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Is x < y ?

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Is x < y ?  [#permalink]

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New post Updated on: 26 Apr 2016, 12:13
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Is x < y ?
1) \(\sqrt{x}\) < y
2) \(x^2\) < y

Experts please explain

Originally posted by Sunil01 on 26 Apr 2016, 11:11.
Last edited by Vyshak on 26 Apr 2016, 12:13, edited 1 time in total.
Formatted the question
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Re: Is x < y ?  [#permalink]

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New post 26 Apr 2016, 12:12
2
2
Is x < y?

St1: sqrt(x) < y --> x < y^2
Since sqrt(x) is positive, y can take only positive values
If x = 1; y = 3; then is x < y? Yes
If x = 5; y = 3; then is x < y? No
Not sufficient

St2: x^2 < y
This statement seems to be fine if x is an integer. But what if x is a fraction?
If x is an integer and x^2 < y then x < y
If x and y are fraction, say, x = 1/2; y = 1/2 then is x < y? No
Not sufficient

Combining St1 and St2:
x^2 < y --> x < y when x is an integer
x < y^2 --> y is positive and x < y when x and y are fraction
Sufficient

Answer: C
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Re: Is x < y ?  [#permalink]

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New post 28 Apr 2016, 04:32
1
Vyshak wrote:
Is x < y?

St1: sqrt(x) < y --> x < y^2
Since sqrt(x) is positive, y can take only positive values
If x = 1; y = 3; then is x < y? Yes
If x = 5; y = 3; then is x < y? No
Not sufficient

St2: x^2 < y
This statement seems to be fine if x is an integer. But what if x is a fraction?
If x is an integer and x^2 < y then x < y
If x and y are fraction, say, x = 1/2; y = 1/2 then is x < y? No
Not sufficient

Combining St1 and St2:
x^2 < y --> x < y when x is an integer
x < y^2 --> y is positive and x < y when x and y are fraction
Sufficient

Answer: C


Hi Vyshak,

Not getting how you combined statement 1 and statement 2.

Thanks & regards,
Sunil01
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Re: Is x < y ?  [#permalink]

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New post 28 Apr 2016, 07:24
Sunil01 wrote:
Vyshak wrote:
Is x < y?

St1: sqrt(x) < y --> x < y^2
Since sqrt(x) is positive, y can take only positive values
If x = 1; y = 3; then is x < y? Yes
If x = 5; y = 3; then is x < y? No
Not sufficient

St2: x^2 < y
This statement seems to be fine if x is an integer. But what if x is a fraction?
If x is an integer and x^2 < y then x < y
If x and y are fraction, say, x = 1/2; y = 1/2 then is x < y? No
Not sufficient

Combining St1 and St2:
x^2 < y --> x < y when x is an integer
x < y^2 --> y is positive and x < y when x and y are fraction
Sufficient

Answer: C


Hi Vyshak,

Not getting how you combined statement 1 and statement 2.

Thanks & regards,
Sunil01


Hi Sunil,

Combining St1 and St2 we can derive the following conclusions:
1. y is positive
2. x^2 < y --> x is definitely less than y when x is an integer. If we check St1 we had a case in which x < y^2, but x > y. Now such cases are not possible.
3. x < y^2 --> x is definitely less than y when x is a fraction. If we check St2 we had a case in which x^2 < y, but x = y. Such cases are not possible now.
Using both the statements we can say that x is definitely less than y irrespective of the nature of x (whether a fraction or an integer)

Hope I am clear now.
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Re: Is x < y ?  [#permalink]

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New post 04 May 2017, 20:37
1
1
Sunil01 wrote:
Is x < y ?
1) \(\sqrt{x}\) < y
2) \(x^2\) < y

Experts please explain


Solving this question is easy

(1) \(\sqrt{x}<y \implies 0\leq x < y^2\)

If \(x=2\) and \(y=2\) then \(x=2<y^2=4\) and \(x=y\)
If \(x=3\) and \(y=2\) then \(x=3<y^2=4\) and \(x>y\).

Hence, insufficient.

(2) \(x^2<y\)

Not the rule that given \(x > 0\), we have
\(\Bigg \{\begin{array}{lr}
x^2 \geq x \quad \text{if } \; x \geq 1 \\
x^2 < x \quad \text{if } \; 0 < x < 1
\end{array}\)

Hence, if \(x=1\) and \(y=2\) then \(x^2=1 < y=2\) and \(x<y\)
If \(x=0.1\) and \(y=0.02\) then \(x^2=0.01 < y=0.02\) and \(x>y\)

Hence, insufficient.


Combine (1) and (2) we have
\(\Bigg \{
\begin{array}{lr}
\sqrt{x}< y \\
x^2 < y
\end{array}
\implies \Bigg \{
\begin{array}{lr}
0< x< y^2 \\
0 < x^2 < y
\end{array}
\implies x \times x^2 < y^2 \times y \implies x^3 < y^3 \implies x < y
\)

Sufficient.

The answer is C.
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Re: Is x < y ?  [#permalink]

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Re: Is x < y ?   [#permalink] 11 Apr 2019, 17:59
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