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# Is x < y ?

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Intern
Joined: 09 Dec 2014
Posts: 32
Is x < y ? [#permalink]

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Updated on: 26 Apr 2016, 12:13
4
00:00

Difficulty:

75% (hard)

Question Stats:

54% (01:49) correct 46% (01:26) wrong based on 81 sessions

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Is x < y ?
1) $$\sqrt{x}$$ < y
2) $$x^2$$ < y

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
SC Moderator
Joined: 13 Apr 2015
Posts: 1695
Location: India
Concentration: Strategy, General Management
GMAT 1: 200 Q1 V1
GPA: 4
WE: Analyst (Retail)
Re: Is x < y ? [#permalink]

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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

Intern
Joined: 09 Dec 2014
Posts: 32
Re: Is x < y ? [#permalink]

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28 Apr 2016, 04:32
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

Hi Vyshak,

Not getting how you combined statement 1 and statement 2.

Thanks & regards,
Sunil01
SC Moderator
Joined: 13 Apr 2015
Posts: 1695
Location: India
Concentration: Strategy, General Management
GMAT 1: 200 Q1 V1
GPA: 4
WE: Analyst (Retail)
Re: Is x < y ? [#permalink]

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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

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.
Senior CR Moderator
Status: Long way to go!
Joined: 10 Oct 2016
Posts: 1393
Location: Viet Nam
Re: Is x < y ? [#permalink]

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

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.

_________________
Re: Is x < y ?   [#permalink] 04 May 2017, 20:37
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