If x and y are integers and y=|x+3| + |4-x|, does y equals 7 : GMAT Data Sufficiency (DS)
Check GMAT Club Decision Tracker for the Latest School Decision Releases http://gmatclub.com/AppTrack

 It is currently 18 Jan 2017, 14:24

### 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

# If x and y are integers and y=|x+3| + |4-x|, does y equals 7

Author Message
TAGS:

### Hide Tags

Intern
Joined: 18 Aug 2010
Posts: 6
Followers: 1

Kudos [?]: 16 [3] , given: 1

If x and y are integers and y=|x+3| + |4-x|, does y equals 7 [#permalink]

### Show Tags

26 Sep 2010, 20:26
3
KUDOS
16
This post was
BOOKMARKED
00:00

Difficulty:

35% (medium)

Question Stats:

66% (02:22) correct 34% (01:31) wrong based on 687 sessions

### HideShow timer Statistics

If x and y are integers and y=|x+3| + |4-x|, does y equals 7?

(1) x < 4
(2) x > -3

Had a hard time solving this, would like to know how to solve this using number picking approach as well as algebraic approach. Thanks.

OA is
[Reveal] Spoiler:
C
[Reveal] Spoiler: OA

_________________

"Learning never exhausts the mind."
--Leonardo da Vinci

Math Expert
Joined: 02 Sep 2009
Posts: 36548
Followers: 7077

Kudos [?]: 93107 [7] , given: 10552

Re: Power Prep - DS - Modulus [#permalink]

### Show Tags

26 Sep 2010, 22:06
7
KUDOS
Expert's post
6
This post was
BOOKMARKED
thirst4edu wrote:
If x & y are integers and y=|x+3| + |4-x|, does y equals 7?

1) x < 4
2) x > -3

Had a hard time solving this, would like to know how to solve this using number picking approach as well as algebraic approach. Thanks.

OA is
[Reveal] Spoiler:
C

$$y=|x+3|+|4-x|$$ two check points: $$x=-3$$ and $$x=4$$ (check point: the value of $$x$$ when expression in || equals to zero), hence three ranges to consider:

A. $$x<{-3}$$ --> $$y=| x + 3| +|4-x| =-x-3+4-x=-2x+1$$, which means that when $$x$$ is in the range {-infinity,-3} the value of $$y$$ is defined by $$x$$ (we would have multiple choices of $$y$$ depending on $$x$$ from the given range);

B. $$-3\leq{x}\leq{4}$$ --> $$y=|x+3|+|4-x|=x+3+4-x=7$$, which means that when $$x$$ is in the range {-3,4} the value of $$y$$ is $$7$$ (value of y does not depend on value of $$x$$, when $$x$$ is from the given range);

C. $$x>{4}$$ --> $$y=|x+3|+|4-x|=x+3-4+x=2x-1$$, which means that when $$x$$ is in the range {4, +infinity} the value of $$y$$ is defined by $$x$$ (we would have multiple choices of $$y$$ depending on $$x$$ from the given range).

Hence we can definitely conclude that $$y=7$$ if $$x$$ is in the range {-3,4}

(1) $$x<4$$ --> not sufficient ($$x<4$$ but we don't know if it's $$\geq{-3}$$);
(2) $$x>-3$$ --> not sufficient ($$x>-3$$ but we don't know if it's $$\leq{4}$$);

(1)+(2) $$-3<x<4$$ exactly the range we needed, so $$y=7$$. Sufficient.

OR: looking at $$y=|x+3|+|4-x|$$ you can notice that $$y=7$$ ($$y$$ doesn't depend on the value of $$x$$) when $$x+3$$ and $$4-x$$ are both positive, in this case $$x-es$$ cancel out each other and we would have $$y=|x+3|+|4-x|=x+3+4-x=7$$. Both $$x+3$$ and $$4-x$$ are positive in the range $$-3<{x}<4$$ ($$x+3>0$$ --> $$x>-3$$ and $$4-x>0$$ --> $$x<4$$).

Hope it's clear.
_________________
Retired Moderator
Joined: 02 Sep 2010
Posts: 805
Location: London
Followers: 105

Kudos [?]: 957 [4] , given: 25

Re: Power Prep - DS - Modulus [#permalink]

### Show Tags

26 Sep 2010, 22:09
4
KUDOS
Let's first solve |x+3|+|4-x|=7 to answer "when is this true ?"

You can solve algebraically but it is much easier to do it using a simple number line approach. Remember |x-a| means distance between x and a on the number line
Here the two points in question are -3 and 4
Now it is easy to imagine the three cases that x is to the left of -3, between -3 and 4 and to the right of 4. The only case when the two distances add up to the distance between -3 and 4, ie, 7 is case two. In case 1 and 3, the sum will exceed 7

1) could mean case 2 or 3. Not sufficient
2) could mean case 1 or 2. Not sufficient
1+2) can only mean case 2. Sufficient to know that y=7

_________________
Manager
Joined: 30 Aug 2010
Posts: 91
Location: Bangalore, India
Followers: 5

Kudos [?]: 160 [3] , given: 27

Re: Power Prep - DS - Modulus [#permalink]

### Show Tags

26 Sep 2010, 23:47
3
KUDOS
1
This post was
BOOKMARKED
thirst4edu wrote:
If x & y are integers and y=|x+3| + |4-x|, does y equals 7?

1) x < 4
2) x > -3

Had a hard time solving this, would like to know how to solve this using number picking approach as well as algebraic approach. Thanks.

OA is
[Reveal] Spoiler:
C

Guys, the be low is my approcah for any modulus qtn in GMAT.

Remember.
The meaning of |x-y| is "On the number line, the distance of X from +Y"
The meaning of |x+y| is "On the number line, the distance of X from -Y"
The meaning of |x| is "On the number line, the distance of X from 0".

Qtn:

for integers X and Y, If y=|x+3| + |4-x|, does y equals 7
==> is the SUM of the distance b/e x and -3 , and x and 4 equals to 7?
==> .........-3......0...........4..... observe that X has to be anywhere b/w -3 and 4 or on any of these points for the total distance to be 7

Stmnt1: X < 4: Answer could be Yes if X is < 4 and b/w -3 and 4 but from the given information (i.e X<4) X could be some where left to -3 in which case the total distance would be > 7 hence insufficient.

......X....-3......0..........4 answer to the qtn: NO
or ............-3...X...0.........4 answer to the qtn: YES

Stmnt2: X > -3: Answer could be Yes if X is > -3 and b/w -3 and 4 but from the given information (i.e X>-3) X could be some where right to 4 in which case the total distance would be > 7 hence insufficient.
..........-3......0..........4...X... answer to the qtn: NO
or ..........-3...X...0.........4 answer to the qtn: YES

1&2

X must be b/w -3 and 4
..........-3.X.X.X...0.X.X.X.X.X...4...... answer is always YES..hence Sufficient.

Hope it helps
Manager
Joined: 06 Aug 2010
Posts: 225
Location: Boston
Followers: 3

Kudos [?]: 183 [2] , given: 5

Re: Power Prep - DS - Modulus [#permalink]

### Show Tags

27 Sep 2010, 10:51
2
KUDOS
thirst4edu wrote:
If x & y are integers and y=|x+3| + |4-x|, does y equals 7?

1) x < 4
2) x > -3

Had a hard time solving this, would like to know how to solve this using number picking approach as well as algebraic approach. Thanks.

OA is
[Reveal] Spoiler:
C

Either choice by itself is clearly insufficient:

(1) If x = 3, y = |3+3| + |4-3| = 7. If x = -100, then y = 97 + 104 = 201.
(2) If x = -2, y = 1 + 6 = 7. If x = 100, then y = 103 + 96 = 199.

Putting them together, you can quickly check every integer value of x from -3 to 4 and see that y = 7 for every one. It's only 6 values to check, you can do it very quickly in your head.

(C)
Manager
Joined: 27 May 2010
Posts: 102
Followers: 2

Kudos [?]: 6 [0], given: 13

Re: Power Prep - DS - Modulus [#permalink]

### Show Tags

28 Sep 2010, 09:58
thanks bunuel. always great explanations!!
SVP
Joined: 05 Jul 2006
Posts: 1743
Followers: 6

Kudos [?]: 316 [0], given: 49

Re: Power Prep - DS - Modulus [#permalink]

### Show Tags

29 Sep 2010, 09:21
Bunuel rocks, cheers mate
Manager
Joined: 27 Dec 2011
Posts: 71
Followers: 1

Kudos [?]: 19 [0], given: 12

Re: If x and y are integers and y=|x+3| + |4-x|, does y equals 7 [#permalink]

### Show Tags

26 May 2012, 07:38
At bunuel,

A. x<{-3} --> y=| x + 3| +|4-x| =-x-3+4-x=-2x+1, which means that when x is in the range {-infinity,-3} the value of y is defined by x (we would have multiple choices of y depending on x from the given range);

B. -3\leq{x}\leq{4} --> y=|x+3|+|4-x|=x+3+4-x=7, which means that when x is in the range {-3,4} the value of y is 7 (value of y does not depend on value of x, when x is from the given range);

C. x>{4} --> y=|x+3|+|4-x|=x+3-4+x=2x-1, which means that when x is in the range {4, +infinity} the value of y is defined by x (we would have multiple choices of y depending on x from the given range).

I understand the how you got the check points -3 and 4 but I am having a hard time understanding how to decide sign for x when you are removing absolute value symbol
for example for
(A) x<-3 y=| x + 3| +|4-x|
how did you decide sign of "x" here ===> -x-3+4-x
-2x+1

Similarly can u also explain for (B) and (C)

thank you!

-K
Math Expert
Joined: 02 Sep 2009
Posts: 36548
Followers: 7077

Kudos [?]: 93107 [0], given: 10552

Re: If x and y are integers and y=|x+3| + |4-x|, does y equals 7 [#permalink]

### Show Tags

28 May 2012, 04:23
kartik222 wrote:
At bunuel,

A. x<{-3} --> y=| x + 3| +|4-x| =-x-3+4-x=-2x+1, which means that when x is in the range {-infinity,-3} the value of y is defined by x (we would have multiple choices of y depending on x from the given range);

B. -3\leq{x}\leq{4} --> y=|x+3|+|4-x|=x+3+4-x=7, which means that when x is in the range {-3,4} the value of y is 7 (value of y does not depend on value of x, when x is from the given range);

C. x>{4} --> y=|x+3|+|4-x|=x+3-4+x=2x-1, which means that when x is in the range {4, +infinity} the value of y is defined by x (we would have multiple choices of y depending on x from the given range).

I understand the how you got the check points -3 and 4 but I am having a hard time understanding how to decide sign for x when you are removing absolute value symbol
for example for
(A) x<-3 y=| x + 3| +|4-x|
how did you decide sign of "x" here ===> -x-3+4-x
-2x+1

Similarly can u also explain for (B) and (C)

thank you!

-K

Absolute value properties:
When $$x\leq{0}$$ then $$|x|=-x$$, or more generally when $$some \ expression\leq{0}$$ then $$|some \ expression|\leq{-(some \ expression)}$$. For example: $$|-5|=5=-(-5)$$;

When $$x\geq{0}$$ then $$|x|=x$$, or more generally when $$some \ expression\geq{0}$$ then $$|some \ expression|\leq{some \ expression}$$. For example: $$|5|=5$$;

So, for example if $$x<-3$$ then $$x+3<0$$ and $$4-x>0$$ which means that $$|x+3|=-(x+3)$$ and $$|4-x|=4-x$$ --> $$|x+3|+|4-x|=-(x+3)+4-x=-2x+1$$.

Similarly for B and C.

Hope it's clear.
_________________
Manager
Joined: 02 Jun 2011
Posts: 159
Followers: 1

Kudos [?]: 75 [0], given: 11

Re: Power Prep - DS - Modulus [#permalink]

### Show Tags

28 May 2012, 08:10
Bunuel wrote:
thirst4edu wrote:
If x & y are integers and y=|x+3| + |4-x|, does y equals 7?

1) x < 4
2) x > -3

Had a hard time solving this, would like to know how to solve this using number picking approach as well as algebraic approach. Thanks.

OA is
[Reveal] Spoiler:
C

$$y=|x+3|+|4-x|$$ two check points: $$x=-3$$ and $$x=4$$ (check point: the value of $$x$$ when expression in || equals to zero), hence three ranges to consider:

A. $$x<{-3}$$ --> $$y=| x + 3| +|4-x| =-x-3+4-x=-2x+1$$, which means that when $$x$$ is in the range {-infinity,-3} the value of $$y$$ is defined by $$x$$ (we would have multiple choices of $$y$$ depending on $$x$$ from the given range);

B. $$-3\leq{x}\leq{4}$$ --> $$y=|x+3|+|4-x|=x+3+4-x=7$$, which means that when $$x$$ is in the range {-3,4} the value of $$y$$ is $$7$$ (value of y does not depend on value of $$x$$, when $$x$$ is from the given range);

C. $$x>{4}$$ --> $$y=|x+3|+|4-x|=x+3-4+x=2x-1$$, which means that when $$x$$ is in the range {4, +infinity} the value of $$y$$ is defined by $$x$$ (we would have multiple choices of $$y$$ depending on $$x$$ from the given range).

Hence we can definitely conclude that $$y=7$$ if $$x$$ is in the range {-3,4}

(1) $$x<4$$ --> not sufficient ($$x<4$$ but we don't know if it's $$\geq{-3}$$);
(2) $$x>-3$$ --> not sufficient ($$x>-3$$ but we don't know if it's $$\leq{4}$$);

(1)+(2) $$-3<x<4$$ exactly the range we needed, so $$y=7$$. Sufficient.

OR: looking at $$y=|x+3|+|4-x|$$ you can notice that $$y=7$$ ($$y$$ doesn't depend on the value of $$x$$) when $$x+3$$ and $$4-x$$ are both positive, in this case $$x-es$$ cancel out each other and we would have $$y=|x+3|+|4-x|=x+3+4-x=7$$. Both $$x+3$$ and $$4-x$$ are positive in the range $$-3<{x}<4$$ ($$x+3>0$$ --> $$x>-3$$ and $$4-x>0$$ --> $$x<4$$).

Hope it's clear.

inequalities are posing problems! - one doubt - when it is said "-3<x<4", in this range shouldn't we check -2, -1 or 2,1 etc and see what is the value for y?
is y independent of x when x<0?
Manager
Joined: 12 May 2012
Posts: 83
Location: India
Concentration: General Management, Operations
GMAT 1: 650 Q51 V25
GMAT 2: 730 Q50 V38
GPA: 4
WE: General Management (Transportation)
Followers: 2

Kudos [?]: 91 [0], given: 14

Re: Power Prep - DS - Modulus [#permalink]

### Show Tags

01 Jun 2012, 04:36
Bunuel wrote:
thirst4edu wrote:
If x & y are integers and y=|x+3| + |4-x|, does y equals 7?

1) x < 4
2) x > -3

Had a hard time solving this, would like to know how to solve this using number picking approach as well as algebraic approach. Thanks.

OA is
[Reveal] Spoiler:
C

$$y=|x+3|+|4-x|$$ two check points: $$x=-3$$ and $$x=4$$ (check point: the value of $$x$$ when expression in || equals to zero), hence three ranges to consider:

A. $$x<{-3}$$ --> $$y=| x + 3| +|4-x| =-x-3+4-x=-2x+1$$, which means that when $$x$$ is in the range {-infinity,-3} the value of $$y$$ is defined by $$x$$ (we would have multiple choices of $$y$$ depending on $$x$$ from the given range);

B. $$-3\leq{x}\leq{4}$$ --> $$y=|x+3|+|4-x|=x+3+4-x=7$$, which means that when $$x$$ is in the range {-3,4} the value of $$y$$ is $$7$$ (value of y does not depend on value of $$x$$, when $$x$$ is from the given range);

C. $$x>{4}$$ --> $$y=|x+3|+|4-x|=x+3-4+x=2x-1$$, which means that when $$x$$ is in the range {4, +infinity} the value of $$y$$ is defined by $$x$$ (we would have multiple choices of $$y$$ depending on $$x$$ from the given range).

Hence we can definitely conclude that $$y=7$$ if $$x$$ is in the range {-3,4}

(1) $$x<4$$ --> not sufficient ($$x<4$$ but we don't know if it's $$\geq{-3}$$);
(2) $$x>-3$$ --> not sufficient ($$x>-3$$ but we don't know if it's $$\leq{4}$$);

(1)+(2) $$-3<x<4$$ exactly the range we needed, so $$y=7$$. Sufficient.

OR: looking at $$y=|x+3|+|4-x|$$ you can notice that $$y=7$$ ($$y$$ doesn't depend on the value of $$x$$) when $$x+3$$ and $$4-x$$ are both positive, in this case $$x-es$$ cancel out each other and we would have $$y=|x+3|+|4-x|=x+3+4-x=7$$. Both $$x+3$$ and $$4-x$$ are positive in the range $$-3<{x}<4$$ ($$x+3>0$$ --> $$x>-3$$ and $$4-x>0$$ --> $$x<4$$).

Hope it's clear.

MOD questions always floor me.
Could you please suggest some good material on MODs?
Math Expert
Joined: 02 Sep 2009
Posts: 36548
Followers: 7077

Kudos [?]: 93107 [1] , given: 10552

Re: Power Prep - DS - Modulus [#permalink]

### Show Tags

01 Jun 2012, 06:37
1
KUDOS
Expert's post
manulath wrote:
MOD questions always floor me.
Could you please suggest some good material on MODs?

Check Absolute Value chapter of Math Book: math-absolute-value-modulus-86462.html

DS questions on absolute value to practice: search.php?search_id=tag&tag_id=37
PS questions on absolute value to practice: search.php?search_id=tag&tag_id=58

Tough absolute value and inequity questions with detailed solutions: inequality-and-absolute-value-questions-from-my-collection-86939.html

Hope it helps.
_________________
Manager
Joined: 24 Mar 2010
Posts: 81
Followers: 1

Kudos [?]: 58 [0], given: 134

Re: If x and y are integers and y=|x+3| + |4-x|, does y equals 7 [#permalink]

### Show Tags

20 Dec 2012, 08:01
2
This post was
BOOKMARKED
Thinking of modulus as distances

|x+3| => distance of x from -3
|x-4| => distance of x from 4

Picture the same on the number line

________-3_____________0_________________4__________

We are given that y is the sum of the distance of x from -3 & of x from 4

Hence y could be anywhere on the number line

For y=7, let us consider the possibilities

Case (1)

_____x______-3_____________0_________________4__________

As you can quickly conclude
Its impossible for the distance to be 7 if x < -3
Take x = -4 and check,
y = 1 + 8 = 9

Case (2)

_________-3_____________0_________________4_____x_____

As you can quickly conclude
Its impossible for the distance to be 7 if x >4
Take x = 5 and check,
y = 8 + 1 = 9

Hence the range for y = 7 has to be in third case

__-3_____________x_________________4_____
i.e. -3<x<4

So we need to find if -3<x<4 ????

(1) x < 4
Insuff

(2) x > -3
Insuff

(3) Combining - -3<x<4

Bang-on.

Hence C
_________________

- Stay Hungry, stay Foolish -

Manager
Joined: 12 Dec 2012
Posts: 230
GMAT 1: 540 Q36 V28
GMAT 2: 550 Q39 V27
GMAT 3: 620 Q42 V33
GPA: 2.82
WE: Human Resources (Health Care)
Followers: 4

Kudos [?]: 78 [0], given: 181

Re: Power Prep - DS - Modulus [#permalink]

### Show Tags

27 Mar 2013, 06:58
Bunuel wrote:
thirst4edu wrote:
If x & y are integers and y=|x+3| + |4-x|, does y equals 7?

1) x < 4
2) x > -3

Had a hard time solving this, would like to know how to solve this using number picking approach as well as algebraic approach. Thanks.

OA is
[Reveal] Spoiler:
C

$$y=|x+3|+|4-x|$$ two check points: $$x=-3$$ and $$x=4$$ (check point: the value of $$x$$ when expression in || equals to zero), hence three ranges to consider:

A. $$x<{-3}$$ --> $$y=| x + 3| +|4-x| =-x-3+4-x=-2x+1$$, which means that when $$x$$ is in the range {-infinity,-3} the value of $$y$$ is defined by $$x$$ (we would have multiple choices of $$y$$ depending on $$x$$ from the given range);

B. $$-3\leq{x}\leq{4}$$ --> $$y=|x+3|+|4-x|=x+3+4-x=7$$, which means that when $$x$$ is in the range {-3,4} the value of $$y$$ is $$7$$ (value of y does not depend on value of $$x$$, when $$x$$ is from the given range);

C. $$x>{4}$$ --> $$y=|x+3|+|4-x|=x+3-4+x=2x-1$$, which means that when $$x$$ is in the range {4, +infinity} the value of $$y$$ is defined by $$x$$ (we would have multiple choices of $$y$$ depending on $$x$$ from the given range).

Hence we can definitely conclude that $$y=7$$ if $$x$$ is in the range {-3,4}

(1) $$x<4$$ --> not sufficient ($$x<4$$ but we don't know if it's $$\geq{-3}$$);
(2) $$x>-3$$ --> not sufficient ($$x>-3$$ but we don't know if it's $$\leq{4}$$);

(1)+(2) $$-3<x<4$$ exactly the range we needed, so $$y=7$$. Sufficient.

OR: looking at $$y=|x+3|+|4-x|$$ you can notice that $$y=7$$ ($$y$$ doesn't depend on the value of $$x$$) when $$x+3$$ and $$4-x$$ are both positive, in this case $$x-es$$ cancel out each other and we would have $$y=|x+3|+|4-x|=x+3+4-x=7$$. Both $$x+3$$ and $$4-x$$ are positive in the range $$-3<{x}<4$$ ($$x+3>0$$ --> $$x>-3$$ and $$4-x>0$$ --> $$x<4$$).

Hope it's clear.

Thanks for the great explanation . I got this answer when I tackled it using the same approach .. but got an E when I tried to tackle it in the +/- (x+3)=+/-(4-x) . I found y = +7 , -7 , 2x-1 , -2x+1 , and on plugging values that satisfy the 2 statements together it turned out to a range of -5 to 5 . Could you please guide ? Thanks a million
_________________

My RC Recipe
http://gmatclub.com/forum/the-rc-recipe-149577.html

My Problem Takeaway Template
http://gmatclub.com/forum/the-simplest-problem-takeaway-template-150646.html

Math Expert
Joined: 02 Sep 2009
Posts: 36548
Followers: 7077

Kudos [?]: 93107 [0], given: 10552

Re: Power Prep - DS - Modulus [#permalink]

### Show Tags

27 Mar 2013, 08:02
TheNona wrote:
Bunuel wrote:
thirst4edu wrote:
If x & y are integers and y=|x+3| + |4-x|, does y equals 7?

1) x < 4
2) x > -3

Had a hard time solving this, would like to know how to solve this using number picking approach as well as algebraic approach. Thanks.

OA is
[Reveal] Spoiler:
C

$$y=|x+3|+|4-x|$$ two check points: $$x=-3$$ and $$x=4$$ (check point: the value of $$x$$ when expression in || equals to zero), hence three ranges to consider:

A. $$x<{-3}$$ --> $$y=| x + 3| +|4-x| =-x-3+4-x=-2x+1$$, which means that when $$x$$ is in the range {-infinity,-3} the value of $$y$$ is defined by $$x$$ (we would have multiple choices of $$y$$ depending on $$x$$ from the given range);

B. $$-3\leq{x}\leq{4}$$ --> $$y=|x+3|+|4-x|=x+3+4-x=7$$, which means that when $$x$$ is in the range {-3,4} the value of $$y$$ is $$7$$ (value of y does not depend on value of $$x$$, when $$x$$ is from the given range);

C. $$x>{4}$$ --> $$y=|x+3|+|4-x|=x+3-4+x=2x-1$$, which means that when $$x$$ is in the range {4, +infinity} the value of $$y$$ is defined by $$x$$ (we would have multiple choices of $$y$$ depending on $$x$$ from the given range).

Hence we can definitely conclude that $$y=7$$ if $$x$$ is in the range {-3,4}

(1) $$x<4$$ --> not sufficient ($$x<4$$ but we don't know if it's $$\geq{-3}$$);
(2) $$x>-3$$ --> not sufficient ($$x>-3$$ but we don't know if it's $$\leq{4}$$);

(1)+(2) $$-3<x<4$$ exactly the range we needed, so $$y=7$$. Sufficient.

OR: looking at $$y=|x+3|+|4-x|$$ you can notice that $$y=7$$ ($$y$$ doesn't depend on the value of $$x$$) when $$x+3$$ and $$4-x$$ are both positive, in this case $$x-es$$ cancel out each other and we would have $$y=|x+3|+|4-x|=x+3+4-x=7$$. Both $$x+3$$ and $$4-x$$ are positive in the range $$-3<{x}<4$$ ($$x+3>0$$ --> $$x>-3$$ and $$4-x>0$$ --> $$x<4$$).

Hope it's clear.

Thanks for the great explanation . I got this answer when I tackled it using the same approach .. but got an E when I tried to tackle it in the +/- (x+3)=+/-(4-x) . I found y = +7 , -7 , 2x-1 , -2x+1 , and on plugging values that satisfy the 2 statements together it turned out to a range of -5 to 5 . Could you please guide ? Thanks a million

Could you please elaborate the red part?
_________________
Manager
Joined: 12 Dec 2012
Posts: 230
GMAT 1: 540 Q36 V28
GMAT 2: 550 Q39 V27
GMAT 3: 620 Q42 V33
GPA: 2.82
WE: Human Resources (Health Care)
Followers: 4

Kudos [?]: 78 [0], given: 181

Re: Power Prep - DS - Modulus [#permalink]

### Show Tags

27 Mar 2013, 08:37
Bunuel wrote:
Could you please elaborate the red part?

I mean that combining both statements x is between -3 and 4 , so I plugged all the values in this range in both 2x-1 and -2x+1 giving the range of Ys -5 to 5
_________________

My RC Recipe
http://gmatclub.com/forum/the-rc-recipe-149577.html

My Problem Takeaway Template
http://gmatclub.com/forum/the-simplest-problem-takeaway-template-150646.html

Math Expert
Joined: 02 Sep 2009
Posts: 36548
Followers: 7077

Kudos [?]: 93107 [0], given: 10552

Re: Power Prep - DS - Modulus [#permalink]

### Show Tags

28 Mar 2013, 10:42
TheNona wrote:
Bunuel wrote:
Could you please elaborate the red part?

I mean that combining both statements x is between -3 and 4 , so I plugged all the values in this range in both 2x-1 and -2x+1 giving the range of Ys -5 to 5

y is equal to 1-2x only when x<-3.
y is equal to 2x-1 only when x>4.

When -3<x<4, then y is equal to 7.
_________________
Intern
Joined: 12 Dec 2011
Posts: 12
Location: Italy
Concentration: Finance, Entrepreneurship
GMAT Date: 04-09-2013
GPA: 4
WE: Management Consulting (Consulting)
Followers: 0

Kudos [?]: 19 [0], given: 5

### Show Tags

07 May 2013, 13:07
Hi guys,
I would like to deeply understand how to deal with absolute value questions like the one attached.
Thank you very much
Attachments

Veritas Preparation.PNG [ 30.01 KiB | Viewed 5577 times ]

Manager
Joined: 24 Apr 2013
Posts: 54
Schools: Duke '16
Followers: 0

Kudos [?]: 10 [0], given: 76

### Show Tags

07 May 2013, 14:33
I got A as statment one alone is sufficient but 2 is a nada
Manager
Status: Pushing Hard
Affiliations: GNGO2, SSCRB
Joined: 30 Sep 2012
Posts: 89
Location: India
Concentration: Finance, Entrepreneurship
GPA: 3.33
WE: Analyst (Health Care)
Followers: 1

Kudos [?]: 83 [0], given: 11

### Show Tags

07 May 2013, 20:08
mario1987 wrote:
Hi guys,
I would like to deeply understand how to deal with absolute value questions like the one attached.
Thank you very much

The Expression given in the Question is Y = lx+3l + l4-xl .. & Question asks Is Y = 7 ????

Statement 1 :: x<4 ..... when we plugin any value of x less than 4 till -3 we will get a result as Y = 7 but below -3 ... Y will not be equal to 7 . Therefore, Insufficient.

Similarly, Statement :: 2 .... is insufficient.......

with 1+2 ..... we get the value of x in between 4 & -3 ...... i.e., -3<x<4 ....... For all the value of x in between -3 & 4 ... the value of

Y is always. 7 .. Therefore, Sufficient....

Hence, C ...........
_________________

If you don’t make mistakes, you’re not working hard. And Now that’s a Huge mistake.

Re: Algebra->Inequalities (Absolute Value)   [#permalink] 07 May 2013, 20:08

Go to page    1   2    Next  [ 27 posts ]

Similar topics Replies Last post
Similar
Topics:
3 If x and y are prime numbers, is y(x-3) odd? 8 07 Jul 2013, 21:40
1 If x and y are prime numbers, is y(x-3) odd? 9 02 Jan 2011, 13:08
5 If xy does not equal 0, what is the value of ((x^2)^3 y^4)/x 7 19 Oct 2010, 11:15
1 If x and y are integers and y = |x + 3| + |4 - x|, does y equal 7? 3 29 Jan 2010, 10:55
12 If x and y are integers and y = | x + 3| + | 4 x|. Does y 12 31 Oct 2009, 22:38
Display posts from previous: Sort by