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02 Mar 2012, 03:54
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A total of 20 amounts are entered on a spreadsheet that has 5 rows and 4 columns; each of the 20 positions in the spreadsheet contains one amount. The average (arithmetic mean) of the amounts in row (i) is R(i) (1≤(i)≤5). The average of the amounts in column (j) is C(j) (1≤(j)≤4). What is the average of all 20 amounts on the spreadsheet?
(1) R1 + R2 + R3 + R4 + R5 = 550
(2) C1 + C2 + C3 + C4 = 440
ID: 700319
(1) R1 + R2 + R3 + R4 + R5 = 550
(2) C1 + C2 + C3 + C4 = 440
ID: 700319
02 Mar 2012, 04:11
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A total of 20 amounts are entered on a spreadsheet that has 5 rows and 4 columns; each of the 20 positions in the spreadsheet contains one amount. The average (arithmetic mean) of the amounts in row (i) is R(i) (1≤(i)≤5). The average of the amounts in column (j) is C(j) (1≤(j)≤4). What is the average of all 20 amounts on the spreadsheet?
X--X--X--X --> R1
X--X--X--X --> R2
X--X--X--X --> R3
X--X--X--X --> R4
X--X--X--X --> R5
__________
C1-C2-C3-C4
(1) R1 + R2 + R3 + R4 + R5 = 550 --> since R1 is the average of row 1, then the sum of the numbers in row 1 is R1*4, the sum of the numbers in row 2 is R2*4, and so on. Hence the sum of all 20 numbers is 4*(R1 + R2 + R3 + R4 + R5)=4*550 --> the average of all 20 numbers is (average)=(sum)/(# of terms)=(4*550)/20=110. Sufficient.
(2) C1 + C2 + C3 + C4 = 440. The same logic here: the sum of the numbers in column 1 is C1*5, the sum of the numbers in column 2 is C2*5, and so on. Hence the sum of all 20 numbers is 5*(C1 + C2 + C3 + C4 )=5*440 --> the average of all 20 numbers is (average)=(sum)/(# of terms)=(5*440)/20=110. Sufficient.
Answer: D.
Hope it's clear.
X--X--X--X --> R1
X--X--X--X --> R2
X--X--X--X --> R3
X--X--X--X --> R4
X--X--X--X --> R5
__________
C1-C2-C3-C4
(1) R1 + R2 + R3 + R4 + R5 = 550 --> since R1 is the average of row 1, then the sum of the numbers in row 1 is R1*4, the sum of the numbers in row 2 is R2*4, and so on. Hence the sum of all 20 numbers is 4*(R1 + R2 + R3 + R4 + R5)=4*550 --> the average of all 20 numbers is (average)=(sum)/(# of terms)=(4*550)/20=110. Sufficient.
(2) C1 + C2 + C3 + C4 = 440. The same logic here: the sum of the numbers in column 1 is C1*5, the sum of the numbers in column 2 is C2*5, and so on. Hence the sum of all 20 numbers is 5*(C1 + C2 + C3 + C4 )=5*440 --> the average of all 20 numbers is (average)=(sum)/(# of terms)=(5*440)/20=110. Sufficient.
Answer: D.
Hope it's clear.
30 Jun 2021, 11:37
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A total of 20 amounts are entered on a spreadsheet that has 5 rows and 4 columns; each of the 20 positions in the spreadsheet contains one amount. The average (arithmetic mean) of the amounts in row (i) is R(i) (1≤(i)≤5). The average of the amounts in column (j) is C(j) (1≤(j)≤4). What is the average of all 20 amounts on the spreadsheet?
(1) R1 + R2 + R3 + R4 + R5 = 550
(2) C1 + C2 + C3 + C4 = 440
Solution:(1) R1 + R2 + R3 + R4 + R5 = 550
(2) C1 + C2 + C3 + C4 = 440
Question Stem Analysis:
We need to determine the average of all 20 numbers on the spreadsheet that contains 5 rows and 4 columns.
Statement One Alone:
Notice that R1 is the average of the 4 numbers in the first row, R2 is the average of the 4 numbers in the second row and so on. The sum of the 4 numbers in the first row is 4*R1, the sum of the 4 numbers in the second row is 4*R2, and so on. Therefore, the sum of all 20 numbers is:
4*R1 + 4*R2 + 4*R3 + 4*R4 + 4*R5 = 4(R1 + R2 + R3 + R4 + R5) = 4(550) = 2200
Thus, the average of the 20 numbers is 2200/20 = 110. Statement one alone is sufficient.
Statement Two Alone:
Notice that C1 is the average of the 5 numbers in the first column, C2 is the average of the 5 numbers in the second column and so on. The sum of the 5 numbers in the first column is 5*C1, the sum of the 5 numbers in the second column is 5*C2, and so on. Therefore, the sum of all 20 numbers is:
5*C1 + 5*C2 + 5*C3 + 5*C4 = 5(C1 + C2 + C3 + C4) = 5(440) = 2200
Thus, the average of the 20 numbers is 2200/20 = 110. Statement two alone is sufficient.
Answer: D
