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M24-18

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Math Expert
Joined: 02 Sep 2009
Posts: 49320

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16 Sep 2014, 01:21
00:00

Difficulty:

5% (low)

Question Stats:

89% (00:30) correct 11% (00:17) wrong based on 114 sessions

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What is $$1 + 2 + 3 + ... + 48 + 49$$ ?

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

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Math Expert
Joined: 02 Sep 2009
Posts: 49320

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16 Sep 2014, 01:21
Official Solution:

What is $$1 + 2 + 3 + ... + 48 + 49$$ ?

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

$$1 + 2 + 3 + ... + 48 + 49 = (1 + 49) + (2 + 48) + (3 + 47) + ... + (24 + 26) + 25 = 50*24 + 25 = 1225$$

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Joined: 21 Jan 2015
Posts: 21

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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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Joined: 02 Sep 2009
Posts: 49320

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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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Joined: 15 Nov 2016
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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

Intern
Joined: 02 Feb 2016
Posts: 27
Location: United States
GMAT 1: 710 Q49 V38
GPA: 3.5

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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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Manager
Joined: 17 Jul 2017
Posts: 116
Location: India
GMAT 1: 690 Q50 V33
GMAT 2: 750 Q50 V40
WE: Engineering (Transportation)

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06 Feb 2018, 10:31
1
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
Intern
Joined: 06 Jan 2018
Posts: 16
GMAT 1: 790 Q50 V51

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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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Joined: 08 Jun 2015
Posts: 503
Location: India
GMAT 1: 640 Q48 V29
GMAT 2: 700 Q48 V38
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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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Intern
Joined: 22 Jan 2018
Posts: 19

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20 Jun 2018, 07:53
This is an arithmetic sequence. Sum = n x 2 first term + (n-1)d / 2 !
Re: M24-18 &nbs [#permalink] 20 Jun 2018, 07:53
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