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# Ax(y) is an operation that adds 1 to y

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03 Jul 2012, 17:17
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95% (hard)

Question Stats:

39% (01:53) correct 61% (02:28) wrong based on 394 sessions

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Ax(y) is an operation that adds 1 to y and then multiplies the result by x. If x = 2/3, then Ax(Ax(Ax(Ax(Ax(x))))) is between

(A) 0 and ½
(B) ½ and 1
(C) 1 and 1½
(D) 1½ and 2
(E) 2 and 2½
[Reveal] Spoiler: OA
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Joined: 14 Apr 2012
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03 Jul 2012, 20:51
If I read the question correctly, AX(Y) = (1+Y)X
Putting X=2/3,
=> AX(X) = (1+ 2/3)*(2/3)
=> AX(X) = (5/3)*(2/3)
=> AX(X) = 10/9 > 1

AX(AX(X) = (1+10/9)(10/9)....The first term in the bracket will be greater than 2 and next term is greater than one...so the product is greater than 2.
The next operation will make the (1+Y) term greater than 3, so basically as per my understanding there is some information missing in the question or I misunderstood.

Regards,
Tushar
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04 Jul 2012, 01:32
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Smita04 wrote:
Ax(y) is an operation that adds 1 to y and then multiplies the result by x. If x = 2/3, then Ax(Ax(Ax(Ax(Ax(x))))) is between

(A) 0 and ½
(B) ½ and 1
(C) 1 and 1½
(D) 1½ and 2
(E) 2 and 2½

Sorry, but I don't have the OA.

$$A_x(y)$$ is an operation that adds 1 to y and then multiplies the result by x. If x = 2/3, then $$A_x(A_x(A_x(A_x(A_x(x)))))$$ is between

(A) 0 and ½
(B) ½ and 1
(C) 1 and 1½
(D) 1½ and 2
(E) 2 and 2½

According to the stem:
$$A_x(x)=(x+1)x=x^2+x$$;
$$A_x(A_x(x))=A_x(x^2+x)=(x^2+x+1)*x=x^3+x^2+x$$;
$$A_x(A_x(A_x(x)))=A_x(x^3+x^2+x)=(x^3+x^2+x+1)*x=x^4+x^3+x^2+x$$;
...

We can see the pattern now, so $$A_x(A_x(A_x(A_x(A_x(x)))))=x^6+x^5+x^4+x^3+x^2+x$$.

For $$x=\frac{2}{3}$$ we'll get: $$(\frac{2}{3})^6+(\frac{2}{3})^5+(\frac{2}{3})^4+(\frac{2}{3})^3+(\frac{2}{3})^2+(\frac{2}{3})$$.

So, we have the sum of the 6 terms of the geometric progression with the first term equal to $$\frac{2}{3}$$ and the common ratio also equal to $$\frac{2}{3}$$.

Now, the sum of infinite geometric progression with common ratio $$|r|<1$$, is $$sum=\frac{b}{1-r}$$, where $$b$$ is the first term. So, if we had infinite geometric progression instead of just 6 terms then its sum would be $$Sum=\frac{\frac{2}{3}}{1-\frac{2}{3}}=2$$. Which means that the sum of this sequence will never exceed 2, also as we have big enough number of terms (6) then the sum will be very close to 2, so we can safely choose answer choice D.

One can also use direct formula.
We have geometric progression with $$b=\frac{2}{3}$$, $$r=\frac{2}{3}$$ and $$n=6$$;

$$S_n=\frac{b(1-r^n)}{(1-r)}$$ --> $$S_{6}=\frac{\frac{2}{3}(1-(\frac{2}{3})^{6})}{(1-\frac{2}{3})}=2*(1-(\frac{2}{3})^{6})$$. Since $$(\frac{2}{3})^{6}$$ is very small number then $$1-(\frac{2}{3})^{6}$$ will be less than 1 but very close to it, hence $$2*(1-(\frac{2}{3})^{6})$$ will be less than 2 but very close to it.

Hope it's clear.
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21 Jun 2013, 02:47
Bumping for review and further discussion*. Get a kudos point for an alternative solution!

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22 Jun 2013, 01:27
Smita04 wrote:
Ax(y) is an operation that adds 1 to y and then multiplies the result by x. If x = 2/3, then Ax(Ax(Ax(Ax(Ax(x))))) is between

(A) 0 and ½
(B) ½ and 1
(C) 1 and 1½
(D) 1½ and 2
(E) 2 and 2½

This one was quite tricky.. lovely explanation by Bunuel using geometric progression.
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15 Jan 2014, 20:56
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tusharacc wrote:
If I read the question correctly, AX(Y) = (1+Y)X
Putting X=2/3,
=> AX(X) = (1+ 2/3)*(2/3)
=> AX(X) = (5/3)*(2/3)
=> AX(X) = 10/9 > 1

AX(AX(X) = (1+10/9)(10/9)....The first term in the bracket will be greater than 2 and next term is greater than one...so the product is greater than 2.
The next operation will make the (1+Y) term greater than 3, so basically as per my understanding there is some information missing in the question or I misunderstood.

Regards,
Tushar

The highlighted portion is incorrect here.

Ax(Ax(X)) = (1+10/9)(2/3)
Note that X is 2/3 only. Ax(X) is 10/9.
Ax(10/9) = (1 + 10/9)(2/3)
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Joined: 22 Apr 2015
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15 May 2015, 10:04
Bunuel

I solved this in about 1:20 by considering that eventually, this progression would stabilize, and so for any number Ax(Ax(Ax(...(Ax(y))) always approaches a number.

So I set up the equation:
y = (y+1) * (2/3)
3/2y = y + 1
1/2y = 1
y = 2

So the equation will always approach the number 2.

Since we can see that the equation is approaching 2 from below, it must be close to 2, but just slightly less, hence D
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27 Apr 2017, 15:25
Bunuel, you never cease to amaze me with the depth of your abstract and theoretical solutions.

The simplicity of the numbers in this question and the depth of recursion in the function make it possible to calculate the value directly. This approach will not always be viable with similar questions.

$$Ax(x) = \frac{2}{3} * (\frac{2}{3} + 1) = \frac{10}{9}$$
$$Ax(10/9) = \frac{2}{3} * (\frac{10}{9} +1) = \frac{38}{27}$$
$$Ax(38/27) = \frac{2}{3} * (\frac{38}{27} + 1) = \frac{130}{81}$$
$$Ax(130/81) = \frac{2}{3} * (\frac{130}{81} +1) = \frac{422}{243}$$
$$Ax(422/243) = \frac{2}{3} * (\frac{422}{243} + 1) = \frac{1330}{729}$$
$$1.5 < \frac{1330}{729} < 2$$

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28 Apr 2017, 07:48
spence11 wrote:
Bunuel, you never cease to amaze me with the depth of your abstract and theoretical solutions.

The simplicity of the numbers in this question and the depth of recursion in the function make it possible to calculate the value directly. This approach will not always be viable with similar questions.

$$Ax(x) = \frac{2}{3} * (\frac{2}{3} + 1) = \frac{10}{9}$$
$$Ax(10/9) = \frac{2}{3} * (\frac{10}{9} +1) = \frac{38}{27}$$
$$Ax(38/27) = \frac{2}{3} * (\frac{38}{27} + 1) = \frac{130}{81}$$
$$Ax(130/81) = \frac{2}{3} * (\frac{130}{81} +1) = \frac{422}{243}$$
$$Ax(422/243) = \frac{2}{3} * (\frac{422}{243} + 1) = \frac{1330}{729}$$
$$1.5 < \frac{1330}{729} < 2$$

You should multiply with the updated number every time and not $$\frac{2}{3}$$
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Re: Ax(y) is an operation that adds 1 to y   [#permalink] 28 Apr 2017, 07:48
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