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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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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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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.

Answer: C.

P.S. Below is an image for better understanding:
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Re: In a numerical table with 10 rows and 10 columns, each entry is either [#permalink]

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

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New post 06 Aug 2017, 02:12
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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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

Answer: 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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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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New post 16 Mar 2018, 11:31
stonecold , could you explain , in the sum section how did you take 1000 - rest
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Re: In a numerical table with 10 rows and 10 columns, each entry is either   [#permalink] 16 Mar 2018, 11:31
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