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In the figure above, points A, B, C, D, and E are evenly spaced along
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02 Sep 2013, 10:33
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In the figure above, points A, B, C, D, and E are evenly spaced along the number line. If E = 9^13 and C = 9^11, what is the distance from point A to point D? A. 9^3 B. 9^9 C. (120)(9^9) D. 9^11 E. (120)(9^11) Attachment:
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Re: In the figure above, points A, B, C, D, and E are evenly spaced along
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02 Sep 2013, 12:48
In the figure above, points A, B, C, D, and E are evenly spaced along the number line. If E = 9^13 and C = 9^11, what is the distance from point A to point D? A. 9^3 B. 9^9 C. (120)(9^9) D. 9^11 E. (120)(9^11) The distance between E and C is \(9^{13}9^{11}=9^{11}*(9^21)=80*9^{11}\); The distance between A and D is 1.5 times the distance between C and E, thus it equal to \(120*9^{11}\) Answer: E. Similar question to practice: Quote: The integers A, B, C, and D shown on the number line above are all equally spaced. If C and D are equal to 5^12 and 5^13, respectively, what is the value of A? A. 5^11 B. 5^10 C. 5^12 D. (7)5^12 E. (12)5^13 Discussed here: theintegersabcanddshownonthenumberlineabovea106968.htmlHope it helps.
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Re: In the figure above, points A, B, C, D, and E are evenly spaced along
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02 Sep 2013, 11:18
First find out the Distance Between E and C i.e 9^13  9^11 = 9^11(9^2  1) = 9^11 * 80 Now this (9^11 * 80) is the distance between C to D and D to E i.e Two Slots distance. As its is given the in the equation that points on the Number line is equal distance so.... 9^11*80/2 = 9^11*40 Now we have to find the distance between A to D  Three slot distance = 3 * 9^11 * 40 = 120 * 9^11




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Re: In the figure above, points A, B, C, D, and E are evenly spaced along
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03 Sep 2013, 22:54
Bunuel, can you help me with this.
I tried to solve it using Arithmetic Progression.
We are given that E = \(9^13\), so a+4d= \(9^13\) and, C = \(9^11\), so a+3d= \(9^11\)
Solving this we get, a=2\(9^11\)\(9^13\) d= {\(9^13\)\(9^11\)}/2
No distance from A to D is 4a+6d. Substituting the values I am getting negative value.
Anything wrong with this?



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Re: In the figure above, points A, B, C, D, and E are evenly spaced along
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In the figure above, points A, B, C, D, and E are evenly spaced along
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09 Sep 2013, 08:45
In the figure above, points A, B, C, D, and E are evenly spaced along the number line. If E = 9^13 and C = 9^11, what is the distance from point A to point D?
If E = 9^13 and C = 9^11 and all points are evenly spaced, then D = 9^12 and A = 9^9.
9^12  9^9 = 9^9*(9^3  1) = 9^9*(80)
I don't understand why the 1.5*80 comes into play. If we know that point D = 9^12 and A = 9^9 then why isn't this a simply subtraction problem?
In other words, let's pretend E=60 and C=40, so D = 50, B = 30, A = 20. The distance between D and A is simply 30. It happens to be 1.5 times the distance between E and C (which is 20) but we don't then multiply 30 by 1.5 to get the distance between D and A.
Any help would be greatly appreciated!
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Re: In the figure above, points A, B, C, D, and E are evenly spaced along
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09 Sep 2013, 16:26
WholeLottaLove wrote: In the figure above, points A, B, C, D, and E are evenly spaced along the number line. If E = 9^13 and C = 9^11, what is the distance from point A to point D?
If E = 9^13 and C = 9^11 and all points are evenly spaced, then D = 9^12 and A = 9^9.
9^12  9^9 = 9^9*(9^3  1) = 9^9*(80)
I don't understand why the 1.5*80 comes into play. If we know that point D = 9^12 and A = 9^9 then why isn't this a simply subtraction problem?
In other words, let's pretend E=60 and C=40, so D = 50, B = 30, A = 20. The distance between D and A is simply 30. It happens to be 1.5 times the distance between E and C (which is 20) but we don't then multiply 30 by 1.5 to get the distance between D and A.
Any help would be greatly appreciated!
Thanks!
A. 9^3 B. 9^9 C. (120)(9^9) D. 9^11 E. (120)(9^11) Taking an example of 2^1, 2^2, 2^3, 2^4, 2^5 i.e 2,4,8,16,32 can not be equally spaced as the distance keeps on increasing between each subsequent point of 2,4,8,16,32. Thus to find the distance between two closest individual points, subtract C from E and then divide the result by 2 i.e. Let us first find the distance between C and E  (9^13  9^11) =9^11 (9^21) = 9^11(811)=9^11(80) Now divide this by 2 to find distance between any two closest points i.e between A&B or B&C or C&D or D&E =9^11(40) Multiply this result by 3 to find distance between A & D =9^11(120)



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Re: In the figure above, points A, B, C, D, and E are evenly spaced along
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16 Sep 2014, 08:04
Bunuel wrote: In the figure above, points A, B, C, D, and E are evenly spaced along the number line. If E = 9^13 and C = 9^11, what is the distance from point A to point D? A. 9^3 B. 9^9 C. (120)(9^9) D. 9^11 E. (120)(9^11) The distance between E and C is \(9^{13}9^{11}=9^{11}*(9^21)=80*9^{11}\); The distance between A and D is 1.5 times the distance between C and E, thus it equal to \(120*9^{11}\) Answer: E. Similar question to practice: Quote: The integers A, B, C, and D shown on the number line above are all equally spaced. If C and D are equal to 5^12 and 5^13, respectively, what is the value of A? A. 5^11 B. 5^10 C. 5^12 D. (7)5^12 E. (12)5^13 Discussed here: theintegersabcanddshownonthenumberlineabovea106968.htmlHope it helps. The distance between A and D is 1.5 times the distance between C and E, thus it equal to \(120*9^{11}\) I have not understood, can you explain the logic?



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Re: In the figure above, points A, B, C, D, and E are evenly spaced along
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16 Sep 2014, 13:06
anu1706 wrote: Bunuel wrote: In the figure above, points A, B, C, D, and E are evenly spaced along the number line. If E = 9^13 and C = 9^11, what is the distance from point A to point D? A. 9^3 B. 9^9 C. (120)(9^9) D. 9^11 E. (120)(9^11) The distance between E and C is \(9^{13}9^{11}=9^{11}*(9^21)=80*9^{11}\); The distance between A and D is 1.5 times the distance between C and E, thus it equal to \(120*9^{11}\) Answer: E. Similar question to practice: Quote: The integers A, B, C, and D shown on the number line above are all equally spaced. If C and D are equal to 5^12 and 5^13, respectively, what is the value of A? A. 5^11 B. 5^10 C. 5^12 D. (7)5^12 E. (12)5^13 Discussed here: theintegersabcanddshownonthenumberlineabovea106968.htmlHope it helps. The distance between A and D is 1.5 times the distance between C and E, thus it equal to \(120*9^{11}\) I have not understood, can you explain the logic? The distance between A and D is 3 units and the distance between C and E is 2 units, hence the distance between A and D is 3/2=1.5 times the distance between C and E. Hope it's clear.
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Re: In the figure above, points A, B, C, D, and E are evenly spaced along
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23 Dec 2015, 12:14
Bunuel wrote: The distance between A and D is 3 units and the distance between C and E is 2 units, hence the distance between A and D is 3/2=1.5 times the distance between C and E.
Hope it's clear. If 2 units if 9^11 * 80 and you want 3 units, why do you only multiply the "80" by 1.5 but not also the "9^11"?
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Re: In the figure above, points A, B, C, D, and E are evenly spaced along
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23 Dec 2015, 12:18
redfield wrote: Bunuel wrote: The distance between A and D is 3 units and the distance between C and E is 2 units, hence the distance between A and D is 3/2=1.5 times the distance between C and E.
Hope it's clear. If 2 units if 9^11 * 80 and you want 3 units, why do you only multiply the "80" by 1.5 but not also the "9^11"? It is getting multiplied by everything. You are doing 1.5*80*9^11 but now as the answer options are all in the form A*9^11, it is better to keep 9^11 as such as you end up with : \ 1.5*80*9^11 = (1.5*80)*9^11 = 120*9^11 . In multiplication, A*B*C = (A*B)*C or A*(B*C) etc. Hope this helps.



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Re: In the figure above, points A, B, C, D, and E are evenly spaced along
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06 Jan 2017, 13:11
crackjack wrote: WholeLottaLove wrote: In the figure above, points A, B, C, D, and E are evenly spaced along the number line. If E = 9^13 and C = 9^11, what is the distance from point A to point D?
Let us first find the distance between C and E  (9^13  9^11) =9^11 (9^21) = 9^11(811)=9^11(80)
Now divide this by 2 to find distance between any two closest points i.e between A&B or B&C or C&D or D&E =9^11(40)
Multiply this result by 3 to find distance between A & D =9^11(120) After 9^11(80) I could not think of divide value by 2 and multiply with 3 to get answer and How The distance between A and D is 1.5 times the distance between C and E, thus it equal to 120∗911 Pls reply. Thanks
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Re: In the figure above, points A, B, C, D, and E are evenly spaced along
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17 May 2017, 05:12
pavan2185 wrote: Attachment: The attachment ABCDE.png is no longer available In the figure above, points A, B, C, D, and E are evenly spaced along the number line. If E = 9^13 and C = 9^11, what is the distance from point A to point D? A. 9^3 B. 9^9 C. (120)(9^9) D. 9^11 E. (120)(9^11) Spoiler Alert: Veritasprep Mock test I could not understand how to solve this question. Did not undertsand theofficial explanation also. The trap in this question is baiting the test taker to think that these lines represent a series of consistently increasing/decreasing exponents (e.x A= 9^9, B =9^10, C=9^11, D=9^12, E =9^13) with answer choice "A." Yet this is an e venly spaced set ; in other words, if we have say E as 2^5 and C as 2^3 then the distance between those two points is 24 so A cannot be 2^1 because 2^324 is 16. More fundamentally, the distance between any two points spaced two letter apart would be 24. In our example, the distance between E and C is 9^139^11 (9^2) 9^11 (1) 9^11  do not forget that there is always an assumed one when subtracting exponents (811) 9^11 (80)9^11 Now the distance between any two lines spaced two letters apart must be (80)9^11  the distance between A and C must also be (80)9^11 as well as the distance between B and D and D and E; however, the distance between and D is 3 spaces which is 1.5 the distance between C and E or moreover  A and C, C and D , B and D, D and E  conceptually, another way of understanding the problem is also that half of (80) 9^11 or alternatively (80) 9^11 /2  would be the distance between two sequential spaces {e.x distance between A, B, or C, B or D,E} so 3 (1/2) (80) 9^11 or (3)(80)9^11/2 would also be another method of calculating the distance. I guess you could also imagine that the spaces between letters in this evenly spaced set form a puzzle piece so the distance between a set of letters would be the length of that puzzle piece ( example: the distance between A and B would be the length of puzzle piece AB)
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Re: In the figure above, points A, B, C, D, and E are evenly spaced along
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pavan2185 wrote: Attachment: ABCDE.png In the figure above, points A, B, C, D, and E are evenly spaced along the number line. If E = 9^13 and C = 9^11, what is the distance from point A to point D? A. 9^3 B. 9^9 C. (120)(9^9) D. 9^11 E. (120)(9^11) We can let the space between each pair of consecutive points = n. Thus, the space from C to E is n + n = 2n. So, we can create the following equation: 9^11 + 2n = 9^13 2n = 9^13  9^11 2n = 9^11(9^2  1) 2n = 9^11(80) n = 9^11(40) Since the distance from A to D = 3n: 3n = 3(9^11)(40) = 120(9^11) Answer: E
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Re: In the figure above, points A, B, C, D, and E are evenly spaced along
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08 Jul 2017, 11:04
pavan2185 wrote: Attachment: ABCDE.png In the figure above, points A, B, C, D, and E are evenly spaced along the number line. If E = 9^13 and C = 9^11, what is the distance from point A to point D? A. 9^3 B. 9^9 C. (120)(9^9) D. 9^11 E. (120)(9^11) Spoiler Alert: Veritasprep Mock test I could not understand how to solve this question. Did not undertsand theofficial explanation also. As the points are equidistant, lets assume, the distance between one unit = \(x\) So Distance between C and E \(= 2x\) \(2x = 9^{13}  9^{11}\) \(2x = 9^{11} (9^2  1)\) \(2x = 9^{11} * (81  1)\) \(2x = 9^{11} * 80\) \(x = 9^{11} * 40\) As the distance between A and D \(= 3x\) \(3x = 9^{11} * 40 * 3\) \(3x = 9^{11} * 120\) Hence, Answer is E
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Re: In the figure above, points A, B, C, D, and E are evenly spaced along
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21 Aug 2017, 18:31
Bunuel wrote: In the figure above, points A, B, C, D, and E are evenly spaced along the number line. If E = 9^13 and C = 9^11, what is the distance from point A to point D? A. 9^3 B. 9^9 C. (120)(9^9) D. 9^11 E. (120)(9^11) The distance between E and C is \(9^{13}9^{11}=9^{11}*(9^21)=80*9^{11}\); The distance between A and D is 1.5 times the distance between C and E, thus it equal to \(120*9^{11}\) Answer: E. Similar question to practice: Quote: The integers A, B, C, and D shown on the number line above are all equally spaced. If C and D are equal to 5^12 and 5^13, respectively, what is the value of A? A. 5^11 B. 5^10 C. 5^12 D. (7)5^12 E. (12)5^13 Discussed here: http://gmatclub.com/forum/theintegers ... 06968.htmlHope it helps. Hi  I have two questions on this problem. 1) how come C does not equal 3*(9^11)*40 2) why can't you just add C + "one distance" unit = 9^11 + (9^11)*40... to arrive at point D? I'm unable to reconcile both 1 and 2 here... Answering my own question here, would appreciate if you could chime in though: the distance between does not equal the distance from zero. C is 9^11 from 0, however we're just interested in calculating the distance between A > D. The distance from A > C indeed does equal 80*9^11, and this implies that A is a negative number since 9^11 < (9^11 * 80)



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Re: In the figure above, points A, B, C, D, and E are evenly spaced along
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23 Sep 2017, 05:32
crackjack wrote: WholeLottaLove wrote: In the figure above, points A, B, C, D, and E are evenly spaced along the number line. If E = 9^13 and C = 9^11, what is the distance from point A to point D?
If E = 9^13 and C = 9^11 and all points are evenly spaced, then D = 9^12 and A = 9^9.
9^12  9^9 = 9^9*(9^3  1) = 9^9*(80)
I don't understand why the 1.5*80 comes into play. If we know that point D = 9^12 and A = 9^9 then why isn't this a simply subtraction problem?
In other words, let's pretend E=60 and C=40, so D = 50, B = 30, A = 20. The distance between D and A is simply 30. It happens to be 1.5 times the distance between E and C (which is 20) but we don't then multiply 30 by 1.5 to get the distance between D and A.
Any help would be greatly appreciated!
Thanks!
A. 9^3 B. 9^9 C. (120)(9^9) D. 9^11 E. (120)(9^11) Taking an example of 2^1, 2^2, 2^3, 2^4, 2^5 i.e 2,4,8,16,32 can not be equally spaced as the distance keeps on increasing between each subsequent point of 2,4,8,16,32. Thus to find the distance between two closest individual points, subtract C from E and then divide the result by 2 i.e. Let us first find the distance between C and E  (9^13  9^11) =9^11 (9^21) = 9^11(811)=9^11(80) Now divide this by 2 to find distance between any two closest points i.e between A&B or B&C or C&D or D&E =9^11(40) Multiply this result by 3 to find distance between A & D =9^11(120) Hi, As per this  D = 9^12 and A = 9^9? If yes then the answer should be arrived at by subtracting the two but this is not the case. Any thoughts on where I am wrong?



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Re: In the figure above, points A, B, C, D, and E are evenly spaced along
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23 Sep 2017, 06:15
Richak91 wrote: In the figure above, points A, B, C, D, and E are evenly spaced along the number line. If E = 9^13 and C = 9^11, what is the distance from point A to point D? A. 9^3 B. 9^9 C. (120)(9^9) D. 9^11 E. (120)(9^11) Hi, As per this  D = 9^12 and A = 9^9? If yes then the answer should be arrived at by subtracting the two but this is not the case. Any thoughts on where I am wrong? The distance between two consecutive points is \(40*9^{11}\), thus: \(A = 79*9^{11}\) \(B = 39*9^{11}\) \(C = 1*9^{11}\) \(D = 41*9^{11}\) \(E = 81*9^{11} = 9^{13}\) Completer solution is here: https://gmatclub.com/forum/inthefigur ... l#p1263035
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