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

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 =

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
C

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

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

stonecold , could you explain , in the sum section how did you take 1000 - rest

Hi loserunderachiever,

https://gmatclub.com/forum/download/file.php?id=36500

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
Hence answer C

Hello Sir Bunuel
Is it okay to say: If we have same number of 9s and 10s then the average would have been 9.5 But since we have more 10s (one entire row without 9s) so the average will be slightly higher than 9.5 and Hence 9.55?

_______________________
Yes, that would be correct.

We'll have a 10*10 matrix.

1st row will have 0 9's and 10 10's. => Sum of 1st row = 100
10th row will have 9 9's and 1 10 => Sum of 10th row = 9*9 + 10*1 = 91

Method 1: Shorter

Avg of 1st row: 10
Avg of 10th row: 9.1

Avg of Avg = (10 + 9.1)/2 = 9.55

Method 2: Longer

If you notice, we have a sequence of consecutive integers (CI).
Mean of CI = (91 + 100)/2 = 95.5

Total sum = 95.5 * 10

Avg = (95.5 * 10)/100 = 9.55

ANSWER: C

I recommend looking for patterns that are easy to visualize and require minimal calculations.

First row adds up to 100.
Second row swaps out a 10 for 9, so it adds up to 99.
Third row swaps out another 10 for a 9, so it adds up to 98.
Etc.

So, if we list our row sums, we have

100
99
98
97
96
95
94
93
92
91

Since they're all evenly spaced, the average is half way between the middle two, so 95.5.

Answer choice C.

Bunuel
Can this question be solved using AP?
If yes, can you please share your approach for the same?

Thank you!

We know we have
We have 10*10 = 100 entries
According to the question, the number of 9s is n-1, where n is the row number.
This means:
10th row the number of 9s is 10 - 1 = 9;
9th row the number of 9s is 9 - 1 = 8;
8th row the number of 9s is 8 - 1 = 7
And so on....
Also, the first row would have 1-1=0 9s

So the number of nines is:

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

So, the number of 10s= 100-45=55 10s

Thus, avg. of 45 nines and 55 tens will be:

(45*9 + 55*10)/100 = 9.55 ie. ANSWER: C

Sum of the first row: 10*10
Sum of the last row: 9*9+10

Since the rows are evenly spaced we can divide the sum of the first and last row by the number of numbers (2*10).

(100+81+10)/20 = 191/20 = 9.55

-> C

Hey stonecold,
Why did you take 1000, can you explain that?

The value 1000 comes from assuming that all entries in the table are initially 10. This is because the table has 10*10 = 100 entries, and if each entry were 10, the total sum would be 10*100 = 1000.

From there, stonecold subtracts the total contribution of the 9s in the table to arrive at the actual sum. Specifically:

• For each 9 in the table, the sum decreases by 1
• Since there are 45 entries that are 9s, the total reduction is 1*45 = 45.

Thus, Actual Sum = 1000 - 45 = 955.

This approach is a shortcut to avoid directly summing the contributions of the 9s and 10s separately, as it assumes everything starts as 10 and adjusts downward for the 9s.