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The reciprocal of x’s non-integer decimal part equals x + 1,

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huda
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The reciprocal of x’s non-integer decimal part equals x + 1, [#permalink] New post   Updated on: 12 Mar 2020, 03:11
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The reciprocal of x’s non-integer decimal part equals x + 1, and x > 0

Quantity A
Quantity B
x
\(\sqrt{2}\)


A) The quantity in Column A is greater.
B) The quantity in Column B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.
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Official Answer
D

Originally posted by huda on 06 Jan 2020, 00:33.
Last edited by Carcass on 12 Mar 2020, 03:11, edited 1 time in total.
Fixed OAs
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Re: The reciprocal of x’s non-integer decimal part equals x + 1, [#permalink] New post   06 Jan 2020, 05:12
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huda wrote:
The reciprocal of x’s non-integer decimal part equals x + 1, and x > 0

Quantity A
Quantity B
x
\(\sqrt{2}\)


A) The quantity in Column A is greater.
B) The quantity in Column B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.


For this type of ques, where not much info is given, try using either QTY B or QTY A

let us take the number is \(\sqrt {2}\) = \(1.41421\)

Now, take the reciprocal of the decimal part = \(\frac{1}{0.41421} = 2.414\), which is \(1 + \sqrt2\)

Therefore, both Quantities are equal
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huda
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Re: The reciprocal of x’s non-integer decimal part equals x + 1, [#permalink] New post   06 Jan 2020, 07:11
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For any fraction \(x\), non integer decimal part = \(x - 1\) (example non-integer decimal part of 1.2 = 1.2 - 1 = 0.2)

The reciprocal part of \( x\)'s non-integer decimal part equals \(x + 1\)

=> Reciprocal of \(( x - 1 )\) = \(x + 1\)

=> \(\frac{1}{(x - 1)}\) = x + 1

cross multiply, we get,

=> 1 = ( \(x\) - 1 ) ( \(x\) + 1 )

Using ( a - b ) * ( a + b ) = \(a^2\) - \(b^2\)

=> 1 = \(x^2\) - \(1^2\) = \(x^2\) - 1

=> \( x^2 \)= 1 + 1 = 2

=> \(x\) = +√2 or \(x\) = -√2

But \(x\) > 0

=> \(x\) = √2

Ans: C
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Re: The reciprocal of x’s non-integer decimal part equals x + 1, [#permalink] New post   07 Jan 2020, 00:14
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huda wrote:
For any fraction \(x\), non integer decimal part = \(x - 1\) (example non-integer decimal part of 1.2 = 1.2 - 1 = 0.2)

The reciprocal part of \( x\)'s non-integer decimal part equals \(x + 1\)

=> Reciprocal of \(( x - 1 )\) = \(x + 1\)

=> \(\frac{1}{(x - 1)}\) = x + 1

cross multiply, we get,

=> 1 = ( \(x\) - 1 ) ( \(x\) + 1 )

Using ( a - b ) * ( a + b ) = \(a^2\) - \(b^2\)

=> 1 = \(x^2\) - \(1^2\) = \(x^2\) - 1

=> \( x^2 \)= 1 + 1 = 2

=> \(x\) = +√2 or \(x\) = -√2

But \(x\) > 0

=> \(x\) = √2

Ans: C


the explanation won't hold for x>2
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Re: The reciprocal of x’s non-integer decimal part equals x + 1, [#permalink] New post   07 Jan 2020, 14:59
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RSQUANT wrote:
huda wrote:
For any fraction \(x\), non integer decimal part = \(x - 1\) (example non-integer decimal part of 1.2 = 1.2 - 1 = 0.2)

The reciprocal part of \( x\)'s non-integer decimal part equals \(x + 1\)

=> Reciprocal of \(( x - 1 )\) = \(x + 1\)

=> \(\frac{1}{(x - 1)}\) = x + 1

cross multiply, we get,

=> 1 = ( \(x\) - 1 ) ( \(x\) + 1 )

Using ( a - b ) * ( a + b ) = \(a^2\) - \(b^2\)

=> 1 = \(x^2\) - \(1^2\) = \(x^2\) - 1

=> \( x^2 \)= 1 + 1 = 2

=> \(x\) = +√2 or \(x\) = -√2

But \(x\) > 0

=> \(x\) = √2

Ans: C


the explanation won't hold for x>2


Did you want to say x > 2 or x > 0? and give us the reason why it not hold (whatever u said)..............?
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Re: The reciprocal of x’s non-integer decimal part equals x + 1, [#permalink] New post   14 Jan 2020, 22:00
In such difficult question, we should ask ourselves WHY a numeric value is given other sector.
Lets play with calculator,
√2=1.4142.......
Then 1.4142.......-1=0.4142.......
then M+
Then 1/MR=2.4142...... which is Exactly √2+1.
Everything matches. So, X=√2.
c
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Re: The reciprocal of x’s non-integer decimal part equals x + 1, [#permalink] New post   11 Mar 2020, 09:07
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huda wrote:
RSQUANT wrote:
huda wrote:
For any fraction \(x\), non integer decimal part = \(x - 1\) (example non-integer decimal part of 1.2 = 1.2 - 1 = 0.2)

The reciprocal part of \( x\)'s non-integer decimal part equals \(x + 1\)

=> Reciprocal of \(( x - 1 )\) = \(x + 1\)

=> \(\frac{1}{(x - 1)}\) = x + 1

cross multiply, we get,

=> 1 = ( \(x\) - 1 ) ( \(x\) + 1 )

Using ( a - b ) * ( a + b ) = \(a^2\) - \(b^2\)

=> 1 = \(x^2\) - \(1^2\) = \(x^2\) - 1

=> \( x^2 \)= 1 + 1 = 2

=> \(x\) = +√2 or \(x\) = -√2

But \(x\) > 0

=> \(x\) = √2

Ans: C


the explanation won't hold for x>2


Did you want to say x > 2 or x > 0? and give us the reason why it not hold (whatever u said)..............?


You are purporting that x can have only 1 possible value, which is true for 1<x<2 (I'm going to assume the phrasing excludes 0<x<1) but as soon as you go above that range, x can have multiple values, which are the positive solutions for:

\(\frac{1}{(x - n)}\) = x + 1

where n={2,3,4,5...}

So, the answer should be D as there is no information allowing us to tell whether x is √2 or some other value.
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Re: The reciprocal of x’s non-integer decimal part equals x + 1, [#permalink] New post   11 Mar 2020, 10:38
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The given answer is wrong.

Assume \(x = N+f\), where \(N\) is a non-negative integer (because \(x\) is positive) and \(f\) is the fractional part ranging between \(0\) and \(1\). According to the question, reciprocal of non-integral part of \(x\) is \(x+1\), which according to how we have defined \(x\) can be written as \(1/f = x+1 = (N+1) + f\)

Multiply both sides by \(f\) to get \(1=(N+1)f+f^2\).

This is a simple quadratic equation whose roots can be found by using the quadratic formula.

The positive root will be \((-(N+1)+\sqrt{(N+1)^2+4})/2\) and this will always lie between \(0\) and \(1\). Thus, we have a value of \(f\) independent of \(N\), which implies value of \(x\) is not fixed.

Hence, D is the correct option. eg - consider the numbers 2.30277563773199464655961063373525 or 0.61803398874989484820458683436564 (this is the golden ratio, btw).

PS:

An easier method is to realize that the function \(1/t\) when \(t\) is between \(0\) and \(1\) has the range from \(1\) to infinity. Thus, we can always create the desired number \(x\) because the number \(x+1\) will itself be in the range of the function \(1/t\)
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Re: The reciprocal of x’s non-integer decimal part equals x + 1, [#permalink] New post   11 Mar 2020, 10:49
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huda wrote:
The reciprocal of x’s non-integer decimal part equals x + 1, and x > 0

Quantity A
Quantity B
x
\(\sqrt{2}\)


A) The quantity in Column A is greater.
B) The quantity in Column B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.



Say x = I + f, where I is the integer part and f is the decimal (fractional) part

\(=> 1/f = x + 1\)

\(=> 1/f = I + 1 + f\)

\(=> 1/f - f = I + 1\)

\(=> 1 - f^2 = f(I + 1)\)

\(=> f^2 + f(I + 1) - 1 = 0\)

\(=> f = [-(I+1) ± \sqrt{(I+1)^2 + 4}]/2\)

If \(I = 1: f = [-2 ± \sqrt{8}]/2 = -1 + \sqrt{2} => x = I + f = \sqrt{2}\)
(note that we took only the positive value of x)
Here, the quantities are equal

If \(I = 2: f = [-3 ± \sqrt{13}]/2 => x = I + f = [1 + \sqrt{13}]/2\)
=> \(x > \sqrt{2}\)
Here, quantity A is greater

Thus, the relationship cannot be determined from the information given

Answer D
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Re: The reciprocal of x’s non-integer decimal part equals x + 1, [#permalink] New post   12 Mar 2020, 03:12
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punindya wrote:
The given answer is wrong.

Assume \(x = N+f\), where \(N\) is a non-negative integer (because \(x\) is positive) and \(f\) is the fractional part ranging between \(0\) and \(1\). According to the question, reciprocal of non-integral part of \(x\) is \(x+1\), which according to how we have defined \(x\) can be written as \(1/f = x+1 = (N+1) + f\)

Multiply both sides by \(f\) to get \(1=(N+1)f+f^2\).

This is a simple quadratic equation whose roots can be found by using the quadratic formula.

The positive root will be \((-(N+1)+\sqrt{(N+1)^2+4})/2\) and this will always lie between \(0\) and \(1\). Thus, we have a value of \(f\) independent of \(N\), which implies value of \(x\) is not fixed.

Hence, D is the correct option. eg - consider the numbers 2.30277563773199464655961063373525 or 0.61803398874989484820458683436564 (this is the golden ratio, btw).

PS:

An easier method is to realize that the function \(1/t\) when \(t\) is between \(0\) and \(1\) has the range from \(1\) to infinity. Thus, we can always create the desired number \(x\) because the number \(x+1\) will itself be in the range of the function \(1/t\)



Thank you. Fixed.

Regards
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Re: The reciprocal of x’s non-integer decimal part equals x + 1, [#permalink] New post   11 Jul 2020, 06:19
If the condition given was 1<x<2, then what would have been the answer? Was it going to be C?
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Re: The reciprocal of x’s non-integer decimal part equals x + 1, [#permalink] New post   23 Jul 2020, 13:07
the fourth root of 40=2.51..
the calc. shows 2.445..
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Re: The reciprocal of xs non-integer decimal part equals x + 1, [#permalink] New post   07 Jul 2023, 03:36
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RSQUANT wrote:
huda wrote:
For any fraction \(x\), non integer decimal part = \(x - 1\) (example non-integer decimal part of 1.2 = 1.2 - 1 = 0.2)

The reciprocal part of \( x\)'s non-integer decimal part equals \(x + 1\)

=> Reciprocal of \(( x - 1 )\) = \(x + 1\)

=> \(\frac{1}{(x - 1)}\) = x + 1

cross multiply, we get,

=> 1 = ( \(x\) - 1 ) ( \(x\) + 1 )

Using ( a - b ) * ( a + b ) = \(a^2\) - \(b^2\)

=> 1 = \(x^2\) - \(1^2\) = \(x^2\) - 1

=> \( x^2 \)= 1 + 1 = 2

=> \(x\) = +√2 or \(x\) = -√2

But \(x\) > 0

=> \(x\) = √2

Ans: C


the explanation won't hold for x>2



Exactly. So how can this be solved algebraically?
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Re: The reciprocal of xs non-integer decimal part equals x + 1, [#permalink] New post   07 Jul 2023, 03:45
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In the previous replies it is solved algebraically such as this one https://gre.myprepclub.com/forum/the-re ... tml#p46710
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