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0 < a < 1 < b < c
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17 Oct 2017, 14:05
17 Oct 2017, 14:05
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\(0 < a < 1 < b < c\)
Quantity A |
Quantity B |
\(\frac{c^2}{a}\) |
\(\frac{bc}{ab}\) |
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: 0 < a < 1 < b < c
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18 Oct 2017, 05:04
18 Oct 2017, 05:04
Quantity A has at the denominator a number lower than 1 but greater than zero, this means it increases when divided by it. Moreover, it has a numerator as the highest number among the three to the power of 2.
Quantity B, instead, has a number greater than 1 at the denominator and a lower numerator, given that b < c and thus bc < c^2.
This leads to the choice of B!
Putting in some numbers as a = 1/2, b = 2 and c = 3; we would get A = 3^2/(1/2) = 18 and B = 3*2/(1/2)*2 = 6.
Quantity B, instead, has a number greater than 1 at the denominator and a lower numerator, given that b < c and thus bc < c^2.
This leads to the choice of B!
Putting in some numbers as a = 1/2, b = 2 and c = 3; we would get A = 3^2/(1/2) = 18 and B = 3*2/(1/2)*2 = 6.
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0 < a < 1 < b < c
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20 Jun 2019, 06:18
20 Jun 2019, 06:18
4
Carcass wrote:
\(0 < a < 1 < b < c\)
Quantity A |
Quantity B |
\(\frac{c^2}{a}\) |
\(\frac{bc}{ab}\) |
We can solve this question using matching operations
Given:
Quantity A: \(\frac{c^2}{a}\)
Quantity B: \(\frac{bc}{ab}\)
Since we're told a is POSITIVE, we can safely multiply both quantities by a to get:
Quantity A: \(c^2\)
Quantity B: \(\frac{bc}{b}\)
Since b ≠ 0, we can cancel out the b's in Quantity B to get:
Quantity A: \(c^2\)
Quantity B: \(c\)
Since we're told c is POSITIVE, we can safely divide both quantities by c to get:
Quantity A: \(c\)
Quantity B: \(1\)
We're told that 1 < c, so the correct answer is A
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