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GMAT Club Olympics 2025 (Day 1): In a store, 40 items are arranged on 30 Jun 2025, 08:00
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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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GMAT Club Olympics 2025 (Day 1): In a store, 40 items are arranged on 30 Jun 2025, 08:30
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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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Experts' Global Explanation:

Average prices of items in row1, row2, row3, row4, and row 5 are X1, X2, X3, X4, and X5 respectively.
Average prices of items in column1, column2.......column7, and column8 are Y1, Y2.......Y7, and Y8 respectively.
The total number of items = 40.
The average price of all items = (Sum of the prices of all items)/(Total number of items).
We need to find whether the value of “Sum of the prices of all items / 40” can be determined.

Statement (1)

Since X1 represents the average price of items in row1, it follows that 8X1 must be the sum of the prices for the 8 items in row1.
Similarly, for row2 through row5, the values 8X2,8X3,8X4, and 8X5 respectively must be the sum of prices for the items in their corresponding rows.

The average price of all items
= Sum of the prices of all items/40
= (Sum of price of items in row1 + Sum of price of items in row2 + ....... + Sum of price of items in row5) / 40
= (8X1 + 8X2 + ....... + 8X5) / 40
= 8(X1 + X2 + ....... + X5) / 40
= 8(850) / 40

It is possible to determine with certainty the average price of all items on the rack. Hence, Statement (1) is sufficient.

Statement (2)

Since Y1 represents the average price of items in column1, it follows that 5Y1 must be the sum of the prices for the 5 items in column1.
Similarly, for column2 through column8, the values 5Y2,5Y3 ......5Y7, and 5Y8 respectively must be the sum of prices for the items in their corresponding columns.

The average price of all items
= Sum of the prices of all items/40
= (Sum of price of items in column1 + Sum of price of items in column2 + ....... + Sum of price of items in column8) / 40
= (5Y1 + 5Y2 + ....... + 5Y8) / 40
= 5(Y1 + Y2 + ....... + Y8) / 40
= 5(1360)/40

It is possible to determine with certainty the average price of all items on the rack. Hence, Statement (2) is sufficient.

D is the correct answer choice.
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(1) X1+X2+...+X5=$850
Average price at m = 850 / 5 = 170
statement (1) is sufficient.
(2) Y1+Y2+...+Y8=$1,360
Average price at n = 1360/8 = 170
statement (1) is sufficient.

Therefore, both statements individually provide information to determine the average price of all items.
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ANSWER IS D

IF WE KNOW TOTAL OF AVERAGES OF EACH ROW OR COLUMN, WE CAN MULTIPLY BY THE NO OF TERMS IN EACH ROW OR COLUMN TO GET TOTAL OF ALL 40 VALUES. THEN SIMPLY DIVIDE BY 40 TO GET AVERAGE OF ALL VALUES.

FOR EX TOTAL X1+X2+X3+X4+X5=850. IF WE MULTIPLY BY 8(NO OF VALUES IN EACH ROW) ON BOTH SIDES, WE GET TOTAL SUM OF ALL VALUES. SIMPLY DIVIDE RESULT BY 40 TO GET AVG OF ALL VALUES
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D. Because (1) 850x8/40 = 170 -> sufficient and (2) 1360x5/40=170 -> 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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To find overall average, we need to find (sum of all elements by 40, 40 is from 5 times 8)

Since we can have sum of all the elements by either multiplying

8*X1 + 8*X2 + 8*X3 + 8*X4 + 8*X5 or doing the same with 5*Y1+5*Y2+.....

so either statements would have been sufficient for finding the average.
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The Answer Should be D....

There are 5 Rows and 8 Columns..Means Each row has 8 Items{ 8 x 5 = 40 } and Each Column has 5 Items ( 5 x 8 } ..Remember ..not to confuse the number of Rows with The Number of Items in 1 Row...1 Row will have items that are no. of Columns..... Given in table below.....

Now Statement 1.... X1 + X2 + X3....X5 ...means the total of averages of Each row is 850...Means the total of all items is (8 x X1) + (8 x X2)....(8 x X5) ...Since 8 is the number of Items in Each row.......Total comes out to be 6800...This is the total of all items..divide. it by 40... We'll get the answer ..hence Sufficient....

Statement 2
Y1 + Y2 + Y3...+Y8 means the total of averages of each column is 1360...Means the total of all items is (5 × Y1) + (5 × Y2) + ... + (5 × 8)...Since 5 is the number of items in each column...Total comes out to be 6800...This is the total of all items...divide it by 40...We'll get the answer...hence Sufficient...

ANSWER 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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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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Here, each of the options will provide the answer.

X1, X2 are all average means of the rows(1-5).

Let x1 be the total value of row 1.
So x1/5 will be X1, where X1 is the average mean of row 1.
Like that, we can change the formula to

x1/5 +x2/5 +x3/5+x4/5+x/5 =X1(that is 850)[x1,x2,x3,x4,x5 are all sum of values in their corresponding rows]
Then x1+x2+x3+x4+x5=850*5
Now we have the sum of the values of all 40 item,s and we can then find the average by dividing the answer we got(850*5) by 40

We will also get the answer from option 2 through this same process.
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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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AM=850*5=4250
AP=1360*8=10880 Average price= 15130/40=$328
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If we think of this 5X8 grid as 8 elements per row or 5 elements per column, we can Step 1: Express the Total and Average

Let S be the total price of all items.
The average price is S/40.

Row Averages
Each row has 8 items, so:

Total price in row m: 8Xm
Total price of all items: S=8(X1+X2+X3+X4+X5)

Column Averages
Each column has 5 items, so:

Total price in column n: 5Yn
Total price of all items: S=5(Y1+Y2+...+Y8)


Step 2: Analyze Each Statement
Statement (1): X_1 + X_2 + X_3 + X_4 + X_5 = $850

S = 8 * 850 = $6,800
Average price = 6,800 / 40 = $170

Statement (1) alone is sufficient.

Statement (2): Y_1 + Y_2 + ... + Y_8 = $1,360

S = 5 * 1,360 = $6,800
Average price = 6,800 / 40 = $170

Statement (2) alone is sufficient.

Hence given that each statement alone is sufficient, the answer to this question is option (D)
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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?


C1C2C3C4C5C6C7C8Average
R1X_1
R2X_2
R3X_3
R4X_4
R5X_5
AverageY_1Y_2Y_3Y_4Y_5Y_6Y_7Y_8

The total price of all items on the rack = 8(X_1+X_2+X_3+X_4+X_5) = 5(Y_1+Y_2+Y_3+Y_4+Y_5+Y_6+Y_7+Y_8)
The average price of all items on the rack = 8(X_1+X_2+X_3+X_4+X_5)/40 = (X_1+X_2+X_3+X_4+X_5)/5 = 5(Y_1+Y_2+Y_3+Y_4+Y_5+Y_6+Y_7+Y_8)/40 = (Y_1+Y_2+Y_3+Y_4+Y_5+Y_6+Y_7+Y_8)/8

(1) \(X_1 + X_2 + ... + X_5 = $850\)
The average price of all items on the rack = (X_1+X_2+X_3+X_4+X_5)/5 = $850/5 = $170
SUFFICIENT

(2) \(Y_1 + Y_2 + ... + Y_8 = $1,360\)
The average price of all items on the rack = (Y_1+Y_2+Y_3+Y_4+Y_5+Y_6+Y_7+Y_8)/8 = $$1360/8 = $170
SUFFICIENT

IMO D
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From Statement 1 we can find Arithmetic mean as 850/5 = 170
From Statement 2 we can find Arithmetic mean as 1360/8 = 170

Thus both statement alone are 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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GMAT Club Olympics 2025 (Day 1): In a store, 40 items are arranged on 30 Jun 2025, 08:40
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We’re told:

There are 5 rows and 8 columns → total = 5 × 8 = 40 items.

Let:
Xm be the average price of the items in row m for 1≤m≤5
Yn be the average price of the items in column n for 1≤n≤8

🎯 Goal:
Find the average of all 40 items =

Total sum of all item prices =40

So we need the total sum of the prices.

🟩 Statement (1):
X1 +X2 +X3 +X4 +X5 =850

Each Xm is the average of 8 items (since there are 8 columns per row).
So the sum of prices in row m is 8*Xm
​
Total sum over all rows:

Total sum=8*(X1 +X2 +X3 +X4 +X5 )=8*850=6,800
Now:

Average price = 6,800/40=170

✅ Sufficient

🟩 Statement (2):
Y1 +Y2 +⋯+Y8 =1,360
Each
Yn is the average of 5 items (since there are 5 rows per column).
So the sum of prices in column n is 5*Yn

Total sum over all columns:

Total sum=5* (Y1 +⋯+Y8 )=5*1,360=6,800
So again:

Average price = 6,800 / 40 =170

✅ Sufficient

✅ Final Answer: (D) — Each statement alone is sufficient.
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My answer here is C.
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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First, let's understand what we're looking for. We need to find the average price of all 40 items on the rack.
Statement 1: X1+X2+..+X5=$850
Each Xm represents the average price of items in row m. Since each row has 8 items, the sum of prices in row m is 8Xm.
The total value of all items = 8-X1 + 8-X2 + ... + 8-X5 = 8(X1 + X2 + ... + X5) = 8$840 = $6,800
Therefore, the average price of all 40 items = $6,800 ÷ 40 = $170
Statement 1 alone is sufficient.
Statement 2: Y1+Y2+....+Y8=$1,360
Each Yn represents the average price of items in column n. Since each column has 5 items, the sum of prices in column n is 5Yn.
The total value of all items = 5-Y1 + 5-Y2 + ... + 5-Y8 = 5(Y1 + Y2 + ... + Y8) = 5-$1,360 = $6,800
Therefore, the average price of all 40 items = $6,800 ÷ 40 = $170
Statement 2 alone is sufficient.
Since both statements independently yield the same answer of $170, either statement alone is sufficient.
The answer is D) Each statement alone is sufficient.
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What is the average price of all items on the rack?

Avg = Total Price/40

We need the total price of al the items

S1
\(X_1 + X_2 + ... + X_5 = $850\)
Each row has 8 items
8*(X_1 + X_2 + ... + X_5) = 850*8 = Total price
Sufficient

S2
\(Y_1 + Y_2 + ... + Y_8 = $1,360\)
Each column has 5 items
5*(Y_1 + Y_2 + ... + Y_8) = 1360*5 = Total price
Sufficient

Answer 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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Statement (1): (Sum Avg Rows) / (Number of Rows) = Average Total (Since the join of all Rows will cover all items). Statement (1) is sufficient. Eiminate choices B, C, and E.

Statement (2): (Sum Avg Columns) / (Number of Columns) = Average Total (Since the join of all columns will cover all items). Statement (2) is sufficient. Eiminate choice A.

Answer: D (Each Statement individually is sufficient)
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Prathu1221
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GMAT Club Olympics 2025 (Day 1): In a store, 40 items are arranged on 30 Jun 2025, 08:50
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This can be found using either of statements alone because-
In 1st statement we have sum of all means of rows so we can use that divide by 5 to get a value.
In second statement we are given data of columns so similarly we can find avg of all items by dividing by 8.
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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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Missinga
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GMAT Club Olympics 2025 (Day 1): In a store, 40 items are arranged on 30 Jun 2025, 08:51
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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 [1 ≤ m ≤5]. The average of the price of items in each column (n) is [1 ≤ n ≤ 8]. What is the average price of all items on the rack?

(1) X1+X2+X3+X4+X5=850
We have average price and number of items in each row.
Sufficient
(2) Y1+Y2+....+Y8=1360
We have average price and number of items in each column.
Sufficient

Answer: D
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arbh77
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GMAT Club Olympics 2025 (Day 1): In a store, 40 items are arranged on 30 Jun 2025, 08:51
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The answer is D.

We have to find Avg = Sum of all prices / 40

Statement I:
Xm is the average of the row so 8 ( X1+ X2+...X5)/ 40 is 170 which is the avg of the set.

Statement II:
Ym is the average of the column so 5 ( Y1+Y2...Y5) / 40 is also 170.

Statement
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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