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Which of the following sets includes ALL of the solutions of
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Which of the following sets includes ALL of the solutions of x that will satisfy the equation: ? x  2  x  3 = x  5 (A) {6, 5, 0, 1, 7, 8} (B) {4, 2, 0, 10/3, 4, 5} (C) {4, 0, 1, 4, 5, 6} (D) {1, 10/3, 3, 5, 6, 8} (E) {2, 1, 1, 3, 4, 5}
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Re: Which of the following sets includes ALL of the solutions of
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17 Sep 2012, 22:50
imhimanshu wrote: Which of the following sets includes ALL of the solutions of x that will satisfy the equation: ?
x2x3=x5
(A) {6, 5, 0, 1, 7, 8} (B) {4, 2, 0, 10/3, 4, 5} (C) {4, 0, 1, 4, 5, 6} (D) {1, 10/3, 3, 5, 6, 8} (E) {2, 1, 1, 3, 4, 5}
Usually, I'm able to solve such questions, however I couldn't solve this one. Can someone run pass me through using number line approach.
Thanks H Responding to a pm: x2x3=x5 First and foremost, 'the difference between the distances' is a concept which is a little harder to work with as compared with 'the sum of distances'. So try to get rid of the negative sign, if possible. x2 = x3+x5 This equation tells you the following: x is a point whose distance from 2 is equal to the sum of its distance from 3 and its distance from 5. Plot the 3 points on the number line and run through the 4 regions. Attachment:
Ques3.jpg [ 3.83 KiB  Viewed 17000 times ]
Green: For every point to the left of 2, the distance of the point from 2 is less than the distance from 3. So there is no way any point here will satisfy the equation. Distance of x from 2 will always be less than the sum of distances from 3 and 5. Blue: Between 2 and 3, 3 is farthest from 2 i.e. at a distance of 1 but 5 is at a distance of 2 from 3. So again, no point here can satisfy the equation. Distance of x from 2 will still be less than the sum of distances from 3 and 5. Black: Now we are going a little farther from 2 and toward 5 so there might be a point where distance from 2 is equal to sum of distances from 3 and 5. Check what happens at 4. You see that 4 satisfies the equation. At 4, the distance of x from 2 is equal to the sum of distances from 3 and 5. After 4, distance of x from 2 will be more than the sum of distances from 3 and 5 till we reach 5. Red: After 5, as x moves to the right, its distance from 3 as well as 5 increases. So it is possible that the distance of x from 2 is again equal to the sum of distances from 3 and 5. At 5, x is 3 units away from 2 and 2 units away from 3 and 5 together. Check what happens at 6. At 6, x is 4 away from 2 and 4 away from 3 and 5 together. Hence 6 satisfies too. As you move one step to the right of 6, distance from 2 increases by 1 unit and distance from 3 and 5 together increases by 2 units. So the distances will never be the same again.
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Re: Which of the following sets includes ALL of the solutions of
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05 Dec 2012, 02:49
1) Get the checkpoints using the absolute value expressions. x2 tells us that x=2 as one checkpoint x3 tells us that x=3 as another checkpoint x5 tells us that x=5 as last checkpoint 2) Let us test for x where x < 2 x2 = x5 + x3 (x2)= (x5)  (x3) x +2 = x +5 x +3 x=6 Invalid since x=6 is not x<2 3) Let us test for x where 2 < x < 3 (x2) = (x5) (x3) x2 = x+5 x +3 3x = 10 x = 10/3 Invalid since x=3.3 is not within x = (2,3) 4) Let us test for x where 3 < x < 5 x2 = x3  (x5) x2 = 2 x = 4 Valid 5) Let us test for x where x > 5 x2 = x3 + x  5 x = 6 x=6 Valid 6) Now let's look for 4 and 6 in the sets of answer choices. (A) {6, 5, 0, 1, 7, 8} (B) {4, 2, 0, 10/3, 4, 5} (C) {4, 0, 1, 4, 5, 6} (D) {1, 10/3, 3, 5, 6, 8} (E) {2, 1, 1, 3, 4, 5} Only C has both x=4 and x=6. Answer: C
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Re: Which of the following sets includes ALL of the solutions of
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17 Sep 2012, 08:20
the equation has 3 intervals 2, 3 , and 5. i.e 4 segments x<2, 2<x<3, 3<x<5, 5<x.
1) When X>5 none of the brackets will change the sign and hence (x2)(x3) = (x5)> x=6 ..We accept this solution as it satisfies the equation.
Now since we know x=6 is one of the solution we know the answer could be C or D
2) When 3<x<5 here (x5) will change the sign (x2)(x3) = (x5) >x=4 ..We accept the solution as 3<4<5
We can stop solving here as the only set consisting both 6 and 4 is set C hence correct answer is C



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Re: Which of the following sets includes ALL of the solutions of
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17 Sep 2012, 08:55
imhimanshu wrote: Which of the following sets includes ALL of the solutions of x that will satisfy the equation: ?
x2x3=x5
(A) {6, 5, 0, 1, 7, 8} (B) {4, 2, 0, 10/3, 4, 5} (C) {4, 0, 1, 4, 5, 6} (D) {1, 10/3, 3, 5, 6, 8} (E) {2, 1, 1, 3, 4, 5}
Usually, I'm able to solve such questions, however I couldn't solve this one. Can someone run pass me through using number line approach.
Thanks H The given equation can be rewritten as \(x2=x3+x5,\) which means that the distance between \(x\) and \(2\) is the sum of the distances between \(x\) and \(3\) plus the distance between \(x\) and 5. Using the number line and visualizing the points, we can easily see that \(x\) cannot be smaller than \(3.\) If \(x>5,\) then \(x2=x3+x5,\) from which \(x=6.\) We have to choose between answers C and D. If \(3<x<5,\) then \(x2=x3x+5,\) which yields \(x=4.\) Answer C
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Re: Which of the following sets includes ALL of the solutions of
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17 Sep 2012, 09:06
EvaJager wrote: imhimanshu wrote: Which of the following sets includes ALL of the solutions of x that will satisfy the equation: ?
x2x3=x5
(A) {6, 5, 0, 1, 7, 8} (B) {4, 2, 0, 10/3, 4, 5} (C) {4, 0, 1, 4, 5, 6} (D) {1, 10/3, 3, 5, 6, 8} (E) {2, 1, 1, 3, 4, 5}
Usually, I'm able to solve such questions, however I couldn't solve this one. Can someone run pass me through using number line approach.
Thanks H The given equation can be rewritten as \(x2=x3+x5,\) which means that the distance between \(x\) and \(2\) is the sum of the distances between \(x\) and \(3\) plus the distance between \(x\) and 5. Using the number line and visualizing the points, we can easily see that \(x\) cannot be smaller than \(3.\) If \(x>5,\) then \(x2=x3+x5,\) from which \(x=6.\) We have to choose between answers C and D. If \(3<x<5,\) then \(x2=x3x+5,\) which yields \(x=4.\) Answer C Hi Eva, If i am not wrong then x will take any value which can satisfy this modulus equation x2x3=x5 i.e. LHS = RHS Now if take x=5 (as 5 is one of the value of x in option C), then LHS> 52  53 = 32 = 32 = 1 RHS> 55 = 0 = 0 In this case LHS is not equal to RHS i.e. how can 1 equals 0 How can C be the correct option if it contains one value which does not satisfy the equation. Waiting for reply
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Re: Which of the following sets includes ALL of the solutions of
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17 Sep 2012, 09:15
fameatop wrote: EvaJager wrote: imhimanshu wrote: Which of the following sets includes ALL of the solutions of x that will satisfy the equation: ?
x2x3=x5
(A) {6, 5, 0, 1, 7, 8} (B) {4, 2, 0, 10/3, 4, 5} (C) {4, 0, 1, 4, 5, 6} (D) {1, 10/3, 3, 5, 6, 8} (E) {2, 1, 1, 3, 4, 5}
Usually, I'm able to solve such questions, however I couldn't solve this one. Can someone run pass me through using number line approach.
Thanks H The given equation can be rewritten as \(x2=x3+x5,\) which means that the distance between \(x\) and \(2\) is the sum of the distances between \(x\) and \(3\) plus the distance between \(x\) and 5. Using the number line and visualizing the points, we can easily see that \(x\) cannot be smaller than \(3.\) If \(x>5,\) then \(x2=x3+x5,\) from which \(x=6.\) We have to choose between answers C and D. If \(3<x<5,\) then \(x2=x3x+5,\) which yields \(x=4.\) Answer C Hi Eva, If i am not wrong then x will take any value which can satisfy this modulus equation x2x3=x5 i.e. LHS = RHS Now if take x=5 (as 5 is one of the value of x in option C), then LHS> 52  53 = 32 = 32 = 1 RHS> 55 = 0 = 0 In this case LHS is not equal to RHS i.e. how can 1 equals 0 How can C be the correct option if it contains one value which does not satisfy the equation. Waiting for reply The question reads "Which of the following sets includes ALL of the solutions of x that will satisfy the equation" it doesn't say that all the elements of the set are solutions of the given equation. When solving, you can find that there are exactly two values only which are solutions: 4 and 6. And they are both included only in the set of answer C.
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Re: Which of the following sets includes ALL of the solutions of
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17 Sep 2012, 09:17
fameatop wrote: EvaJager wrote: imhimanshu wrote: Which of the following sets includes ALL of the solutions of x that will satisfy the equation: ?
x2x3=x5
(A) {6, 5, 0, 1, 7, 8} (B) {4, 2, 0, 10/3, 4, 5} (C) {4, 0, 1, 4, 5, 6} (D) {1, 10/3, 3, 5, 6, 8} (E) {2, 1, 1, 3, 4, 5}
Usually, I'm able to solve such questions, however I couldn't solve this one. Can someone run pass me through using number line approach.
Thanks H The given equation can be rewritten as \(x2=x3+x5,\) which means that the distance between \(x\) and \(2\) is the sum of the distances between \(x\) and \(3\) plus the distance between \(x\) and 5. Using the number line and visualizing the points, we can easily see that \(x\) cannot be smaller than \(3.\) If \(x>5,\) then \(x2=x3+x5,\) from which \(x=6.\) We have to choose between answers C and D. If \(3<x<5,\) then \(x2=x3x+5,\) which yields \(x=4.\) Answer C Hi Eva, If i am not wrong then x will take any value which can satisfy this modulus equation x2x3=x5 i.e. LHS = RHS Now if take x=5 (as 5 is one of the value of x in option C), then LHS> 52  53 = 32 = 32 = 1 RHS> 55 = 0 = 0 In this case LHS is not equal to RHS i.e. how can 1 equals 0 How can C be the correct option if it contains one value which does not satisfy the equation. Waiting for reply We need the set that INCLUDES all the solutions (which are 4 and 6), it's not necessary that all numbers of the set to satisfy the equation. Only option C contains all of the solutions, so C is the correct answer. Hope it's clear.
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Re: Which of the following sets includes ALL of the solutions of
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23 Sep 2012, 00:03
EvaJager wrote: imhimanshu wrote: Which of the following sets includes ALL of the solutions of x that will satisfy the equation: ?
x2x3=x5
(A) {6, 5, 0, 1, 7, 8} (B) {4, 2, 0, 10/3, 4, 5} (C) {4, 0, 1, 4, 5, 6} (D) {1, 10/3, 3, 5, 6, 8} (E) {2, 1, 1, 3, 4, 5}
Usually, I'm able to solve such questions, however I couldn't solve this one. Can someone run pass me through using number line approach.
Thanks H The given equation can be rewritten as \(x2=x3+x5,\) which means that the distance between \(x\) and \(2\) is the sum of the distances between \(x\) and \(3\) plus the distance between \(x\) and 5. Using the number line and visualizing the points, we can easily see that \(x\) cannot be smaller than \(3.\) If \(x>5,\) then \(x2=x3+x5,\) from which \(x=6.\) We have to choose between answers C and D. If \(3<x<5,\) then \(x2=x3x+5,\) which yields \(x=4.\) Answer C Hi, Please explain in detail why x cannot be smaller than 3... I am not able to understand this. thanks



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Re: Which of the following sets includes ALL of the solutions of
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23 Sep 2012, 01:08
abhi398 wrote: EvaJager wrote: imhimanshu wrote: Which of the following sets includes ALL of the solutions of x that will satisfy the equation: ?
x2x3=x5
(A) {6, 5, 0, 1, 7, 8} (B) {4, 2, 0, 10/3, 4, 5} (C) {4, 0, 1, 4, 5, 6} (D) {1, 10/3, 3, 5, 6, 8} (E) {2, 1, 1, 3, 4, 5}
Usually, I'm able to solve such questions, however I couldn't solve this one. Can someone run pass me through using number line approach.
Thanks H The given equation can be rewritten as \(x2=x3+x5,\) which means that the distance between \(x\) and \(2\) is the sum of the distances between \(x\) and \(3\) plus the distance between \(x\) and 5. Using the number line and visualizing the points, we can easily see that \(x\) cannot be smaller than \(3.\) If \(x>5,\) then \(x2=x3+x5,\) from which \(x=6.\) We have to choose between answers C and D. If \(3<x<5,\) then \(x2=x3x+5,\) which yields \(x=4.\) Answer C Hi, Please explain in detail why x cannot be smaller than 3... I am not able to understand this. thanks The equation is \(x2=x3+x5.\) Draw the number line and place 2, 3 and 5 on it. If \(x\) is between 2 and 3, the distance between \(x\) and 2, which is \(x2,\) is less than 1, while the distance between \(x\) and 5, which is \(x5,\) is greater than 2. So, the left hand side is for sure less than the right hand side of the equality. If \(x\) is less than 2, the distance between \(x\) and 2 is the smallest, is both less than distance between \(x\) and 3 and less than the distance between \(x\) and 5. Again, equality cannot hold.
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Which of the following sets includes ALL of the solutions of
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21 Apr 2013, 09:02
Usually, plugging in numbers will be the best option but things are a little complicated here. Let me point out one thing: "Which of the following sets includes ALL of the solutions of x that will satisfy the equation: ?" means you need to find ALL the solutions and some members of the set may not be a solution. So say, you check for 8 and see that it doesn't satisfy the equation, it doesn't mean that options (A) and (D) are out of the running. It may be one of the numbers which do not satisfy the equation but the set (A) or (D) may still contain all the solutions. With a lot of different numbers, it could get confusing and complicated. Anyway, it all depends on your comfort with the various methods.
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Re: Which of the following sets includes ALL of the solutions of
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22 Apr 2013, 02:41
VeritasPrepKarishma wrote: This equation tells you the following: x is a point whose distance from 2 is equal to the sum of its distance from 3 and its distance from 5. Plot the 3 points on the number line and run through the 4 regions. Attachment: Ques3.jpg Green: For every point to the left of 2, the distance of the point from 2 is less than the distance from 3. So there is no way any point here will satisfy the equation. Distance of x from 2 will always be less than the sum of distances from 3 and 5. Blue: Between 2 and 3, 3 is farthest from 2 i.e. at a distance of 1 but 5 is at a distance of 2 from 3. So again, no point here can satisfy the equation. Distance of x from 2 will still be less than the sum of distances from 3 and 5. Black: Now we are going a little farther from 2 and toward 5 so there might be a point where distance from 2 is equal to sum of distances from 3 and 5. Check what happens at 4. You see that 4 satisfies the equation. At 4, the distance of x from 2 is equal to the sum of distances from 3 and 5. After 4, distance of x from 2 will be more than the sum of distances from 3 and 5 till we reach 5. Red: After 5, as x moves to the right, its distance from 3 as well as 5 increases. So it is possible that the distance of x from 2 is again equal to the sum of distances from 3 and 5. At 5, x is 3 units away from 2 and 2 units away from 3 and 5 together. Check what happens at 6. At 6, x is 4 away from 2 and 4 away from 3 and 5 together. Hence 6 satisfies too. As you move one step to the right of 6, distance from 2 increases by 1 unit and distance from 3 and 5 together increases by 2 units. So the distances will never be the same again. I didnt get this part at all I am sorry, could you please elaborate a bit more, or give a link regarding this method ?
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Re: Which of the following sets includes ALL of the solutions of
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22 Apr 2013, 22:39
LalaB wrote: VeritasPrepKarishma wrote: This equation tells you the following: x is a point whose distance from 2 is equal to the sum of its distance from 3 and its distance from 5. Plot the 3 points on the number line and run through the 4 regions. Attachment: Ques3.jpg Green: For every point to the left of 2, the distance of the point from 2 is less than the distance from 3. So there is no way any point here will satisfy the equation. Distance of x from 2 will always be less than the sum of distances from 3 and 5. Blue: Between 2 and 3, 3 is farthest from 2 i.e. at a distance of 1 but 5 is at a distance of 2 from 3. So again, no point here can satisfy the equation. Distance of x from 2 will still be less than the sum of distances from 3 and 5. Black: Now we are going a little farther from 2 and toward 5 so there might be a point where distance from 2 is equal to sum of distances from 3 and 5. Check what happens at 4. You see that 4 satisfies the equation. At 4, the distance of x from 2 is equal to the sum of distances from 3 and 5. After 4, distance of x from 2 will be more than the sum of distances from 3 and 5 till we reach 5. Red: After 5, as x moves to the right, its distance from 3 as well as 5 increases. So it is possible that the distance of x from 2 is again equal to the sum of distances from 3 and 5. At 5, x is 3 units away from 2 and 2 units away from 3 and 5 together. Check what happens at 6. At 6, x is 4 away from 2 and 4 away from 3 and 5 together. Hence 6 satisfies too. As you move one step to the right of 6, distance from 2 increases by 1 unit and distance from 3 and 5 together increases by 2 units. So the distances will never be the same again. I didnt get this part at all I am sorry, could you please elaborate a bit more, or give a link regarding this method ? Go through this link: http://www.veritasprep.com/blog/2011/01 ... spartii/It might help.
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13 Jun 2013, 09:52
Why can we not plug in the answer choices into the question stem to solve that way?



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13 Jun 2013, 09:56
WholeLottaLove wrote: Why can we not plug in the answer choices into the question stem to solve that way? You can, but you have to be extremely cautious. (and the process would be quite time counsiming) Which of the following sets includes ALL of the solutions of x that will satisfy the equation: ?The question asks for ALL the solutions, even if a statement gives you valid options, you cannot be 100% sure that they are ALL the possible options.
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14 Jun 2013, 04:27
WholeLottaLove wrote: Looking at x2x3=x5
in Answer C, one of the choices is 5. So, plugging five in x2x3=x5 gets:
5253=55
32=0
32=0
That appears to be incorrect.
Read the question carefully again: Which of the following sets includes ALL of the solutions of x that will satisfy the equation: ? You need to find the set that includes ALL the solutions. The set needn't have ONLY the solutions. The solution set for x is 4 and 6. The only set that includes both 4 and 6 is (C) Note that (C) has other elements as well but it doesn't matter. It is the only set that INCLUDES all the solutions. And that is the reason why number plugging can be exhausting.
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01 Jul 2013, 14:35
Which of the following sets includes ALL of the solutions of x that will satisfy the equation: ?
x2x3=x5
Three checkpoints: 2, 3 and 5 meaning there are four ranges to test:
x<2, 2<x<3, 3<x<5, x>5
x<2: (x2)  (x3) = (x5) (x+2)  (x+3) = (x+5) x+2 +x3 = x+5 x=6 INVALID as 6 doesn't fall within the range of x<2
2<x<3: (x2)  (x3) = (x5) (x2)  (x+3) = (x+5) x2 + x3 = x+5 3x=10 x=10/3 INVALID as 10/3 does not fall within the range of 2<x<3
3<x<5: (x2)  (x3)= (x5) x2 x+3 =x+5 x=4 VALID as 4 falls in the range of 3<x<5
x>5: (x2)  (x3) = (x5) x2x+3 = x5 x=6 VALID as x falls within the range of x>5
(A) {6, 5, 0, 1, 7, 8} (B) {4, 2, 0, 10/3, 4, 5} (C) {4, 0, 1, 4, 5, 6} (D) {1, 10/3, 3, 5, 6, 8} (E) {2, 1, 1, 3, 4, 5}
The only answer choice with both valid answers (x=4, x=6) is (C)



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01 Jul 2013, 19:38
WholeLottaLove wrote: Why can we not plug in the answer choices into the question stem to solve that way? Idk why plugging in answers doesnt work. When I first saw this question, I started plugging in the numbers from the left of the sets in the answer choices and clearly none of the ve values work. So I just started plugging in the values from the right and luckily only Option C worked. So I got the correct answer but I think I spent more than 2 minutes by using brute force method. Maybe I am not understanding the exact meaning of the question, but from what I understand, all the values in each of the sets should meet the requirements of the problem. I know GMAT problems are desgined to trick you, but I think the phrasing of the question here is debatable and I think GMAT questions usually dont leave room for interpretations. just IMO. Please correct me



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01 Jul 2013, 20:53
jjack0310 wrote: WholeLottaLove wrote: Why can we not plug in the answer choices into the question stem to solve that way? Idk why plugging in answers doesnt work. When I first saw this question, I started plugging in the numbers from the left of the sets in the answer choices and clearly none of the ve values work. So I just started plugging in the values from the right and luckily only Option C worked. So I got the correct answer but I think I spent more than 2 minutes by using brute force method. Maybe I am not understanding the exact meaning of the question, but from what I understand, all the values in each of the sets should meet the requirements of the problem. I know GMAT problems are desgined to trick you, but I think the phrasing of the question here is debatable and I think GMAT questions usually dont leave room for interpretations. just IMO. Please correct me Ok, so how did you plug in? I an ignoring options (A) and (B) and focusing on just (C) Option (C) {4, 0, 1, 4, 5, 6} x2x3=x5 You put x = 4. It didn't satisfy You put x = 0. It didn't satisfy You put x = 1. It didn't satisfy You put x = 4. It satisfied You put x = 5. It didn't satisfy You put x = 6. It satisfied How can you say there is no other value of x that satisfies this equation? How can you say that all solutions of this equation are included in (C)?
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Re: Which of the following sets includes ALL of the solutions of
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02 Jul 2013, 16:34
VeritasPrepKarishma wrote: Ok, so how did you plug in? I an ignoring options (A) and (B) and focusing on just (C)
Option (C) {4, 0, 1, 4, 5, 6}
x2x3=x5 You put x = 4. It didn't satisfy You put x = 0. It didn't satisfy You put x = 1. It didn't satisfy You put x = 4. It satisfied You put x = 5. It didn't satisfy You put x = 6. It satisfied
How can you say there is no other value of x that satisfies this equation? How can you say that all solutions of this equation are included in (C)?
Calm down. Dont get excited. I meant I plugged in all the values in all the options (a,b,c,d and e) Option C had two values that satisfy the equations and the other options didnt have two values. So I just went with C and got the correct answer. I did say I got lucky  maybe/maybe not. "How can you say that all solutions of this equation are included in (C)" lol Pretty ironic  all solutions Precisely my point that the wordings of this problem can be easily misinterpreted. Just like you did mine.




Re: Which of the following sets includes ALL of the solutions of
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02 Jul 2013, 16:34



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