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

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.

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.

Hi adkikani ,

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.

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 PKN. Yes. I didn't read it correctly. Thanks for pointing that out!

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))

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

Answer: A

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 what`s wring with it ?

thanks and have a great day !:-)

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.

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.

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

Answer: A

Hi PKN

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





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

Hi Jeff, these were the numbers that I picked as well, only to realize that C also holds true.

√a = √4 = 2
√b = √9 = 3
√ab = √(4x9) = √36 = 6

2 < 3 < 6

I solved the issue by testing a = 1 and b = 100 for the two remaining options.

How do I determine which numbers are best to test in such cases?

I solved this algebraically as following: