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805+ Level|   Absolute Values|   Algebra|      
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On the number line shown, the distance between 0 and a . . . Updated on: 07 Aug 2018, 04:27
Originally posted by EgmatQuantExpert on 08 Nov 2016, 05:04.
Last edited by EgmatQuantExpert on 07 Aug 2018, 04:27, edited 7 times in total.
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On the number line shown, the distance between 0 and a, a and b, and a and c is in the ratio of 1:2:3. If the distance of point b from 15 is twice the distance of point a from 15, what is the value of |c|?

    A. 9
    B. 27
    C. 36
    D. 45
    E. 54


Take a stab at this fresh question from e-GMAT. Post your analysis below.

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C
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On the number line shown, the distance between 0 and a . . . 09 Nov 2016, 11:41
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Let Point a-'x" units away from 0
Point b- 3x units away from 0
Point C- 4x units away from 0.

15 has to be between points a and b such that
3x-15=2(15-x)
=> x=9
Point c=4x=4*9=36.

Hence C

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On the number line shown, the distance between 0 and a . . . 27 Nov 2016, 21:49
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From: "The distance between 0 and a, a and b, and a and c is in the ratio of 1:2:3"
Set a= 1, we have b = 3 and c =4
=> a : b : c is in the ratio of 1 : 3 : 4
=> c is divisible by 4
=> c is 36
C
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On the number line shown, the distance between 0 and a . . . 08 Nov 2016, 05:49
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Let dist between a and 0 be x,then that between a and b would be 2x,and between b and C would be x.
Mod b-15 = 2* Mod 15-a
So, x= 9
Therefore mod c = 27

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On the number line shown, the distance between 0 and a . . . Updated on: 04 Dec 2016, 23:50
Originally posted by EgmatQuantExpert on 28 Nov 2016, 06:25.
Last edited by EgmatQuantExpert on 04 Dec 2016, 23:50, edited 4 times in total.
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Alright, let's look at the detailed solution of the above question. Let's use the concept of absolute value \(|p-q|\) as the distance between the point p and q on the number line to solve this question.

Given
The following information is given in the question.
    • Points a, b and c such that \(c > b > a\)
    • \(a: (b – a) : (c – a) = 1: 2: 3\)
    • \(|b – 15| = 2 * |a-15|\) (We are taking the absolute values here, as we do not know the position of 15 with respect to a and b. For example, if \(b > 15\), then \(|b – 15| = b – 15\) but if b < 15, then\(|b – 15| = (15 - b)\)

Let’s assume the value of a to be x.
Hence, \(b-a = 2x\) and substituting the value of a, we get the value of \(b = 3x\)
Similarly, the value of \(c = 4x\)

To find
As, we need to find |c|, which is the distance of c from the origin, we can say that \(|c| = 4x\)

Approach and Working Out
Now, we are given the relationship between the distance of points a and b with respect to 15. However, we do not know the position of 15 with respect to a and b on the number line. So, the following cases can occur depending upon the position of 15 on the number line:



    o Distance between 15 and \(a = a – 15 = x – 15\) (as a is greater than 15)
    o Distance between 15 and b = \(b – 15 = 3x – 15\) (as b is greater than 15)
    o \(3x – 15 = 2*(x – 15)\) {as \(|b – 15| = 2 * |a-15|\)}
    o \(3x – 15 = 2x – 30\)
    o \(x = -15\), which is not possible as x is the coordinate of a point that lies on the right hand side of 0, i.e. it is positive. So, \(x > 0\).
    o Thus, we can reject this case.



    o Distance between 15 and a = \(15 – a = 15 – x\) (as 15 is greater than a)
    o Distance between 15 and b = \(b – 15 = 3x – 15\) (as b is greater than 15)
    o \(3x – 15 = 2*(15 – x)\){as \(|b – 15| = 2 * |a-15|\)}
    o \(3x – 15 = 30 – 2x\)
    o \(5x = 45\), i.e. \(x = 9\).
    o In this case we have assumed that \(a < 15\). So, \(x < 15\) (As the coordinate of point a is x). Also \(x > 0\), since it is the coordinate of a point that lies on the right hand side of 0.
    o So, \(0 < x< 15\). As we have calculated \(x = 9\), this is a possible case
    o We have calculated above, the value of \(c = 4x\). So, we have \(4x = 36\)



    o Distance between 15 and a = \(15 – a = 15 – x\) (as 15 is greater than a)
    o Distance between 15 and b = \(15 – b = 15 – 3x\) (as 15 is greater than b)
    o \(15 – 3x = 2*( 15 – x)\) {as \(|b – 15| = 2 * |a-15|\) }
    o \(15 – 3x = 30 – 2x\), i.e. \(x = -15\),
    o We know that \(x > 0\). So, \(x = -15\) is not a valid value.
    o Thus, we can reject this case.

Therefore, the value of |c|is 36
Correct Answer: C

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On the number line shown, the distance between 0 and a . . . 28 Nov 2016, 07:02
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nailin16
Let dist between a and 0 be x,then that between a and b would be 2x,and between b and C would be x.
Mod b-15 = 2* Mod 15-a
So, x= 9
Therefore mod c = 27

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You have correctly calculated the value of x i.e. the distance between point a and 0 on the number line.
However, the value of |c| is 4x, and hence the answer would be 36.

Kindly review our solution above to understand where you went wrong.
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On the number line shown, the distance between 0 and a . . . 29 Nov 2016, 05:53
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From: "The distance between 0 and a, a and b, and a and c is in the ratio of 1:2:3"
Set a= 1, we have b = 3 and c =4
=> a : b : c is in the ratio of 1 : 3 : 4
=> c is divisible by 4
=> c is 36
C
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There is a small flaw in your approach. You seem to have assumed c as a multiple of 4, which may not be the case always. For instance, let's look at a slightly different version of the same question.

On the number line shown, the distance between 0 and a, a and b, and a and c is in the ratio of 1:2:3. If the distance of point b from 11.25 is twice the distance of point a from 15, what is the value of |c|?

    A. 9
    B. 27
    C. 36
    D. 45
    E. 54


Solve this question using the approach mentioned by you and also using the solution provided by us. If you use the approach mentioned by you in your response, you will still the answer as 36 (Choice C). However, the answer to this modified version would be 27 (choice B), which is clearly not a multiple of 4.

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On the number line shown, the distance between 0 and a . . . 08 Dec 2016, 07:04
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EgmatQuantExpert

If the distance of point b from 15 is twice the distance of point a from 15
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In your previous explanation you have also placed the point "15" to the left of A and to the right of B, but would this even be possible given the above limitation? Given the order of the letters, the only way I can see this condition being met is if 15 is 1/3 on the line between A and B.

I would appreciate your response.
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On the number line shown, the distance between 0 and a . . . 08 Dec 2016, 22:38
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koenh
EgmatQuantExpert

If the distance of point b from 15 is twice the distance of point a from 15

In your previous explanation you have also placed the point "15" to the left of A and to the right of B, but would this even be possible given the above limitation? Given the order of the letters, the only way I can see this condition being met is if 15 is 1/3 on the line between A and B.

I would appreciate your response.
Show more

Yes, you're absolutely right. :)

We have taken these cases as Case I (15 to the left of a or a>15) and Case III (15 to the right of B, 15>b). However, if you go through our explanation in details, you will understand that we have also concluded that these cases are not possible. Hence, the only case possible is Case II, in which 15 lies between a and b. The same has been explained in details in the solution.

The reason why we have discussed all possibilities is to ensure that we critically analyze all cases before arriving at a conclusion. Doing so, is extremely important while solving GMAT questions.

Regards,
Saquib
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On the number line shown, the distance between 0 and a . . . 18 Sep 2018, 10:35
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EgmatQuantExpert

On the number line shown, the distance between 0 and a, a and b, and a and c is in the ratio of 1:2:3. If the distance of point b from 15 is twice the distance of point a from 15, what is the value of |c|?

    A. 9
    B. 27
    C. 36
    D. 45
    E. 54
Show more
Using info given (figure included) we have DATA and FOCUS as below:

\(0 < a < b < c\,\,\left( * \right)\)

\({\text{dist}}\left( {0,a} \right) = a\)

\(\left. \begin{gathered}\\
{\text{dist}}\left( {a,b} \right) = 2a\, \hfill \\\\
{\text{dist}}\left( {a,c} \right) = 3a \hfill \\ \\
\end{gathered} \right\}\,\,\,\mathop \Rightarrow \limits^{\left( * \right)} \,\,\,{\text{dist}}\left( {b,c} \right) = 3a - 2a = a\)

\(? = c = 4a\)

From the fact that \({\text{dist}}\left( {b,15} \right) = 2 \cdot {\text{dist}}\left( {a,15} \right)\) we have three possible scenarios:

\(b < 15:\)

\(15 - b = 2\left( {15 - a} \right)\,\,\,\, \Rightarrow \,\,\,\,15 = 2a - b = 2a - 3a = - a\,\,\,\, \Rightarrow \,\,\,a < 0\,\,\,\,{\text{impossible}}\,\,\)


\(a < 15 \leqslant b:\)

\(b - 15 = 2\left( {15 - a} \right)\,\,\,\, \Rightarrow \,\,\,\,3 \cdot 15 = 2a + b = 2a + 3a = 5a\,\,\,\, \Rightarrow \,\,? = 4a = 4 \cdot 9 = 36\)

And from the fact that we have already found (one viable) correct alternative choice, no need to evaluate the third scenario!


This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.
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On the number line shown, the distance between 0 and a . . . 18 Sep 2018, 11:27
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EgmatQuantExpert
koenh
EgmatQuantExpert

If the distance of point b from 15 is twice the distance of point a from 15

In your previous explanation you have also placed the point "15" to the left of A and to the right of B, but would this even be possible given the above limitation? Given the order of the letters, the only way I can see this condition being met is if 15 is 1/3 on the line between A and B.

I would appreciate your response.

Yes, you're absolutely right. :)

We have taken these cases as Case I (15 to the left of a or a>15) and Case III (15 to the right of B, 15>b). However, if you go through our explanation in details, you will understand that we have also concluded that these cases are not possible. Hence, the only case possible is Case II, in which 15 lies between a and b. The same has been explained in details in the solution.

The reason why we have discussed all possibilities is to ensure that we critically analyze all cases before arriving at a conclusion. Doing so, is extremely important while solving GMAT questions.

Regards,
Saquib
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Hello!! With all respect, I request you to explain how you got C=4x ?? I am getting a lil bit confused here.
Thank you in advance.
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On the number line shown, the distance between 0 and a . . . 01 Nov 2019, 19:03
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EgmatQuantExpert
Alright, let's look at the detailed solution of the above question. Let's use the concept of absolute value \(|p-q|\) as the distance between the point p and q on the number line to solve this question.

Given
The following information is given in the question.
    • Points a, b and c such that \(c > b > a\)
    • \(a: (b – a) : (c – a) = 1: 2: 3\)
    • \(|b – 15| = 2 * |a-15|\) (We are taking the absolute values here, as we do not know the position of 15 with respect to a and b. For example, if \(b > 15\), then \(|b – 15| = b – 15\) but if b < 15, then\(|b – 15| = (15 - b)\)

Let’s assume the value of a to be x.
Hence, \(b-a = 2x\) and substituting the value of a, we get the value of \(b = 3x\)
Similarly, the value of \(c = 4x\)

To find

As, we need to find |c|, which is the distance of c from the origin, we can say that \(|c| = 4x\)

Approach and Working Out
Now, we are given the relationship between the distance of points a and b with respect to 15. However, we do not know the position of 15 with respect to a and b on the number line. So, the following cases can occur depending upon the position of 15 on the number line:



    o Distance between 15 and \(a = a – 15 = x – 15\) (as a is greater than 15)
    o Distance between 15 and b = \(b – 15 = 3x – 15\) (as b is greater than 15)
    o \(3x – 15 = 2*(x – 15)\) {as \(|b – 15| = 2 * |a-15|\)}
    o \(3x – 15 = 2x – 30\)
    o \(x = -15\), which is not possible as x is the coordinate of a point that lies on the right hand side of 0, i.e. it is positive. So, \(x > 0\).
    o Thus, we can reject this case.



    o Distance between 15 and a = \(15 – a = 15 – x\) (as 15 is greater than a)
    o Distance between 15 and b = \(b – 15 = 3x – 15\) (as b is greater than 15)
    o \(3x – 15 = 2*(15 – x)\){as \(|b – 15| = 2 * |a-15|\)}
    o \(3x – 15 = 30 – 2x\)
    o \(5x = 45\), i.e. \(x = 9\).
    o In this case we have assumed that \(a < 15\). So, \(x < 15\) (As the coordinate of point a is x). Also \(x > 0\), since it is the coordinate of a point that lies on the right hand side of 0.
    o So, \(0 < x< 15\). As we have calculated \(x = 9\), this is a possible case
    o We have calculated above, the value of \(c = 4x\). So, we have \(4x = 36\)



    o Distance between 15 and a = \(15 – a = 15 – x\) (as 15 is greater than a)
    o Distance between 15 and b = \(15 – b = 15 – 3x\) (as 15 is greater than b)
    o \(15 – 3x = 2*( 15 – x)\) {as \(|b – 15| = 2 * |a-15|\) }
    o \(15 – 3x = 30 – 2x\), i.e. \(x = -15\),
    o We know that \(x > 0\). So, \(x = -15\) is not a valid value.
    o Thus, we can reject this case.

Therefore, the value of |c|is 36
Correct Answer: C

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after we establish c = 4x...can we just look at the answer choices for multiples of 4 and narrow down to choice C
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On the number line shown, the distance between 0 and a . . . 25 Jun 2021, 23:37
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Rather then going through each algebraic case, I found it much easier and faster to use the answer choices

(1st) understand the concept that the Absolute Value of C is given by the distance on the Number Line from variable C to 0


(2nd) we are dealing with all positive values here. So the quantity inside the Modulus of:

[a]
Absolute value of b
[c]

Will be positive, so the Modulus is not needed.

[a] = a
Absolute value of b = b
[c] = c

(3rd)
We will name the unknown ratio multiplier = M

Distance from 0 to a = M
Distance from a to b = 2M
Distance from a to c = 3M

Now, once we set these up on the number line, we can see that:

Distance from b to c = (Dist. from a to c) - (Dist. from a to b)

= 3M - 2M = M


distance from b to c = M

And finally

Distance from c to the origin 0 = M + 2M + M = 4M



[c] = c = 4M = ????


Answer (a) 9

If c = 9 ———-> 4M = 9 ——-> M = 9/4

[a] = M = 9/4 = 2.25

= 3M = 27/4 = 6.75

a is further from 15 than c is, so this can’t be the answer


Answer (b) 27

If [c] = 4M = 27 ———> M = 27/4

[a] = M = 27/4 = 6.75

[b] = 3M = 81/4 = 20.25

b is closer to 15 than a is on the number line (can be done by simple inspection)

Answer (c) 36

If [c] = 36 ———> 4M = 36 ——-> M = 9

[a] = M = 9

[b] = 3M = 27

Distance from a to 15 ——- [15 - 9] = 6

Distance from b to 15 ——- [27 - 15] = 12

Since 12 is TWICE the value of 6, when the absolute value of c is 36, the constraints given are satisfied.


Answer (c)

Seems as if taking the algebraic road takes too long on this problem….

[b]Posted from my mobile device
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On the number line shown, the distance between 0 and a . . . 20 Nov 2024, 04:35
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