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

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If x and y are integers and y=|x+3| + |4-x|, does y equals 7 [#permalink] New post 26 Sep 2010, 20:26
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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

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Re: Power Prep - DS - Modulus [#permalink] New post 26 Sep 2010, 22:06
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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.

Answer: C.

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.
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Re: Power Prep - DS - Modulus [#permalink] New post 26 Sep 2010, 22:09
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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

Answer is (c)
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Re: Power Prep - DS - Modulus [#permalink] New post 26 Sep 2010, 23:47
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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.

Answer C.

Hope it helps
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Re: Power Prep - DS - Modulus [#permalink] New post 27 Sep 2010, 10:51
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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)
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Re: Power Prep - DS - Modulus [#permalink] New post 28 Sep 2010, 09:58
thanks bunuel. always great explanations!!
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Re: Power Prep - DS - Modulus [#permalink] New post 29 Sep 2010, 09:21
Bunuel rocks, cheers mate
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Re: If x and y are integers and y=|x+3| + |4-x|, does y equals 7 [#permalink] New post 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
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Re: If x and y are integers and y=|x+3| + |4-x|, does y equals 7 [#permalink] New post 28 May 2012, 04:23
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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.
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Re: Power Prep - DS - Modulus [#permalink] New post 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.

Answer: C.

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?
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Re: Power Prep - DS - Modulus [#permalink] New post 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.

Answer: C.

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?
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Re: Power Prep - DS - Modulus [#permalink] New post 01 Jun 2012, 06:37
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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.
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Re: If x and y are integers and y=|x+3| + |4-x|, does y equals 7 [#permalink] New post 20 Dec 2012, 08:01
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
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Re: Power Prep - DS - Modulus [#permalink] New post 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.

Answer: C.

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 :)
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Re: Power Prep - DS - Modulus [#permalink] New post 27 Mar 2013, 08:02
Expert's post
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.

Answer: C.

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?
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Re: Power Prep - DS - Modulus [#permalink] New post 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
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Re: Power Prep - DS - Modulus [#permalink] New post 28 Mar 2013, 10:42
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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.
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Algebra->Inequalities (Absolute Value) [#permalink] New post 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
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Re: Algebra->Inequalities (Absolute Value) [#permalink] New post 07 May 2013, 14:33
I got A as statment one alone is sufficient but 2 is a nada
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Re: Algebra->Inequalities (Absolute Value) [#permalink] New post 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 ...........
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Re: Algebra->Inequalities (Absolute Value)   [#permalink] 07 May 2013, 20:08
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