In the set of positive integers from 1 to 500, what is the sum of all

Question

In the set of positive integers from 1 to 500, what is the sum of all the odd multiples of 5?

A. 10,000 
B. 12,500
C. 17,500
D. 22,500
E. 25,000

mehrdadtaheri92 · Accepted Answer

ANSWER IS B....

AT the first step , lets define the set : ODD MULTIPLES OF 5 ARE : 5 , 15,25,..... as here we have finite numbers between 1 and 500 , so we have to define

the first and the last element of the set. here we have : 5 as the first element and 495 as the last element of the set.

so we have a set of these numbers like this : 5,15,25,35,....,495 .here we are asked to find the sum of these numbers

in order to find the sum of a set we must to find THE NUMBER OF SET AND ALSO FIND THE AVERAGE OF THE SET and then MULTIPLE THESE TO NUMBER .

here we have evenly spaced set of number (as every element is 10 more than the previous element) so we have a nice formula to find the number of elements in the set

the formula is :   +1 , so here we have : ( 495-5 ) /10 +1 = 490/10 +1 = 49 +1 =50 ( the number of elements in the given set)

TO find the average in the evenly spaced set we should use this formula : the last element + the first one / 2 = 495 +5 / 2 = 500/2 =250

FINAL STEP : multiple these to numbers : 250 * 50=12500 ( the sum of elements in the set ) , SO answer is B...

with the best regards,

LaxAvenger · Answer

reduce 1 - 500 to 1 - 50

5 - 15 - 25 - 35 - 45 are valid multiples. Add them --> 125

to get from 1 - 50 to 1 - 500 one has just to change the dimension, thus answer B

chetan2u · Answer

hi ,
there are two methods discussed above..
one more which is pure mathematical based on formula (can be very useful in saving time on some ocassions) is..
sum=5+15+25+.....+495..=5(1+3+5+7+.....99)..
sum of first n odd numbers is  n^2 , here the terms are 50...
so sum=5( 50^2 )=5*2500=12500..
ans B..

EMPOWERgmatRichC · Answer

Hi All,

These type of 'large sum' questions can be approached in a variety of ways, but there will always be a pattern involved. We can use BUNCHING to solve this problem.

Here, we're asked for all of the ODD multiples of 5 from 1 to 500....

To start, take the 1st multiple of 5 involved and add it to the last multiple of 5 involved.

5+495 = 500

Next, take the 2nd multiple of 5 involved and the second-to-last multiple of 5 involved....

15 + 485 = 500

This proves that there are a certain number of "500s" in this sequence. We have to figure out how many and if there is an "unpaired" term in the middle. Given the restrictions in this prompt, we're taking 1 number out of every 10 consecutive numbers....

The '5' out of 1 through 10
The '15' out of 11 through 20
The '25' out of 21 through 30,
Etc.

Since we're dealing with the numbers from 1 to 500, we're dealing with 50 sets of 10. This means that we are taking the sum of 50 numbers....

50 numbers --> 25 pairs (and each pair totals 500)

25(500) = 12,500

Final Answer:  B

GMAT assassins aren't born, they're made,
Rich

Bunuel · Answer

MAGOOSH OFFICIAL SOLUTION: Let’s think about the terms in this sequence: 5, 15, 25, 35, …., 485, 495 The first term is 5 and the last is 495. There are 100 multiples of 5 from 1 to 500, so there are 50 odd multiples and 50 even multiple. The sum is: gpp-s_img13.png Answer = (B).

Lucky2783 · Answer

each set 1-100, 101-200, 201-30, 301-400, 401-500 will have 20 numbers which will be multiple of 5 . half of them will be odd multiples and half of them will be even . 
so we have total 50 numbers which are odd and 50 which are even .
Lets take odd sequence as that's what the question ask for

First term = 5 
Last term = 495 
common difference = 10 
SUM of a AP =  N/2 (2A+ (N-1)*D)  
 50/2 * (2*5 + (50 -1) *10 ) 
 = 50*250 = 12500

ScottTargetTestPrep · Answer

The odd multiples of 5 from 1 to 500 are 5, 15, 25, …, 495.

Because this is an evenly-spaced set, the average of these multiples of 5 is (495 + 5)/2 = 250.

The number of these multiples of 5 is (495 - 5)/10 + 1 = 50.

So the sum is 250 x 50 = 12,500.

Answer: B

ueh55406 · Answer

it's a direct formula plug-in question:

Sn=n/2* {2a1+ (n-1)*d}

n=500
a1=5
d=10

Odd multiples of 5= 5*1, 5*3, 5*5.... difference b/w each number is ten.

bumpbot · Answer

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In the set of positive integers from 1 to 500, what is the sum of all

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