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a^2<a,b^2<b,c^2<c and d^2<d

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a^2<a,b^2<b,c^2<c and d^2<d [#permalink] New post   05 Jan 2020, 03:05
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\(a^2<a,\)
\(b^2<b,\)
\(c^2<c\)
and \(d^2<d\)



Quantity A
Quantity B
\(\frac{1}{a}\) + \(\frac{b}{1 + \frac{1}{c+\frac{1}{d}}}\)
The reciprocal of \(\frac{1}{a}\) + \(\frac{b}{1+\frac{1}{c+\frac{1}{d}}}\)



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.

Kudos for the right answer and explanation
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A
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Re: a^2<a,b^2<b,c^2<c and d^2<d [#permalink] New post   05 Jan 2020, 07:55
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\(a^2<a\) is true only when a is a positive fraction. That meas 0<a<1

Let's take Quantity A

\(\frac{1}{a}\) + \(\frac{b}{1 + \frac{1}{c+\frac{1}{d}}}\)

\(c+\frac{1}{d}\)
Now d is a positive fraction that means 1/d will become more than 1 and c, which is fraction, + 1/d (Which is more than 1) will become more than 1
Now, \(1 + \frac{1}{c+\frac{1}{d}}\)
will become 1 + 1/something more than 1 => 1 + fraction => greater than 1
\(\frac{1}{a}\) will become more than 1 as 1/positive fraction is more than 1
and \(\frac{b}{1 + \frac{1}{c+\frac{1}{d}}}\) will become b/something more than 1 so fraction/more than 1 => fraction
=> \(\frac{1}{a}\) + \(\frac{b}{1 + \frac{1}{c+\frac{1}{d}}}\)
will become Something more than 1 + fraction => something more than 1

Quantity B is reciprocal of Quantity A => 1/something more than 1 => Positive fraction
So, Quantity A > Quantity B.
So, Answer is A
Hope it helps!
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Re: a^2<a,b^2<b,c^2<c and d^2<d [#permalink] New post   14 Jan 2020, 21:44
a^2<a,
b^2<b,
c^2<c
and d^2<d.
Indicates a,b,c,d are positive proper fraction.
Let's take a=b=c=d=0.5

Here, both quantity has same equation. Only difference is Quantity A asks for the value of the fraction and quantity B asks for the value of RECIPROCAL of the fraction . . If we can find out the value of the fraction, we can easily find out the value of RECIPROCAL of the fraction.

Let's USE the CALCULATOR.
Begin from the denominator
1/0.5=2>>>>>>
0.5+2=2.5>>>>>
1/2.5=0.04>>>>>
1+2.5=3.5>>>>>
0.5/3.5=0.14>>>
1/0.5+0.14=2.14
So, Value of the fraction=2.14 (Quantity A)
So value of the reciprocal of the fraction= 1/2.14=0.46 (Quantity B).
So, Quantity A is greater.
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Re: a^2<a,b^2<b,c^2<c and d^2<d [#permalink] New post   24 May 2021, 07:09
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Carcass wrote:
\(a^2<a,\)
\(b^2<b,\)
\(c^2<c\)
and \(d^2<d\)



Quantity A
Quantity B
\(\frac{1}{a}\) + \(\frac{b}{1 + \frac{1}{c+\frac{1}{d}}}\)
The reciprocal of \(\frac{1}{a}\) + \(\frac{b}{1+\frac{1}{c+\frac{1}{d}}}\)



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.

Kudos for the right answer and explanation



Put the value of a,b,c,d , 0.2 as square of number is less than original. Do a small calculation. You will get 5.smtg in quantity a and its reciprocal is small hence quantity a is greater.
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Re: a^2<a,b^2<b,c^2<c and d^2<d [#permalink] New post   25 Apr 2024, 22:36
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Re: a^2<a,b^2<b,c^2<c and d^2<d [#permalink]
25 Apr 2024, 22:36
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