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# If a and b are both positive integers such that a < b, which of the fo

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If a and b are both positive integers such that a < b, which of the fo  [#permalink]

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07 Jul 2018, 09:25
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If a and b are both positive integers such that a < b, which of the following must be true?

(A) $$a < \sqrt{ab} <b$$

(B) $$\sqrt{a} < \sqrt{ab} < \sqrt{b}$$

(C) $$\sqrt{a} < \sqrt{b} < \sqrt{ab}$$

(D) $$\sqrt{ab}<a<b$$

(D) $$a<\sqrt{ab}<\sqrt{b}$$

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If a and b are both positive integers such that a < b, which of the fo  [#permalink]

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07 Jul 2018, 10:15
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Bunuel wrote:
If a and b are both positive integers such that a < b, which of the following must be true?

(A) $$a < \sqrt{ab} <b$$

(B) $$\sqrt{a} < \sqrt{ab} < \sqrt{b}$$

(C) $$\sqrt{a} < \sqrt{b} < \sqrt{ab}$$

(D) $$\sqrt{ab}<a<b$$

(D) $$a<\sqrt{ab}<\sqrt{b}$$

$$a=\sqrt{aa}$$ and b=$$\sqrt{bb}$$since a<b...... $$\sqrt{aa}<\sqrt{ab}<\sqrt{bb}.......................a<\sqrt{ab}<b$$

A
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Re: If a and b are both positive integers such that a < b, which of the fo  [#permalink]

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07 Jul 2018, 10:35
Bunuel wrote:
If a and b are both positive integers such that a < b, which of the following must be true?

(A) $$a < \sqrt{ab} <b$$

(B) $$\sqrt{a} < \sqrt{ab} < \sqrt{b}$$

(C) $$\sqrt{a} < \sqrt{b} < \sqrt{ab}$$

(D) $$\sqrt{ab}<a<b$$

(D) $$a<\sqrt{ab}<\sqrt{b}$$

Let a=4 & b=16 (a<b)
(A) $$a < \sqrt{ab} <b$$: 4<8<16 (True)
(B) $$\sqrt{a} < \sqrt{ab} < \sqrt{b}$$: 2<8<4 (False)
(C) $$\sqrt{a} < \sqrt{b} < \sqrt{ab}$$: 2<4<8 (False)
(D) $$\sqrt{ab}<a<b$$: 8<4<16 (False)
(E) $$a<\sqrt{ab}<\sqrt{b}$$: 2<8<4 (False)

Ans. (A)

P.S.- Perfect square are chosen for checking purpose since square roots are involved in the answer choices.
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Re: If a and b are both positive integers such that a < b, which of the fo  [#permalink]

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25 Jul 2018, 15:48
If I choose a = 1 and b = 2 (both positive integers), then sqrt(b) = sqrt(ab).
So no option "must be true". Did I miss anything?
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If a and b are both positive integers such that a < b, which of the fo  [#permalink]

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25 Jul 2018, 20:17
1
rohithtv89 wrote:
If I choose a = 1 and b = 2 (both positive integers), then sqrt(b) = sqrt(ab).
So no option "must be true". Did I miss anything?

Hi rohithtv89,
Let's check options A to E with your chosen (a,b), i.e. (1,2)

(A) $$a < \sqrt{ab} <b$$: $$1<\sqrt{2}=1.41(approx.)<2$$ (True)
(B) $$\sqrt{a} < \sqrt{ab} < \sqrt{b}$$: $$1<\sqrt{2}<\sqrt{2}$$ (False)
(C) $$\sqrt{a} < \sqrt{b} < \sqrt{ab}$$: $$1<\sqrt{2}<\sqrt{2}$$ (False)
(D) $$\sqrt{ab}<a<b$$: $$\sqrt{2}<1<2$$ (False)
(E) $$a<\sqrt{ab}<\sqrt{b}$$: $$1<\sqrt{2}<\sqrt{2}$$(False)

You might have checked option A with $$\sqrt{b}$$ but it is 'b' only.
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Re: If a and b are both positive integers such that a < b, which of the fo  [#permalink]

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25 Jul 2018, 20:48
Hi PKN. Yes. I didn't read it correctly. Thanks for pointing that out!
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Re: If a and b are both positive integers such that a < b, which of the fo  [#permalink]

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25 Jul 2018, 21:04
Why option C is not correct. If I take a as 4 and b as 9.
Than √4=2 less than √9=3 less than √4*9=6.

What I am missing here.

Posted from my mobile device
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If a and b are both positive integers such that a < b, which of the fo  [#permalink]

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26 Jul 2018, 00:34
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Orionstar wrote:
Why option C is not correct. If I take a as 4 and b as 9.
Than √4=2 less than √9=3 less than √4*9=6.

What I am missing here.

Posted from my mobile device

Hi Orionstar,
First of all, this is a MUST BE TRUE question. We have to find out the correct relationship among a,b, $$\sqrt{a}$$,$$\sqrt{b}$$, and $$\sqrt{ab}$$ in terms of magnitude. And that relationship MUST HOLD TRUE at all the possible positive integer values of a and b, available in the universe. (with the given limitation:- a<b)

With the chosen pairing (4,9), the relationship in C is true. Is the given relationship in option 'C" true for all positive integer values of a and b? Answer is No (Say a=1 , b=2, (a<b))
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Re: If a and b are both positive integers such that a < b, which of the fo  [#permalink]

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26 Jul 2018, 16:25
Bunuel wrote:
If a and b are both positive integers such that a < b, which of the following must be true?

(A) $$a < \sqrt{ab} <b$$

(B) $$\sqrt{a} < \sqrt{ab} < \sqrt{b}$$

(C) $$\sqrt{a} < \sqrt{b} < \sqrt{ab}$$

(D) $$\sqrt{ab}<a<b$$

(D) $$a<\sqrt{ab}<\sqrt{b}$$

If we let a = 4 and b = 9, we see that only A is true, since √(4 x 9) = √36 = 6.

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Re: If a and b are both positive integers such that a < b, which of the fo  [#permalink]

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27 Jul 2018, 04:12
PKN wrote:
Bunuel wrote:
If a and b are both positive integers such that a < b, which of the following must be true?

(A) $$a < \sqrt{ab} <b$$

(B) $$\sqrt{a} < \sqrt{ab} < \sqrt{b}$$

(C) $$\sqrt{a} < \sqrt{b} < \sqrt{ab}$$

(D) $$\sqrt{ab}<a<b$$

(D) $$a<\sqrt{ab}<\sqrt{b}$$

Let a=4 & b=16 (a<b)
(A) $$a < \sqrt{ab} <b$$: 4<8<16 (True)
(B) $$\sqrt{a} < \sqrt{ab} < \sqrt{b}$$: 2<8<4 (False)
(C) $$\sqrt{a} < \sqrt{b} < \sqrt{ab}$$: 2<4<8 (False)
(D) $$\sqrt{ab}<a<b$$: 8<4<16 (False)
(E) $$a<\sqrt{ab}<\sqrt{b}$$: 2<8<4 (False)

Ans. (A)

P.S.- Perfect square are chosen for checking purpose since square roots are involved in the answer choices.

PKN hey there, can you please explain how can option C be wrong. 2<4<8 if A=2 B= 4 o correct 2<4 or a<b so whats wring with it ?

thanks and have a great day !:-)
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Re: If a and b are both positive integers such that a < b, which of the fo  [#permalink]

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27 Jul 2018, 04:31
1
dave13 wrote:
PKN wrote:
Bunuel wrote:
If a and b are both positive integers such that a < b, which of the following must be true?

(A) $$a < \sqrt{ab} <b$$

(B) $$\sqrt{a} < \sqrt{ab} < \sqrt{b}$$

(C) $$\sqrt{a} < \sqrt{b} < \sqrt{ab}$$

(D) $$\sqrt{ab}<a<b$$

(D) $$a<\sqrt{ab}<\sqrt{b}$$

Let a=4 & b=16 (a<b)
(A) $$a < \sqrt{ab} <b$$: 4<8<16 (True)
(B) $$\sqrt{a} < \sqrt{ab} < \sqrt{b}$$: 2<8<4 (False)
(C) $$\sqrt{a} < \sqrt{b} < \sqrt{ab}$$: 2<4<8 (False)
(D) $$\sqrt{ab}<a<b$$: 8<4<16 (False)
(E) $$a<\sqrt{ab}<\sqrt{b}$$: 2<8<4 (False)

Ans. (A)

P.S.- Perfect square are chosen for checking purpose since square roots are involved in the answer choices.

PKN hey there, can you please explain how can option C be wrong. 2<4<8 if A=2 B= 4 o correct 2<4 or a<b so whats wring with it ?

thanks and have a great day !:-)

Hi dave13 ,
Have a great day!!

First of all, this is a MUST BE TRUE question. We have to find out the correct relationship among a,b, a√a,b√b, and ab−−√ab in terms of magnitude. And that relationship MUST HOLD TRUE at all the possible positive integer values of a and b, available in the universe. (with the given limitation:- a<b)

With the chosen pairing (2,4), the relationship in C is true(1.41<2<2.82). Is the given relationship in option 'C" true for all positive integer values of a and b? Answer is No (Say a=1 , b=2, (a<b))

The correct answer option or relationship must hold true for all positive integers a and b (a<b), not on a piecemeal basis:-), true in some cases and false in other cases. Hence, options C is discarded.

Thanking You.
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Re: If a and b are both positive integers such that a < b, which of the fo  [#permalink]

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27 Jul 2018, 08:46
Hi PKN

In this question, is there a chance that SQRT(A*B) is negetive?
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Re: If a and b are both positive integers such that a < b, which of the fo  [#permalink]

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27 Jul 2018, 10:15
ETLim wrote:
Hi PKN

In this question, is there a chance that SQRT(A*B) is negetive?

$$\sqrt{}$$ denotes a function. Mathematically the square root function cannot give negative result. When the GMAT provides the square root sign for an even root, such as $$\sqrt{x}$$ or $$\sqrt[4]{x}$$, then the only accepted answer is the positive root.

That is, $$\sqrt{25}=5$$, NOT +5 or -5. In contrast, the equation $$x^2=25$$ has TWO solutions, +5 and -5. Even roots have only a positive value on the GMAT.

OFFICIAL GUIDE:
$$\sqrt{n}$$ denotes the positive number whose square is n.
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If a and b are both positive integers such that a < b, which of the fo  [#permalink]

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27 Jul 2018, 10:28
ETLim wrote:
Hi PKN

In this question, is there a chance that SQRT(A*B) is negetive?

Hi ETLim,

Since it is mentioned in the question stem that 'a' and 'b' are positive integers, hence multiplication of two positive integers will always yield a positive integer. Therefore, a*b=positive. Hence, $$\sqrt{ab}=\sqrt{{positive}}=Positive$$ in GMAT.
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If a and b are both positive integers such that a < b, which of the fo  [#permalink]

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28 Jul 2018, 05:35
Bunuel wrote:
If a and b are both positive integers such that a < b, which of the following must be true?

(A) $$a < \sqrt{ab} <b$$

(B) $$\sqrt{a} < \sqrt{ab} < \sqrt{b}$$

(C) $$\sqrt{a} < \sqrt{b} < \sqrt{ab}$$

(D) $$\sqrt{ab}<a<b$$

(E) $$a<\sqrt{ab}<\sqrt{b}$$

Let's think to DISAPPROVE the choices

a=1 & b=2

(A) $$a < \sqrt{ab} <b$$.............(A) $$1 < \sqrt{2} <2$$.............Keep

(B) $$\sqrt{1} < \sqrt{2} < \sqrt{2}$$.................Eliminate

(C) $$\sqrt{1} < \sqrt{2} < \sqrt{2}$$.................Eliminate

(D) $$\sqrt{2}<1<2$$..................................................Eliminate

(E) $$1<\sqrt{2}<\sqrt{2}$$...............................................Eliminate

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If a and b are both positive integers such that a < b, which of the fo  [#permalink]

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28 Jul 2018, 05:43
Hi PKN

Can you please explain the highlighted text in chetan2u 's approach?

Quote:
If a and b are both positive integers such that a < b, which of the following must be true?

Quote:
$$a=\sqrt{aa}$$ and b=$$\sqrt{bb}$$since a<b...... $$\sqrt{aa}<\sqrt{ab}<\sqrt{bb}$$......................$$.a<\sqrt{ab}<b$$

I could not proceed from $$\sqrt{a}$$ $$\sqrt{b}$$ = $$\sqrt{ab}$$
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If a and b are both positive integers such that a < b, which of the fo  [#permalink]

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28 Jul 2018, 06:53
1
Hi PKN

Can you please explain the highlighted text in chetan2u 's approach?

Quote:
If a and b are both positive integers such that a < b, which of the following must be true?

Quote:
$$a=\sqrt{aa}$$ and b=$$\sqrt{bb}$$since a<b...... $$\sqrt{aa}<\sqrt{ab}<\sqrt{bb}$$......................$$.a<\sqrt{ab}<b$$

I could not proceed from $$\sqrt{a}$$ $$\sqrt{b}$$ = $$\sqrt{ab}$$

First of all, you must be knowing the following property of exponent:-
$$(a^m)*(a^n)=a^{m+n}$$
Let $$m=n=\frac{1}{2}$$, now $$a^{\frac{1}{2}} * a^{\frac{1}{2}}=a^{\frac{1}{2}+\frac{1}{2}}$$=a----(1)
Also, you know, $$a^{1/2}=\sqrt{a}$$-------(b)

Now moving to the question:-

Given a<b, --------------(c)
I am sure we can write $$a=\sqrt{a}*\sqrt{a}$$ & $$b=\sqrt{b}*\sqrt{b}$$
Substituting in (c), we have
$$\sqrt{a}*\sqrt{a} < \sqrt{b}*\sqrt{b}$$-----(d)

Again, a<b (You know we can multiply positive numbers on both sides of the inequality)(Given a and b are positive integers)
Or, $$a*b<b^2$$ (multiplying 'b' both sides)
Or, $$\sqrt{ab}<\sqrt{b^2}$$ (we can take square root on both sides of inequalities when they are positive)
Or, $$\sqrt{ab}< b$$------------------(e)
Similarly, b>a (You know we can multiply positive numbers on both sides of the inequality)(Given a and b are positive integers)
Or, $$b*a>a^2$$ (multiplying 'a' both sides)
Or, $$\sqrt{ab}>\sqrt{a^2}$$ (we can take square root on both sides of inequalities when they are positive)
Or, $$\sqrt{ab}> a$$------------------(f)
Combining (d),(e), and (f), we have $$a<\sqrt{ab}<b$$

We have derived it.
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