Based on the inequality y must be greater than x, right?

\(0<\frac{x}{y}<1\)
then
\(0<x<y\)

Based on this, C should be a solution as well. I believe your example for C is incorrect because in that case, x>y.

For option A, multiplying by y and dividing x on both sides, we get,

\(\frac{x}{y}<1\) - true

For option B,

Since we do not know the individual value of X, this will not hold true.

For option C,

if x/y is less than 1 but greater than 0, then adding or subtracting won't change anything. It would be still less than 1. - true.

Hence, A and C.

\(\frac{x}{y}\) must be a Fraction where \(x < y\) (when both are positive) and \(x > y\) (when both are negative)

Let us say \(x = -3\) and \(y = -4\)

Is \(0 < \frac{x}{y} < 1\)

Yes - \(0 < \frac{3}{4} < 1\)

C. \(\frac{x+5}{y+5} < 1\)

i.e. \(\frac{-3+5}{-4+5} < 1\)

\(\frac{2}{1} < 1\) - NO

Great debate! Maybe it's the phrasing of the question, but is it implied that x and y are positive?

0<x<y

The inequality direction won't change by assigning x and y negative values, and if we do assign negative values, the inequality no longer holds true.

The answer shows both A and C as correct, and this is the only way I can explain it.

BTW this is a MAC

Multiple-choice Questions — Select One or More Answer Choices

The question can be solved in 10 seconds conceptually

Now, the stem is a fraction positive between 0 and one such as 1/2

A. we do know by math rule that a square of a fraction , then the result is always <

suppose 1/2, squared is 1/4 and 1/4 <1/2. A must be true

B. not always true because we do have just the x^2

So if our fraction is 1/2 >>> then x^2 in the numerator is still 1 so we do have >.

But if the x=9 >>>> then squared is 81 and LHS > RHS

Not always true

C.

we just add 5 to both the numerator and denominator of the fraction. nothing change it is just a +5

Therefore the fraction will be always <1

A and C are always true

x and y must be positive being in the range between 0 and 1

The answer to this question should be only A.
The question specifies the fraction x/y between 0 and 1 which holds true in negative cases as well.
0<x/y<1 this equation cannot be entirely multiplied by y to get 0<x<y unless and until y is given as positive number because if y is negative number, inequality changes.

If \(0<\frac{x}{y}<1\), then which of the following must be true ?


A. \(\frac{x^2}{y^2}<\frac{x}{y}\) --> divide by \(\frac{x^2}{y^2}\) to get 1<y/x, 1>y/x true as can't be false

B. \(\frac{x^2}{y }> \frac{x}{y}\) --> divide by \(\frac{x^2}{y^2}\) to get y<y/x, y>y/x not true as can be false

C. \(\frac{(x+5)}{(y+5)}<1\) --> it's true according to arithmetic conventions

answer is A, C

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