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\(y=|x+3|+|4-x|\) 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 \(4-x>0\) --> \(x<4\)).
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
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.
(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.
Re: If x and y are integers and y=|x+3| + |4-x|, does y equals 7 [#permalink]
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
Re: If x and y are integers and y=|x+3| + |4-x|, does y equals 7 [#permalink]
28 May 2012, 04:23
Expert's post
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 \(|4-x|=4-x\) --> \(|x+3|+|4-x|=-(x+3)+4-x=-2x+1\).
\(y=|x+3|+|4-x|\) 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?
\(y=|x+3|+|4-x|\) 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?
\(y=|x+3|+|4-x|\) 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 _________________
\(y=|x+3|+|4-x|\) 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? _________________
Re: Power Prep - DS - Modulus [#permalink]
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 _________________
Re: Power Prep - DS - Modulus [#permalink]
28 Mar 2013, 10:42
Expert's post
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. _________________
Re: Algebra->Inequalities (Absolute Value) [#permalink]
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.
gmatclubot
Re: Algebra->Inequalities (Absolute Value)
[#permalink]
07 May 2013, 20:08
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