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

Question

What is the approximate arithmetic mean of $199879805$, $199879817$, $199879831$, $199879833$, $199879839$, $199879869$

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

Kudos for a correct Solution

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Official Answer

C

broall · Answer

Oh, there are a lot of big numbers :-D

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 :-)

vitaliyGMAT · Answer

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.

broall · Answer

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

vitaliyGMAT · Answer

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

broall · Answer

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

$Sum=6 \times 199,879,800 + 5+17+31+33+39+69=6 \times 199,879,800 +194$
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

Abhishek009 · Answer

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)...

stonecold · Answer

Looks like i missed the Word=> "Approximate" Kudos for the solution. Great Work guys Abhishek009 vitaliyGMAT :shock:

broall · Answer

Yeah

Tks for stonecold's Kudos :twisted:

vitaliyGMAT · Answer

Thanks. Doing great job with posting interesting questions!
Regards

JANISHAR · Answer

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.

premprabhs111 · Answer

c is the correct answer just add the last numbers of each number to get 194 as the sum 
194/6= approx 32

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