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The sum of the first 100 even positive integers – divided by the sum o

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The sum of the first 100 even positive integers – divided by the sum o  [#permalink]

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New post 23 Jan 2016, 07:09
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The sum of the first 100 even positive integers – divided by the sum of the next 100 even positive integers – is equal to which of the following?

A. 1/3
B. 101/ 301
C. 51/151
D. 1/2
E. 1

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Re: The sum of the first 100 even positive integers – divided by the sum o  [#permalink]

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New post 09 Dec 2016, 07:01
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Sum of the first even integers: n(n+1)

Sum of the first 100-even integers: 100(101) = 10100
Sum of the first 200-even integers: 200(201) = 40200
Sum of the the even integers between the first 100 and the first 200: 40200-10100 = 30100

--> 10100/30100
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Re: The sum of the first 100 even positive integers – divided by the sum o  [#permalink]

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New post 23 Jan 2016, 08:40
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The first 100 positive numbers are
2, 4, 6, ....198, 200.
If we make 50 pairs made by the first and the last, second and the second last, we get:
2+200 + 4+198 + 6+196 + ... + 100+102
=50*(202)
=10100

same
202 - 400 = 30100

so 101/301

Ans B
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The sum of the first 100 even positive integers – divided by the sum o  [#permalink]

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New post 23 Jan 2016, 09:03
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shasadou wrote:
The sum of the first 100 even positive integers – divided by the sum of the next 100 even positive integers – is equal to which of the following?

A. 1/3
B. 101/ 301
C. 51/151
D. 1/2
E. 1


Hi,
i will tell you a simpler way of doing these Qs..

Remember sum of consecutive numbers can be found by finding average of these numbers * total number..

I..... sum of the first 100 even positive integers=:-
first even integer=2, and 100th even integer=200..
\(sum = \frac{2+200}{2}*100=101*100\)..

II....sum of the next 100 even positive integers:-
first even integer=202, and 100th even integer=400..
\(sum = \frac{202+400}{2}*100=301*100\)..

answer \(=\frac{101*100}{301*100}= \frac{101}{301}\)

B
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The sum of the first 100 even positive integers – divided by the sum o  [#permalink]

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New post 28 Jul 2017, 20:27
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First 100 even integers: 2...200
Sum of first 100 even integers: (a1+an)n/2 = (2+200)*100/2=202*50
Next 100 even integers: 202~400
Sum of next 100 even integers: (a1+an)n/2 = (202+400)*100/2=602*50
Ratio of the sums: 202/402 = 101/301
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Re: The sum of the first 100 even positive integers – divided by the sum o  [#permalink]

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New post 28 Jul 2017, 22:52
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shasadou wrote:
The sum of the first 100 even positive integers – divided by the sum of the next 100 even positive integers – is equal to which of the following?

A. 1/3
B. 101/ 301
C. 51/151
D. 1/2
E. 1


Sum of first 100 even positive integers = 2 + 4 +6 +8 +.... + 200 = 2(1+2+3+4+..... +100) = 2* (100*101/2) = 100*101
Sum of next 100 even positive integers = 202 + 204 +206 +..... +400 = 2 (101 + 102 + ... + 200) = 2 ((1+2+3+.....+200)-(1+2+3+...+100))
= 2 [200* 201/2 - 100*101/2] = 200*201 - 100*101

Ratio = 100*101/ (200*201 - 100*101)= 101/(402 - 101) = 101/301

Answer B
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Re: The sum of the first 100 even positive integers – divided by the sum o  [#permalink]

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New post 28 Jul 2017, 23:25
shasadou wrote:
The sum of the first 100 even positive integers – divided by the sum of the next 100 even positive integers – is equal to which of the following?

A. 1/3
B. 101/ 301
C. 51/151
D. 1/2
E. 1


Ans(B) 101/301

Given that, p = (2+4+6.....+200 ) [First 100 Terms] / (202+204....+400) [Next 100 Terms]
p= 2(1+2+...+100) / 2 ( 101+ 102+200)
P= (1+2+....+100) / (101+102+...+200) ( Cancelling out 2)
After this we can, we can solve this using 2 methods.
Method1:
To calculate the sum of a continuous series, with a same difference between any two continuous numbers, simply take the average of first & last number and multiple it by the number of terms. For ex : 4+7+10+13= (average of 13&4, multiplied by 4) = 8.5*4 = 34
Now in the original problem,
p = Average of (1&100)*100 / Average of (101&200)*100
p=(1+100)/(101+200) (Cancelling out 2, not shown here)
p=101/301 Answer.

Method2: [This is a formula dependent approach and lengthy one also]
For Equation, P= (1+2+....+100) / (101+102+...+200)
We can use the formula for the sum of first "n" natural numbers : S= n(n+1)/2. But this formula works when the series starts from 1.
So, using this formula, we can calculate the sum for the numerator part of the above equation.
But for denominator, first we need to calculate the some of 200 numbers and then subtract the sum of first 100.
And this double calculation makes this approach slightly lengthy.

Thanks!
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Re: The sum of the first 100 even positive integers – divided by the sum o  [#permalink]

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New post 02 Nov 2017, 14:55
paidlukkha wrote:
The first 100 positive numbers are
2, 4, 6, ....198, 200.
If we make 50 pairs made by the first and the last, second and the second last, we get:
2+200 + 4+198 + 6+196 + ... + 100+102
=50*(202)
=10100

same
202 - 400 = 30100

so 101/301

Ans B



why don't we use 0?
isn't it even integer??
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Re: The sum of the first 100 even positive integers – divided by the sum o  [#permalink]

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New post 02 Nov 2017, 21:28
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soodia wrote:
paidlukkha wrote:
The first 100 positive numbers are
2, 4, 6, ....198, 200.
If we make 50 pairs made by the first and the last, second and the second last, we get:
2+200 + 4+198 + 6+196 + ... + 100+102
=50*(202)
=10100

same
202 - 400 = 30100

so 101/301

Ans B



why don't we use 0?
isn't it even integer??


"The sum of the first 100 even positive integers..."

Yes, 0 is an even integer but it's neither positive nor negative.
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The sum of the first 100 even positive integers – divided by the sum o  [#permalink]

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New post 05 May 2018, 11:14
Here I would try to solve this with the help of a formula.
'Sum of the first N even natural numbers = N(N+1)'

So, Sum of 1st 100 even numbers = 100(101), and Sum of the 1st 200 even numbers =200(201)
Therefore, the sum of the Next 100 even numbers 'Minus' the 1st 100 even numbers = 200(201)-100(101) which is equal to 30100 OR 100(301).

So, The sum of the first 100 even positive integers – divided by the sum of the next 100 even positive integers equal to

100(101)/ 100(301) =101/301.


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Re: The sum of the first 100 even positive integers – divided by the sum o  [#permalink]

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New post 02 Oct 2018, 10:06
dassamik89 wrote:
Here I would try to solve this with the help of a formula.
'Sum of the first N even natural numbers = N(N+1)'

So, Sum of 1st 100 even numbers = 100(101), and Sum of the 1st 200 even numbers =200(201)
Therefore, the sum of the Next 100 even numbers 'Minus' the 1st 100 even numbers = 200(201)-100(101) which is equal to 30100 OR 100(301).

So, The sum of the first 100 even positive integers – divided by the sum of the next 100 even positive integers equal to

100(101)/ 100(301) =101/301.


Kudos would be just fine!


Thanks for the explanation.
Just sharing another way to look at this same question:

The first 100 even numbers are 2, 4, 6, ... 200.
The next 100 even numbers are 202, 204, 206, ... 400.

Since the numerator and denominator have exactly the same number of addends, we can simply divide the average of those terms in the numerator by the average of those terms in the denominator.
(In other words, the numerator sum will equal 100 times the average of its addends, and the denominator sum will equal 100 times the average of its addends, and the 100s will cancel.)

Since the terms in both the numerator and the denominator constitute evenly spaced sets, these averages can be found simply by averaging the first and last terms of each set.

So the average of the terms in the numerator is (2+200)/2 = 101, and the average of the terms in the denominator is (202+400)/2 = 301.

So the fraction simplifies to 101/301.
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Re: The sum of the first 100 even positive integers – divided by the sum o  [#permalink]

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New post 03 Oct 2018, 17:54
shasadou wrote:
The sum of the first 100 even positive integers – divided by the sum of the next 100 even positive integers – is equal to which of the following?

A. 1/3
B. 101/ 301
C. 51/151
D. 1/2
E. 1


The first 100 positive even numbers constitute an evenly-spaced set with the first number 2 and last number 200. The average of any evenly-spaced set is (smallest + largest)/2. We can multiply this average by the number of numbers in the set to get the sum:
(2 + 200)/2 x 100 = 101 x 100

The next set - the next 100 even positive integers - is also an evenly-spaced set, with first number 202 and last number 400. Using the same procedure, we calculate the sum of that set as:

(202 + 400)/2 x 100 = 301 x 100

So the sum of the first set divided by the sum of the second set is:

(101 x 100)/(301 x 100) = 101/301

Answer: B
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Re: The sum of the first 100 even positive integers – divided by the sum o   [#permalink] 03 Oct 2018, 17:54
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