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The arithmetic mean of a, b, c, and d is 14
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21 Mar 2019, 11:03
21 Mar 2019, 11:03
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The arithmetic mean of a, b, c, and d is 14.
Quantity A is greater.
Quantity B is greater.
The two quantities are equal.
The relationship cannot be determined from the information given.
Quantity A |
Quantity B |
\(32\) |
The arithmetic mean of \(a + 3b + 2d\) and \(a - b + 2c - 48\) |
Quantity A is greater.
Quantity B is greater.
The two quantities are equal.
The relationship cannot be determined from the information given.
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The arithmetic mean of a, b, c, and d is 14
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22 Mar 2019, 07:20
22 Mar 2019, 07:20
1
Carcass wrote:
The arithmetic mean of a, b, c, and d is 14.
Quantity A is greater.
Quantity B is greater.
The two quantities are equal.
The relationship cannot be determined from the information given.
Quantity A |
Quantity B |
\(32\) |
The arithmetic mean of \(a + 3b + 2d\) and \(a - b + 2c - 48\) |
Quantity A is greater.
Quantity B is greater.
The two quantities are equal.
The relationship cannot be determined from the information given.
GIVEN: The arithmetic mean of a, b, c, and d is 14
In other words: (a + b + c + d)/4 = 14
Multiply both sides by 4 to get: a + b + c + d = 56
The mean of a + 3b + 2d and a - b + 2c - 48 = [(a + 3b + 2d) + (a - b + 2c - 48)]/2
= [2a + 2b + 2c + 2d - 48]/2
= a + b + c + d - 24
= 56 - 24
= 32
We get:
QUANTITY A: 32
QUANTITY B: 32
Answer: C
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The arithmetic mean of a, b, c, and d is 14 [#permalink]
22 Mar 2019, 07:20
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