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.
May 0510:00 AM PDT
-11:00 AM PDT
GMAT Inequalities is a high-frequency topic in GMAT Quant, but many students struggle because the concepts behave differently from standard algebra. Understanding the right rules, patterns, and edge cases can significantly improve both speed and accuracy.- May 06
08:30 AM PDT
-09:30 AM PDT
In Episode 3 of our GMAT Ninja Critical Reasoning series, we tackle Discrepancy, Paradox, and Explain an Oddity questions. You know the feeling: the passage gives you two facts that seem completely contradictory.... - May 08
08:00 AM PDT
-08:30 AM PDT
Join the special YouTube live-stream for selecting the winners of GMAT Club MBA Scholarships sponsored by Juno live. Watch who gets these coveted MBA scholarships offered by GMAT Club and Juno.
03 Dec 2024, 09:00
Kudos
Bookmarks
\(H_{6}\): 85
Flowers: 97
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
55%
(hard)
Question Stats:
70% (03:29) correct
30%
(03:31)
wrong
based on 487
sessions
History
Date
Time
Result
Not Attempted Yet
A species of tree grows in a unique pattern such that its height in a given year follows a sequence defined as:
where \(H_{1}\) = 3 meters and \(H_{2}\) = 5 meters, and \(H_{n}\) is the height of the tree in meters at the end of year n.
The tree produces flowers in odd years (e.g., 1, 3, 5, 7, ...). The number of flowers produced in year 2n+1 is equal to the value of \(H_{2n+1} * n\).
Based on the above information, select for \(H_{6}\) the height of the tree in year 6 and for Flowers the total amount of flowers produced from year 1 to 6.
\(H_{n} = H_{n-1} + 2 * H_{n-2}\)
where \(H_{1}\) = 3 meters and \(H_{2}\) = 5 meters, and \(H_{n}\) is the height of the tree in meters at the end of year n.
The tree produces flowers in odd years (e.g., 1, 3, 5, 7, ...). The number of flowers produced in year 2n+1 is equal to the value of \(H_{2n+1} * n\).
Based on the above information, select for \(H_{6}\) the height of the tree in year 6 and for Flowers the total amount of flowers produced from year 1 to 6.
| \(H_{6}\) | Flowers | |
| 67 | ||
| 79 | ||
| 85 | ||
| 86 | ||
| 97 | ||
| 100 |
ShowHide Answer
Official Answer
\(H_{6}\): 85
Flowers: 97
03 Dec 2024, 12:11
Kudos
Bookmarks
Can someone explain how they got 97 flowers
03 Dec 2024, 14:19
Kudos
Bookmarks
cdrectenwald
Flowers are produced only in odd years (n=1,3,5...). For odd year 2n+1, the number of flowers is:
Flowers in year 2n+1 = H2n+1 * n
1. Year 1 (2n+1=1, solving n we get n=0):
Flowers = H1*0 = 3*0 = 0
2. Year 3 (2n+1=3, solving n we get n=1):
Flowers = H3*1 = 11*1 = 11
3. Year 5 (2n+1=5, solving n we get n=2):
Flowers = H5*2 = 43*2 = 86
Total Flowers = 0+11+86 = 97.
