GMAT Question of the Day: Daily via email | Daily via Instagram New to GMAT Club? Watch this Video

It is currently 01 Apr 2020, 23:34

Close

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
Your Progress

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

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

Close

Request Expert Reply

Confirm Cancel

The larger of two negative consecutive even integers 2t and 2(t-1) is

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
TAGS:

Hide Tags

Find Similar Topics 
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 62469
The larger of two negative consecutive even integers 2t and 2(t-1) is  [#permalink]

Show Tags

New post 06 Sep 2016, 03:49
00:00
A
B
C
D
E

Difficulty:

  5% (low)

Question Stats:

90% (01:30) correct 10% (01:32) wrong based on 134 sessions

HideShow timer Statistics

Retired Moderator
avatar
G
Joined: 26 Nov 2012
Posts: 554
The larger of two negative consecutive even integers 2t and 2(t-1) is  [#permalink]

Show Tags

New post 06 Sep 2016, 04:46
Bunuel wrote:
The larger of two negative consecutive even integers 2t and 2(t-1) is multiplied by 3 and then added to the smaller of the two original integers. Which of the following represents this operation?

A. 6t-2
B. 8t-2
C. -2
D. 3
E. -2-4t^2


lets take t = -1, since it is given consecutive even integers and given that larget is mutiplied by 3.

For t =-1 case we get -2 and -4 and the larger value is -2 among them and this value is multiplied by 3 , we get -6.

and this value is added to smaller of two, i.e. -6 is added -4 we get -10.

Only option B is correct...i.e. 8(-1) - 2 = -10.

IMO option B.
Intern
Intern
avatar
B
Joined: 05 Sep 2015
Posts: 43
Re: The larger of two negative consecutive even integers 2t and 2(t-1) is  [#permalink]

Show Tags

New post 07 Sep 2016, 03:12
larger of 2t-2, 2t (when they are negative) is 2t. the smallest of 2t and 2t-2 when they are positive is 2t-2.
therefore 3* 2t + 2t-2 = 8t-2.
option b.
Current Student
User avatar
V
Joined: 19 Mar 2012
Posts: 4426
Location: India
GMAT 1: 760 Q50 V42
GPA: 3.8
WE: Marketing (Non-Profit and Government)
The larger of two negative consecutive even integers  [#permalink]

Show Tags

New post 05 May 2018, 06:07

GST Week 4 Day 4 Economist GMAT Tutor Question 4


Give your best shot at writing a top notch explanation and you will have the chance to win GMAT Club tests daily and GMAT Basic Prep Plan by The Economist GMAT Tutor. See the GMAT Spring Training Thread for all details


The larger of two negative consecutive even integers \(2t\) and \(2(t-1)\) is multiplied by 3 and then added to the smaller of the two original integers. Which of the following represents this operation?

A. \(6t-2\)

B. \(8t-2\)

C. \(-2\)

D. \(3\)

E. \(-2-4t^2\)
_________________
Senior Manager
Senior Manager
User avatar
G
Joined: 29 Dec 2017
Posts: 365
Location: United States
Concentration: Marketing, Technology
GMAT 1: 630 Q44 V33
GMAT 2: 690 Q47 V37
GMAT 3: 710 Q50 V37
GPA: 3.25
WE: Marketing (Telecommunications)
Re: The larger of two negative consecutive even integers  [#permalink]

Show Tags

New post 05 May 2018, 06:23
souvik101990 wrote:
[textarea][header3]

The larger of two negative consecutive even integers \(2t\) and \(2(t-1)\) is multiplied by 3 and then added to the smaller of the two original integers. Which of the following represents this operation?

A. \(6t-2\)
B. \(8t-2\)
C. \(-2\)
D. \(3\)
E. \(-2-4t^2\)


It is clear that 2(t-1)= 2t-2 - is smaller than 2t (or standing 2 grades lefter on a number line)

2t*3+2t-2 = 8t-2. Answer (B)
Manager
Manager
avatar
S
Joined: 13 Apr 2010
Posts: 86
Reviews Badge
Re: The larger of two negative consecutive even integers  [#permalink]

Show Tags

New post 05 May 2018, 06:27
souvik101990 wrote:

GST Week 4 Day 4 Economist GMAT Tutor Question 4


Give your best shot at writing a top notch explanation and you will have the chance to win GMAT Club tests daily and GMAT Basic Prep Plan by The Economist GMAT Tutor. See the GMAT Spring Training Thread for all details


The larger of two negative consecutive even integers \(2t\) and \(2(t-1)\) is multiplied by 3 and then added to the smaller of the two original integers. Which of the following represents this operation?

A. \(6t-2\)

B. \(8t-2\)

C. \(-2\)

D. \(3\)

E. \(-2-4t^2\)



Out of 2t and 2(t - 1) , 2t is the largest even consecutive integer . you can verify by substituting " t "with positive , negative and zero

t= 0 ; 2t = 0 and 2(t - 1) = -2
t= -5 ; 2t = -10 and 2(t - 1) = -12
t = 6 ; 2t = 12 and 2(t - 1) = 10

coming to question , 2t multiplied by 3 equals 6t
added to smaller of two orginal integers , that is, 2(t - 1) is equal to 8t - 2
Answer is B
RC Moderator
User avatar
V
Joined: 24 Aug 2016
Posts: 785
GMAT 1: 540 Q49 V16
GMAT 2: 680 Q49 V33
GMAT ToolKit User Reviews Badge CAT Tests
Re: The larger of two negative consecutive even integers  [#permalink]

Show Tags

New post 05 May 2018, 06:39
souvik101990 wrote:

GST Week 4 Day 4 Economist GMAT Tutor Question 4


Give your best shot at writing a top notch explanation and you will have the chance to win GMAT Club tests daily and GMAT Basic Prep Plan by The Economist GMAT Tutor. See the GMAT Spring Training Thread for all details


The larger of two negative consecutive even integers \(2t\) and \(2(t-1)\) is multiplied by 3 and then added to the smaller of the two original integers. Which of the following represents this operation?

A. \(6t-2\)

B. \(8t-2\)

C. \(-2\)

D. \(3\)

E. \(-2-4t^2\)


The negative consecutive even integers are : -2, -4, -6, -8,-10 ........................etc
Given,two negative consecutive even integers \(2t\) and \(2(t-1)\) ..... thus the max possible value of t=-1, as the boundary conditions
(1) when t=0, 2t=0 .... is not negative integer and (2) when t=-0.5 , 2t =-1.... is not negative even integer.
Thus between \(2t\) and \(2(t-1)\) , the larger is \(2t\) and smaller is \(2(t-1)\)
hence , \(3 *2t\) + \(2(t-1)\) =\(6t+2t-2\) = \(8t-2\) ....Hence I would go for answer B
_________________
Please let me know if I am going in wrong direction.
Thanks in appreciation.
Intern
Intern
avatar
B
Joined: 27 Nov 2016
Posts: 41
Location: India
Schools: ISB '19
GPA: 2.75
The larger of two negative consecutive even integers  [#permalink]

Show Tags

New post 05 May 2018, 07:05
souvik101990 wrote:

GST Week 4 Day 4 Economist GMAT Tutor Question 4


Give your best shot at writing a top notch explanation and you will have the chance to win GMAT Club tests daily and GMAT Basic Prep Plan by The Economist GMAT Tutor. See the GMAT Spring Training Thread for all details


The larger of two negative consecutive even integers \(2t\) and \(2(t-1)\) is multiplied by 3 and then added to the smaller of the two original integers. Which of the following represents this operation?

A. \(6t-2\)

B. \(8t-2\)

C. \(-2\)

D. \(3\)

E. \(-2-4t^2\)


Since the given expressions are negative, we have to consider "t" as negative for "2t" to be negative. This means that 2(t-1) has a higher absolute value, which further means that 2(t-1) has a smaller value than 2t.
Hence, 2t*3 + 2(t-1) = 8t-2 i.e. B
Ask GMAT Experts Forum Moderator
User avatar
V
Status: Preparing for GMAT
Joined: 25 Nov 2015
Posts: 1043
Location: India
GPA: 3.64
GMAT ToolKit User Reviews Badge
Re: The larger of two negative consecutive even integers  [#permalink]

Show Tags

New post 05 May 2018, 07:11
souvik101990 wrote:

GST Week 4 Day 4 Economist GMAT Tutor Question 4


Give your best shot at writing a top notch explanation and you will have the chance to win GMAT Club tests daily and GMAT Basic Prep Plan by The Economist GMAT Tutor. See the GMAT Spring Training Thread for all details


The larger of two negative consecutive even integers \(2t\) and \(2(t-1)\) is multiplied by 3 and then added to the smaller of the two original integers. Which of the following represents this operation?

A. \(6t-2\)

B. \(8t-2\)

C. \(-2\)

D. \(3\)

E. \(-2-4t^2\)


Larger of two consecutive negative even integers among 2t and 2t-2 is 2t (for e.g. -2>-4)
2t multiplied by 3 = 6t
Adding to the smaller of the two original integer = 6t+2t-2 = 8t-2
Answer B.
Manager
Manager
avatar
B
Joined: 02 Jul 2017
Posts: 63
Re: The larger of two negative consecutive even integers  [#permalink]

Show Tags

New post 05 May 2018, 07:12
Two consecutive negative integers are 2t and 2t-2 , It is clear 2t is > 2t-2 ( (as 2t -(2t-2) =2 is positive ))

so , larger of the two consecutive integers is 2t ..
as question asks ..let us proceed ..
multiplying larger with 3 and adding to smaller will give -> 3 * (2t) +2t-2 =6t+2t -2 =8t-2 ..

I hope all steps are clear
Manager
Manager
User avatar
P
Joined: 07 Apr 2018
Posts: 99
Reviews Badge
Re: The larger of two negative consecutive even integers  [#permalink]

Show Tags

New post 05 May 2018, 08:08
Two consecutive negative even integers are 2t and 2(t-1)

Larger negative integer = 2t

Mutiplying 2t by 3= 6t....................(1)
Adding (1) to the smaller negative integer= 6t+ 2(t-1)
= 8t-2

Correct Answer= B
Intern
Intern
avatar
B
Joined: 02 Jan 2018
Posts: 16
Location: Australia
WE: Information Technology (Computer Software)
Premium Member
Re: The larger of two negative consecutive even integers  [#permalink]

Show Tags

New post 05 May 2018, 08:25
souvik101990 wrote:

GST Week 4 Day 4 Economist GMAT Tutor Question 4


Give your best shot at writing a top notch explanation and you will have the chance to win GMAT Club tests daily and GMAT Basic Prep Plan by The Economist GMAT Tutor. See the GMAT Spring Training Thread for all details


The larger of two negative consecutive even integers \(2t\) and \(2(t-1)\) is multiplied by 3 and then added to the smaller of the two original integers. Which of the following represents this operation?

A. \(6t-2\)

B. \(8t-2\)

C. \(-2\)

D. \(3\)

E. \(-2-4t^2\)



Whenever there are questions of this type, assume numbers. Let t=-1. So, options will be

A. -8
B. -10
C. -2
D. 3
E. -6

Now, coming back to question

2 Negative consecutive even integers are

-2 and -4. Larger is -2, it is multiplied by 3 so -6 and then added to smaller -6+ (-4) . So, answer is -10. Hence B.


If we are not supposing numbers, then

Since 2t will be greater out of these 2, Hence 2t * 3= 6t. Then 6t + 2(t-1) = 8t-2
Intern
Intern
avatar
B
Joined: 19 Dec 2017
Posts: 5
Re: The larger of two negative consecutive even integers 2t and 2(t-1) is  [#permalink]

Show Tags

New post 18 May 2018, 05:27
Bunuel wrote:
The larger of two negative consecutive even integers 2t and 2(t-1) is multiplied by 3 and then added to the smaller of the two original integers. Which of the following represents this operation?

A. 6t-2
B. 8t-2
C. -2
D. 3
E. -2-4t^2


Since t<0, 2t> 2(t-1)

2 x t x 3 + 2(t-1)
6t + 2t - 2
8t - 2

B is the answer
GMAT Club Bot
Re: The larger of two negative consecutive even integers 2t and 2(t-1) is   [#permalink] 18 May 2018, 05:27
Display posts from previous: Sort by

The larger of two negative consecutive even integers 2t and 2(t-1) is

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  





Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne