GMAT Changed on April 16th - Read about the latest changes here

 It is currently 28 May 2018, 01:00

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# M24-18

Author Message
TAGS:

### Hide Tags

Math Expert
Joined: 02 Sep 2009
Posts: 45498

### Show Tags

16 Sep 2014, 01:21
00:00

Difficulty:

5% (low)

Question Stats:

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

### HideShow timer Statistics

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

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

_________________
Math Expert
Joined: 02 Sep 2009
Posts: 45498

### Show Tags

16 Sep 2014, 01:21
Expert's post
1
This post was
BOOKMARKED
Official Solution:

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

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

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

_________________
Intern
Joined: 21 Jan 2015
Posts: 21

### Show Tags

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
Math Expert
Joined: 02 Sep 2009
Posts: 45498

### Show Tags

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.
_________________
Manager
Joined: 15 Nov 2016
Posts: 132

### Show Tags

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: 25

### Show Tags

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

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

Sent from my iPhone using GMAT Club Forum
Manager
Joined: 17 Jul 2017
Posts: 111
Location: India
WE: Engineering (Transportation)

### Show Tags

06 Feb 2018, 10:31
1
KUDOS
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: 15
GMAT 1: 790 Q50 V51

### Show Tags

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".
Director
Joined: 08 Jun 2015
Posts: 504
Location: India
GMAT 1: 640 Q48 V29

### Show Tags

01 May 2018, 05:41
+1 for option E. Use sum of consecutive numbers formula. The answer comes to 1225. Option E it is !!
_________________

" The few , the fearless "

Re: M24-18   [#permalink] 01 May 2018, 05:41
Display posts from previous: Sort by

# M24-18

Moderators: chetan2u, Bunuel

 Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.