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M24-18

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M24-18 [#permalink]

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New post 16 Sep 2014, 01:21
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A
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C
D
E

Difficulty:

  5% (low)

Question Stats:

88% (00:32) correct 12% (00:17) wrong based on 100 sessions

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Expert Post
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Re M24-18 [#permalink]

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New post 16 Sep 2014, 01:21
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Re: M24-18 [#permalink]

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New post 23 Aug 2015, 08:54
I think this is a poor-quality question and I agree with explanation. I was confused by the stem. Is it possible to make it more obvious that we are supposed to find consecutive integers from 1 to 49 and not some fancy exponents ?
sorry, I might be tired from GMAT :)
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Re: M24-18 [#permalink]

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New post 23 Aug 2015, 10:32
Barnal wrote:
I think this is a poor-quality question and I agree with explanation. I was confused by the stem. Is it possible to make it more obvious that we are supposed to find consecutive integers from 1 to 49 and not some fancy exponents ?
sorry, I might be tired from GMAT :)


Edited as suggested. Hope it's better now.
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Re: M24-18 [#permalink]

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New post 10 Oct 2017, 05:26
What is 1+2+3+...+..+48+49 ?

A. 1168
B. 1176
C. 1195
D. 1215
E. 1225

An alternate approach:

Average = sum of integers x No. of integers

Average = Last term + First term / 2

Average = 25
No of terms = 49

25x49=1225

E is the answer.
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Re: M24-18 [#permalink]

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New post 10 Oct 2017, 06:00
Sum of consecutive integers (starting at 1): n(n+1)/2

49(49+1)/2 = 49*50/2 = 1225


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Re: M24-18 [#permalink]

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New post 06 Feb 2018, 10:31
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Hi,

1,2,3..49 is an AP series.

So for that we can take the sum as averageXno of integers

Here average = (first term+last term)/2--> 1+49/2=25

Sum =25x49 = 1225
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Re: M24-18 [#permalink]

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New post 03 Mar 2018, 02:00
Agree with previous calculations.

You also don't really need to count how many multiples of 50 there are (e.g. (1+49) + (2+48) etc).

You can realise very quickly that 25 will be "on its own" (as 25+25 would equal 50 but obviously we can't have 2 copies of the same number).

Therefore the final answer will either end with 25 (if there are an even number of multiples of 50) or 75 (if there are an odd number of multiples of 50).

Only 1225 satisfies this requirement so you can almost do this one "at a glance".
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Re: M24-18 [#permalink]

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New post 01 May 2018, 05:41
+1 for option E. Use sum of consecutive numbers formula. The answer comes to 1225. Option E it is !!
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Re: M24-18   [#permalink] 01 May 2018, 05:41
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