In the figure above, points A, B, C, D, and E are evenly spaced along

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

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 \(9^{13}-9^{11}=9^{11}*(9^2-1)=80*9^{11}\);

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:


Discussed here: the-integers-a-b-c-and-d-shown-on-the-number-line-above-a-106968.html

Hope it helps.
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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?

Since you don't get the correct answer then obviously there is something wrong.

The distance from A to D should be simply 3d.
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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?

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!

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 (9^2-1) = 9^11(81-1)=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)

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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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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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

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 evenly 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 2^3-24 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

9^13-9^11
(9^2) 9^11- (1) 9^11 - do not forget that there is always an assumed one when subtracting exponents
(81-1) 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)

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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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

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)

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/in-the-figur ... l#p1263035
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