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Re: a + b + c/2 = 60 –a – b + c/2 = –10 [#permalink]
Carcass wrote:
\(a + b + \frac{c}{2} = 60\)

\(–a – b + \frac{c}{2} = –10\)

Quantity A
Quantity B
b
c


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.



pls explain how the answer is D and not B ?
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Re: a + b + c/2 = 60 –a – b + c/2 = –10 [#permalink]
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Carcass wrote:
\(a + b + \frac{c}{2} = 60\)

\(–a – b + \frac{c}{2} = –10\)

Quantity A
Quantity B
b
c


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.



We have:
\(a+b+c/2 = 60\) ... (i)
\(-a-b+c/2 = -10\) ... (ii)

Adding the 2 equations, we have: \(c = 50\)

Plugging in \(c=50\) in (i):

\(a + b = 35\)
\(=> b = 35-a\)

Apparently, it seems that \(b\) is less than 35, and hence is less than \(c\) (which equals 50) implying Quantity A is less than Quantity B

However, the value of \(a\) could be negative:

# If \(a = -15\), then \(b = 50 = c\) => Quantity A equals Quantity B

# If \(a\) is less than -15, say \(a = -16\), then \(b = 51 > c\) => Quantity A is greater than Quantity B

Thus, there is no relation

Answer D
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Re: a + b + c/2 = 60 a b + c/2 = 10 [#permalink]
Underdetermined: three unknowns and two equations... no way to solve the system.
This annoying "a"...
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Re: a + b + c/2 = 60 a b + c/2 = 10 [#permalink]
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