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Given that a is the average (arithmetic mean) of the first nine

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kingflo
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Given that a is the average (arithmetic mean) of the first nine [#permalink] New post   22 Jul 2013, 11:12
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Given that a is the average (arithmetic mean) of the first nine positive multiples of six and b is the median of the first twelve positive multiples of six, what is the ratio of a to b?

A. 3:4
B. 10:13
C. 5:6
D. 13:10
E. 4:3
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B
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Bunuel
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Re: Given that a is the average (arithmetic mean) of the first nine [#permalink] New post   22 Jul 2013, 11:20
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kingflo wrote:
Given that a is the average (arithmetic mean) of the first nine positive multiples of six and b is the median of the first twelve positive multiples of six, what is the ratio of a to b?

A. 3:4
B. 10:13
C. 5:6
D. 13:10
E. 4:3


The first nine positive multiples of six are {6, 12, 18, 24, 30, 36, 42, 48, 54}
The first twelve positive multiples of six are {6, 12, 18, 24, 30, 36, 42, 48, 54, 60, 66, 72}

Both sets are evenly spaced, thus their median=mean:
a=30 and b=(36+42)/2=39 --> a/b=30/39=10/13.

Answer: B.
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Re: Given that a is the average (arithmetic mean) of the first nine [#permalink] New post   01 Aug 2013, 09:26
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In both cases median=mean
The first nine positive multiples of six =median of that 9 numbers=6*5
The first twelve positive multiples of six=6(6+7)/2=6*13/2

A/b=6*5/(6*13/2)=10/13
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Re: Given that a is the average (arithmetic mean) of the first nine [#permalink] New post   20 Aug 2013, 07:21
kingflo wrote:
Given that a is the average (arithmetic mean) of the first nine positive multiples of six and b is the median of the first twelve positive multiples of six, what is the ratio of a to b?

A. 3:4
B. 10:13
C. 5:6
D. 13:10
E. 4:3


Both sets are evenly spaced, thus their median=mean:
Median
Set 1 – 5th number
Set 2 – Average of 6th and 7th position number
Formula to find the Nth number value:
A + (N-1)D
A = first number of the set
N = Total number of multiples
D = Difference
Median for set 1 i.e. 5th number of the set
6 + (5-1)*6 = 30
Similarly median for set 2 = 39
Ratio: 30/39 = 10/13
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Re: Given that a is the average (arithmetic mean) of the first nine [#permalink] New post   06 Oct 2013, 11:21
Is there anything wrong with this approach. I didn't recognize that the numbers were evenly spaced and decided to use a factor method. I made a careless addition mistake and got the question wrong.

\(a=\frac{6(1+2+3+4+5+6+7+8+9)}{9}=\frac{2}{3}*45=30\)

\(b=\frac{6(1+2+3+4+5+6+7+8+9+10+11+12)}{12}\frac{6}{12}*78=39\)

\(\frac{30}{39}=\frac{10}{13}\)
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Re: Given that a is the average (arithmetic mean) of the first nine [#permalink] New post   06 Oct 2013, 11:28
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avohden wrote:
Is there anything wrong with this approach. I didn't recognize that the numbers were evenly spaced and decided to use a factor method. I made a careless addition mistake and got the question wrong.

\(a=\frac{6(1+2+3+4+5+6+7+8+9)}{9}=\frac{2}{3}*45=30\)

\(b=\frac{6(1+2+3+4+5+6+7+8+9+10+11+12)}{12}\frac{6}{12}*78=39\)

\(\frac{30}{39}=\frac{10}{13}\)


Nothing wrong: b is the median of evenly spaced set, so it equal to the mean, thus you can find it the way you did.
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Re: Given that a is the average (arithmetic mean) of the first nine [#permalink] New post   16 Dec 2016, 15:05
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Here is my solution to this one =>
Both the sets are AP series with D=6

For any Ap series => Mean = Median = Average of the last and first term


a=6+54/2 => 30
b=6+72/2 = 39

Hence a/b = 30/39 = 10/13

Hence B
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Re: Given that a is the average (arithmetic mean) of the first nine [#permalink] New post   13 Jul 2017, 15:30
6 x 8 =54

9/2 [54 + 6] =270

270/9= 30

36 +42 /2 =39

30/39

10/13
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Re: Given that a is the average (arithmetic mean) of the first nine [#permalink] New post   16 Jul 2017, 17:47
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kingflo wrote:
Given that a is the average (arithmetic mean) of the first nine positive multiples of six and b is the median of the first twelve positive multiples of six, what is the ratio of a to b?

A. 3:4
B. 10:13
C. 5:6
D. 13:10
E. 4:3


The first multiple of 6 is 6 and the ninth multiple of 6 is 54. Thus, the average of those numbers is (54 + 6)/2 = 30.

The twelfth multiple of 6 is 72, so the median of the first 12 multiples of 6 is (72 + 6)/2 = 78/2 = 39.

So, a = 30 and b = 39, and the ratio of a/b = 30/39 = 10/13.

Answer: B
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Re: Given that a is the average (arithmetic mean) of the first nine [#permalink] New post   19 Dec 2022, 00:54
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