Bunuel wrote:
Integers a and b are consecutive multiples of 6. Integers x and y are consecutive multiples of 8. In terms of a, b, x, and y, what is the ratio of the average of a and b and the average of x and y?
A. \(\frac{(x+y)}{(a+b)}\)
B. \(\frac{(a+b)}{(x+y)}\)
C. \((a+\frac{b}{4})/(y-\frac{x}{4})\)
D. \((a+\frac{b}{4})/(y-\frac{x}{2})\)
E. \(\frac{1}{2}\frac{(a+b)}{(x+y)}\)
a = 6*1 = 6
b= 6 *2 = 12
Total = 12 + 6 = 18
Average : 18 / 2 = 9
x = 8*1 = 8
y = 8*2 = 16
total = 18 + 6 = 24
Average = 24 / 2 = 12
Ratio: a + b/2 : x + y/2 = 9 : 12 = 3 : 4.
Option A) a + b / x + y = 18 / 24 = 3: 4.
Thus A is the best answer.