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Retired Moderator
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WE:Education (Education)
If X_k > Y_{k+2}, then what is the smallest value that k can take?
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11 Mar 2022, 23:04
11 Mar 2022, 23:04
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Question Stats:
50% (02:53) correct
50% (03:21) wrong based on 10 sessions
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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?
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25 Mar 2022, 02:10
25 Mar 2022, 02:10
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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.
