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


I. \(\frac{x^2}{y^2}<\frac{x}{y}\)

II. \(\frac{x^2}{y }> \frac{x}{y}\)

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

A. I only
B. II only
C. III only
D. I and III only
E. I, II, and III

\(0<\frac{x}{y}<1\)
Analyzing the inequality:
1) Both x and y have same sign.
2) If both positive x<y. for example 2<3
3) If both negative, x>y. for example -2>-3

I. \(\frac{x^2}{y^2}<\frac{x}{y}\)
Any fraction in range 0 to 1 will decrease as the power is increased.
TRUE

II. \(\frac{x^2}{y }> \frac{x}{y}\)
If x and y are negative, say -2 and -3
\(\frac{(-2)^2}{-3}>\frac{-2}{-3}......-\frac{4}{3}>\frac{2}{3}\)...NO

III. \(\frac{(x+5)}{(y+5)}<1\)
x=2, and y=3......\(\frac{x+5}{y+5}=\frac{7}{8}<1\)..Yes
x=-2, and y=-3......\(\frac{x+5}{y+5}=\frac{3}{2}<1\)..No
Need not be TRUE

Only I

A

I know a property which states that if you increase the numerator and the denominator of a fraction < 1 by the same number, the fraction approaches 1. Is that only applicable when the numbers are positive?
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chetan2u
Bunuel
If \(0<\frac{x}{y}<1\), then which of the following must be true ?


I. \(\frac{x^2}{y^2}<\frac{x}{y}\)

II. \(\frac{x^2}{y }> \frac{x}{y}\)

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

A. I only
B. II only
C. III only
D. I and III only
E. I, II, and III

\(0<\frac{x}{y}<1\)
Analyzing the inequality:
1) Both x and y have same sign.
2) If both positive x<y. for example 2<3
3) If both negative, x>y. for example -2>-3

I. \(\frac{x^2}{y^2}<\frac{x}{y}\)
Any fraction in range 0 to 1 will decrease as the power is increased.
TRUE

II. \(\frac{x^2}{y }> \frac{x}{y}\)
If x and y are negative, say -2 and -3
\(\frac{(-2)^2}{-3}>\frac{-2}{-3}......-\frac{4}{3}>\frac{2}{3}\)...NO

III. \(\frac{(x+5)}{(y+5)}<1\)
x=2, and y=3......\(\frac{x+5}{y+5}=\frac{7}{8}<1\)..Yes
x=-2, and y=-3......\(\frac{x+5}{y+5}=\frac{3}{2}<1\)..No
Need not be TRUE

Only I

A

I know a property which states that if you increase the numerator and the denominator of a fraction < 1 by the same number, the fraction approaches 1. Is that only applicable when the numbers are positive?


Yes, you are correct.
The following link has further details on the same topic.
https://gmatclub.com/forum/effects-of-arithmetic-operations-on-fractions-269413.html
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beeunoia
According to the condition 0<x/y<1 we can say that x<y

1. x^2/y^2 < x/y => Let's pick numbers 4/9 = 2/3, but 1/4 < 1/2 - not sufficient
2. x^2/y > x/y => Let's pick numbers 4/3 > 2/3, but 1/2 = 1/2 - not sufficient
3. (x+5)/(y+5)<1 => Let's pick numbers (2+5)/(3+5)<1 and (1+5)/(2+5)<1 - sufficient

Option C is correct!

1. calc error here; 4/9 = 2/4.5, not 2/3

1 works

3. use -.5 x and -1 y: -.5+5 = 4.5, -1+5 = 4; 4.5/4 not less than 1.

3 doesnt work
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astiles67
beeunoia
According to the condition 0<x/y<1 we can say that x<y

1. x^2/y^2 < x/y => Let's pick numbers 4/9 = 2/3, but 1/4 < 1/2 - not sufficient
2. x^2/y > x/y => Let's pick numbers 4/3 > 2/3, but 1/2 = 1/2 - not sufficient
3. (x+5)/(y+5)<1 => Let's pick numbers (2+5)/(3+5)<1 and (1+5)/(2+5)<1 - sufficient

Option C is correct!

1. calc error here; 4/9 = 2/4.5, not 2/3

1 works

3. use -.5 x and -1 y: -.5+5 = 4.5, -1+5 = 4; 4.5/4 not less than 1.

3 doesnt work


As for 3, yes, thanks, did not consider the negative number. But as for 1 - I considered it as \(\frac{\sqrt{4}}{\sqrt{9}}\)
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The answer should be D (I and III). Numbers are lying in the range of 0 to 1 will always have a lower numerator and higher denominator.

I: This is a property. The more you square this number, the smaller it will become.

II: Is not true for single case 1/2. (Should be enough to negate this in a 'must be true' answer)

III: We are adding a constant to the numerator and denominator. Hence the distance travelled by the numbers will be the same on the number line. So Denominator will still be larger than the numerator. Hence it will always be less than 1.
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