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abc > 0
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24 Sep 2018, 03:35
24 Sep 2018, 03:35
Expert Reply
00:00
Question Stats:
88% (00:48) correct
11% (00:52) wrong based on 59 sessions
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\(abc > 0\)
A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.
Quantity A |
Quantity B |
c(a + b) |
abc |
A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.
Re: abc > 0
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11 Jan 2019, 23:51
11 Jan 2019, 23:51
1
Given:
abc > 0 which implies that,
Out of a,b,c either two of the variables can be negative or all of the three variables can be positive.
Suppose a=1, b=2 and c=3.
We get, A > B ( A=9 and B=6)
Suppose a= -1, b=-2 and c=3
We get, B > A ( A=-9 and B=3)
Hence, the answer is D.
Please rectify if the process is wrong.
Posted from my mobile device
abc > 0 which implies that,
Out of a,b,c either two of the variables can be negative or all of the three variables can be positive.
Suppose a=1, b=2 and c=3.
We get, A > B ( A=9 and B=6)
Suppose a= -1, b=-2 and c=3
We get, B > A ( A=-9 and B=3)
Hence, the answer is D.
Please rectify if the process is wrong.
Posted from my mobile device
Re: abc > 0
[#permalink]
12 Jan 2019, 04:36
12 Jan 2019, 04:36
Expert Reply
Carcass wrote:
\(abc > 0\)
A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.
Quantity A |
Quantity B |
c(a + b) |
abc |
A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.
A very straight forward solution, even if we take all as positive..
(1) Let \(a=b=c=3\)
A=(c(a + b))=\(3(3+3)=18\)
B=(abc)=\(3*3*3=27\)
so B>A
(2) Let \(a=b=c=1\)
A=(c(a + b))=\(1(1+1)=2\)
B=(abc)=\(1*1*1=1\)
so A>B
D
