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Bunuel
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


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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..
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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:
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Bunuel
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


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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:
Attachment:
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gpp-s_img13.png [ 2.59 KiB | Viewed 12469 times ]
Answer = (B).
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Bunuel
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


Kudos for a correct solution.

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 = \(N/2 (2A+ (N-1)*D)\)
\(50/2 * (2*5 + (50 -1) *10 )\)
\(= 50*250 = 12500\)
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Bunuel
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


Kudos for a correct solution.


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
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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.
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