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if x and y are integers and y=|x+3| +|4-x|, does y equal 7

If we carefully look at the question we can see that "does y equal 7" actually means that y doesn't depend on x and then you can see that if we open two moduli with same signs (++ or --), x and -x disappear and 3+4 is 7. So, let's see when we can open moduli ++ or --:

++) x>-3 and x<4 or x e (-3,4)
--) x<-3 and x>4 - it can't be.

So, if x between -3 and 4, y =7. Now, look at our statements: it is obvious that we need two statements.

C.
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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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thirst4edu
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

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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thanks bunuel. always great explanations!!
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Bunuel rocks, cheers mate
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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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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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thirst4edu
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

\(y=|x+3|+|4-x|\) 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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manulath

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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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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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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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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Hey nishitraj,

Little confused here could you explain by picking numbers for statement 1 and 2.
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Amerendra
Hey nishitraj,

Little confused here could you explain by picking numbers for statement 1 and 2.


Hi...

y=|x+3|+|4-x|

Without solving you can say for all values between -3 and 4, y will be 7 as any increase in x in one mod will be negated with decrease in other mod..

Let's see the statements
1) x<4...
Take x as 0..
y=|0+3|+|4-0|=3+4=7
Take x as -5
y=|-5+3|+|4-(-5)|=|-2|+|9|=2+9=11
So different answers
Insufficient
2) x>-3
Take x as -2
y=|-2+3|+|4-(-2)|=|1|+|4+2|=1+6=7
Take x as 7
y=|3+7|+|4-7|=10+3=13
Different answers
Insufficient

Combined sufficient

C
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I got everything right and was on my way to choose option C but then I tried using x = -3 and it worked. But the option says x > -3, whereas -3 is also a valid option. That is why I though of choosing E because C does not contain the whole range.
zaarathelab
If x and y are integers and y = |x + 3| + |4 - x|, does y equal 7?

(1) x < 4
(2) x > -3
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Ishaan30
I got everything right and was on my way to choose option C but then I tried using x = -3 and it worked. But the option says x > -3, whereas -3 is also a valid option. That is why I though of choosing E because C does not contain the whole range.
zaarathelab
If x and y are integers and y = |x + 3| + |4 - x|, does y equal 7?

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

We got that if −3 ≤ x ≤ 4, then |x + 3| + |4 - x| = 7. When combining the statements, we get −3 < x < 4, so x can be any number from the desired range except −3 and 4. However, the values x can take are still within the desired range, so |x + 3| + |4 - x| = 7.

P.S. Pure algebraic questions are no longer a part of the DS syllabus of the GMAT.

DS questions in GMAT Focus encompass various types of word problems, such as:

  • Word Problems
  • Work Problems
  • Distance Problems
  • Mixture Problems
  • Percent and Interest Problems
  • Overlapping Sets Problems
  • Statistics Problems
  • Combination and Probability Problems

While these questions may involve or necessitate knowledge of algebra, arithmetic, inequalities, etc., they will always be presented in the form of word problems. You won’t encounter pure "algebra" questions like, "Is x > y?" or "A positive integer n has two prime factors..."

Check GMAT Syllabus for Focus Edition

You can also visit the Data Sufficiency forum and filter questions by OG 2024-2025, GMAT Prep (Focus), and Data Insights Review 2024-2025 sources to see the types of questions currently tested on the GMAT.

So, you can ignore this question.

Hope it helps.­
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Ishaan30
I got everything right and was on my way to choose option C but then I tried using x = -3 and it worked. But the option says x > -3, whereas -3 is also a valid option. That is why I though of choosing E because C does not contain the whole range.
zaarathelab
If x and y are integers and y = |x + 3| + |4 - x|, does y equal 7?

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

We got that if −3 ≤ x ≤ 4, then |x + 3| + |4 - x| = 7. When combining the statements, we get −3 < x < 4, so x can be any number from the desired range except −3 and 4. However, the values x can take are still within the desired range, so |x + 3| + |4 - x| = 7.

P.S. Pure algebraic questions are no longer a part of the DS syllabus of the GMAT.

DS questions in GMAT Focus encompass various types of word problems, such as:

  • Word Problems
  • Work Problems
  • Distance Problems
  • Mixture Problems
  • Percent and Interest Problems
  • Overlapping Sets Problems
  • Statistics Problems
  • Combination and Probability Problems

While these questions may involve or necessitate knowledge of algebra, arithmetic, inequalities, etc., they will always be presented in the form of word problems. You won’t encounter pure "algebra" questions like, "Is x > y?" or "A positive integer n has two prime factors..."

Check GMAT Syllabus for Focus Edition

You can also visit the Data Sufficiency forum and filter questions by OG 2024-2025, GMAT Prep (Focus), and Data Insights Review 2024-2025 sources to see the types of questions currently tested on the GMAT.

So, you can ignore this question.

Hope it helps.­
Oh okay! This makes sense. They're not asking for every value, they're asking if the values within the given range verifies.
And thanks for the additional wisdom. I feel lost sometimes so this really helped.
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