Option III (5 and 6):
This pair is valid both physically (they're on the same sheet) and mathematically — removing them results in a total remaining sum of 10,000.

So, only Option III satisfies both the physical sheet structure and the math condition. That’s why I believe the question is flawed unless it clearly states that the torn pages are just "any two consecutive page numbers" — not necessarily on the same sheet.
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You're absolutely correct and this question is flawed.

Nice catch.
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Archiving the topic. Thank you!

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