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Set S contains consecutive natural numbers from 1 to 100, in increasin

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Set S contains consecutive natural numbers from 1 to 100, in increasin  [#permalink]

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New post 08 Sep 2018, 07:10
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e-GMAT Question of the Week #13


Set S contains consecutive natural numbers from 1 to 100, in increasing order. \(S_1\) and \(S_2\) denote the sum of first n even numbers and last n odd numbers of set S, respectively. What is the value of n, if \(S_1: S_2\) = 16: 85?

    A. 5
    B. 10
    C. 15
    D. 20
    E. 25


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Re: Set S contains consecutive natural numbers from 1 to 100, in increasin  [#permalink]

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New post 08 Sep 2018, 08:13
The sum of the first n even numbers is n*(n+1) and the ratio of S1:S2 = 16: 85 thus S1 has to be a multiple of 16. Substituting the values of n available we get n=15.
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Re: Set S contains consecutive natural numbers from 1 to 100, in increasin  [#permalink]

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New post 08 Sep 2018, 09:27
EgmatQuantExpert wrote:
e-GMAT Question of the Week #13


Set S contains consecutive natural numbers from 1 to 100, in increasing order. \(S_1\) and \(S_2\) denote the sum of first n even numbers and last n odd numbers of set S, respectively. What is the value of n, if \(S_1: S_2\) = 16: 85?

    A. 5
    B. 10
    C. 15
    D. 20
    E. 25


[(2+2n)/2]n/[99+(101-2n)/2]n=16/85➡
202n=3030➡
n=15
C
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Re: Set S contains consecutive natural numbers from 1 to 100, in increasin  [#permalink]

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New post 08 Sep 2018, 09:51
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EgmatQuantExpert wrote:
Set S contains consecutive natural numbers from 1 to 100, in increasing order. \(S_1\) and \(S_2\) denote the sum of first n even numbers and last n odd numbers of set S, respectively. What is the value of n, if \(S_1: S_2\) = 16: 85?

    A. 5
    B. 10
    C. 15
    D. 20
    E. 25


Sum of first n natural numbers is \frac{n*(n+1)}{2}.
Sum of first N even natural numbers can be calculated below:
2+4+6+.....+2n
take 2 common
2(1+2+3+....+n)
=n(n+1)

Now, calculating sum of first n even nos is easy.
However, calculating the sum of last n odd numbers is not
99+97+95+93.It can be done but takes a lot of time.
Let us denote the sum of last n odd numbers by So

Given,
\(\frac{Se}{So}=\frac{16}{85}\)
\(\frac{n(n+1)}{so}=\frac{16}{85}\)

\(85*n(n+1)=16*So\)
\(16=2^4\)
this implies, \(85*n(n+1)=2^4*So\)
since 85 is odd, and So must be an integer. therefore RHS must contain atleast \(2^4\)
This implies that n(n+1) should be such that it provides at least \(2^4\)

Checking Answer choices. for n:
A. 5 - \(5*6\). Drop it.
B. 10 - \(10*11\). Drop it
C. 15 - \(15*16\). Bingo. Should be our answer, but just for precaution, we'll check the rest of the answer choices.
D. 20 - \(21*26\). Drop it.
E. 25 - \(26*26\). Drop it.

Hence our answer: C

Please note that we can calculate the sum of last n odd numbers the following way:
99+97+95+93+.......nth odd no
=(100-1)+(100-3)+(100-5)+.......[100-(2n-1)]
=[100*n] - (1+2+3+......nth term(ie 2n-1)....................a

(1+2+3+......nth term(ie 2n-1)
= sum of first 2n natural no - sum of first n even no.
\(= \frac{2n(2n+1)}{2}-n(n+1)\)
\(=n(2n+1)-n(n+1)\)
\(=2n^2 + n - n^2 -n\)
\(=n^2\)

Using a,
we get, sum of last n odd numbers = 100n-n^2
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Set S contains consecutive natural numbers from 1 to 100, in increasin  [#permalink]

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New post 08 Sep 2018, 13:57
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Took a while to figure out the method.

Sum of any arithmetic progression of n numbers = \(n/2*(first number + last number)\)
S1 = Arithmetic progression of first n even numbers = \(n/2*(2 + 2n)\) = \(n(n+1)\)
S2 = Arithmetic progression of last n odd numbers = \(n/2*(99 + (100 -2n + 1)\) = \(n(100-n)\) where 99 is last odd number & (100-2n + 1) is the first odd number

\(S1/S2 = n(n+1)/n(100-n) = 16/85\)
\(85n+85=1600-16n\)
\(101n=1515\)
\(n=15\)

Answer: C
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Re: Set S contains consecutive natural numbers from 1 to 100, in increasin  [#permalink]

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New post 14 Sep 2018, 00:25

Solution


Given:
    • A set, S, which contains all the natural numbers from 1 to 100, in increasing order.
    • \(S_1\) denotes the sum of first n even numbers of the set, S.
    • \(S_2\) denotes the sum of last n odd numbers of the set, S.
    • \(S_1: S_2\) = 16 : 85


To find:
    • The value of n

Approach and Working:
\(S_1\) = sum of first n even numbers = 2 + 4 + 6 + 8 + 10 + …. + 2n

    • Now, if we take 2 common from all the terms, we get,
      o \(S_1 = 2(1 + 2 + 3 + 4 + 5 + …. + n)\)
      o The sum of first ‘n’ natural numbers, \((1+ 2 + 3 + 4 + 5 + …. + n) = \frac{n(n+1)}{2}\)
      o Which implies, \(S_1 = 2*n*\frac{(n+1)}{2} = n(n+1)\) ………………… (1)

As we have total 100 consecutive numbers in S, out of those 100 numbers, 50 will be even and rest 50 will be odd.

\(S_2\) = sum of last n odd numbers = sum of all odd numbers from 1 to 100 – sum of the first (50 – n) odd numbers

    • The sum of first n odd natural numbers = \(n^2\)
      o Implies, the sum of all 50 odd numbers from 1 to \(100 = 50^2\) (Since, the total number of odd numbers from 1 to 100 = 50)
    • Similarly, the sum of the first (50 – n) odd numbers = \((50 – n)^2\)
    • Thus, \(S_2 = 50^2 - (50 – n)^2\) …………………………… (2)

Substituting, (1) and (2), in \(S_1: S_2 = 16 : 85\), we get,

    • \(\frac{n(n + 1)}{[50^2 - (50 – n)^2]} = \frac{16}{85}\)
      o If we observe the denominator is in the form of \(a^2 – b^2\)
    • Implies, \(\frac{[n *(n + 1)]}{[(100 - n) * (n)]}= 16/85\)
      o \(\frac{(n + 1)}{(100 - n)} = \frac{16}{85} = \frac{(15 + 1)}{(100 - 15)}\)

Thus, the value of n = 15

Hence, the correct answer is option C.

Answer: C

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Re: Set S contains consecutive natural numbers from 1 to 100, in increasin  [#permalink]

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New post 14 Sep 2018, 00:51
EgmatQuantExpert wrote:
e-GMAT Question of the Week #13


Set S contains consecutive natural numbers from 1 to 100, in increasing order. \(S_1\) and \(S_2\) denote the sum of first n even numbers and last n odd numbers of set S, respectively. What is the value of n, if \(S_1: S_2\) = 16: 85?

    A. 5
    B. 10
    C. 15
    D. 20
    E. 25


Image


Here we know set includes first 100 natural numbers.

Now \(S_1\) = Sum of first n Natural even numbers in set
and, \(S_2\) = Sum of last n Natural odd numbers in set

Using Arithmetic Progression here,

\(S_n = \frac{n}{2} * [2a + (n-1)d]\)

Therefore, For \(S_1\) we have "a = 2" and "d = 2" (Series will be like this -> 2,4,6,.....100)
\(S_1 = \frac{n}{2} * [2*2 + (n-1)*2 ]\)
On Solving,
\(S_1 = n * (n+1)\) --------- (1)

For \(S_2\) we have "a = 99" and "d = -2" (Series will be like this -> 99,97,95,.....3,1)
\(S_2 = \frac{n}{2} * [2*99 + (n-1)*(-2) ]\)
On Solving,
\(S_2 = n * (100-n)\) --------- (2)

Now from question we know,
\(\frac{S_1}{S_2} = \frac{16}{85}\) --------- (3)

Using equations 1, 2 and 3
\(\frac{n*(n + 1)}{n*(100-n)}\) = \(\frac{16}{85}\)
On solving,
\(n = 15\)

Answer is C
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