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If x and y are integers and y=x+3 + 4x, does y equals 7 [#permalink]
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If x and y are integers and y=x+3 + 4x, 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
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Re: Power Prep  DS  Modulus [#permalink]
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thirst4edu wrote: If x & y are integers and y=x+3 + 4x, 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 \(y=x+3+4x\) 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. \(x<{3}\) > \(y= x + 3 +4x =x3+4x=2x+1\), 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. \(3\leq{x}\leq{4}\) > \(y=x+3+4x=x+3+4x=7\), 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. \(x>{4}\) > \(y=x+3+4x=x+34+x=2x1\), 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 \(y=x+3+4x\) you can notice that \(y=7\) (\(y\) doesn't depend on the value of \(x\)) when \(x+3\) and \(4x\) are both positive, in this case \(xes\) cancel out each other and we would have \(y=x+3+4x=x+3+4x=7\). Both \(x+3\) and \(4x\) are positive in the range \(3<{x}<4\) (\(x+3>0\) > \(x>3\) and \(4x>0\) > \(x<4\)). Hope it's clear.
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Re: Power Prep  DS  Modulus [#permalink]
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Let's first solve x+3+4x=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 xa 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]
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thirst4edu wrote: If x & y are integers and y=x+3 + 4x, 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 Guys, the be low is my approcah for any modulus qtn in GMAT. Remember. The meaning of xy 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 + 4x, 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]
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thirst4edu wrote: If x & y are integers and y=x+3 + 4x, 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 + 43 = 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]
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28 Sep 2010, 10:58
thanks bunuel. always great explanations!!



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Re: Power Prep  DS  Modulus [#permalink]
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29 Sep 2010, 10:21
Bunuel rocks, cheers mate



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Re: If x and y are integers and y=x+3 + 4x, does y equals 7 [#permalink]
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26 May 2012, 08:38
At bunuel,
A. x<{3} > y= x + 3 +4x =x3+4x=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+4x=x+3+4x=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+4x=x+34+x=2x1, 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 +4x how did you decide sign of "x" here ===> x3+4x 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 + 4x, does y equals 7 [#permalink]
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28 May 2012, 05:23
kartik222 wrote: At bunuel,
A. x<{3} > y= x + 3 +4x =x3+4x=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+4x=x+3+4x=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+4x=x+34+x=2x1, 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 +4x how did you decide sign of "x" here ===> x3+4x 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 \(4x>0\) which means that \(x+3=(x+3)\) and \(4x=4x\) > \(x+3+4x=(x+3)+4x=2x+1\). Similarly for B and C. Hope it's clear.
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Re: Power Prep  DS  Modulus [#permalink]
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28 May 2012, 09:10
Bunuel wrote: thirst4edu wrote: If x & y are integers and y=x+3 + 4x, 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 \(y=x+3+4x\) 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. \(x<{3}\) > \(y= x + 3 +4x =x3+4x=2x+1\), 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. \(3\leq{x}\leq{4}\) > \(y=x+3+4x=x+3+4x=7\), 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. \(x>{4}\) > \(y=x+3+4x=x+34+x=2x1\), 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 \(y=x+3+4x\) you can notice that \(y=7\) (\(y\) doesn't depend on the value of \(x\)) when \(x+3\) and \(4x\) are both positive, in this case \(xes\) cancel out each other and we would have \(y=x+3+4x=x+3+4x=7\). Both \(x+3\) and \(4x\) are positive in the range \(3<{x}<4\) (\(x+3>0\) > \(x>3\) and \(4x>0\) > \(x<4\)). 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]
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01 Jun 2012, 05:36
Bunuel wrote: thirst4edu wrote: If x & y are integers and y=x+3 + 4x, 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 \(y=x+3+4x\) 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. \(x<{3}\) > \(y= x + 3 +4x =x3+4x=2x+1\), 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. \(3\leq{x}\leq{4}\) > \(y=x+3+4x=x+3+4x=7\), 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. \(x>{4}\) > \(y=x+3+4x=x+34+x=2x1\), 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 \(y=x+3+4x\) you can notice that \(y=7\) (\(y\) doesn't depend on the value of \(x\)) when \(x+3\) and \(4x\) are both positive, in this case \(xes\) cancel out each other and we would have \(y=x+3+4x=x+3+4x=7\). Both \(x+3\) and \(4x\) are positive in the range \(3<{x}<4\) (\(x+3>0\) > \(x>3\) and \(4x>0\) > \(x<4\)). 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]
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Re: If x and y are integers and y=x+3 + 4x, does y equals 7 [#permalink]
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Thinking of modulus as distances x+3 => distance of x from 3 x4 => 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 Bangon. Hence C
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Re: Power Prep  DS  Modulus [#permalink]
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27 Mar 2013, 07:58
Bunuel wrote: thirst4edu wrote: If x & y are integers and y=x+3 + 4x, 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 \(y=x+3+4x\) 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. \(x<{3}\) > \(y= x + 3 +4x =x3+4x=2x+1\), 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. \(3\leq{x}\leq{4}\) > \(y=x+3+4x=x+3+4x=7\), 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. \(x>{4}\) > \(y=x+3+4x=x+34+x=2x1\), 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 \(y=x+3+4x\) you can notice that \(y=7\) (\(y\) doesn't depend on the value of \(x\)) when \(x+3\) and \(4x\) are both positive, in this case \(xes\) cancel out each other and we would have \(y=x+3+4x=x+3+4x=7\). Both \(x+3\) and \(4x\) are positive in the range \(3<{x}<4\) (\(x+3>0\) > \(x>3\) and \(4x>0\) > \(x<4\)). 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)=+/(4x) . I found y = +7 , 7 , 2x1 , 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]
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27 Mar 2013, 09:02
TheNona wrote: Bunuel wrote: thirst4edu wrote: If x & y are integers and y=x+3 + 4x, 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 \(y=x+3+4x\) 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. \(x<{3}\) > \(y= x + 3 +4x =x3+4x=2x+1\), 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. \(3\leq{x}\leq{4}\) > \(y=x+3+4x=x+3+4x=7\), 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. \(x>{4}\) > \(y=x+3+4x=x+34+x=2x1\), 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 \(y=x+3+4x\) you can notice that \(y=7\) (\(y\) doesn't depend on the value of \(x\)) when \(x+3\) and \(4x\) are both positive, in this case \(xes\) cancel out each other and we would have \(y=x+3+4x=x+3+4x=7\). Both \(x+3\) and \(4x\) are positive in the range \(3<{x}<4\) (\(x+3>0\) > \(x>3\) and \(4x>0\) > \(x<4\)). 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)=+/(4x) . I found y = +7 , 7 , 2x1 , 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]
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27 Mar 2013, 09:37
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 2x1 and 2x+1 giving the range of Ys 5 to 5
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Algebra>Inequalities (Absolute Value) [#permalink]
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07 May 2013, 14:07
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]
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07 May 2013, 15:33
I got A as statment one alone is sufficient but 2 is a nada



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Re: Algebra>Inequalities (Absolute Value) [#permalink]
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07 May 2013, 21:08
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 + l4xl .. & 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)
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07 May 2013, 21:08



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