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 = 1Given expression: a + 1/(b + 1/c) =
1 + 1/(
1 + 1/
1) =
= 1 + 1/2
=
3/2So, when
a = b = c = 1, the given expression evaluates to be
3/2So, the correct answer choice must also evaluate to be
3/2 when we plug in
a = b = c = 1Check 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
=
5/3So, when
a = b = 1 and c = 2 , the given expression evaluates to be
5/3So, 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
_________________
Brent Hanneson – Creator of gmatprepnow.com
Before you spend another second preparing for the GMAT, check out my article series, Are you doing it wrong?.
You’ll learn what the GMAT actually tests, and why memorizing a ton of formulas actually makes you less effective.