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17 Jul 2020, 01:58
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Q1, Q2, Q3, …, Qn is a sequence of n consecutive multiples of positive integer k. If n > 4, then what is the average (arithmetic mean) of the sequence?
(1) k = 4
(2) The sum of the second greatest and the second least terms of the series is 24.
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17 Jul 2020, 02:56
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(1) k = 4
So the sequence will be
4(1), 4(2), 4(3), 4(4),4(5).........
Average of set till 4(5) = 12
Average of set till 4(6) = 14
Average of set till 4(7) = 16
The average is varying..
Hence Insufficient.
(2) The sum of the second greatest and the second least terms of the series is 24
K(2) + K(n-1) = 24
K{2+n-1} = 24
K{n+1} = 24
Factors of 24 = 2*2*2*3
Since n > 4
N+1 can only be 8 or 12
Then n = 7 or 11
If N = 11
K = 2
Sequence will be 2(1), 2(2).......2(11)
And average 2(6) = 12
If N = 7
K = 3
Sequence will be 3(1), 3(2).......3(7)
And average 3(4) = 12
(Sufficient)
IMO B
Posted from my mobile device
So the sequence will be
4(1), 4(2), 4(3), 4(4),4(5).........
Average of set till 4(5) = 12
Average of set till 4(6) = 14
Average of set till 4(7) = 16
The average is varying..
Hence Insufficient.
(2) The sum of the second greatest and the second least terms of the series is 24
K(2) + K(n-1) = 24
K{2+n-1} = 24
K{n+1} = 24
Factors of 24 = 2*2*2*3
Since n > 4
N+1 can only be 8 or 12
Then n = 7 or 11
If N = 11
K = 2
Sequence will be 2(1), 2(2).......2(11)
And average 2(6) = 12
If N = 7
K = 3
Sequence will be 3(1), 3(2).......3(7)
And average 3(4) = 12
(Sufficient)
IMO B
Posted from my mobile device
17 Jul 2020, 03:23
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Q1, Q2, Q3, …, Qn is a sequence of n consecutive multiples of positive integer k. If n > 4, then what is the average (arithmetic mean) of the sequence?
\(Avg. = \frac{Q_1 + Q_2 + Q_3 + Q_4 ..... Q_{n-1} + Q_n}{n}\) = ?
(1) k = 4
Case1: Let \(Q_1 = k, Q_2 = 2k, Q_3 = 3k ... Q_{10} = 10k\)
\(Avg. = \frac{Q_1 + Q_2 + Q_3 + Q_4 ..... Q_{n-1} + Q_n}{n}\)
\(Avg. = \frac{k + 2k + 3k + 4k + .... 10k}{10}\)
\(Avg. = \frac{11}{2}*k = 22\)
Case 2: Let \(Q_1 = 5k, Q_2 = 6k, Q_3 = 7k ... Q_5 = 9k\)
\(Avg. = \frac{Q_1 + Q_2 + Q_3 + Q_4 ..... Q_{n-1} + Q_n}{n}\)
\(Avg. = \frac{5k + 6k + 7k + 8k + 9k}{5}\)
Avg. = 7k = 28
INSUFFICIENT.
(2) The sum of the second greatest and the second least terms of the series is 24.
\(Q_2 + Q_{n-1} = 24\)
Case 1: Let \(Q_1 = k, Q_2 = 2k, Q_3 = 3k ... + Q_{10} = 10k\)
2k + 9k = 24
\(k = \frac{24}{11}\)
\(Avg. = \frac{Q_1 + Q_2 + Q_3 + Q_4 ..... Q_{n-1} + Q_n}{n}\)
\(Avg. = \frac{k + 2k + 2k + 4k + ..... + 10k}{10}\)
\(Avg. = \frac{11}{2}k = \frac{11}{2}*\frac{24}{11} = 12\)
Case 2: Let \(Q_1 = 5k, Q_2 = 6k, Q_3 = 7k ... Q_5 = 9k\)
6k + 8k = 24
\(k = \frac{24}{14}\)
\(Avg. = \frac{Q_1 + Q_2 + Q_3 + Q_4 + Q_5}{5}\)
\(Avg. = \frac{5k + 6k + 7k + 8k + 9k}{5}\)
\(Avg. = 7k = 7*\frac{24}{14} = 12 \)
SUFFICIENT.
Answer B.
\(Avg. = \frac{Q_1 + Q_2 + Q_3 + Q_4 ..... Q_{n-1} + Q_n}{n}\) = ?
(1) k = 4
Case1: Let \(Q_1 = k, Q_2 = 2k, Q_3 = 3k ... Q_{10} = 10k\)
\(Avg. = \frac{Q_1 + Q_2 + Q_3 + Q_4 ..... Q_{n-1} + Q_n}{n}\)
\(Avg. = \frac{k + 2k + 3k + 4k + .... 10k}{10}\)
\(Avg. = \frac{11}{2}*k = 22\)
Case 2: Let \(Q_1 = 5k, Q_2 = 6k, Q_3 = 7k ... Q_5 = 9k\)
\(Avg. = \frac{Q_1 + Q_2 + Q_3 + Q_4 ..... Q_{n-1} + Q_n}{n}\)
\(Avg. = \frac{5k + 6k + 7k + 8k + 9k}{5}\)
Avg. = 7k = 28
INSUFFICIENT.
(2) The sum of the second greatest and the second least terms of the series is 24.
\(Q_2 + Q_{n-1} = 24\)
Case 1: Let \(Q_1 = k, Q_2 = 2k, Q_3 = 3k ... + Q_{10} = 10k\)
2k + 9k = 24
\(k = \frac{24}{11}\)
\(Avg. = \frac{Q_1 + Q_2 + Q_3 + Q_4 ..... Q_{n-1} + Q_n}{n}\)
\(Avg. = \frac{k + 2k + 2k + 4k + ..... + 10k}{10}\)
\(Avg. = \frac{11}{2}k = \frac{11}{2}*\frac{24}{11} = 12\)
Case 2: Let \(Q_1 = 5k, Q_2 = 6k, Q_3 = 7k ... Q_5 = 9k\)
6k + 8k = 24
\(k = \frac{24}{14}\)
\(Avg. = \frac{Q_1 + Q_2 + Q_3 + Q_4 + Q_5}{5}\)
\(Avg. = \frac{5k + 6k + 7k + 8k + 9k}{5}\)
\(Avg. = 7k = 7*\frac{24}{14} = 12 \)
SUFFICIENT.
Answer B.
