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Re: a and b are positive integers such that 4 < a + b < 6 and [#permalink]
kloppismydad wrote:
1st equation:

a+b = 5 (1)

2nd equation:

|a-b| = 1

a-b = 1 (2)
a-b = -1 (3)

Solving (1) and (2) gives us,
a = 3, b = 2

Solving (2) and (3) gives us,
a = 2, b = 3

Hence, volume of cylinder 1 = πr^2h = π*(3)^2*2 = 18π

Volume of cylinder 2 = π*(2)^2*3 = 12π

However, since radius a has two volumes, the answer is D.


Thank you, this explained it well!
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Re: a and b are positive integers such that 4 < a + b < 6 and [#permalink]
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a and b are positive integers such that 4<a+b<6 and |a−b|=1

A+B has to be 5 since A & B Are integers and the absolute value equation has to be equal to 1.

Two numbers that satisfy this are 2 and 3.

Now:
The volume of a cylinder with radius a and height b
The volume of a cylinder with radius b and height a

We don't know which is 2 and which is 3. The one with Radius 3 will be greater than Radius 2.

Answer D.

For completeness:

Pi(3^2)*2 = 18pi
Pi(2^2)*3= 12pi
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a and b are positive integers such that 4 < a + b < 6 and [#permalink]
Carcass wrote:
a and b are positive integers such that \(4 < a + b < 6\) and \(| a   −   b | = 1\)


Quantity A
Quantity B
The volume of a cylinder with radius \(a\) and height \(b\)
The volume of a cylinder with radius \(b\) and height \(a\)


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.


\(a + b = 5\)

So, we can have \(a = 1, 2, 3,\) or \(4\) and \(b = 4, 3, 2,\) or \(1\)

Also, \(|a - b| = 1\)

i.e. \(|1 - 4| = 1\)
\(|2 - 3| = 1\)
\(|3 - 2| = 1\)
\(|4 - 1| = 1\)

Col. A: \(πa^2b\)
Col. B: \(πb^2a\)

Dividing both sides by \(πab\):

Col. A: \(a\)
Col. B: \(b\)

When \(a = 2\) and \(b = 3\), Col. A < Col. B
When \(a = 3\) and \(b = 2\), Col. A > Col. B

Hence, option D
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a and b are positive integers such that 4 < a + b < 6 and [#permalink]
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