GRE Prep 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.
Retired Moderator
Joined: 16 Apr 2020
Status:Founder & Quant Trainer
Affiliations: Prepster Education
Posts: 1546
Given Kudos: 172
Location: India
WE:Education (Education)
If X_k > Y_{k+2}, then what is the smallest value that k can take?
[#permalink]
11 Mar 2022, 23:04
11 Mar 2022, 23:04
1
1
Bookmarks
00:00
Question Stats:
44% (03:06) correct
55% (03:21) wrong based on 9 sessions
Hide Show timer Statistics
Sequence X is defined by \(X_n = X_{n-1} + 3\), \(X_1 = 11\)
Sequence Y is defined as \(Y_n = Y_{n-1} – 4\), \(Y_3 = 103\)
If \(X_k > Y_{k+2}\), then what is the smallest value that \(k\) can take?
A. 6
B. 9
C. 11
D. 14
E. 15
Sequence Y is defined as \(Y_n = Y_{n-1} – 4\), \(Y_3 = 103\)
If \(X_k > Y_{k+2}\), then what is the smallest value that \(k\) can take?
A. 6
B. 9
C. 11
D. 14
E. 15
Retired Moderator
Joined: 16 Apr 2020
Status:Founder & Quant Trainer
Affiliations: Prepster Education
Posts: 1546
Given Kudos: 172
Location: India
WE:Education (Education)
If X_k > Y_{k+2}, then what is the smallest value that k can take?
[#permalink]
25 Mar 2022, 02:10
25 Mar 2022, 02:10
1
1
Bookmarks
OA Explanation:
Sequence X is in AP with common difference 3 and first term as 11
\(X_2 = X_1 + 3 = 14\)
\(X_3 = X_2 + 3 = 17\)
\(X_4 = X_3 + 3 = 20\)....
Sequence Y is in AP too with common difference -4 and first term as 111
\(Y_n = Y_{n-1} – 4\)
\(Y_3 = Y_2 – 4\)
\(Y_2 = Y_3 + 4 = 103 + 4 = 107\)
\(Y_2 = Y_1 – 4\)
\(Y_1 = Y_2 + 4 = 107 + 4 = 111\)
Given: \(X_k > Y_{k+2}\)
Let us check the option choices;
A. 6
\(X_6 > Y_8\)
\(11 + (6-1)3 > 111 + (8-1)(-4)\)
\(11 + 15 > 111 - 28\)
\(26 > 83\)
B. 9
\(X_9 > Y_{11}\)
\(11 + (9-1)3 > 111 + (11-1)(-4)\)
\(11 + 24 > 111 - 40\)
\(35 > 71\)
C. 11
\(X_{11} > Y_{13}\)
\(11 + (11-1)3 > 111 + (13-1)(-4)\)
\(11 + 30 > 111 - 48\)
\(41 > 63\)
D. 14
\(X_{14} > Y_{16}\)
\(11 + (14-1)3 > 111 + (16-1)(-4)\)
\(11 + 39 > 111 - 60\)
\(50 > 51\)
E. 15
\(X_{15} > Y_{17}\)
\(11 + (15-1)3 > 111 + (17-1)(-4)\)
\(11 + 42 > 111 - 64\)
\(53 > 47\)
Hence, option E
Sequence X is in AP with common difference 3 and first term as 11
\(X_2 = X_1 + 3 = 14\)
\(X_3 = X_2 + 3 = 17\)
\(X_4 = X_3 + 3 = 20\)....
Sequence Y is in AP too with common difference -4 and first term as 111
\(Y_n = Y_{n-1} – 4\)
\(Y_3 = Y_2 – 4\)
\(Y_2 = Y_3 + 4 = 103 + 4 = 107\)
\(Y_2 = Y_1 – 4\)
\(Y_1 = Y_2 + 4 = 107 + 4 = 111\)
Given: \(X_k > Y_{k+2}\)
Let us check the option choices;
A. 6
\(X_6 > Y_8\)
\(11 + (6-1)3 > 111 + (8-1)(-4)\)
\(11 + 15 > 111 - 28\)
\(26 > 83\)
B. 9
\(X_9 > Y_{11}\)
\(11 + (9-1)3 > 111 + (11-1)(-4)\)
\(11 + 24 > 111 - 40\)
\(35 > 71\)
C. 11
\(X_{11} > Y_{13}\)
\(11 + (11-1)3 > 111 + (13-1)(-4)\)
\(11 + 30 > 111 - 48\)
\(41 > 63\)
D. 14
\(X_{14} > Y_{16}\)
\(11 + (14-1)3 > 111 + (16-1)(-4)\)
\(11 + 39 > 111 - 60\)
\(50 > 51\)
E. 15
\(X_{15} > Y_{17}\)
\(11 + (15-1)3 > 111 + (17-1)(-4)\)
\(11 + 42 > 111 - 64\)
\(53 > 47\)
Hence, option E
Re: If X_k > Y_{k+2}, then what is the smallest value that k can take?
[#permalink]
14 Jul 2022, 02:53
14 Jul 2022, 02:53
confused with the method.
