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 . The average of the price of items in each column (n) is Y_n . 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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GMAT Club Olympics 2025 (Day 1): In a store, 40 items are arranged on

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Either of the Sum of Xm or Ym gives the sum of of items. Average = Sum of all items / no of items .
Since we can find using each statement that any one option is sufficient.

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1.
X1, X2...X5 -> The average value for each single item in that row.

8(X1+X2+X3+X4+X5) -> Total price of all items in the grid

8(X1+X2+X3+X4+X5) / 40 -> Average price of each item in the grid

8 * 850 / 40 = 170

Sufficient

2.
Y1, Y2...Y8 -> The average value for each single item in that column
5( Y1+ Y2+ Y3+ Y4+Y5+Y6+Y7+Y8) -> Total price of all items in the grid
5( Y1+ Y2+ Y3+ Y4+Y5+Y6+Y7+Y8) / 40 -> Average price of each item in the grid

5 * 1360 / 40 = 170

Sufficient

Answer.D

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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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Rows = 5
Columns = 8

Average Row price = $X_m$ [1 ≤ m ≤5]
Average Row price = $Y_n$ [1 ≤ n ≤ 8]

We need to find the average price of all items on the rack

(1) $X_1 + X_2 + ... + X_5 = $850$

Lets take the first row
Average price = $X_1$
Number of items in the first row = 8
Total price of items in the first row = $X_1$ x 8

Similarly we have the total price of the second row = $X_2$ x 8
Similarly we have the total price of the third row = $X_3$ x 8
Similarly we have the total price of the fourth row = $X_4$ x 8
Similarly we have the total price of the fifth row = $X_5$ x 8

Total price of the rack = Total price of all rows combined =
$X_1$ x 8 + $X_2$ x 8 + $X_3$ x 8 + $X_4$ x 8 + $X_5$ x 8
=> ($X_1$ + $X_2$ + $X_3$ + $X_4$ + $X_5$) x 8
=> $850 * 8 = $6800
Average price of all items = $6800/40 = 170

(2) $Y_1 + Y_2 + ... + Y_8 = $1,360$
Lets take the first column
Average price = $Y_1$
Number of items in the first column = 5
Total price of items in the first column = $Y_1$ x 5

Similarly we have the total price of the second column = $Y_2$ x 5
Similarly we have the total price of the third column = $Y_3$ x 5
Similarly we have the total price of the fourth column = $Y_4$ x 5
Similarly we have the total price of the fifth column = $Y_5$ x 5
......
.....
Similarly we have the total price of the eight column = $Y_8$ x 5

Total price of the rack = Total price of all rows combined =
$Y_1$ x 5 + $Y_2$ x 5 + $Y_3$ x 5 + ........................... + $Y_8$ x 5
=> ($Y_1$ + $Y_2$ + $Y_3$ + ............ + $Y_8$) x 5
=> $1360 * 5 = $6800
Average price of all items = $6800/40 = 170

IMHO Option D

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Answer: D

5 rows: $x_1$, $x_2$, $x_3$, $x_4$, $x_5$
8 columns: $y_1$, $y_2$, $y_3$, $y_4$, $y_5$, $y_6$, $y_7$, $y_8$

(1) For $x_1$, we have the average price of 8 columns on first row. For $x_2$, the average price of 8 columns on second row. And so on and so forth.
Therefore, (8 * $x_1$ + $x_2$ + $x_3$ + $x_4$ + $x_5$) / (5 * 8) = x
x = (8* $850) / 40 = average price of all items.
x = $850 / 5 = $ 170

Sufficient.

(2) Same logic applied. $y_1$ is the average price of 5 rows on first column.
Therefore, (5 * $y_1$ + $y_2$ + ... + $y_8$) / (5 * 8) = x
x = $ 1,360 / 8 = $ 170

Sufficient.

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The Goal: Find the average price of all 40 items. To do this, we need the total price of all 40 items.

Statement (1): $X_1 + X_2 + ... + X_5 = $850

There are 5 rows, and each X_m is the average of 8 items in that row.
Total price from rows = (Sum of row averages) \times (Items per row) = $850 \times 8 = 6800.
Since we have the total price and total items (40), we can find the average. Sufficient.

Statement (2): $Y_1 + Y_2 + ... + Y_8 = $1360
There are 8 columns, and each Y_n is the average of 5 items in that column.
Total price from columns = (Sum of column averages) \times (Items per column) = $1360 \times 5 = 6800.
Since we have the total price and total items (40), we can find the average. Sufficient.
Answer: D (Each statement alone is sufficient).

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Bunuel

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To find the average of all items, we have to find the total of all items. Since we know the number of items= 40. We can easily find the average.

Consider statement 1,
The total of each row can be calculated using 8*Xm. (1<=Xm<=5) As each row has 8 items.
Therefore, the total of all items: 8*X1+ 8*X2 +... + 8*X5.
= 8(X1+X2+.....+X5).
= 8*850 = 6800. Total items=40.
Hence, the average of all items is 6800/40= 170.
Sufficient.

Similarly,
Consider statement 2,
The total of each column can be calculated using 5*Yn. (1<=Yn<=8) As each column has 5 items.
Therefore, the total of all items: 5*Y1+ 5*Y2 +... + 5*Y5.
= 5(Y1+Y2+.....+Y5).
= 5*1360 = 6800. Total items=40.
Hence, the average of all items is 6800/40= 170.
Sufficient.

The correct answer: D
Each Statement Alone is Sufficient.

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01 Jul 2025, 00:15

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Item=40, row=m=5,column=n=8;1<=m<=5 ,1<=n<=8;

(1) $X_1 + X_2 + ... + X_5 = $850$
total=8(850);Avg=8*850/40=170
(2) $Y_1 + Y_2 + ... + Y_8 = $1,360$
total=5(1360);Avg=5*1360/40=170
Ans:D

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D. EACH statement ALONE is sufficient.

The sum of all prices for the products = 8(X1 + X2 +...+ Xn) = 5(Y1 + Y2 +...+ Yn) = S --> A

Total products = 40

Given the formula for mean = S/40, either statement will suffice.

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01 Jul 2025, 00:22

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1) since we have sum of average of all rows and all rows have same number of items we can find the total of the rack and hence the average of the rack
Suff
2) since we have sum of average of all columns and they have same number of items we can find the total of rack and hence the average
Suff

Ans D

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(D) Both statements are sufficient and the avg price is 170

Bunuel

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01 Jul 2025, 02:37

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There are 40 items on the rack arranged in a 5×8 grid:

That means each row has 8 items, and each column has 5 items.

You're given:

The average price in each row is X1,X2,...X5
The average price in each column is Y1,Y2,...,Y8

We want the overall average price of all 40 items.
Statement (1): X1+...+X5=850
This sum is over row averages. Since each row has 8 items, the total price of all items is: Total=8×(X1+X2+...+X5)=8×850=6800
Average price per item = 6800/40=170
So, statement (1) is sufficient.
Statement (2):
Y1+Y2+...+Y8=1360
This sum is over column averages. Each column has 5 items, so: Total=5×(Y1+Y2+..+Y8)=5×1360=6800
Average = 680040=170
So, statement (2) is also sufficient.
Final Answer: (D) Each statement alone is sufficient

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Answer is option D. each statements is enough

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S1 provides the average price for each row meaning from it we can get the total price of all items hence sufficient
S2 provides the average price for each column meaning from it we can get the total price of all items hence sufficient
Ans D

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Bunuel

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X - rows Y -Columns

1. We know in each row there are 8 columns and 8 items are arranged. so X= total sum of price in the row/ total no. items.
Total sum of price in the row = X * Total no. of Items
8 ( X_1 + X_2 + ... + X_5) = $850*8 = total Price of all the item.

Average required = total Price of the items/ 40

2. Similarly,

5 ( Y_1 + Y_2 + ... + Y_8) = $1,360 *5 -> Average price of all items can be determined.

Hence, D is the answer

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C, each sufficient , because sum can be found be multiplying the eqns by 8 and 5 respectively, and you have the total no. of terms which is 40, avg can be calculated

Bunuel

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We’re told there are 40 items in total, arranged in 5 rows and 8 columns.
The average price of items in each row is given as X1,X2,X3,X4,X5
and the average price of items in each column is given as Y1,Y2,...,Y8

[hr]
Statement (1): X1+X2+X3+X4+X5=850
Total price of all 40 items =
8(X1+X2+X3+X4+X5)=8×850=6800
Average price =
6800÷40=170
So, Statement (1) alone is sufficient to find the average.
[hr]
Statement (2): Y1+Y2+...+Y8=1360
Total price =
5(Y1+Y2+...+Y8)=5×1360=6800
Average price =
6800÷40=170
So, Statement (2) alone is also sufficient.
[hr]
Final Answer: (D)
Each statement alone is sufficient to answer the question.

Bunuel

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01 Jul 2025, 05:40

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The answer is option D).

Statement 1 says ,
(X1 + x2 + x3 + x4 + x5)* 8 => total price
Hence average price = total price / 40 = (850*8)/40 = 170
Sufficient

Statement 2 says ,
(y1 + Y2 + y3+ Y4 + y5) * 8 => total price
Hence , average price = total price / 40 = 1360 * 5 ) / 40 = 170
Sufficient

Hence D is the answer.

Bunuel

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IMO, answer is D.

Statement 1.
X1 + X2 + ... + X5 = $850

Average of elements in row 1, that is X1 = (X1C1 +X1C2+X1C3+X1C4+....+X1C8)/8
So, X1C1 +X1C2+X1C3+X1C4+....+X1C8 = 8X1
similarly,
X2c1+..............................+X2C8=8X2

Following this
sum of elements of all the rows = 8X1 + 8X2 + ........ + 8X5
Multiplying the statement 1 with 8 on both sides:
sum of elements of all the rows = 8X1 + 8X2 + ........ + 8X5 = 6800
Avg of all 40 elements is = (8X1 + 8X2 + ........ + 8X5)/40 = 6800/40 = 170

Statement 2:
Applying the above logic statement 2 is also sufficient.

Avg of all 40 elements is = (5Y1 + 5Y2 + ........ + 5Y8)/40 = 6800/40 = 170

Bunuel

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From prompt/question:
We get the average of the first row is X1, and the second row is X2, and so on till X5. And average of the first column in Y1, the second column is Y2, and so on till Y8.

Now, the average price of all items = sum of prices / 40.

We can get the average price of all items either if we have the sum of (X1 + X2 +... + X5) or (Y1 + Y2 + ... + Y8).
Why?
Because what is X1?
X1 = sum of the prices of items in the first row / 5,
which gives, sum of the prices of items in the first row = 5*X1, the same goes for other rows
Now, the sum of prices of all items would be 5*X1 + 5*X2 +... + 5*X5
Then the average would be 5( X1 + X2 + .... + X5) / 40.

The same goes for column-wise as well.
Hence, each statement alone is sufficient. Option D

Bunuel

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We can test this by 2 rows x 3 columns:

C1 C2 C3
R1: | a b c |
R2: | d e f |

average of R1, X1: (a+b+c)/3
average of R2, X2: (d+e+f)/3

Average of all 6 items = (a+b+c+d+e+f)/6
we can write this as: 0.5 * [(a+b+c)/3 +(d+e+f)/3]
or, : 0.5* [X1+X2]

so if we know X1+X2, we can find the overall average

Similarly, it works for Columns

1 2 3 4 5 6 7 8

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