Events & Promotions
|
|
GMAT Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
Track
Your Progress
Practice
Pays
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Sep 1507:00 PM EDT
-09:00 PM EST
Elite scores are possible when you enroll in a powerful GMAT course, taught live online + 6 months access to TTP OnDemand video courses included! Class starts Tues/Thurs Sept. 15, 2026 - Nov. 15, 2027, 7:00pm-9:00pm EST
Sep 2107:00 PM EDT
-10:00 PM EDT
Boost your GMAT score in less than one month in a live online class + 6 months access to TTP OnDemand video courses included! Class starts Mon, Tues, Wed, Thur, Fri Sept. 21, 2026 - Oct. 9, 2027, 7:00pm-10:00pm EST
21 Mar 2019, 12:50
Kudos
Bookmarks
E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
75%
(hard)
Question Stats:
58% (02:31) correct
42%
(02:55)
wrong
based on 194
sessions
History
Date
Time
Result
Not Attempted Yet
GMATH practice exercise (Quant Class 3)
If \({\left( {r + {1 \over r}} \right)^2} = 5\), what is the value of \(\,{\left( {{r^3} + {1 \over {{r^3}}}} \right)^2}\,\) ?
\(\eqalign{\\
& \left( A \right)\,\,7.75 \cr \\
& \left( B \right)\,\,10 \cr \\
& \left( C \right)\,\,12.25 \cr \\
& \left( D \right)\,\,15 \cr \\
& \left( E \right)\,\,20 \cr}\)
If \({\left( {r + {1 \over r}} \right)^2} = 5\), what is the value of \(\,{\left( {{r^3} + {1 \over {{r^3}}}} \right)^2}\,\) ?
\(\eqalign{\\
& \left( A \right)\,\,7.75 \cr \\
& \left( B \right)\,\,10 \cr \\
& \left( C \right)\,\,12.25 \cr \\
& \left( D \right)\,\,15 \cr \\
& \left( E \right)\,\,20 \cr}\)
Updated on: 21 Mar 2019, 17:53
Originally posted by GMATGuruNY on 21 Mar 2019, 13:54.
Last edited by GMATGuruNY on 21 Mar 2019, 17:53, edited 1 time in total.
Last edited by GMATGuruNY on 21 Mar 2019, 17:53, edited 1 time in total.
Kudos
Bookmarks
fskilnik
(r + 1/r)² = 5
r + 1/r = √5
(r + 1/r)² = 5
r² + 1/r² + 2(r)(1/r) = 5
r² + 1/r² + 2 = 5
r² + 1/r² = 3
Multiplying the blue equation and the red equation, we get:
(r² + 1/r²)(r + 1/r) = 3√5
r³ + (r²)(1/r) + (1/r²)(r) + 1/r³ = 3√5
r³ + (r + 1/r) + 1/r³ = 3√5
Substituting r + 1/r = √5 into the green equation above, we get:
r³ + √5 + 1/r³ = 3√5
r³ + 1/r³ = 2√5
(r³ + 1/r³)² = (2√5)²
(r³ + 1/r³)² = 20
.
21 Mar 2019, 14:23
Kudos
Bookmarks
fskilnik
\({\left( {r + {1 \over r}} \right)^2} = 5\,\,\,\,\,\left( * \right)\)
\(? = \,{\left( {{r^3} + {1 \over {{r^3}}}} \right)^2}\)
\(?\,\,\, = \,\,\,{\left[ {\left( {r + {1 \over r}} \right)\left( {{r^2} - r \cdot {1 \over r} + {1 \over {{r^2}}}} \right)} \right]^{\,2}} = {\left( {r + {1 \over r}} \right)^2}{\left( {{r^2} - 1 + {1 \over {{r^2}}}} \right)^2}\,\,\,\mathop = \limits^{\left( * \right)} \,\,\,5 \cdot {\left( {{r^2} - 1 + {1 \over {{r^2}}}} \right)^2}\,\,\,\mathop = \limits^{\left( {**} \right)} \,\,\,20\)
\(\left( {**} \right)\,\,{r^2} - 1 + {1 \over {{r^2}}} = \left( {{r^2} + 2 + {1 \over {{r^2}}}} \right) - 3 = {\left( {r + {1 \over r}} \right)^2} - 3\,\,\,\mathop = \limits^{\left( * \right)} \,\,\,2\)
The correct answer is (E).
We follow the notations and rationale taught in the GMATH method.
Regards,
Fabio.
