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Posts: 112
Updated on: 27 Jul 2015, 13:57
Originally posted by zaarathelab on 29 Jan 2010, 10:55.
Last edited by Bunuel on 27 Jul 2015, 13:57, edited 1 time in total.
Last edited by Bunuel on 27 Jul 2015, 13:57, edited 1 time in total.
Renamed the topic, edited the question and added the OA.
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
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If x and y are integers and y = |x + 3| + |4 - x|, does y equal 7?
(1) x < 4
(2) x > -3
(1) x < 4
(2) x > -3
Solving this question helps.
Taking a timed set of similar questions in
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is even better.
If x and y are integers and y = |x + 3| + |4 - x|, does y equal 7?
\(y=|x+3|+|4-x|\) 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.
_________________
\(y=|x+3|+|4-x|\) 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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thirst4edu
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
General Discussion
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.
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.
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)
_________________
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)
_________________
thirst4edu
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)
Posts: 53
thanks bunuel. always great explanations!!
Bunuel rocks, cheers mate
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
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
kartik222
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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Bunuel
MOD questions always floor me.
Could you please suggest some good material on MODs?
manulath
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
|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
TheNona
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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
TheNona
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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Amerendra
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
_________________
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
Ishaan30
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.
Bunuel
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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