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a, b, c, and d are consecutive integers such that a < b < c
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30 Aug 2018, 16:27
30 Aug 2018, 16:27
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a, b, c, and d are consecutive integers such that \(a < b < c < d\).
A)The quantity in Column A is greater.
B)The quantity in Column B is greater.
C)The two quantities are equal.
D)The relationship cannot be determined from the information given.
Quantity A |
Quantity B |
The average (arithmetic mean) of a, b, c, and d |
The average (arithmetic mean)of b and c |
A)The quantity in Column A is greater.
B)The quantity in Column B is greater.
C)The two quantities are equal.
D)The relationship cannot be determined from the information given.
Re: a, b, c, and d are consecutive integers such that a < b < c
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01 Sep 2018, 10:10
01 Sep 2018, 10:10
1
sandy wrote:
a, b, c, and d are consecutive integers such that \(a < b < c < d\).
A)The quantity in Column A is greater.
B)The quantity in Column B is greater.
C)The two quantities are equal.
D)The relationship cannot be determined from the information given.
Quantity A |
Quantity B |
The average (arithmetic mean) of a, b, c, and d |
The average (arithmetic mean)of b and c |
A)The quantity in Column A is greater.
B)The quantity in Column B is greater.
C)The two quantities are equal.
D)The relationship cannot be determined from the information given.
Let the series be 4<3<2<1 as consecutive integers.
avg of four number = 10/4 = 2.5
b+c avg = b+c/2 = 5/2 = 2.5.
both same . C.
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a, b, c, and d are consecutive integers such that a < b < c
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20 Aug 2019, 10:33
20 Aug 2019, 10:33
1
sandy wrote:
a, b, c, and d are consecutive integers such that \(a < b < c < d\).
A)The quantity in Column A is greater.
B)The quantity in Column B is greater.
C)The two quantities are equal.
D)The relationship cannot be determined from the information given.
Quantity A |
Quantity B |
The average (arithmetic mean) of a, b, c, and d |
The average (arithmetic mean)of b and c |
A)The quantity in Column A is greater.
B)The quantity in Column B is greater.
C)The two quantities are equal.
D)The relationship cannot be determined from the information given.
Here's an algebraic approach:
If a, b, c, and d are consecutive integers such that \(a < b < c < d\), then we can conclude the following:
\(a = a\) (no duh :-D )
\(b = a + 1\)
\(c = a + 2\)
\(d = a + 3\)
We can now replace b, c and d with their equivalent values to get:
Quantity A: Average \(= \frac{a+(a+1)+(a+2)+(a+3)}{4}= \frac{4a+6}{4}=\frac{4a}{4}+\frac{6}{4}=a+1.5\)
Quantity B: Average \(=\frac{(a+1)+(a+2)}{2}=\frac{2a+3}{2}=\frac{2a}{2}+\frac{3}{2}=a+1.5\)
Answer: C
Cheers,
Brent
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