Last visit was: 23 May 2024, 09:31 It is currently 23 May 2024, 09:31
Toolkit
GMAT Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

# On the number line shown, the distance between 0 and a . . .

SORT BY:
Tags:
Show Tags
Hide Tags
e-GMAT Representative
Joined: 04 Jan 2015
Posts: 3723
Own Kudos [?]: 17017 [82]
Given Kudos: 165
Intern
Joined: 23 Jul 2015
Status:I am not giving up. Not yet.
Posts: 35
Own Kudos [?]: 71 [32]
Given Kudos: 392
Intern
Joined: 19 Nov 2016
Posts: 6
Own Kudos [?]: 16 [14]
Given Kudos: 14
General Discussion
Manager
Joined: 05 Apr 2014
Posts: 111
Own Kudos [?]: 93 [0]
Given Kudos: 255
Location: India
Schools: ISB '19 Fox"19
GMAT 1: 660 Q48 V33
GPA: 3
Re: On the number line shown, the distance between 0 and a . . . [#permalink]
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

Sent from my HM 1SW using GMAT Club Forum mobile app
e-GMAT Representative
Joined: 04 Jan 2015
Posts: 3723
Own Kudos [?]: 17017 [14]
Given Kudos: 165
Re: On the number line shown, the distance between 0 and a . . . [#permalink]
7
Kudos
7
Bookmarks
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

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.
e-GMAT Representative
Joined: 04 Jan 2015
Posts: 3723
Own Kudos [?]: 17017 [1]
Given Kudos: 165
Re: On the number line shown, the distance between 0 and a . . . [#permalink]
1
Kudos
nailin16 wrote:
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

Sent from my HM 1SW using GMAT Club Forum mobile app

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.
e-GMAT Representative
Joined: 04 Jan 2015
Posts: 3723
Own Kudos [?]: 17017 [3]
Given Kudos: 165
Re: On the number line shown, the distance between 0 and a . . . [#permalink]
3
Kudos
khiemchii wrote:
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

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.

Intern
Joined: 15 Nov 2015
Posts: 23
Own Kudos [?]: 12 [0]
Given Kudos: 14
Location: Netherlands
GMAT 1: 730 Q47 V44
GPA: 3.99
Re: On the number line shown, the distance between 0 and a . . . [#permalink]
EgmatQuantExpert wrote:

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.

e-GMAT Representative
Joined: 04 Jan 2015
Posts: 3723
Own Kudos [?]: 17017 [0]
Given Kudos: 165
Re: On the number line shown, the distance between 0 and a . . . [#permalink]
koenh wrote:
EgmatQuantExpert wrote:

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.

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
Tutor
Joined: 12 Oct 2010
Status:GMATH founder
Posts: 892
Own Kudos [?]: 1382 [1]
Given Kudos: 56
Re: On the number line shown, the distance between 0 and a . . . [#permalink]
1
Bookmarks
EgmatQuantExpert wrote:
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

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.
Attachments

18Set18_3w.gif [ 1.92 KiB | Viewed 12839 times ]

Intern
Joined: 27 Apr 2018
Posts: 17
Own Kudos [?]: 1 [0]
Given Kudos: 144
Location: India
GMAT 1: 600 Q44 V29
GPA: 3.9
Re: On the number line shown, the distance between 0 and a . . . [#permalink]
EgmatQuantExpert wrote:
koenh wrote:
EgmatQuantExpert wrote:

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.

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

Hello!! With all respect, I request you to explain how you got C=4x ?? I am getting a lil bit confused here.
Intern
Joined: 12 Aug 2019
Posts: 18
Own Kudos [?]: 11 [0]
Given Kudos: 400
Location: United States
GMAT 1: 690 Q49 V33
Re: On the number line shown, the distance between 0 and a . . . [#permalink]
EgmatQuantExpert wrote:
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

after we establish c = 4x...can we just look at the answer choices for multiples of 4 and narrow down to choice C
VP
Joined: 10 Jul 2019
Posts: 1390
Own Kudos [?]: 549 [1]
Given Kudos: 1656
On the number line shown, the distance between 0 and a . . . [#permalink]
1
Kudos
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 = ????

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

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)

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.

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

[b]Posted from my mobile device
Non-Human User
Joined: 09 Sep 2013
Posts: 33151
Own Kudos [?]: 829 [0]
Given Kudos: 0
Re: On the number line shown, the distance between 0 and a . . . [#permalink]
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
Re: On the number line shown, the distance between 0 and a . . . [#permalink]
Moderator:
Math Expert
93419 posts