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# A little difficult math problem, pls help me

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A little difficult math problem, pls help me  [#permalink]

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24 Jul 2011, 03:40
00:00

Difficulty:

35% (medium)

Question Stats:

88% (01:04) correct 12% (02:30) wrong based on 25 sessions

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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

I have no idea how to solve this ,pls help~~~

the explanation says that suppose the sum of x is m, then 2m-1=1+2+3+.......+10.
M=28, so x = 7

why x = 7 can be concluded from M=28?

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Re: A little difficult math problem, pls help me  [#permalink]

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24 Jul 2011, 05:58
1
tracyyahoo wrote:
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

I have no idea how to solve this ,pls help~~~

the explanation says that suppose the sum of x is m, then 2m-1=1+2+3+.......+10.
M=28, so x = 7

why x = 7 can be concluded from M=28?

sum of first x terms where 1<x<10 is x(x+1)/2 --- Eq 1
Now we know that sum of terms in Arithmetic Progression is n/2[2a +d(n-1)]
where n = number of terms; here n= 10-x.
a = first number of the seq. here first term will be x+1
d = difference between 2 consecutive numbers; 1 in this case

Hence the new expression is (10-x)/2[2(x+1) + (10-x+1)] --- Eq 2

Now the question says Sum of X terms = 1 + sum of remaining terms of (10-X)

therefore putting eq1 and eq2.

We get :

x(x+1)/2 = 1+ (10-x)/2[2(x+1) + (10-x+1)]
...
solving eq we will get:

x^2 +x -56 = 0
or (x+8)(x-7) = 0.
there for X = 7 as 1< x < 10

Hence D
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Re: A little difficult math problem, pls help me  [#permalink]

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24 Jul 2011, 17:10
1
We know that the sum of the integers from 1 to 10 inclusive is $$(\frac{10+1}{2})*(10-1+1) = 5.5*10 = 55$$

$$55/2 = 27.5$$, and the sum of the first x digits is one more than the sum of the rest, so the sum of the first x digits is 28 and the sum of the rest is 27.

List the numbers from 1 to 10:

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

We can start adding the consecutive integers from 1 until we reach the integer x that brings the sum to 28:

$$1+2+3+4=10$$

$$10+5+6=21$$

$$21+7=28$$

$$x=7$$
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Re: A little difficult math problem, pls help me  [#permalink]

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25 Jul 2011, 09:36
Both the replies given so far are correct, but I think there's a better way to approach this - use the answers. The answers are there to help you - you should always read them as part of the question.

One way to solve this problem would be to guess and check, then adjust your guess based on whether it was too high or too low. In these situation you should pick C as your first guess and check number. Because the GMAT always puts numerical answers in order, picking C will either get you the correct answer or eliminate 3 choices immediately. Then you only have to check one more answer

Guess C: 1 + 2 + 3 + 4 + 5 + 6 = 21 and 7 + 8 + 9 + 10 = 34. Too low

Now, obviously A and B can't be the answers, so we'll check D, and if that doesn't work E will be our answer.

Guess D: 21 + 7 = 28 and 34 - 7 = 27. Correct

Note that I used calculations I'd already done for my first guess in my second guess.

Most questions on the GMAT can be answered with a little common sense - students spending hours relearning algebra and geometry that they've forgotten since high school or college often overlook this approach.
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Re: A little difficult math problem, pls help me  [#permalink]

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25 Jul 2011, 10:29
I want to use the AP sums, same as Sudhanshuacharya, but explain a bit differently,
Use of the formal to sum the AP with first term "a", last term "l" and number of terms "n" is given as shown next, = n/2 (a + l)

We can sum the series as " 1 + 2 + ......+ x + (x+1) + .....+10

The sum of the first x terms = x(1 + x)/2

Now we have "10 - x" more terms starting with "x+1" and last term is 10, we can find the sum of these too; as = (10 - x )(x + 1 + 10)/2 = (10 - x)(x + 11)/2

But x(x + 1)/2 - 1= (10 - x)(x + 11)/2

Now solving above equation we get a quadratic equation x^2 + x - 56 = 0 and we can factor it as

(x + 8) (x - 7) = 0
Hence x = -8, 7 Reject -8 as it is not in page numbers and our answer is x = 7 which is given by choice d.
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Re: A little difficult math problem, pls help me  [#permalink]

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25 Jul 2011, 18:26
Thank u . ur post is most comprehensable.

uote="2ndgrademath"]I want to use the AP sums, same as Sudhanshuacharya, but explain a bit differently,
Use of the formal to sum the AP with first term "a", last term "l" and number of terms "n" is given as shown next, = n/2 (a + l)

We can sum the series as " 1 + 2 + ......+ x + (x+1) + .....+10

The sum of the first x terms = x(1 + x)/2

Now we have "10 - x" more terms starting with "x+1" and last term is 10, we can find the sum of these too; as = (10 - x )(x + 1 + 10)/2 = (10 - x)(x + 11)/2

But x(x + 1)/2 - 1= (10 - x)(x + 11)/2

Now solving above equation we get a quadratic equation x^2 + x - 56 = 0 and we can factor it as

(x + 8) (x - 7) = 0
Hence x = -8, 7 Reject -8 as it is not in page numbers and our answer is x = 7 which is given by choice d.[/quote]

Posted from my mobile device
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Re: A little difficult math problem, pls help mew  [#permalink]

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25 Jul 2011, 22:09
why x(x-1)/2 - 1? I don"t follow this. Pls help me to explaim this

I want to use the AP sums, same as Sudhanshuacharya, but explain a bit differently,
Use of the formal to sum the AP with first term "a", last term "l" and number of terms "n" is given as shown next, = n/2 (a + l)

We can sum the series as " 1 + 2 + ......+ x + (x+1) + .....+10

The sum of the first x terms = x(1 + x)/2

Now we have "10 - x" more terms starting with "x+1" and last term is 10, we can find the sum of these too; as = (10 - x )(x + 1 + 10)/2 = (10 - x)(x + 11)/2

But x(x + 1)/2 - 1= (10 - x)(x + 11)/2

Now solving above equation we get a quadratic equation x^2 + x - 56 = 0 and we can factor it as

(x + 8) (x - 7) = 0
Hence x = -8, 7 Reject -8 as it is not in page numbers and our answer is x = 7 which is given by choice d.

Posted from my mobile device
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Re: A little difficult math problem, pls help me  [#permalink]

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26 Jul 2011, 00:34
I went directly into the answers and started eliminating.

the only answer choice that satisfies the given condition is 7
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Re: A little difficult math problem, pls help me  [#permalink]

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27 Jul 2011, 09:23
Hi tracyyahoo

Thank you very much for your appreciation.

And we take away one from the left side because of the following reason:

Because the sum of first "x" terms, [which is x(1 + x)/2], is 1 more than the sum of 10-x terms. So, to make them equal we need to subtract 1 from the greater. Or you can add one to the smaller sum.

Hope it will help.
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Re: A little difficult math problem, pls help me  [#permalink]

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27 Jul 2011, 22:26
tracyyahoo wrote:
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

I have no idea how to solve this ,pls help~~~

the explanation says that suppose the sum of x is m, then 2m-1=1+2+3+.......+10.
M=28, so x = 7

why x = 7 can be concluded from M=28?

The moment I look at the question, I find myself jumping to option C to see if it works. Notice that adding numbers between 1 to 10 is really easy and quick. I wouldn't bother using Algebra in this question. But if we change this to pages numbered from 1 to 100 or something, I wouldn't bother with the options and go to Algebra instead. So knowing both the techniques is important. Let me discuss both of them.

Case 1: Using Options
If I add 1 to 6, I get 6*7/2 = 21
Adding 7+8+9+10 gives me 34 (You can simply add these OR say that the sum will be (3+2+1) less than 40 i.e. 6 less than 40 OR multiply the average (8.5) of the numbers by 4 i.e. 8.5*4 = 8*4 + .5*4 = 34)
Nope. We need to go to D and do the same there. We see that it works.

Case 2: Using Algebra
Sum of first x pages = x(x+1)/2
Sum of all pages = 10*11/2 = 55
Sum of all pages except first x pages = 55 - x(x+1)/2

Given: x(x+1)/2 = 1 + 55 - x(x+1)/2
x(x+1) = 56 = 7*8
x must be 7.
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Re: A little difficult math problem, pls help me  [#permalink]

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27 Jul 2011, 22:45
1
This is how I would solve:

S(n) = n*(n+1)/2 ... S = Sum
S(10) = 10*11/2 = 55

We need to find "largest" n: S(n) > S(10) - S(n)
=> S(n) > 55 - S(n)
=> 2*S(n) > 55
=> n*(n+1) > 55
=> n*(n+1) >= 56 = 7*8
=> n=7

Note: Here the solution came easily as 56=7*8. But you might have something like n*(n+1) < 45 normally in a question ... then you need to just do some mental math to find largest n, such that n*(n+1) < 45... This is how I would start thinking... 5*6 = 30, 6*7 = 42, so n = 6. So be prepared to do some easy mental math in the Test. From all the tests I have given, I have seen at-least one problem that has Sum of an arithmetic progression (S(n) is the simplest form of Arithmetic pregression) combined with inequality.

Hope this helps.
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Re: A little difficult math problem, pls help me  [#permalink]

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28 Jul 2011, 05:52
VeritasPrepKarishma wrote:
tracyyahoo wrote:
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

I have no idea how to solve this ,pls help~~~

the explanation says that suppose the sum of x is m, then 2m-1=1+2+3+.......+10.
M=28, so x = 7

why x = 7 can be concluded from M=28?

The moment I look at the question, I find myself jumping to option C to see if it works. Notice that adding numbers between 1 to 10 is really easy and quick. I wouldn't bother using Algebra in this question. But if we change this to pages numbered from 1 to 100 or something, I wouldn't bother with the options and go to Algebra instead. So knowing both the techniques is important. Let me discuss both of them.

Case 1: Using Options
If I add 1 to 6, I get 6*7/2 = 21
Adding 7+8+9+10 gives me 34 (You can simply add these OR say that the sum will be (3+2+1) less than 40 i.e. 6 less than 40 OR multiply the average (8.5) of the numbers by 4 i.e. 8.5*4 = 8*4 + .5*4 = 34)
Nope. We need to go to D and do the same there. We see that it works.

Case 2: Using Algebra
Sum of first x pages = x(x+1)/2
Sum of all pages = 10*11/2 = 55
Sum of all pages except first x pages = 55 - x(x+1)/2

Given: x(x+1)/2 = 1 + 55 - x(x+1)/2
x(x+1) = 56 = 7*8
x must be 7.

yep picking numbers is what helped me ... good that this question was for lesser number of pages
Else Probly we might have to resort to formulae
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Re: A little difficult math problem, pls help me  [#permalink]

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28 Jul 2011, 11:00

Quote:
the explanation says that suppose the sum of x is m, then 2m-1=1+2+3+.......+10.
M=28, so x = 7

Sum of x = m which is x(x+1)/2 = m

and M = 28 => x (x+1)/2 = 28
=> x(x+1) = 56. since x is positive we get x = 7.
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Re: A little difficult math problem, pls help me  [#permalink]

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28 Jul 2011, 12:56
I felt substitution was easy ..

sum of consecutive 10 numbers = 55

try substituting answers to satisfy the condition ..
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Re: A little difficult math problem, pls help me  [#permalink]

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09 Aug 2011, 13:07
Can't we just guess and check? I guessed 7 and got it right by chance, but even if we guessed 6 first it wouldn't be hard.

Guess: 6
1+6+2+5+3+4 = 7*3 = 21
7+8+9+10 = 19 + 16 = 35
Wrong, too small

Guess 7:
1+7+2+6+3+5+4 = 3*8+4= 28
8+9+10 = 27
Correct
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Re: A little difficult math problem, pls help me  [#permalink]

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09 Aug 2011, 19:36
ssarkar wrote:
How you getting 7 if you use case1 - using options-

If we add 1 to 7, then we get 7*8/2=28
Adding 8+9+10, we get 27. ( average = 27/3=9)
Then how we get 7. Is the fact that 10*11/2=55 and 28+27=55.

The condition that the answer should satisfy is this: 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

If x = 7, the sum of page number up to and including x is 1+2+3..+6+7 = 7*8/2 = 28 (The sum of n consecutive numbers starting from 1 is n(n+1)/2)
The sum of the page numbers following page number x = 8+9+10 = 27
Since 28 is one more than 27, our condition is satisfied. So answer is 7.
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Re: A little difficult math problem, pls help me  [#permalink]

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09 Aug 2011, 21:33
tracyyahoo wrote:
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

I have no idea how to solve this ,pls help~~~

the explanation says that suppose the sum of x is m, then 2m-1=1+2+3+.......+10.
M=28, so x = 7

why x = 7 can be concluded from M=28?

Another approach:

Sum of the page numbers from 1 to 10 = 55

The sum of the page numbers up to and including page number x of the report = a
The sum of the page numbers following page number x = b

a + b = 55 ..........i
a - b = 1.............ii

Add i and ii: a = 28 and b = 27.

what makes of b = 27? 10+9? no. 10+9+8? yes.

so x has to be 7.
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Re: A little difficult math problem, pls help me  [#permalink]

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21 Aug 2011, 19:09
1+2...+x = 1+(x+1)...10

1+2+..+x = 1 + (1+2...10)- (1+2+...x)

2(1+2...x) = 1+55

x(x+1) = 56

x=7
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Re: A little difficult math problem, pls help me  [#permalink]

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21 Aug 2011, 21:21
Acc. to the question, the page numbers are between 1 and 10

Let x be the point where we divide the page numbers in two parts

so no. of terms between 1 & x inclusive is x
and between 10 & x is 10-x

first term of series from 1 to x is 1 & last term is x
using sum of series formula s1=x(x+1)/2

similarly for the next set
first term is x+1 and last term is 10
s2=(10-x)(10+x+1)/2=(10-x)*(11+x)/2

acc. to given conditions
(10-x)*(11+x)+2=x(x+1)

112-x-x2=x2+x
2(x2+x-56)=0
x=7 or -8
x can't be negative as it's a page no.
so x-7

Hope it helps
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Re: A little difficult math problem, pls help me  [#permalink]

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23 Aug 2011, 09:06
Substituting the given values of X would be easy in this case.
Re: A little difficult math problem, pls help me   [#permalink] 23 Aug 2011, 09:06

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