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What is the sum of odd integers from 35 to 85, inclusive?

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What is the sum of odd integers from 35 to 85, inclusive?  [#permalink]

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03 Jul 2016, 10:27
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What is the sum of odd integers from 35 to 85, inclusive?

A) 1,560
B) 1,500
C) 1,240
D) 1,120
E) 1,100

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Re: What is the sum of odd integers from 35 to 85, inclusive?  [#permalink]

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03 Jul 2016, 11:08
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Bunuel wrote:
What is the sum of odd integers from 35 to 85, inclusive?

A) 1,560
B) 1,500
C) 1,240
D) 1,120
E) 1,100

Number of odd integers = (85-35)/2 + 1
= 50/2 + 1
= 26
Sum of odd integers = (35+85)/2 * 26
= 60 * 26
= 1560
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Re: What is the sum of odd integers from 35 to 85, inclusive?  [#permalink]

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03 Jul 2016, 23:18
2
A

take average and multiply by number of terms

here number of odd terms=26 and average is 60.

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Re: What is the sum of odd integers from 35 to 85, inclusive?  [#permalink]

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14 Apr 2017, 10:00
1
Skywalker18 wrote:
Bunuel wrote:
What is the sum of odd integers from 35 to 85, inclusive?

A) 1,560
B) 1,500
C) 1,240
D) 1,120
E) 1,100

Number of odd integers = (85-35)/2 + 1
= 50/2 + 1
= 26
Sum of odd integers = (35+85)/2 * 26
= 60 * 26
= 1560

Hi Skywalker18,

I do understand the following on this exercise:

1. We're dealing with consecutive integers so I can find the average by: $$\cfrac { First\quad Number+Last\quad Number }{ 2 }$$
2. To find the range in a set of numbers for which both end points are inclusive, we: $$\left( Hightest\quad Number-Lowest\quad Number \right) +1$$

I don't understand why you can simply divide the range by 2 in $$\cfrac { \left( Hightest\quad Number-Lowest\quad Number \right) }{ 2 } +1$$ to find the number of odd numbers. Would it be the same if I wanted to find the number of even numbers?

Thanks'
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Re: What is the sum of odd integers from 35 to 85, inclusive?  [#permalink]

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02 Jun 2017, 20:53
He has divided by2 as we need to find out the number of odd numbers.(85-35)+1 is total number which includes even and odd
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Re: What is the sum of odd integers from 35 to 85, inclusive?  [#permalink]

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01 Jul 2017, 11:18
1
Notice that by adding pairs of numbers we get a sum of 120. We need to find the number of integers between 35 and 85 inclusive.
The number of integers is $$\frac{(85-35)}{2}+1$$ = 26

So we add 120 , 26 times to get 120*26, dividing this by 2 , we get 1560.
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What is the sum of odd integers from 35 to 85, inclusive?  [#permalink]

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01 Jul 2017, 12:20
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Bunuel wrote:
What is the sum of odd integers from 35 to 85, inclusive?

A) 1,560
B) 1,500
C) 1,240
D) 1,120
E) 1,100

Different approach (for kicks )

We want: 35 + 37 + 39 + 41 + . . . . 79 + 81 + 83 + 85

Add the numbers in pairs, starting from the outside and working towards the middle, we get:
35 + 37 + 39 + 41 + . . . . 79 + 81 + 83 + 85 = (35 + 85) + (37 + 83) + (39 + 81) + . . .
= (120) + (120) + (120) + (120) + .....
= 120 x some integer

So, the desired sum must be DIVISIBLE by 120
Only one answer choice is divisible by 120

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Re: What is the sum of odd integers from 35 to 85, inclusive?  [#permalink]

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26 May 2018, 22:44
Hi Skywalker18 niks18 gmatbusters

What is the sum of odd integers from 35 to 85, inclusive?

Quote:
Number of odd integers = (85-35)/2 + 1
= 50/2 + 1
= 26

Can you explain why did you divide by 2?
If I want to count no. of terms from 1 to 10,
I would perform (Last term - First term) + 1.
But is not above approach only valid for consecutive terms
and sequence starting with 1?

Quote:
Sum of odd integers = (35+85)/2 * 26
= 60 * 26
= 1560

I guess you used that for consecutive numbers, mean = median = (First term + last term) / 2.
Let me know if we are on same page
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Re: What is the sum of odd integers from 35 to 85, inclusive?  [#permalink]

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26 May 2018, 22:58
1
1

Your query: Can you explain why did you divide by 2?
Actually we divide by 2 because the difference between the consecutive odd numbers = 2.

In fact we have to count the multiple of any number say 7, multiple of 7 between 7 and 98 both inclusive.
it will be (98-7)/7 +1

Always divide by common difference
( this is as per Arithmetic progression , AP)

As per you "If I want to count no. of terms from 1 to 10, I would perform (Last term - First term) + 1."

you are right here, the common difference = 1. so dividing by 1 is not necessary.

Your Query :I guess you used that for consecutive numbers, mean = median = (First term + last term) / 2.

yes you are right, in fact, it is true for any set of equally spaced numbers (AP)

Hi Skywalker18 niks18 gmatbusters

What is the sum of odd integers from 35 to 85, inclusive?

Quote:
Number of odd integers = (85-35)/2 + 1
= 50/2 + 1
= 26

Can you explain why did you divide by 2?
If I want to count no. of terms from 1 to 10,
I would perform (Last term - First term) + 1.
But is not above approach only valid for consecutive terms
and sequence starting with 1?

Quote:
Sum of odd integers = (35+85)/2 * 26
= 60 * 26
= 1560

I guess you used that for consecutive numbers, mean = median = (First term + last term) / 2.
Let me know if we are on same page

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What is the sum of odd integers from 35 to 85, inclusive?  [#permalink]

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27 May 2018, 01:55
GMATPrepNow wrote:
Bunuel wrote:
What is the sum of odd integers from 35 to 85, inclusive?

A) 1,560
B) 1,500
C) 1,240
D) 1,120
E) 1,100

Quote:
Different approach (for kicks )

We want: 35 + 37 + 39 + 41 + . . . . 79 + 81 + 83 + 85

Add the numbers in pairs, starting from the outside and working towards the middle, we get:
35 + 37 + 39 + 41 + . . . . 79 + 81 + 83 + 85 = (35 + 85) + (37 + 83) + (39 + 81) + . . .
= (120) + (120) + (120) + (120) + .....
= 120 x some integer

So, the desired sum must be DIVISIBLE by 120
Only one answer choice is divisible by 120

This observation is interesting but I would take a separate approach after the first two steps:-

Add the numbers in pairs, starting from the outside and working towards the middle, we get:
35 + 37 + 39 + 41 + . . . . 79 + 81 + 83 + 85 = (35 + 85) + (37 + 83) + (39 + 81) + . . .
= (120) + (120) + (120) + (120) + .....
= 120 x some integer

Some integer is 13 because the average of extreme values is 60. So for every odd number from 35 to 60 there will be a pair for it from 60 to 85. Its easy to count that the number of pairs will be 13. therefore the answer is 120*13. [Only reason I mentioned this approach was because sometimes dividing 5 numbers could be time consuming. Counting the number of pairs to 13 was simple and straight forward)

Edit: Formating

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Re: What is the sum of odd integers from 35 to 85, inclusive?  [#permalink]

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15 Sep 2018, 04:51
Check this out hope it helps
https://gmatclub.com/forum/how-many-mul ... ml#p730075
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Re: What is the sum of odd integers from 35 to 85, inclusive?  [#permalink]

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15 Sep 2018, 04:56
gmatbusters wrote:

Your query: Can you explain why did you divide by 2?
Actually we divide by 2 because the difference between the consecutive odd numbers = 2.

In fact we have to count the multiple of any number say 7, multiple of 7 between 7 and 98 both inclusive.
it will be (98-7)/7 +1

Always divide by common difference
( this is as per Arithmetic progression , AP)

As per you "If I want to count no. of terms from 1 to 10, I would perform (Last term - First term) + 1."

you are right here, the common difference = 1. so dividing by 1 is not necessary.

Your Query :I guess you used that for consecutive numbers, mean = median = (First term + last term) / 2.

yes you are right, in fact, it is true for any set of equally spaced numbers (AP)

Hi Skywalker18 niks18 gmatbusters

What is the sum of odd integers from 35 to 85, inclusive?

Quote:
Number of odd integers = (85-35)/2 + 1
= 50/2 + 1
= 26

Can you explain why did you divide by 2?
If I want to count no. of terms from 1 to 10,
I would perform (Last term - First term) + 1.
But is not above approach only valid for consecutive terms
and sequence starting with 1?

Quote:
Sum of odd integers = (35+85)/2 * 26
= 60 * 26
= 1560

I guess you used that for consecutive numbers, mean = median = (First term + last term) / 2.
Let me know if we are on same page

I think you are missing one point here
Always divide by common difference ( this is as per Arithmetic progression , AP)
this is not true we have to divide by common difference of first and last multiple of the integer whose multiples are being found out in this case 7 and 98 are by chance multiples of 7 lets say multiples b/w 7 and 99 even then we do 98-7
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Re: What is the sum of odd integers from 35 to 85, inclusive?  [#permalink]

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23 Oct 2018, 05:41
Bunuel wrote:
What is the sum of odd integers from 35 to 85, inclusive?

A) 1,560
B) 1,500
C) 1,240
D) 1,120
E) 1,100

Because the numbers are evenly spaced we can find the average.

85+ 35 = 120/2 = 60

# of terms = 85 - 35 = 50/2 = 25 + 1 = 26

26 * 60 = 1,560

Re: What is the sum of odd integers from 35 to 85, inclusive? &nbs [#permalink] 23 Oct 2018, 05:41
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