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

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e-GMAT Representative
Joined: 04 Jan 2015
Posts: 3230
Set S contains consecutive natural numbers from 1 to 100, in increasin  [#permalink]

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08 Sep 2018, 07:10
00:00

Difficulty:

45% (medium)

Question Stats:

69% (03:01) correct 31% (02:33) wrong based on 72 sessions

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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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Joined: 02 Sep 2015
Posts: 40
Set S contains consecutive natural numbers from 1 to 100, in increasin  [#permalink]

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08 Sep 2018, 13:57
2
4
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

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

##### General Discussion
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Joined: 29 Jun 2018
Posts: 5
Location: India
GMAT 1: 720 Q49 V39
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Re: Set S contains consecutive natural numbers from 1 to 100, in increasin  [#permalink]

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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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Joined: 07 Dec 2014
Posts: 1230
Re: Set S contains consecutive natural numbers from 1 to 100, in increasin  [#permalink]

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08 Sep 2018, 09:27
EgmatQuantExpert wrote:
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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Joined: 11 Aug 2016
Posts: 364
Re: Set S contains consecutive natural numbers from 1 to 100, in increasin  [#permalink]

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08 Sep 2018, 09:51
3
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 $$2^4$$

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.

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
e-GMAT Representative
Joined: 04 Jan 2015
Posts: 3230
Re: Set S contains consecutive natural numbers from 1 to 100, in increasin  [#permalink]

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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.

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Joined: 11 Mar 2018
Posts: 157
Re: Set S contains consecutive natural numbers from 1 to 100, in increasin  [#permalink]

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14 Sep 2018, 00:51
1
EgmatQuantExpert wrote:
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

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,
$$n = 15$$

Re: Set S contains consecutive natural numbers from 1 to 100, in increasin   [#permalink] 14 Sep 2018, 00:51
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