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If a > 0, b > 0, and c > 0, a + 1/(b + 1/c) =

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Bunuel
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If a > 0, b > 0, and c > 0, a + 1/(b + 1/c) = [#permalink] New post   30 Jul 2018, 00:43
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If \(a > 0\), \(b > 0\) , and \(c > 0\) , \(a + \frac{1}{b + \frac{1}{c}}\) =


A. \(\frac{a + b}{c}\)

B. \(\frac{ac + bc + 1}{c}\)

C. \(\frac{abc + b + c}{bc}\)

D. \(\frac{a + b + c}{abc + 1}\)

E. \(\frac{abc + a + c}{bc + 1}\)
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E
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If a > 0, b > 0, and c > 0, a + 1/(b + 1/c) = [#permalink] New post   30 Jul 2018, 01:44
Bunuel wrote:
If \(a > 0\), \(b > 0\) , and \(c > 0\) , \(a + \frac{1}{b + \frac{1}{c}}\) =


A. \(\frac{a + b}{c}\)

B. \(\frac{ac + bc + 1}{c}\)

C. \(\frac{abc + b + c}{bc}\)

D. \(\frac{a + b + c}{abc + 1}\)

E. \(\frac{abc + a + c}{bc + 1}\)



It's better to solve this one part to part:

= \(b + \frac{1}{c}\)

=\(\frac{bc + 1}{c}\)

Then

=\(\frac{1}{(bc + 1) / c}\)

= \(\frac{c}{bc + 1}\)

last part :

= a + \(\frac{c}{bc +1}\)

=\(\frac{c + abc + a}{bc +1}\)

The best answer is E.
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Re: If a > 0, b > 0, and c > 0, a + 1/(b + 1/c) = [#permalink] New post   30 Aug 2018, 13:25
assume values for a,b,c and put values of abc in options
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BrentGMATPrepNow
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Re: If a > 0, b > 0, and c > 0, a + 1/(b + 1/c) = [#permalink] New post   30 Aug 2018, 17:25
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Bunuel wrote:
If \(a > 0\), \(b > 0\) , and \(c > 0\) , \(a + \frac{1}{b + \frac{1}{c}}\) =


A. \(\frac{a + b}{c}\)

B. \(\frac{ac + bc + 1}{c}\)

C. \(\frac{abc + b + c}{bc}\)

D. \(\frac{a + b + c}{abc + 1}\)

E. \(\frac{abc + a + c}{bc + 1}\)


We're looking for an expression that is equivalent to the given expression.
So, if we find the value of the given expression for certain values of a, b and c, then the correct answer choice must also equal the same value for the same value of values of a, b and c.

So, let's see what happens when a = b = c = 1

Given expression: a + 1/(b + 1/c) = 1 + 1/(1 + 1/1) =
= 1 + 1/2
= 3/2

So, when a = b = c = 1, the given expression evaluates to be 3/2
So, the correct answer choice must also evaluate to be 3/2 when we plug in a = b = c = 1

Check the answer choices....
A. (a + b)/c = (1 + 1)/1 = 2. No good. We need 3/2. ELIMINATE.

B. (ac + bc + 1)/c = (1 + 1 + 1)/1 = 3. No good. We need 3/2. ELIMINATE.

C. (abc + b + c)/bc = (1 + 1 + 1)/1 = 1. No good. We need 3/2. ELIMINATE.

D. (a + b + c)/(abc + 1) = (1 + 1 + 1)/(1 + 1) = 3/2. Perfect!! KEEP

E. (abc + a + c)/(bc + 1) = (1 + 1 + 1)/(1 + 1) = 3/2. Perfect!! KEEP


Okay, we have two possible answers: D or E.
So, we must test one more set of values.



Let's see what happens when a = b = 1 and c = 2
Here, the given expression: a + 1/(b + 1/c) = 1 + 1/(1 + 1/2) =
= 1 + 1/(3/2)
= 1 + 2/3
= 5/3

So, when a = b = 1 and c = 2 , the given expression evaluates to be 5/3
So, the correct answer choice must also evaluate to be 5/3 when we plug in a = b = 1 and c = 2

Check the REMAINING answer choices....
D. (a + b + c)/(abc + 1) = (1 + 1 + 2)/(2 + 1) = 4/3. No good. We need 5/3. ELIMINATE.

E. (abc + a + c)/(bc + 1) = (2 + 1 + 2)/(2 + 1) = 5/3. Perfect!! KEEP

Answer: E

Cheers,
Brent
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Re: If a > 0, b > 0, and c > 0, a + 1/(b + 1/c) = [#permalink] New post   30 Mar 2021, 01:45
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Re: If a > 0, b > 0, and c > 0, a + 1/(b + 1/c) = [#permalink]
30 Mar 2021, 01:45
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