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29 Jul 2024, 01:50
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The ratio of two positive integers, \(m\) and \(n\), is 4 to 3. If the values of \(m\) and \(n\) are increased in a ratio of 1 to 1, which of the following cannot be the resulting integers?
I. 8 and 5
II. 25 and 21
III. 13 and 11
A. I only
B. II only
C. III only
D. I and II only
E. I and III only
The ratio of two positive integers, \(m\) and \(n\), is 4 to 3. If the values of \(m\) and \(n\) are increased in a ratio of 1 to 1, which of the following cannot be the resulting integers?
I. 8 and 5
II. 25 and 21
III. 13 and 11
A. I only
B. II only
C. III only
D. I and II only
E. I and III only
29 Jul 2024, 01:50
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Official Solution:
The ratio of two positive integers, \(m\) and \(n\), is 4 to 3. If the values of \(m\) and \(n\) are increased in a ratio of 1 to 1, which of the following cannot be the resulting integers?
I. 8 and 5
II. 25 and 21
III. 13 and 11
A. I only
B. II only
C. III only
D. I and II only
E. I and III only
When two numbers are increased in a ratio of 1:1, it implies that the same positive number was added to both of them.
If a fraction is between 0 and 1, adding the same positive number to both the numerator and denominator increases the value of the fraction. For instance, if we add 2 to the numerator and denominator of \(\frac{1}{3}\), we get \(\frac{1 + 2}{3 + 2} = \frac{3}{5}\), which is greater than the initial fraction of \(\frac{1}{3}\).
If a fraction is more than 1, adding the same positive number to both the numerator and denominator decreases the value of the fraction. For instance, if we add 2 to the numerator and denominator of \(\frac{3}{2}\), we get \(\frac{3 + 2}{2 + 2} = \frac{5}{4} = 1.25\), which is less than the initial fraction of \(\frac{3}{2} = 1.5\).
For the question at hand, the original ratio was 4:3, which is more than 1 (≈1.3), thus adding the same positive number to both the numerator and the denominator will decrease the value of the new ratio, bringing it closer to, but still above 1. Therefore, both 25 and 21 (\(\frac{25}{21} \approx 1.2\)), and 13 and 11 (\(\frac{13}{11} \approx 1.2\)), are possible values for the resulting integers since their ratio is less than 1.3, while 8 to 5 cannot be the resulting integers, as their ratio is 1.6, which is greater than 1.3.
Answer: A
The ratio of two positive integers, \(m\) and \(n\), is 4 to 3. If the values of \(m\) and \(n\) are increased in a ratio of 1 to 1, which of the following cannot be the resulting integers?
I. 8 and 5
II. 25 and 21
III. 13 and 11
A. I only
B. II only
C. III only
D. I and II only
E. I and III only
When two numbers are increased in a ratio of 1:1, it implies that the same positive number was added to both of them.
If a fraction is between 0 and 1, adding the same positive number to both the numerator and denominator increases the value of the fraction. For instance, if we add 2 to the numerator and denominator of \(\frac{1}{3}\), we get \(\frac{1 + 2}{3 + 2} = \frac{3}{5}\), which is greater than the initial fraction of \(\frac{1}{3}\).
If a fraction is more than 1, adding the same positive number to both the numerator and denominator decreases the value of the fraction. For instance, if we add 2 to the numerator and denominator of \(\frac{3}{2}\), we get \(\frac{3 + 2}{2 + 2} = \frac{5}{4} = 1.25\), which is less than the initial fraction of \(\frac{3}{2} = 1.5\).
For the question at hand, the original ratio was 4:3, which is more than 1 (≈1.3), thus adding the same positive number to both the numerator and the denominator will decrease the value of the new ratio, bringing it closer to, but still above 1. Therefore, both 25 and 21 (\(\frac{25}{21} \approx 1.2\)), and 13 and 11 (\(\frac{13}{11} \approx 1.2\)), are possible values for the resulting integers since their ratio is less than 1.3, while 8 to 5 cannot be the resulting integers, as their ratio is 1.6, which is greater than 1.3.
Answer: A
Moh12345
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17 Nov 2024, 14:08
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I'm a bit confused. If they are added to 1:1 ratio, wouldn't the resulting num & denom always be 1 apart? I understand the fraction value decreases, but shouldn't the num and denom value difference stay constant?
Bunuel
:shh:... 
