Bunuel
Which of the following fractions is closest to 1 given that a > b > 1?
(A) a/b
(B) (a + 2)/(b + 2)
(C) (a + 1)/(b + 1)
(D) (a + 1)/b
(E) (a - 1)/(b - 1)
Given: \(a > b > 1\).
Let's choose simple values that satisfy this condition:
Let \(a = 3\) and \(b = 2\).
Now, let's calculate the value of each option to see which is closest to 1:
(A) \(\frac{a}{b} = \frac{3}{2} = 1.5\)
Distance from 1 is \(0.5\).
(B) \(\frac{a + 2}{b + 2} = \frac{3 + 2}{2 + 2} = \frac{5}{4} = 1.25\)
Distance from 1 is \(0.25\).
(C) \(\frac{a + 1}{b + 1} = \frac{3 + 1}{2 + 1} = \frac{4}{3} \approx 1.33\)
Distance from 1 is approximately \(0.33\).
(D) \(\frac{a + 1}{b} = \frac{3 + 1}{2} = \frac{4}{2} = 2\)
Distance from 1 is \(1\).
(E) \(\frac{a - 1}{b - 1} = \frac{3 - 1}{2 - 1} = \frac{2}{1} = 2\)
Distance from 1 is \(1\).
Comparing the results, \(1.25\) is the value closest to 1 (with the smallest distance of 0.25).
Conceptual Approach:Since \(a > b\), the fraction \(\frac{a}{b}\) is greater than 1 (an improper fraction).
For any fraction greater than 1,
adding the same positive number to both the numerator and the denominator pulls the value
down closer to 1. The larger the number you add, the closer it gets to 1.
Conversely,
subtracting from both (as in option E) pushes the value
up, further away from 1.
Since option (B) adds the largest number (+2) to both numerator and denominator, it must be the closest to 1.
Answer: B