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

Question

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

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e-GMAT Question of the Week #13

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

e-GMAT         Question of the Week #13

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magbot · Accepted Answer

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

adstudy · Answer

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} *

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} *   
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} *   
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

prashant17p · Answer

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.

gracie · Answer

n/ n=16/85➡
202n=3030➡
n=15
C

GmatDaddy · Answer

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)+....... 
=  - (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

EgmatQuantExpert · Answer

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)}{ } = \frac{16}{85} 
 o If we observe the denominator is in the form of  a^2 – b^2  
• Implies,  \frac{ }{ }= 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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MFMarwat · Answer

Hi
I tried solving this Qs multiple times and also read the solutions being provided but I couldn't understand what approach is being followed. Can you please elaborate it stepwise  EgmatQuantExpert  @Bunnel
Regards

Nonbiryani · Answer

Since we know the ratio is 16:85 and the sum of first n even numbers is n(n+1), 16k = n(n+1)

We need a multiple of 16 such that it is also consecutive. The only option is 15 such that 15*16 also ensures it is a multiple of 16 while being consecutive.

adityaprateek15 · Answer

Bunuel   Krunaal  I believe the value in Option A got removed inadvertently.

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

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