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If you average each row and average each column, then you end up with the same number. Each statement says the same thing so (d).
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honestly not sure on this one: the total should be the sum of the rows + the sum of the columns and then divide that total by 40 items to get the average of all the items on the rectangular rack.

but I'm not sure how to deduce any further information from the statements, and I know guessing E generally isn't best practice, but I'm going to do that in this case. Answer: E
Bunuel
In a store, 40 items are arranged on a rack in 5 rows and 8 columns. The average (arithmetic mean) of the price of items in each row (m) is \(X_m\) [1 ≤ m ≤5]. The average of the price of items in each column (n) is \(Y_n\) [1 ≤ n ≤ 8]. What is the average price of all items on the rack?

(1) \(X_1 + X_2 + ... + X_5 = $850\)
(2) \(Y_1 + Y_2 + ... + Y_8 = $1,360\)


 


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For any matrix, average of mean of all rows or average of mean of all column is equal to the mean of the matrix.

Also

Sum of all rows = sum of all columns.
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From the perspective of mean of row
Sum of all the items in a particular row
Row 1 = Average * #elements = X1 * 8 (total number of columns is 8)
Row 2 = X2 * 8

and so on.

That is sum of all the elements = (X1 * 8) + (X2 * 8) + (X3 * 8) + (X4 * 8) + (X5 * 8)= 8 * (X1 + X2 + X3 + X4 +X5)

This sum is given in the first statement hence it satisfies.



From perspective of mean of Column

Sum of all the items in a particular Column
Column 1 = Average * #elements = Y1 * 5 (total number of row is 5)
Column 2 = Y2 * 5
And so on

That is sum of all the elements = (Y1 * 5) + (Y2 * 5) .... (Y5 * 5)= 5 * (Y1 + Y2 .... Y8)
This sum is given in the first statement, hence it satisfies.
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Bunuel
In a store, 40 items are arranged on a rack in 5 rows and 8 columns. The average (arithmetic mean) of the price of items in each row (m) is \(X_m\) [1 ≤ m ≤5]. The average of the price of items in each column (n) is \(Y_n\) [1 ≤ n ≤ 8]. What is the average price of all items on the rack?

(1) \(X_1 + X_2 + ... + X_5 = $850\)
(2) \(Y_1 + Y_2 + ... + Y_8 = $1,360\)


 


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Somehow it was more difficult to get the writing of the variables than getting the answer right.

We are looking for any numbers that allow us to use them for multiplication to calculate the average for all given items.

We can imagine a matrix:

X_1
X_2
X_3
X_4
X_5
Y_1 .... Y_8

We could fill it like:

1 2 3 4 5 6 7 8
7 6 5 7 5 3 5 7
1 1 1 1 1 1 1 1
2 2 2 2 2 2 2 2
3 6 7 3 1 5 7 5

We have a total of: 151
But how do we get there with averages?

X1: 5,625 ((1+2+3+4+5+6+7+8+9)/8)
X2: 5,625
X3: 1
X4: 2
X5: 4,625
The summ of all those is: 18,875
Multiplied by 8, we get: 151.

Why? Because we did nothing else than reversing the calculation.
We had to divide by 8 to get the average and multiply by 8 to get the total again.

Vice versa the same logic counts for Y.

Therefore: D) (Both alone are sufficient)
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Since there are 8 columns, each row has 8 items. Similarly, since there are 5 rows, each column has 5 items.

Average price of all items = Total of all items/No. of items

Total of all items = X1*8 + X2*8....X5*8 = 8(X1+X2+X3+X4+X5)

Alternatively, Total = 5*(Y1 + Y2 + ... + Y8)

St. 1 we have, the value of X1+X2+X3+X4+X5

Average price of all items = 850*8/40 = 170. Sufficient

St. 2 (Y1 + Y2 + ... + Y8) = 1,360

Total = 5*(Y1 + Y2 + ... + Y8)

= 5*1360

Average = 5*1360/40 = 170. Sufficient.

So, each statement alone is sufficient. Option D
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The total price of all items can be found by summing up the average price of each row, then multiplying by the number of items per row. Alternatively, sum the average price of each column and multiply by the number of items per column. Both statements provide one of these sums, allowing you to calculate the total price, and thus the overall average. Therefore, each statement alone is sufficient. D

Bunuel
In a store, 40 items are arranged on a rack in 5 rows and 8 columns. The average (arithmetic mean) of the price of items in each row (m) is \(X_m\) [1 ≤ m ≤5]. The average of the price of items in each column (n) is \(Y_n\) [1 ≤ n ≤ 8]. What is the average price of all items on the rack?

(1) \(X_1 + X_2 + ... + X_5 = $850\)
(2) \(Y_1 + Y_2 + ... + Y_8 = $1,360\)


 


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S1. We have mean across each row => we get sum across each row = 8* mean of row. => 8*(sum of means of all rows) = sum of all items = sufficient to find avg of all items

S2: similarly we have sum of means of all columns * 5 = sum of all items. sufficient to find avg


Bunuel
In a store, 40 items are arranged on a rack in 5 rows and 8 columns. The average (arithmetic mean) of the price of items in each row (m) is \(X_m\) [1 ≤ m ≤5]. The average of the price of items in each column (n) is \(Y_n\) [1 ≤ n ≤ 8]. What is the average price of all items on the rack?

(1) \(X_1 + X_2 + ... + X_5 = $850\)
(2) \(Y_1 + Y_2 + ... + Y_8 = $1,360\)


 


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Total price can be written in both terms of Xm and Yn, which would be:
8X1+8X2 +...+ 8X5 (8 items in each row) = 8(X1+X2+...+X5)
5Y1+5Y2+...+5Y8 (5 items in each column) = 5(Y1+Y2+...+Y8)
Thus, average price can be written in either way
Avg = 8(X1+X2+...+X5) / 40 = 5(Y1+Y2+...+Y8) / 40

(1) We can get an average of 8*$850 / 40 = $170 => Sufficient
(2) We can get an average of 5*$1,360 / 40 $170 => Sufficient

Answer: Each statement alone is sufficient
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Each Statement alone is sufficient.

1- We are able to draw out the total for all 40 items in all 5 rows (8 items in each row)
2- We are able to draw out the total for all 40 items in all 8 columns (5 items in each column)
Bunuel
In a store, 40 items are arranged on a rack in 5 rows and 8 columns. The average (arithmetic mean) of the price of items in each row (m) is \(X_m\) [1 ≤ m ≤5]. The average of the price of items in each column (n) is \(Y_n\) [1 ≤ n ≤ 8]. What is the average price of all items on the rack?

(1) \(X_1 + X_2 + ... + X_5 = $850\)
(2) \(Y_1 + Y_2 + ... + Y_8 = $1,360\)


 


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(1)
X1+X2+X3+X4+X5 = 850

Total sum = 8X1+8X2+8X3+8X4+8X5 = 8(X1+X2+X3+X4+X5) = 6800

Average = 6800/40 = 170

SUFFICIENT

(2)
Y1+Y2+Y3+Y4+Y5+Y6+Y7+Y8 = 1360

Total sum = 5Y1+5Y2+5Y3+5Y4+5Y5+5Y6+5Y7+5Y8 = 5(Y1+Y2+Y3+Y4+Y5+Y6+Y7+Y8) = 6800

Average = 6800/40 = 170

SUFFICIENT

IMO D
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40 items arranged in 5 rows and 8 columns, price of each item is not known and we need to find the average price of all 40 items.

Statement 1: X1+X2+X3+X4+X5 = 850
X1 is the average of 8 items in row 1, X2 is the average of 8 items in row 2, and on it goes.
Hence, total price of items in row 1 = 8X1.
Similarly, we can do it for the other 4 rows.
Therefore, Price of all 40 items = 8(X1+X2+X3+X4+X5) = 8*850 = $6,800
Average price of all 40 items = $6,800/40 = $170
Statement 1 is sufficient.

Statement 2: Y1+Y2+...+Y8=$1,360
Similar to the last statement, Y1 is the average of 5 items in column 1, Y2 is the average of the 5 items in column 2, and so on.
Hence, total price of items in column 1 = 5Y1.
Similarly, we can do it for the other 7 columns.
Therefore, Price of all 40 items = 5(Y1+Y2+...+Y8) = 5*1,360 = $6,800
Average price of all 40 items = $6,800/40 = $170
Statement 2 is also sufficient.

Answer = D
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We need both mean to define the total value. letter C
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Each statement is the sum of the averages. Since they all have the same weight, by dividing each one by its respective number of rows or columns, we will have the average of the total rack.

Bunuel
In a store, 40 items are arranged on a rack in 5 rows and 8 columns. The average (arithmetic mean) of the price of items in each row (m) is \(X_m\) [1 ≤ m ≤5]. The average of the price of items in each column (n) is \(Y_n\) [1 ≤ n ≤ 8]. What is the average price of all items on the rack?

(1) \(X_1 + X_2 + ... + X_5 = $850\)
(2) \(Y_1 + Y_2 + ... + Y_8 = $1,360\)


 


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Statement 1:
X_1 is the mean for row 1
Thus the total value for row 1 is X_1 * 8 (This is because there are 8 columns, thus 8 items in row 1)

Total for all the rows
= X_1*8 + X_2*8 + ... + X_8 * 8
= 8 (X_1 + X_2 + ... + X_8)
= 8 (850)
= 6800

6800 is the price of all 40 items. Thus the average price is 6800/4 = 1700
Statement 1 is sufficient

Statement 2 is sufficient as well. Just do the above but column-wise :)
Bunuel
In a store, 40 items are arranged on a rack in 5 rows and 8 columns. The average (arithmetic mean) of the price of items in each row (m) is \(X_m\) [1 ≤ m ≤5]. The average of the price of items in each column (n) is \(Y_n\) [1 ≤ n ≤ 8]. What is the average price of all items on the rack?

(1) \(X_1 + X_2 + ... + X_5 = $850\)
(2) \(Y_1 + Y_2 + ... + Y_8 = $1,360\)


 


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IMO should be D

If we have the sum total of either the rows or the colums, that should be enough for us to solve the total average
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Statement 1: \(X_1+X_2+...+X_5=$850\)

\(X_m\) gives the average price of each row. Sum of avg row prices is \($850\).

Therefore average row price \(= \frac{$850}{5}=$170\)

Since each row has 8 items (columns), the avg price of one item \(= \frac{$170}{8}=$21.25\)

Sufficient

Statement 2: \(Y_1+Y_2+...+Y_5=$1360\)

Avg column price \(=\frac{$1360}{8}=$170\)

Since each column has 5 items (rows), the avg price of one item \(= \frac{$170}{5}=$34\)

Sufficient

Answer: D
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Statement 1 provides the sum of prices of all items in every column. To calculate the average, we can divide this number by 5. Therefore, S1 alone is sufficient. Likewise, S2 provides the sum of prices of all items in every row, which then can be divided by 8 to arrive at the average price. Therefore, S2 alone is also sufficient.
Bunuel
In a store, 40 items are arranged on a rack in 5 rows and 8 columns. The average (arithmetic mean) of the price of items in each row (m) is \(X_m\) [1 ≤ m ≤5]. The average of the price of items in each column (n) is \(Y_n\) [1 ≤ n ≤ 8]. What is the average price of all items on the rack?

(1) \(X_1 + X_2 + ... + X_5 = $850\)
(2) \(Y_1 + Y_2 + ... + Y_8 = $1,360\)


 


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Bunuel
In a store, 40 items are arranged on a rack in 5 rows and 8 columns. The average (arithmetic mean) of the price of items in each row (m) is \(X_m\) [1 ≤ m ≤5]. The average of the price of items in each column (n) is \(Y_n\) [1 ≤ n ≤ 8]. What is the average price of all items on the rack?

(1) \(X_1 + X_2 + ... + X_5 = $850\)
(2) \(Y_1 + Y_2 + ... + Y_8 = $1,360\)
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Since there are 8 columns, therefore each of the 5 rows will have 8 items (A quick diagram of the setup will help)
Similarly, each column will have 5 items.

Statement 1:
If we multiply the average of the row \(X_1\) with the number of items in the row, which is 8, we would get the sum of the items in row 1.
Doing the same for the other rows and adding all together would give the sum of all the 40 items:
\(8X_1 + 8X_2 + ... + 8X_5 = 8(X_1 + X_2 + ... + X_5)= $8*850\)

Similarly for statement 2:
we can multiply the number of items in a column, which is 5, with the average of the column \(Y_1\) to get the sum of the items in the column
\(5Y_1 + 5Y_2 + ... + 5Y_8= 5(Y_1 + Y_2 + ... + Y_8)= $5*1,360\)

Therefore, D.
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If we make 5x8 we see that both the options are enough to find the answers:
For example. Let's take the first option and see, x1 denotes the average sum of all elements in row1. That means 8*(x1) gives the sum of all elements in row 1. We can check for other rows, and we will come to see that sum of all the 40 items come to 8{x1+x2+x3+x4}. since we have the sum of x1+x2..x5 given hence option 1 is sufficient.

For option 2 we see that, y1 is average price of 5 items in a column so total price in 1 column will be 5y1. For other 8 columns it will 5y1+5y2+5y3 ..5y8.

Since we have y1+y2+y3 ..+y8 given, hence this option is also sufficient to answer the average of all items.

Hence option D. Both options are sufficient is the correct answer
Bunuel
In a store, 40 items are arranged on a rack in 5 rows and 8 columns. The average (arithmetic mean) of the price of items in each row (m) is \(X_m\) [1 ≤ m ≤5]. The average of the price of items in each column (n) is \(Y_n\) [1 ≤ n ≤ 8]. What is the average price of all items on the rack?

(1) \(X_1 + X_2 + ... + X_5 = $850\)
(2) \(Y_1 + Y_2 + ... + Y_8 = $1,360\)


 


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