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If x+3  12 < 13, what is the range of x?
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Updated on: 13 Sep 2018, 08:20
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Different methods to solve absolute value equations and inequalities Exercise Question #5If x+3  12 < 13, what is the range of x? Options a) (4, 22) b) (28, 2) c) (28, 22) d) (28, 2) U (2, 22) e) (28, 2) U (4, 22) Previous Question To read the article: Different methods to solve absolute value equations and inequalities
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If x+3  12 < 13, what is the range of x?
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Updated on: 17 Sep 2018, 04:04
Solution Given: We are given an absolute value inequality, x+3  12 < 13 Note that we do not have any constraint on x. So, x can be an integer or a decimal To find: We need to find the range of values of x, that satisfies the given inequality Approach and Working: First, let us substitute the inner modulus function, x + 3, with t, which gives, t  12 < 13 We know that, the range of x, for x < a, is a < x < a So, the range of x, for which t  12 < 13, is
13 < t – 12 < 13, Adding 12 on all the sides, we get, 1 < t < 25 Now, substituting, back the value of t as x + 3, we get, 1 < x + 3 < 25
Since, the value of x + 3 is always greater than or equal to 0 Hence, 0 ≤ x + 3 < 25 Now, solving the inequality, x + 3 < 25, we get the range of x as (28, 22) Therefore, the range of x, that satisfies the inequality, x+3  12 < 13, is (28, 22) Hence, the correct answer is option C. Answer: C
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If x+3  12 < 13, what is the range of x?
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Updated on: 09 Oct 2018, 06:36
My Approach: x+3 will always be positive. Therefore x+3<25 in order for x+3 12< 13 to be true! x+3<25 > 28<x<22 Answer: C Please correct me if I went wrong somewhere!!!
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Originally posted by T1101 on 13 Sep 2018, 08:35.
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If x+3  12 < 13, what is the range of x?
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13 Sep 2018, 14:47
Hi Payal: Is my process okay for this problem? I solved the problem this way: Since the equation is pretty messy(by my standards), I decided to use substitution. x+3  12 < 13
1.) I let x+3 = a and then rephrased the eqn a12 < 13 a  12 < 13 and (a25) < 13 > Since x+3 > 0 I didn't bother to solve the 2nd eqn a < 25
2.) I plugged in the original value of a to find the range x+3 < 25 x+3 < 25 and (x+3) < 25 x < 22 x> 28
Answer C



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Re: If x+3  12 < 13, what is the range of x?
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14 Sep 2018, 13:15
EgmatQuantExpert wrote: If x+3  12 < 13, what is the range of x? Options a) (4, 22) b) (28, 2) c) (28, 22) d) (28, 2) U (2, 22) e) (28, 2) U (4, 22) \(?\,\,\,:\,\,\,{\text{solution}}\,\,{\text{set}}\) The geometric interpretation of (some) absolute values is essential to solve this sort of question quickly and safely: \(\left {z  w} \right = {\text{dist}}\left( {z,w} \right)\) \(\left {k + m} \right = \left {k  \left( {  m} \right)} \right = {\text{dist}}\left( {k,  m} \right)\) The application of these ideas is used in the particular case: \(a = \left {x + 3} \right\) \(?\,\,\,:\,\,\,\,\left {a  12} \right < 13\,\,\,\,\,\,\,\left( {{\text{variable}}\,\,x} \right)\,\,\,\,\,\) See image attached. Right answer: (C) This solution follows the notations and rationale taught in the GMATH method. Regards, Fabio.
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If x+3  12 < 13, what is the range of x?
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Updated on: 23 Sep 2018, 04:17
EgmatQuantExpert wrote: Solution Given: We are given an absolute value inequality, x+3  12 < 13 Note that we do not have any constraint on x. So, x can be an integer or a decimal To find: We need to find the range of values of x, that satisfies the given inequality Approach and Working: First, let us substitute the inner modulus function, x + 3, with t, which gives, t  12 < 13 We know that, the range of x, for x < a, is a < x < a So, the range of x, for which t  12 < 13, is
13 < t – 12 < 13, Adding 12 on all the sides, we get, 1 < t < 25 Now, substituting, back the value of t as x + 3, we get, 1 < x + 3 < 25
Since, the value of x + 3 is always greater than or equal to 0 Hence, 0 ≤ x + 3 < 25 Now, solving the inequality, x + 3 < 25, we get the range of x as (28, 22) Therefore, the range of x, that satisfies the inequality, x+3  12 < 13, is (28, 22) Hence, the correct answer is option C. Answer: CHi Payal & EgmatQuantExpert , thanks for explanation but I have a doubt which may be stupid . Normally for x < a we do have answer as a < x < a , then why for 1 < x + 3 < 25 we made it 0 ≤ x + 3 < 25. Again in our final answer we had negative range inclusive > (28, 22) . Please clear this doubt . Thanks
Originally posted by KARISHMA315 on 23 Sep 2018, 03:56.
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Re: If x+3  12 < 13, what is the range of x?
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23 Sep 2018, 04:17
In this problem, the easiest way would to mind answers for information. If x+3  12 < 13, what is the range of x? Our bounderis should satisfy x+312 = 13 equation. a) (4, 22) Let's put 4 in our equation: 4+312 = 11 => A out b) (28, 2) 28+312 = 13 2+312 = 11 => B out c) (28, 22) 28  good 22+312 = 13 good, but we need to check others d) (28, 2) U (2, 22) 28  good 2+312 = 7 => D out e) (28, 2) U (4, 22) 28  good 2  bad => E out
So, we know that only C



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Re: If x+3  12 < 13, what is the range of x?
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26 Sep 2018, 13:45
KARISHMA315, sorry that I can't quote but here is the answer: Hi! That is because the absolute value can't be negative, so zero is the lesser value that x+3 can have. Then, you can rewrite the equation: \(0 ≤ x + 3 < 25\) \(x + 3 < 25\) The second equation is analog because an inequation with absolute value is nonnegative by definition. Thus, the solution is straightforward: \(25 < x + 3 < 25\) \(28 < x < 22\) Hope it will be useful



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Re: If x+3  12 < 13, what is the range of x?
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09 Oct 2018, 06:33
You can do this by substitution: E+D+B > Try x = 2 > the equation is still correct > only A, C left A,C > Try x = 5 > the equation is still correct > C left > C is the correct answer



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If x+3  12 < 13, what is the range of x?
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14 Oct 2018, 03:44
EgmatQuantExpert wrote: Solution Given: We are given an absolute value inequality, x+3  12 < 13 Note that we do not have any constraint on x. So, x can be an integer or a decimal To find: We need to find the range of values of x, that satisfies the given inequality Approach and Working: First, let us substitute the inner modulus function, x + 3, with t, which gives, t  12 < 13 We know that, the range of x, for x < a, is a < x < a So, the range of x, for which t  12 < 13, is
13 < t – 12 < 13, Adding 12 on all the sides, we get, 1 < t < 25 Now, substituting, back the value of t as x + 3, we get, 1 < x + 3 < 25
Since, the value of x + 3 is always greater than or equal to 0 Hence, 0 ≤ x + 3 < 25 Now, solving the inequality, x + 3 < 25, we get the range of x as (28, 22) Therefore, the range of x, that satisfies the inequality, x+3  12 < 13, is (28, 22) Hence, the correct answer is option C. Answer: CHi EgmatQuantExpertbased on which formula/rule do you assume that the value of x + 3 is always greater than or equal to 0 we always have two cases positive and negative.... isn't it so ? can you please explain hey Are you there pushpitkc maybe you know answer to my beautiful question



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Re: If x+3  12 < 13, what is the range of x?
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14 Oct 2018, 07:11
Hey dave13The cardinal rule for an absolute value expression is that it can never be negative. For example 2 = +2 = 2 (whatever comes outside the is ALWAYS positive) In the problem we have the expression x+3  12 < 13. The minimum value for the part x+3 is 0, when x = 3. For anything else, the value will be greater than 0. For instance, if x = 4, x+3 = 1 = 1(which is again positive) Hope that helps!
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If x+3  12 < 13, what is the range of x?
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Updated on: 14 Oct 2018, 13:37
pushpitkc wrote: Hey dave13The cardinal rule for an absolute value expression is that it can never be negative. For example 2 = +2 = 2 (whatever comes outside the is ALWAYS positive) In the problem we have the expression x+3  12 < 13. The minimum value for the part x+3 is 0, when x = 3. For anything else, the value will be greater than 0. For instance, if x = 4, x+3 = 1 = 1(which is again positive) Hope that helps! Hi pushpitkc, thanks for nice explanation, if so then why in the example below when we get 2 < x  3 < 14 consider two cases negative and positive ? ... whereas in this question when we get, 1 < x + 3 < 25 we don't consider TWO cases ? Any idea generis ? who knows, may be you know Since, the value of x + 3 is always greater than or equal to 0 Hence, 0 ≤ x + 3 < 25 Now, solving the inequality, x + 3 < 25, we get the range of x as (28, 22)Example 5If x  3  8 < 6, what is the range of x?SolutionWe solved a question of double modulus in absolute value equations. So, we will follow the similar approach. First, substitute x  3 as y, and write the inequality as, • y  8 < 6 • Now, this is in the form of x < a • Therefore, the range of x is a < x < a.
o Similarly, we can find the range of y if we substitute x by y8 and a by 6.
Thus, 6 < y8< 6 Adding 8 on all the sides of the inequality, we get: 2< y< 14.
Now, we can substitute back the value of y as x  3 and get: • 2 < x  3 < 14 • Here, we need to consider two cases, when x – 3 is positive and when x  3 is negative
o Considering first case, x – 3 is positive, we have:
2 < x  3 < 14 Simplifying this inequality, we get
o Now, considering second case, x – 3 is negative, we have:
2 <  (x – 3) < 14 So, let us now multiply by 1 on all the sides, we get
• 2 > x – 3 > 14 or 14 < x – 3 < 2 Note: Did you notice that, the sign of the inequality is changed when you multiply the whole expression by 1?
So, always keep in mind to change the signs when you multiply an inequality by a negative number. o Simplifying this inequality, we get
Therefore, the range of x is (1) ⋃ (2), which is equal to (11, 1) ⋃ (5, 17). https://gmatclub.com/forum/differentme ... l#p2130767
Originally posted by dave13 on 14 Oct 2018, 07:22.
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Re: If x+3  12 < 13, what is the range of x?
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14 Oct 2018, 07:40
EgmatQuantExpert wrote: Different methods to solve absolute value equations and inequalities Exercise Question #5If x+3  12 < 13, what is the range of x? Options a) (4, 22) b) (28, 2) c) (28, 22) d) (28, 2) U (2, 22) e) (28, 2) U (4, 22) There are 2 possible cases: (i) \((x+3)12 < 13\) (ii) \((x+3)12 < 13\) Case (i): \(x+312 < 13\) \(x9 < 13\) \(x < 22\) Case (ii): \(x312 < 13\) \(x15 < 13\) \(x < 28\) \(x > 28\) Range of \(x: 28 < x < 22\).



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Re: If x+3  12 < 13, what is the range of x?
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14 Oct 2018, 08:01
dave13 wrote: pushpitkc wrote: Hey dave13The cardinal rule for an absolute value expression is that it can never be negative. For example 2 = +2 = 2 (whatever comes outside the is ALWAYS positive) In the problem we have the expression x+3  12 < 13. The minimum value for the part x+3 is 0, when x = 3. For anything else, the value will be greater than 0. For instance, if x = 4, x+3 = 1 = 1(which is again positive) Hope that helps! Hi pushpitkc, thanks for nice explanation, if so then why in the example below when we get 2 < x  3 < 14 consider two cases negative and positive ? ... whereas in this question when we get, 1 < x + 3 < 25 we don't consider TWO cases ? Since, the value of x + 3 is always greater than or equal to 0 Hence, 0 ≤ x + 3 < 25 Now, solving the inequality, x + 3 < 25, we get the range of x as (28, 22)Example 5If x  3  8 < 6, what is the range of x?SolutionWe solved a question of double modulus in absolute value equations. So, we will follow the similar approach. First, substitute x  3 as y, and write the inequality as, • y  8 < 6 • Now, this is in the form of x < a • Therefore, the range of x is a < x < a.
o Similarly, we can find the range of y if we substitute x by y8 and a by 6.
Thus, 6 < y8< 6 Adding 8 on all the sides of the inequality, we get: 2< y< 14.
Now, we can substitute back the value of y as x  3 and get: • 2 < x  3 < 14 • Here, we need to consider two cases, when x – 3 is positive and when x  3 is negative
o Considering first case, x – 3 is positive, we have:
2 < x  3 < 14 Simplifying this inequality, we get
o Now, considering second case, x – 3 is negative, we have:
2 <  (x – 3) < 14 So, let us now multiply by 1 on all the sides, we get
• 2 > x – 3 > 14 or 14 < x – 3 < 2 Note: Did you notice that, the sign of the inequality is changed when you multiply the whole expression by 1?
So, always keep in mind to change the signs when you multiply an inequality by a negative number. o Simplifying this inequality, we get
Therefore, the range of x is (1) ⋃ (2), which is equal to (11, 1) ⋃ (5, 17). https://gmatclub.com/forum/differentme ... l#p2130767Why x is written as x, when absolute value represents the positive value of the number. • However, if you pay attention to the constraint that “if x is negative”, then x is equal to x. • And, we all know that negative of a negative number is always positive. o Hence, x is always positive. Hope this helps. For more on inequality please refer: https://gmatclub.com/forum/differentme ... l#p2130767 Please hit Kudos if it helps!



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If x+3  12 < 13, what is the range of x?
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14 Oct 2018, 22:55
EgmatQuantExpert wrote: Different methods to solve absolute value equations and inequalities Exercise Question #5If x+3  12 < 13, what is the range of x? Options a) (4, 22) b) (28, 2) c) (28, 22) d) (28, 2) U (2, 22) e) (28, 2) U (4, 22) dave13  whenever I see "nested" absolute value inequalities, I break the problem into pieces or stages. That process often includes considering negative and positive cases. I think that this approach is easier and more basic than the theoretical references I see. Essentially, we simplify one part of the LHS for part of the time that we solve. We have to find the range of solutions for a "nested" absolute value inequality. That situation is complicated. We want to make it simpler. 1) Make the LHS simpler.Change the inside absolute value expression with a placeholder for a little while. I will let \(u\) stand for \(x+3\) \(x+3\) becomes \(u\). That is, \(u=x+3\) I will "put back" \(x+3\) later. Change the original: \(x+3  12 < 13\) Changed: NOW we have: \(u  12 < 13\)I use \(u\) without brackets to solve more easily BUT I do keep in mind that the brackets are "in the background" \(u\) stands for an absolute value expression, x + 3 \(u\) is "really" an absolute value expression One way to think about absolute value: it measures distance from 0. By definition, distances cannot be negative. Absolute value must be positive or 0. Absolute value can never be negative. 3) Using \(u  12 < 13\) proceed as usual Case 1: the value inside \(u12\) is positive. (LHS is positive) Simply remove the absolute value bars \(u12<13\) \(u<25\)That result is okay. RHS is positive. Case 2: \(u12\) is negative* (LHS is negative. I used a shortcut to get from \((u12)<25\) to \(u12>25\) . The shortcut is in the footnote*) \(u12>25\)\(u>13\)No. Cannot use this case. RHS is negative. \(u\) is actually an absolute value expression (is really x+3). Absolute values cannot be negative. Ignore that case. Toss it in the trash. Case 1 works. Case 2 does not work. Next stage. 5) Use \(u<25\) from Case 1. "Put back" \(x+3\). Proceed as usual We have yet another inequality: \(x+3<25\) Case 1: LHS is positive \(x+3<25\) \(x<22\)Case 2: LHS is negative \(x+3>25\) \(x>28\), which is the same as \(28<x\)Those two values give us our range of solutions \(28<x<22\) End values that specify this range are \((28,22)\) I hope that helps. Answer C *Finding the negative case in an absolute value inequality. I took a shortcut. Use x + 3 < 25, from #5, above. The case in which the value inside the brackets is negative is written x + 3 > 25 That inequality does not appear to be "LHS is negative." It is. I used a shortcut / mechanical rule.
Mechanical Rule  In an absolute value inequality, to find Case 2 in which LHS is negative: Remove absolute value brackets, add a negative sign to RHS, and flip the sign. Start: x + 3 < 25 , negative case • EITHER  (x + 3) < 25 (1) (x + 3) < 25 Divide by negative 1. Dividing an inequality by a negative number flips the sign.\((\frac{1)(x+3)}{(1)}> \frac{25}{1}\)x + 3 > 25 Then simplify to x > 28 • OR  (x + 3) < 25 x  3 < 25 x < 28 (1) * (x) < 28\(\frac{(1)(x)}{(1)}>\frac{28}{1}\)x > 28 **Regarding the theoretical references (I am not sure they're helpful, but . . .) In the statement "if x is less than 0, then x = x " The right hand side is NOT a negative number. It is the negative of a negative number: RHS is positive. Try ANY negative number: 3 = 3 3 = (3) The absolute value of 3 is 3, and that absolute value, 3, is the opposite of negative 3. The negative sign in variable x "hides." You may want to take a look at Bunuel 's post starting HERE. My post follows: Approaches: If x is negative, then x = x . Finally, just below that post Bunuel here: finishes the discussion.
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If x+3  12 < 13, what is the range of x?
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14 Oct 2018, 23:08
elincopi wrote: KARISHMA315, sorry that I can't quote but here is the answer: Hi! That is because the absolute value can't be negative, so zero is the lesser value that x+3 can have. Then, you can rewrite the equation: \(0 ≤ x + 3 < 25\) \(x + 3 < 25\) The second equation is analog because an inequation with absolute value is nonnegative by definition. Thus, the solution is straightforward: \(25 < x + 3 < 25\) \(28 < x < 22\) Hope it will be useful elincopi your explanation was good, and more importantly, you went out of your way to help quite awhile ago. I think it's time you were thanked. +1 for both the explanation (which will indeed be useful) and the graciousness P. S. In order to make sure that a person sees a post, make sure that the person is tagged. To tag:  type the '@' sign  put the person's username right next to it, NO space, but  leave ONE space afterwards. (In other words, if you want to call on KARISHMA315 and put a comma after the tag, you have to leave a space after the last letter of the tag, then put the comma.)
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If x+3  12 < 13, what is the range of x?
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14 May 2019, 23:57
dave13 wrote: pushpitkc wrote: Hey dave13The cardinal rule for an absolute value expression is that it can never be negative. For example 2 = +2 = 2 (whatever comes outside the is ALWAYS positive) In the problem we have the expression x+3  12 < 13. The minimum value for the part x+3 is 0, when x = 3. For anything else, the value will be greater than 0. For instance, if x = 4, x+3 = 1 = 1(which is again positive) Hope that helps! Hi pushpitkc, thanks for nice explanation, if so then why in the example below when we get 2 < x  3 < 14 consider two cases negative and positive ? ... whereas in this question when we get, 1 < x + 3 < 25 we don't consider TWO cases ? Any idea generis ? who knows, may be you know Since, the value of x + 3 is always greater than or equal to 0 Hence, 0 ≤ x + 3 < 25 Now, solving the inequality, x + 3 < 25, we get the range of x as (28, 22)Example 5If x  3  8 < 6, what is the range of x?SolutionWe solved a question of double modulus in absolute value equations. So, we will follow the similar approach. First, substitute x  3 as y, and write the inequality as, • y  8 < 6 • Now, this is in the form of x < a • Therefore, the range of x is a < x < a.
o Similarly, we can find the range of y if we substitute x by y8 and a by 6.
Thus, 6 < y8< 6 Adding 8 on all the sides of the inequality, we get: 2< y< 14.
Now, we can substitute back the value of y as x  3 and get: • 2 < x  3 < 14 • Here, we need to consider two cases, when x – 3 is positive and when x  3 is negative
o Considering first case, x – 3 is positive, we have:
2 < x  3 < 14 Simplifying this inequality, we get
o Now, considering second case, x – 3 is negative, we have:
2 <  (x – 3) < 14 So, let us now multiply by 1 on all the sides, we get
• 2 > x – 3 > 14 or 14 < x – 3 < 2 Note: Did you notice that, the sign of the inequality is changed when you multiply the whole expression by 1?
So, always keep in mind to change the signs when you multiply an inequality by a negative number. o Simplifying this inequality, we get
Therefore, the range of x is (1) ⋃ (2), which is equal to (11, 1) ⋃ (5, 17). https://gmatclub.com/forum/differentme ... l#p2130767As generis explained the absolute value is the measure of the distance from 0. that is why the absolute value can never be negative. Eg.1 x=3 which can also be written as x0=3 means the distance from zero is 3 units. i.e. x can be 3 or 3 Eg.2 x2=5 means the distance from 2 is 5 units. i.e. 7 or 3 Eg.3 x+5=13 can be written as x(5)=13 means the distance from 5 is 13 units i.e. 18 or 8 Eg.4 x+3<13 we generally solve the equation as follows 13<x+3<13 16< x<10 but when it comes to this question it says the absolute value lies between 1 < x + 3 < 25 Since, the absolute value couldn't be less than zero, we write the inequality as follows 0 ≤ x + 3 < 25 Hence solving for x + 3 < 25 gives the answer as (28,22)
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If x+3  12 < 13, what is the range of x?
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