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In a numerical table with 10 rows and 10 columns, each entry is either

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In a numerical table with 10 rows and 10 columns, each entry is either  [#permalink]

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25 Jul 2017, 10:27
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In a numerical table with 10 rows and 10 columns, each entry is either a 9 or a 10. If the number of 9s in the nth row is n – 1 for each n from 1 to 10, what is the average (arithmetic mean) of all the numbers in the table?

A. 9.45

B. 9.50

C. 9.55

D. 9.65

E. 9.70

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In a numerical table with 10 rows and 10 columns, each entry is either  [#permalink]

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25 Jul 2017, 22:10
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carcass wrote:
In a numerical table with 10 rows and 10 columns, each entry is either a 9 or a 10. If the number of 9s in the nth row is n – 1 for each n from 1 to 10, what is the average (arithmetic mean) of all the numbers in the table?

A. 9.45

B. 9.50

C. 9.55

D. 9.65

E. 9.70

We have 10*10 = 100 entries.

"The number of 9s in the nth row is n – 1 for each n from 1 to 10" means that:
In the 10th row the number of 9s is 10 - 1 = 9;
In the 9th row the number of 9s is 9 - 1 = 8;
In the 8th row the number of 9s is 8 - 1 = 7;
...
In the 1st row the number of 9s is 1 - 1 = 0.

Thus, the total number of 9s is 9 + 8 + 7 + 6 + 5 + 4 + 3 + 2 + 1 = 9*(9 + 1)/2 = 45.

The number of 10s is therefore 100 - 45 = 55.

The average of all numbers = (45*9 + 55*10)/100 = 9.55.

P.S. Below is an image for better understanding:

Attachment:

Untitled.png [ 8.57 KiB | Viewed 11971 times ]

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Re: In a numerical table with 10 rows and 10 columns, each entry is either  [#permalink]

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05 Aug 2017, 22:13
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Painful question.

I feel this question is just pure lengthy and nothing else.

Here is what I did -->

Sum => 1000-(1+2+3+4+5+6+7+9) => 1000-(45) => 955

Total spaces => 100

Mean => 955/100 => 9.55

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Re: In a numerical table with 10 rows and 10 columns, each entry is either  [#permalink]

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01 Aug 2017, 18:39
1
Key formula to note: Avg of an evenly spaced set = $$\frac{(First + Last)}{2}$$

Yellow part sum: 9 ( 1 + 2 + 3 ..... 9) = 9 * (Avg * 9) = 9 * [(9+1)/2]*9 = 45 * 9 = 405
This also hints there are 9 9s

Green part sum: 10 ( 1 + 2 + 3 .... 10) = 10 * 5.5*10 = 550
There are 10 10s

Total elements = 10 * 10 = 100

Thus Avg =
(550+405)/100 = 9.55 (C)
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Re: In a numerical table with 10 rows and 10 columns, each entry is either  [#permalink]

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06 Aug 2017, 01:12
1
2
We can do it following way also to be quick -
Assume al lnumbers were 10, then the sum would have been 10*10(rows)*10(columns) = 1000
Now, this sum is diminshed by including some 9s in the rows in a specific pattern. So remove the extra counted 1 (10-9 = 1) from above sum.
There are 0+1+2+3+4+5+6+7+8+9 = 45 9s. Hence, subtract 45 from 1000.
This gives 955 which is the sum of all numbers on the grid.
Hence, average = 955/100 = 9.55
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Re: In a numerical table with 10 rows and 10 columns, each entry is either  [#permalink]

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09 Aug 2017, 12:34
1
1
carcass wrote:
In a numerical table with 10 rows and 10 columns, each entry is either a 9 or a 10. If the number of 9s in the nth row is n – 1 for each n from 1 to 10, what is the average (arithmetic mean) of all the numbers in the table?

A. 9.45

B. 9.50

C. 9.55

D. 9.65

E. 9.70

Since the table has 10 rows and 10 columns there are 100 spots in the table.

So we see there are:

For row 1: n = 1 and there are (n - 1) = (1 - 1) = 0. Thus, there are 0 nines in row 1

For row 2: n = 2 and there are (n - 1) = (2 - 1) = 1. Thus, there is 1 nine in row 2.

For row 3: n = 3 and there are (n - 1) = (3 - 1) = 2. Thus, there are 2 nines in row 3, and so on.

We can see that a pattern emerges.

So the number of nines is:

0 + 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 = 45

Because there are 100 numbers in the entire table, we have (100 - 45) = 55 tens.

Thus, the average of the 45 nines and 55 tens is:

[9(45) + 10(55)]/100 = (405 + 550)/100 = 9.55

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Re: In a numerical table with 10 rows and 10 columns, each entry is either  [#permalink]

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21 Aug 2017, 07:17
1
1st row has 0 9's
2nd row has 1 9's
.....
10th row has 9 9's

No of 9's = 1+2.....+9 = 45.

=> No of 10's = 55

So total sum of numbers = (45*9)+(55*10) = 955

Mean = 955/100 = 9.55
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Re: In a numerical table with 10 rows and 10 columns, each entry is either  [#permalink]

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16 Mar 2018, 10:31
stonecold , could you explain , in the sum section how did you take 1000 - rest
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In a numerical table with 10 rows and 10 columns, each entry is either  [#permalink]

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15 Aug 2018, 20:58
1
Bunuel wrote:
carcass wrote:
In a numerical table with 10 rows and 10 columns, each entry is either a 9 or a 10. If the number of 9s in the nth row is n – 1 for each n from 1 to 10, what is the average (arithmetic mean) of all the numbers in the table?

A. 9.45

B. 9.50

C. 9.55

D. 9.65

E. 9.70

We have 10*10 = 100 entries.

"The number of 9s in the nth row is n – 1 for each n from 1 to 10" means that:
In the 10th row the number of 9s is 10 - 1 = 9;
In the 9th row the number of 9s is 9 - 1 = 8;
In the 8th row the number of 9s is 8 - 1 = 7;
...
In the 1st row the number of 9s is 1 - 1 = 0.

Thus, the total number of 9s is 9 + 8 + 7 + 6 + 5 + 4 + 3 + 2 + 1 = 9*(9 + 1)/2 = 45.

The number of 10s is therefore 100 - 45 = 55.

The average of all numbers = (45*9 + 55*10)/100 = 9.55.

P.S. Below is an image for better understanding:

Attachment:
Untitled.png

Hi loserunderachiever,

May be i can try to explain. View the figure above
Let's assume that each value in 10 rows and 10 columns is "10". Then then total sum of each columns would be 100 and there are 10 columns= 100*10= 100
Now since in every row there is one less 10 than previous and we have assumed the value of it 1 each more than actual ,we will subtract it from total.
In the ninth coloumn our assumed total is 100 but actually its 99 so we have taken 1 more than assumed value .
Now in column number 8 number our assumed total is 100 but actually it is 98 so its two more than assumed. Why because we assumed two "10's" in place of 9 so each contributes additional 1.
similarly up-to column number 1 where our assumed total is 100, but actually it has to be 91 . each of the 9 assumed 10's are contributing 1 each to total .

So 1000-(9+8+7+6+5+4+3+2+1+0)
is 1000-45
955
total number observations are 100

so avg = 955/100
9.55
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Re: In a numerical table with 10 rows and 10 columns, each entry is either  [#permalink]

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02 Sep 2018, 18:30
Total entries of 9 and 10 are 100 in the table.
In first row number of 9’s are 0 as according to n-1
In second row number of 9’s are 1
In third, fourth, fifth,....,tenth rows are 2,3,4,5,6,7,8,9 respectively.
So total 9’s are 0+1+2+3+4+5+6+7+8+9=45
Total 10’s are =100-45=55
So total sum of entries is 45*9 + 55*10 = 955
Average is=Sum/total number of entries
Average=955/100=9.55

Posted from my mobile device
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Re: In a numerical table with 10 rows and 10 columns, each entry is either  [#permalink]

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16 Oct 2018, 11:09
Bunuel wrote:
carcass wrote:
In a numerical table with 10 rows and 10 columns, each entry is either a 9 or a 10. If the number of 9s in the nth row is n – 1 for each n from 1 to 10, what is the average (arithmetic mean) of all the numbers in the table?

A. 9.45

B. 9.50

C. 9.55

D. 9.65

E. 9.70

Attachment:
Untitled.png

Lengthy question for sure; not sure I'd be able to stick with it during test-like conditions.
Anyways, if you note that in rows 2-10 there is an equal number of 9s and 10s, you can calculate it as a weighted average between those rows and the first row:

$$\frac{((9x10)(9.50) + (10)(10))}{(10x10)}$$ =8.55 + 1 = 9.55
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In a numerical table with 10 rows and 10 columns, each entry is either  [#permalink]

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18 Nov 2018, 23:01
As per question, in 1 row the number of 9's are row number-1, and the remaining numbers in the row are 10s.
Hence, quickest way is to list number 9s by row number and 10s would be 10-9s (because only 10 numbers in each row)

Ans.= c
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In a numerical table with 10 rows and 10 columns, each entry is either &nbs [#permalink] 18 Nov 2018, 23:01
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