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

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

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06 Sep 2016, 04:49
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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

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

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06 Sep 2016, 05:46
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.
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Re: The larger of two negative consecutive even integers 2t and 2(t-1) is  [#permalink]

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07 Sep 2016, 04:12
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.
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The larger of two negative consecutive even integers  [#permalink]

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05 May 2018, 07:07

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. $$-2-4t^2$$
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Re: The larger of two negative consecutive even integers  [#permalink]

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05 May 2018, 07:23
souvik101990 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. $$-2-4t^2$$

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

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Re: The larger of two negative consecutive even integers  [#permalink]

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05 May 2018, 07:27
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
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Re: The larger of two negative consecutive even integers  [#permalink]

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05 May 2018, 07:39
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
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The larger of two negative consecutive even integers  [#permalink]

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05 May 2018, 08:05
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
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Re: The larger of two negative consecutive even integers  [#permalink]

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05 May 2018, 08:11
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
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Re: The larger of two negative consecutive even integers  [#permalink]

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05 May 2018, 08:12
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
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Re: The larger of two negative consecutive even integers  [#permalink]

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05 May 2018, 09:08
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

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Re: The larger of two negative consecutive even integers  [#permalink]

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05 May 2018, 09:25
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
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Re: The larger of two negative consecutive even integers 2t and 2(t-1) is  [#permalink]

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18 May 2018, 06:27
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

Re: The larger of two negative consecutive even integers 2t and 2(t-1) is   [#permalink] 18 May 2018, 06:27
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