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.
Answer: C.
P.S. Below is an image for better understanding:
Hi
loserunderachiever,
https://gmatclub.com/forum/download/file.php?id=36500May 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
Hence answer C
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