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If 3 < x < 7 and 4 > y > − 2, which of the following must
[#permalink]
09 Apr 2019, 02:19

Expert Reply

6

Bookmarks

Question Stats:

If \(3 < x < 7\) and \(4 > y > − 2\), which of the following must be true?

Indicate all possible choices.

A. \(x − y > 0\)

B. \(x + y > 0\)

C. \(x > y\)

D. \(2y − x > 0\)

E. \(2x − y > 1\)

_________________

Indicate all possible choices.

A. \(x − y > 0\)

B. \(x + y > 0\)

C. \(x > y\)

D. \(2y − x > 0\)

E. \(2x − y > 1\)

_________________

If 3 < x < 7 and 4 > y > − 2, which of the following must
[#permalink]
24 Apr 2019, 09:41

3

1

Bookmarks

Carcass wrote:

If \(3 < x < 7\) and \(4 > y > − 2\), which of the following must be true?

Indicate all possible choices.

A. \(x − y > 0\)

B. \(x + y > 0\)

C. \(x > y\)

D. \(2y − x > 0\)

E. \(2x − y > 1\)

Indicate all possible choices.

A. \(x − y > 0\)

B. \(x + y > 0\)

C. \(x > y\)

D. \(2y − x > 0\)

E. \(2x − y > 1\)

If this question had only 1 correct answer, I might try to find cases where 4 of the 5 answer choices are incorrect, and then choose the remaining answer choice.

However, this strategy won't work if there's more than 1 correct answer. So, I'm going to take a HYBRID approach....

A. \(x − y > 0\)

If x = 3.1 and y = 3.9, then x - y = 3.1 - 3.9 = -0.8

So, statement A need not be true

ELIMINATE

-----------------------------

B. \(x + y > 0\)

Given:

3 < x < 7

-2 < y < 4

Since the inequality symbols are facing the SAME DIRECTION, we can ADD them to get: 1 < x + y < 11

Since x+y is greater than 1, we know that x+y is greater than 0

So, statement B is TRUE

---------------------

C. \(x > y\)

If x = 3.1 and y = 3.9, then x < y

So, statement C need not be true

ELIMINATE

----------------------

D. \(2y − x > 0\)

If y = 0 and x = 5, then x - y = 2y - x = 2(0) - 5 = -5

So, statement C need not be true

ELIMINATE

-----------------------

E. \(2x − y > 1\)[/quote]

GIVEN:

3 < x < 7

-2 < y < 4

Take the top inequality and multiply all sides by 2. Also, take the bottom equation and multiply all sides by -1. We get:

6 < 2x < 14

-4 < -y < 2

Since the inequality symbols are facing the SAME DIRECTION, we can ADD them to get: 2 < 2x - y < 16

Since 2x - y is greater than 2, we know that 2x - y is greater than 1

So, statement E is TRUE

Answer: B, E

_________________

Re: If 3 < x < 7 and 4 > y > − 2, which of the following must
[#permalink]
21 Aug 2019, 19:13

we can ADD them to get: 2 < 2x - y < 165

I guess its 14+2 =16

Re: If 3 < x < 7 and 4 > y > − 2, which of the following must
[#permalink]
01 Jul 2020, 22:09

I understood the question but failed to take non integers into consideration.

Re: If 3 < x < 7 and 4 > y > − 2, which of the following must
[#permalink]
07 Feb 2022, 06:21

1

Instead of plugging in numbers, can I add / subtract the two inequalities for the given options? Will I arrive at the right option?

If this question had only 1 correct answer, I might try to find cases where 4 of the 5 answer choices are incorrect, and then choose the remaining answer choice.

However, this strategy won't work if there's more than 1 correct answer. So, I'm going to take a HYBRID approach....

A. \(x − y > 0\)

If x = 3.1 and y = 3.9, then x - y = 3.1 - 3.9 = -0.8

So, statement A need not be true

ELIMINATE

-----------------------------

B. \(x + y > 0\)

Given:

3 < x < 7

-2 < y < 4

Since the inequality symbols are facing the SAME DIRECTION, we can ADD them to get: 1 < x + y < 11

Since x+y is greater than 1, we know that x+y is greater than 0

So, statement B is TRUE

---------------------

C. \(x > y\)

If x = 3.1 and y = 3.9, then x < y

So, statement C need not be true

ELIMINATE

----------------------

D. \(2y − x > 0\)

If y = 0 and x = 5, then x - y = 2y - x = 2(0) - 5 = -5

So, statement C need not be true

ELIMINATE

-----------------------

E. \(2x − y > 1\)

GIVEN:

3 < x < 7

-2 < y < 4

Take the top inequality and multiply all sides by 2. Also, take the bottom equation and multiply all sides by -1. We get:

6 < 2x < 14

-4 < -y < 2

Since the inequality symbols are facing the SAME DIRECTION, we can ADD them to get: 2 < 2x - y < 16

Since 2x - y is greater than 2, we know that 2x - y is greater than 1

So, statement E is TRUE

Answer: B, E[/quote]

GreenlightTestPrep wrote:

Carcass wrote:

If \(3 < x < 7\) and \(4 > y > − 2\), which of the following must be true?

Indicate all possible choices.

A. \(x − y > 0\)

B. \(x + y > 0\)

C. \(x > y\)

D. \(2y − x > 0\)

E. \(2x − y > 1\)

Indicate all possible choices.

A. \(x − y > 0\)

B. \(x + y > 0\)

C. \(x > y\)

D. \(2y − x > 0\)

E. \(2x − y > 1\)

If this question had only 1 correct answer, I might try to find cases where 4 of the 5 answer choices are incorrect, and then choose the remaining answer choice.

However, this strategy won't work if there's more than 1 correct answer. So, I'm going to take a HYBRID approach....

A. \(x − y > 0\)

If x = 3.1 and y = 3.9, then x - y = 3.1 - 3.9 = -0.8

So, statement A need not be true

ELIMINATE

-----------------------------

B. \(x + y > 0\)

Given:

3 < x < 7

-2 < y < 4

Since the inequality symbols are facing the SAME DIRECTION, we can ADD them to get: 1 < x + y < 11

Since x+y is greater than 1, we know that x+y is greater than 0

So, statement B is TRUE

---------------------

C. \(x > y\)

If x = 3.1 and y = 3.9, then x < y

So, statement C need not be true

ELIMINATE

----------------------

D. \(2y − x > 0\)

If y = 0 and x = 5, then x - y = 2y - x = 2(0) - 5 = -5

So, statement C need not be true

ELIMINATE

-----------------------

E. \(2x − y > 1\)

GIVEN:

3 < x < 7

-2 < y < 4

Take the top inequality and multiply all sides by 2. Also, take the bottom equation and multiply all sides by -1. We get:

6 < 2x < 14

-4 < -y < 2

Since the inequality symbols are facing the SAME DIRECTION, we can ADD them to get: 2 < 2x - y < 16

Since 2x - y is greater than 2, we know that 2x - y is greater than 1

So, statement E is TRUE

Answer: B, E[/quote]

Re: If 3 < x < 7 and 4 > y > − 2, which of the following must
[#permalink]
07 Feb 2022, 06:49

1

Chaithraln2499 wrote:

Instead of plugging in numbers, can I add / subtract the two inequalities for the given options? Will I arrive at the right option?

Subtracting inequalities will yield new inequalities that are not necessarily true.

Consider this example.

We know the following two inequalities are true:

8 < 9

2 < 4

However, if we subtract the bottom inequality from the top inequality we get: 6 < 5, which is not true.

_________________

Moderator

Joined: **02 Jan 2020 **

Status:**GRE Quant Tutor**

Posts: **997**

Location: **India**

Concentration: **General Management**

Schools: **XLRI Jamshedpur, India - Class of 2014**

GMAT 1: **700 Q51 V31**

GPA: **2.8**

WE:**Engineering (Computer Software)**

Re: If 3 < x < 7 and 4 > y > 2, which of the following must
[#permalink]
21 May 2022, 11:38

1

Given that \(3 < x < 7\) and \(4 > y > − 2\) and we need to find out which of the answer choices MUST be true

Let's solve it using two methods

Method 1: Substitution

Since it's a must be a true question so we need to prove the answer choices wrong

\(3 < x < 7\) and \(4 > y > − 2\)

A. \(x − y > 0\)

=> So we need to take value of x smaller than y to prove this one wrong

=> x = 3.5 and y = 3.6

making x - y < 0 => FALSE

B. \(x + y > 0\)

Now, the value of x is positive and minimum value of y can only be very close to -2. Even if we take that value then also x+y will be positive as minimum value of x is very close to 3

So, this is ALWAYS TRUE

C. \(x > y\)

=> So we need to take value of x smaller than y to prove this one wrong

=> x = 3.5 and y = 3.6

making x - y < 0 => FALSE

D. \(2y − x > 0\)

Now, any negative value of y and a positive value of x will be sufficient to prove this one wrong

y = -1 and x = 4

=> 2y - x = 2*-1 - 4 < 0 => FALSE

E. \(2x − y > 1\)

We need to take minimum value of x and maximum value of y to try to prove this one wrong

=> x = 3.1, y = 3.9

=> 2x - y = 2*3.1 - 3.9 = 6.2 - 3.9 > 1 => ALWAYS TRUE

So, Answer will be B and E

Method 2: Algebra

\(3 < x < 7\) and \(4 > y > − 2\)

A. \(x − y > 0\)

\(3 < x < 7\) ...(1)

\(-2 < y < 4\) ...(2)

Multiply with -1 we get

-4 < -y < 2 ...(3)

Add (1) and (3) we get

3-4 < x - y < 7+2

=> -1 < x-y < 9

Clearly x-y can be < 0 also => FALSE

B. \(x + y > 0\)

Add (1) and (2) we get

3-2 < x + y < 7+4

=> 1 < x-y < 11

Clearly x+y will always be > 0 => TRUE

C. \(x > y\)

\(3 < x < 7\)

\(-2 < y < 4\)

Clearly y can be < x as y can be negative also but x is always positive => FALSE

D. \(2y − x > 0\)

Multiply (1) by -1 we get

-7 < -x < 3 ...(4)

Multiple (2) by 2 we get

\(-4 < 2y < 8\) ..(5)

Adding (4) and (5) we get

-7-4 < 2y -x < 3+8

=> -11 < 2y-x < 11

Clearly 2y - x can be < 0 => FALSE

E. \(2x − y > 1\)

Multiply (1) by 2 we get

6 < 2x < 14 ...(6)

Multiple (2) by -1 we get

\(-4 < y < 2\) ..(7)

Adding (6) and (7) we get

6-4 < 2x-y < 14+2

=> 2 < 2x-y < 16

Clearly, 2x-y > 1 => TRUE

So, Answer will be B and E

Hope it helps!

Watch the following video to learn the Basics of Inequalities

_________________

Let's solve it using two methods

Method 1: Substitution

Since it's a must be a true question so we need to prove the answer choices wrong

\(3 < x < 7\) and \(4 > y > − 2\)

A. \(x − y > 0\)

=> So we need to take value of x smaller than y to prove this one wrong

=> x = 3.5 and y = 3.6

making x - y < 0 => FALSE

B. \(x + y > 0\)

Now, the value of x is positive and minimum value of y can only be very close to -2. Even if we take that value then also x+y will be positive as minimum value of x is very close to 3

So, this is ALWAYS TRUE

C. \(x > y\)

=> So we need to take value of x smaller than y to prove this one wrong

=> x = 3.5 and y = 3.6

making x - y < 0 => FALSE

D. \(2y − x > 0\)

Now, any negative value of y and a positive value of x will be sufficient to prove this one wrong

y = -1 and x = 4

=> 2y - x = 2*-1 - 4 < 0 => FALSE

E. \(2x − y > 1\)

We need to take minimum value of x and maximum value of y to try to prove this one wrong

=> x = 3.1, y = 3.9

=> 2x - y = 2*3.1 - 3.9 = 6.2 - 3.9 > 1 => ALWAYS TRUE

So, Answer will be B and E

Method 2: Algebra

\(3 < x < 7\) and \(4 > y > − 2\)

A. \(x − y > 0\)

\(3 < x < 7\) ...(1)

\(-2 < y < 4\) ...(2)

Multiply with -1 we get

-4 < -y < 2 ...(3)

Add (1) and (3) we get

3-4 < x - y < 7+2

=> -1 < x-y < 9

Clearly x-y can be < 0 also => FALSE

B. \(x + y > 0\)

Add (1) and (2) we get

3-2 < x + y < 7+4

=> 1 < x-y < 11

Clearly x+y will always be > 0 => TRUE

C. \(x > y\)

\(3 < x < 7\)

\(-2 < y < 4\)

Clearly y can be < x as y can be negative also but x is always positive => FALSE

D. \(2y − x > 0\)

Multiply (1) by -1 we get

-7 < -x < 3 ...(4)

Multiple (2) by 2 we get

\(-4 < 2y < 8\) ..(5)

Adding (4) and (5) we get

-7-4 < 2y -x < 3+8

=> -11 < 2y-x < 11

Clearly 2y - x can be < 0 => FALSE

E. \(2x − y > 1\)

Multiply (1) by 2 we get

6 < 2x < 14 ...(6)

Multiple (2) by -1 we get

\(-4 < y < 2\) ..(7)

Adding (6) and (7) we get

6-4 < 2x-y < 14+2

=> 2 < 2x-y < 16

Clearly, 2x-y > 1 => TRUE

So, Answer will be B and E

Hope it helps!

Watch the following video to learn the Basics of Inequalities

_________________

Ankit

to BrushMyQuant YouTube Channel to get WEEKLY new VIDEOS!!!

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to BrushMyQuant YouTube Channel to get WEEKLY new VIDEOS!!!

How to Solve :

Similar Triangles || Exponents || Work Rate || Probability: Coin Toss || Probability: Dice Roll || Prime Numbers || Circles || Equilateral Triangle in a Circle || Arithmetic and Geometric Progression || Divisibility Rules || Absolute Value || Inequalities || Statistics || Functions and Custom Characters || Remainders || LCM and GCD || Stem & Leaf Plot || Box And Whisker Plot

GRE Club Topic Wise Problems

GRE Quant - 2019 Ed

GRE Verbal - 2019 Ed

The Princeton Review Quantitative Directory

The 5 lb. Book of GRE Practice Problems (2nd Ed.) - QUANT

GRE - PowerPrep® The FREE Practice Tests Explained 2019

Re: If 3 < x < 7 and 4 > y > 2, which of the following must
[#permalink]
17 Feb 2023, 09:33

Think option like E will not be feasible under exam condition?

Is there any strategy here Brent GreenlightTestPrep? Maybe after choosing B, elimiate rest and choose E? As it can't be just one answer choice in this type of GRE MC questions?

Is there any strategy here Brent GreenlightTestPrep? Maybe after choosing B, elimiate rest and choose E? As it can't be just one answer choice in this type of GRE MC questions?

Re: If 3 < x < 7 and 4 > y > 2, which of the following must
[#permalink]
17 Feb 2023, 09:46

Expert Reply

I am not sure I got what you meant

However, this is a question conceived to test every single answer choice. There is no a real shortcut

Also keep in mind that in a MAC question, the option could be ALL correct or just one. It depends

_________________

However, this is a question conceived to test every single answer choice. There is no a real shortcut

Also keep in mind that in a MAC question, the option could be ALL correct or just one. It depends

_________________

gmatclubot

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