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What is the arithmetic mean of 199879805, 199879817, 99879831, 998798

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What is the arithmetic mean of 199879805, 199879817, 99879831, 998798  [#permalink]

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09 Dec 2016, 07:58
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What is the approximate arithmetic mean of $$199879805$$, $$199879817$$, $$199879831$$, $$199879833$$, $$199879839$$, $$199879869$$

A) $$199879822$$
B) $$199879828$$
C) $$199879832$$
D) $$199879842$$
E) $$199879845$$

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Re: What is the arithmetic mean of 199879805, 199879817, 99879831, 998798  [#permalink]

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09 Dec 2016, 09:08
stonecold wrote:
What is the arithmetic mean of $$199879805,199879817,99879831,99879833,199879839,199879869$$

A)$$99879822$$
B)$$99879828$$
C)$$99879832$$
D)$$99879842$$
E)$$99879845$$

Oh, there are a lot of big numbers

Note that all given numbers have the same form as $$x998798xx$$
Hence we have
$$199,879,805=99,879,800 + 100,000,000 +5$$
$$199,879,817=99,879,800+100,000,000 +17$$
$$099,879,831=99,879,800 +31$$
$$099,879,833=99,879,800+33$$
$$199,879,839=99,879,800+100,000,000+39$$
$$199,879,869=99,879,800+100,000,000+69$$

The sum of all these numbers is
$$6 \times 99,879,800 + 4 \times 100,000,000 + 5 +17+31+33+39+69$$
$$6 \times 99,879,800 + 4 \times 100,000,000+194$$
And the arithmetic mean is

$$\frac{Sum}{6}=\frac{6 \times 99,879,800 + 4 \times 100,000,000+194}{6}=99,879,800 + \frac{400,000,194}{6}=99,879,800 + 66,666,699=166,546,499$$

I think the given numbers are missing some digits
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Re: What is the arithmetic mean of 199879805, 199879817, 99879831, 998798  [#permalink]

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Updated on: 09 Dec 2016, 09:17
I can't get answer from the options

199879805 = X

$$\frac{X + 12 + X – 100000000 + 26 + X – 100000000 + 28 + X + 34 + X + 64}{6}$$

$$\frac{6X + 164 – 200000000}{6}$$

166546499

Now let’s pretend that 1 should not be in front.

99879805 = X

We’ll get

$$\frac{X + X+ 12 + X + 26 + X + 28 + X + 34 + X + 64}{6} = X + \frac{164}{6} = X + 27$$,… not an integer but very close to 99879832 (C)

Please stonecold check your question and confirm if everything is right.

Originally posted by vitaliyGMAT on 09 Dec 2016, 09:13.
Last edited by vitaliyGMAT on 09 Dec 2016, 09:17, edited 2 times in total.
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Re: What is the arithmetic mean of 199879805, 199879817, 99879831, 998798  [#permalink]

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09 Dec 2016, 09:15
2
stonecold wrote:
What is the arithmetic mean of 199,879,805, 199,879,817, 99,879,831, 99,879,833, 199,879,839, 199,879,869

A)$$99879822$$
B)$$99879828$$
C)$$99879832$$
D)$$99879842$$
E)$$99879845$$

I think all numbers highlighted need to be decreased by 100,000,000. Hence we have

$$99,879,805=99,879,800+05$$
$$99,879,817=99,879,800+17$$
$$99,879,831=99,879,800+31$$
$$99,879,833=99,879,800+33$$
$$99,879,839=99,879,800+39$$
$$99,879,869=99,879,800+69$$

the arithmetic mean is
$$99,879,800 + \frac{5+17+31+33+39+69}{6}=99,879,800 +\frac{194}{6}$$

Since 194 is not divisible by 6, the arithmetic mean is not an integer. Hence none of choices is the correct answer.

In fact, the arithmetic mean is 99,879,832.3333
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Re: What is the arithmetic mean of 199879805, 199879817, 99879831, 998798  [#permalink]

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09 Dec 2016, 09:25
1
stonecold wrote:
What is the arithmetic mean of $$199879805,199879817,199879831,199879833,199879839,199879869$$

A)$$199879822$$
B)$$199879828$$
C)$$199879832$$
D)$$199879842$$
E)$$199879845$$

Kudos for a correct Solution

Same answer but not an integer

199879805 = X

$$\frac{X + X+ 12 + X + 26 + X + 28 + X + 34 + X + 64}{6} = X + \frac{164}{6} = X + 27$$,… 199879832 (C)

C
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Re: What is the arithmetic mean of 199879805, 199879817, 99879831, 998798  [#permalink]

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Updated on: 09 Dec 2016, 09:42
stonecold wrote:
What is the approximate arithmetic mean of $$199879805,199879817,199879831,199879833,199879839,199879869$$

A)$$199879822$$
B)$$199879828$$
C)$$199879832$$
D)$$199879842$$
E)$$199879845$$

$$199,879,805=199,879,800+05$$
$$199,879,817=199,879,800+17$$
$$199,879,831=199,879,800+31$$
$$199,879,833=199,879,800+33$$
$$199,879,839=199,879,800+39$$
$$199,879,869=199,879,800+69$$

$$Sum=6 \times 199,879,800 + 5+17+31+33+39+69=6 \times 199,879,800 +194$$
the arithmetic mean $$=199,879,800 +\frac{194}{6}=199,879,800+32.33=199,879,832.33 \approx 199,879,832$$

The correct answer is C
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Originally posted by broall on 09 Dec 2016, 09:26.
Last edited by broall on 09 Dec 2016, 09:42, edited 1 time in total.
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Re: What is the arithmetic mean of 199879805, 199879817, 99879831, 998798  [#permalink]

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09 Dec 2016, 09:30
1
stonecold wrote:
What is the arithmetic mean of $$199879805,199879817,199879831,199879833,199879839,199879869$$

A)$$199879822$$
B)$$199879828$$
C)$$199879832$$
D)$$199879842$$
E)$$199879845$$

199879805 = 199879805 + 0
199879817 = 199879805 + 12
199879831 = 199879805 + 26
199879833 = 199879805 + 28
199879839 = 199879805 + 34
199879869 = 199879805 + 64

So, The arithmatic Mean will be 199879805 + ( 12 + 26 + 38 + 64 )/6

Or, Mean = 199879805 + 27.33 => $$199879832$$

Answer will be (C)...

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Re: What is the arithmetic mean of 199879805, 199879817, 99879831, 998798  [#permalink]

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09 Dec 2016, 09:36
nguyendinhtuong wrote:
stonecold wrote:
What is the arithmetic mean of $$199879805,199879817,199879831,199879833,199879839,199879869$$

A)$$199879822$$
B)$$199879828$$
C)$$199879832$$
D)$$199879842$$
E)$$199879845$$

$$199,879,805=199,879,800+05$$
$$199,879,817=199,879,800+17$$
$$199,879,831=199,879,800+31$$
$$199,879,833=199,879,800+33$$
$$199,879,839=199,879,800+39$$
$$199,879,869=199,879,800+69$$

$$Sum=6 \times 199,879,800 + 5+17+31+33+39+69=6 \times 199,879,800 +194$$
the arithmetic mean $$=199,879,800 +\frac{194}{6}$$

Since 194 is not divisible by 6, the arithmetic mean is not an integer.

Hence none of these choices is the correct answer.

In fact, the arithmetic mean is 199,879,832.3333

Looks like i missed the Word=> "Approximate"

Kudos for the solution.
Great Work guys Abhishek009 vitaliyGMAT

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Re: What is the arithmetic mean of 199879805, 199879817, 99879831, 998798  [#permalink]

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09 Dec 2016, 09:39
stonecold wrote:
nguyendinhtuong wrote:
stonecold wrote:
What is the arithmetic mean of $$199879805,199879817,199879831,199879833,199879839,199879869$$

A)$$199879822$$
B)$$199879828$$
C)$$199879832$$
D)$$199879842$$
E)$$199879845$$

$$199,879,805=199,879,800+05$$
$$199,879,817=199,879,800+17$$
$$199,879,831=199,879,800+31$$
$$199,879,833=199,879,800+33$$
$$199,879,839=199,879,800+39$$
$$199,879,869=199,879,800+69$$

$$Sum=6 \times 199,879,800 + 5+17+31+33+39+69=6 \times 199,879,800 +194$$
the arithmetic mean $$=199,879,800 +\frac{194}{6}$$

Since 194 is not divisible by 6, the arithmetic mean is not an integer.

Hence none of these choices is the correct answer.

In fact, the arithmetic mean is 199,879,832.3333

Looks like i missed the Word=> "Approximate"

Kudos for the solution.
Great Work guys Abhishek009 vitaliyGMAT

Yeah Tks for stonecold's Kudos
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Re: What is the arithmetic mean of 199879805, 199879817, 99879831, 998798  [#permalink]

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09 Dec 2016, 09:44
1
Thanks. Doing great job with posting interesting questions!
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Joined: 03 Aug 2016
Posts: 1
Re: What is the arithmetic mean of 199879805, 199879817, 99879831, 998798  [#permalink]

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23 Aug 2017, 05:02
1
I think below is the easiest method to solve this.
Take any number for reference
Lets take 199879831.
Find out the difference of other numbers from this number.
1)199879805-199879831=-26
2)199879817-199879831=-14
3)199879833-199879831=+2
4)199879839-199879831=+8
5)199879869-199879831=+38
Net=+8
Avg=199879831+8/6=199879831+1.33=199879832.33
Closest answer is C.
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What is the arithmetic mean of 199879805, 199879817, 99879831, 998798  [#permalink]

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27 Aug 2017, 04:39
2
c is the correct answer just add the last numbers of each number to get 194 as the sum
194/6= approx 32
What is the arithmetic mean of 199879805, 199879817, 99879831, 998798   [#permalink] 27 Aug 2017, 04:39
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