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a * b = a/b - b/a
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04 Apr 2018, 12:37
04 Apr 2018, 12:37
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\(a * b = \frac{a}{b} - \frac{b}{a}\) , \(m > n > 0\)
Quantity A |
Quantity B |
\(\frac{1}{m} * \frac{1}{n}\) |
\(\frac{1}{n} * \frac{1}{m}\) |
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 = a/b - b/a
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04 Apr 2018, 15:19
04 Apr 2018, 15:19
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Answer: B
a*b = a/b - b/a
A: 1/m * 1/n a=1/m and b = 1/n
1/m * 1/n = [1/m / 1/n] - [1/n / 1/m] = n/m -m/n = (n^2 - m^2) / m*n
B: 1/n * 1/m a=1/n and b = 1/m
1/n * 1/m = [1/n / 1/m] - [1/m / 1/n] = m/n -n/m = (m^2 - n^2) / m*n
As we see A equals (n^2 - m^2) / m*n and B equals (m^2 - n^2) / m*n. As m is more than n and they are both positive, m^2 is also bigger than n^2. So m^2 - n^2 is positive and n^2-m^2 is negative. So A is a negative value and B is a positive value and thus B is bigger than A.
Answer is B
a*b = a/b - b/a
A: 1/m * 1/n a=1/m and b = 1/n
1/m * 1/n = [1/m / 1/n] - [1/n / 1/m] = n/m -m/n = (n^2 - m^2) / m*n
B: 1/n * 1/m a=1/n and b = 1/m
1/n * 1/m = [1/n / 1/m] - [1/m / 1/n] = m/n -n/m = (m^2 - n^2) / m*n
As we see A equals (n^2 - m^2) / m*n and B equals (m^2 - n^2) / m*n. As m is more than n and they are both positive, m^2 is also bigger than n^2. So m^2 - n^2 is positive and n^2-m^2 is negative. So A is a negative value and B is a positive value and thus B is bigger than A.
Answer is B
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a * b = a/b - b/a
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11 Jan 2021, 08:06
11 Jan 2021, 08:06
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Carcass wrote:
\(a * b = \frac{a}{b} - \frac{b}{a}\) , \(m > n > 0\)
Quantity A |
Quantity B |
\(\frac{1}{m} * \frac{1}{n}\) |
\(\frac{1}{n} * \frac{1}{m}\) |
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
\(\frac{1}{m} * \frac{1}{n}=\frac{\frac{1}{m}}{\frac{1}{n}} - \frac{\frac{1}{n}}{\frac{1}{m}} = (\frac{1}{m})(\frac{n}{1})- (\frac{1}{n})(\frac{m}{1}) = \frac{n}{m}-\frac{m}{n}\)
\(\frac{1}{n} * \frac{1}{m}=\frac{\frac{1}{n}}{\frac{1}{m}} - \frac{\frac{1}{m}}{\frac{1}{n}} = (\frac{1}{n})(\frac{m}{1})- (\frac{1}{m})(\frac{n}{1}) = \frac{m}{n}-\frac{n}{m}\)
We now have:
Quantity A: \(\frac{n}{m}-\frac{m}{n}\)
Quantity B: \(\frac{m}{n}-\frac{n}{m}\)
Add \(\frac{m}{n}\) and add \(\frac{n}{m}\) to both quantities to get:
Quantity A: \(2(\frac{n}{m})\)
Quantity B: \(2(\frac{m}{n})\)
Divide both quantities by 2 to get:
Quantity A: \(\frac{n}{m}\)
Quantity B: \(\frac{m}{n}\)
Since we are told \(m > n > 0\), we can see that \(\frac{n}{m}\) is LESS THAN 1 and \(\frac{m}{n}\) is GREATER THAN 1
Answer: B
