If |a/b| and |x/y| are reciprocals and a/b*x/y < 0, which of the follo

the modulus fractions are reciprocals of each other : \(|\frac{a}{b}|\) * \(|\frac{x}{y}|\) =1
\(\frac{a}{b}*\frac{x}{y} < 0\) : the fractions have opposite signs (since product is negative)
thus \(\frac{a}{b}*\frac{x}{y} =-1 \)
\(\frac{a}{b} = \frac{-y}{x}\)
D

question mentions that a/b and x/y are reciprocals i.e reciprocals of each other, so product will be 1, As product is less than < 0, so a/b*x/y = -1.

Hope i am able to make it clear

I think something is missed. The facts given in the question are not enough to go further. (such as signs of x and y or correlation between variables)

Bunuel
It is given that |a/b| and |x/y| are reciprocals (implies a=y and b=x) therefore, there product should always be 1, because if one of them a (or b) is negative we will always have one more negative i.e y (or x) and product of two negative is positive.

Therefore, the condition given (a/b)*(x/y) < 0 is not a valid condition. As reciprocals are true irrespective of absolute values.

Or please let me know if I am missing something here.

my line of reasoning was pretty complex, but i guess it was ok.

we know abs(a/b) = 1/(abs(x/y))
and a/b * x/y <0

from the second condition we are sure either a/b is positive and x/y is negative or vice versa (in general there should be an odd number of negative signs to be distributed among a,b,x,y).

from the first equation we know that, by opening the absolute value:
a/b > 0 and 1/x/y > 0 --> a/b = 1/x/y --> a/b = y/x [EXCLUDE because of the second condition]
a/b < 0 and 1/x/y > 0 --> -a/b = 1/x/y --> -a/b = y/x
a/b > 0 and 1/x/y < 0 --> a/b = -1/x/y --> a/b = -y/x
a/b < 0 and 1/x/y < 0 --> -a/b = -1/x/y --> -a/b = -y/x [EXCLUDE because of the second condition]

we are left with either
-a/b = y/x
or
a/b = -y/x

therefore a/b must be equal to -y/x