What is the value of a^(-2)*b^(-3)?

Question

What is the value of  a^{(-2)}*b^{(-3)} ?

(1)  a^{(-3)}*b^{(-2)}=36^{(-1)}

(2)  a*b^{(-1)}=6^{(-1)}

Bunuel · Accepted Answer

What is the value of a^-2*b^-3?

Note that we are not told that  a  and  b  are integers.

a^{-2}*b^{-3}=\frac{1}{a^2b^3}=?  So, basically we need to find the value of  a^2b^3 .

(1)  a^{-3}*b^{-2}=36^{-1}  -->  a^3b^2=36 . Not sufficient.
(2)  ab^{-1}=6^{-1}  -->  \frac{b}{a}=6 . Not sufficient.

(1)+(2) Multiply (1) by (2):  a^3b^2*\frac{b}{a}=a^2b^3=36*6 . Sufficient.

Answer: C.

Sarang · Answer

Statement 1 alone is clearly insufficient since we have an additional ab^-1 present.
Statement 2 alone is also not adequate to calculate a^-2*b^-3

Multiplying both statements above- (a^-3*b^-2) *(a*b^-1) = a^-2*b^-3

which is 6/36 = 1/6,

Answer:- C

ulm · Answer

1) or 2) alone ins

Consider 1)+2)
from 2) we have  6a=b 
plug it in 1) and solve for "a". Than plug it into 2).
Now with known "a" and "b" you can find the value of  (a^(-2))(b^(-3)) 
Remember that you don't need to solve it, you need to know that the equation could be solved.
(C)

fluke · Answer

What is  a^{-2}b^{-3}=\frac{1}{a^2b^3}=\frac{1}{(ab)^2b}

1) a^-3b^-2 = 36^-1
 a^{-3}b^{-2} = 36^{-1} 
 \frac{1}{a^3b^2} = \frac{1}{36} 
 \frac{1}{a^3b^2} = \frac{1}{36} 
 \frac{1}{(ab)^2a} = \frac{1}{36} 
 \frac{1}{(ab)^2} = \frac{a}{36} 
 \frac{1}{(ab)^2b}=\frac{a}{36b}  --------------------1

a^3b^2=36 
Possible values of a and b;
 a=1; b=6 
 a=2; b=\sqrt{\frac{36}{8}} 
 a=0.1; b=\sqrt{\frac{36}{0.001}}

Not Sufficient.

2) ab^-1 = 6^-1
 ab^{-1} = 6^{-1} 
 \frac{a}{b} = \frac{1}{6}  ------------------------2

a=1; b=6
a=2; b=12
a=3; b=18
a=3.2; b=19.2
Not Sufficient.

Combining both and using 1 and 2:

\frac{1}{(ab)^2b}=\frac{a}{36b}=\frac{1}{36*6}=\frac{1}{216} 
Sufficient.

Ans: "C"

amit2k9 · Answer

a. gives values (1,6) and (1,-6) 
b gives values (1,6) and (-1,-6)

a+b gives (1,6) hence C

KarishmaB · Answer

Putting the question in the right form can help you quickly arrive at the answer.
What is  \frac{1}{a^2b^3}?

1.  \frac{1}{a^3b^2} = \frac{1}{36} 
Since a and b are real numbers so they can occur in many combinations to give 1/36. This statement alone is not sufficient.

2.  \frac{a}{b} = \frac{1}{6} 
Again, a and b are real numbers and they can take many different values to give 1/6 (e.g. a = 1, b = 6 or a = 2, b = 12 etc). This statement alone is not sufficient.

You can easily get the value of  \frac{1}{a^2b^3}  by combining the two statements. 
 \frac{1}{a^2b^3} = \frac{1}{a^3b^2} * \frac{a}{b} = \frac{1}{36} * \frac{1}{6} 
Hence they are sufficient together. Answer (C)

wizard · Answer

I am a bit confused on (1)

Here is the way I thought:
(a^-3)(b^-2)=(36^-1)
(1/a^3)(1/b^2)=1/36
(a^3)(b^2)=36
(a^3)(b^2)=(3^2)(2^2)
(a^3)(b^2)=(1^3)(6^2)
a= 1 ; b = 6

How using fractions would make this reasoning wrong?

Bunuel · Answer

You are missing a point there: we are NOT told that a and b are integers, hence from a^3*b^2=36 you cannot say for sure that a=1 and b=6 . Because for ANY a there will exist some b which will satisfy a^3*b^2=36 (and vise-versa). For example if a=2 then b^3=9 and b=\sqrt {9} . Similar question to practice: if-3-a-4-b-c-what-is-the-value-of-b-1-5-a-25-2-c-106047.html Hope it helps.

Versatile720 · Answer

Answer is C .

From 1: 1/a^3 *1/b^2= 1/36 Not sufficient 
From 2 : 1/ab =1/6 means a, b can be :2,3 ; 3,2 ;6,1;1,6 . not sufficient

Combining both statements only a=1 and b=6 fulfills the statement 1 so answer is C
 
Try and fail but never fail to try.

fameatop · Answer

Hi all, I would like to add to the explanation given by Bunuel.
(1)as explained by bunuel  a^3b^2=36  - Insufficient ----> Why? 
 Because we can be sure that a=1 but we b can be either +6 or -6
(2) b = 6a----Insufficient because this is just a ratio & nothing is mentioned about their values.

Hope it helps.

Bunuel · Answer

Let me correct you: we cannot be sure that a=1 from (1). Since we are not told that a and b are integers, then a could, for example be 2 and b could be -\frac{3}{\sqrt{2}} or \frac{3}{\sqrt{2}} . Hope it's clear. P.S. This post might also help: what-is-the-value-of-a-2-b-104673.html#p1047996

alphabeta1234 · Answer

Bunuel,

Even if we were told a&b are postive integers does knowing a^3*b^2=constant ever gaurentee knowing the value of a^2*a^3??

Posted from my mobile device

Bunuel · Answer

Well, if we were told that a and b are  positive  integers, then from a^3b^2=36 it follows that a=1 and b=6.

cg0588 · Answer

Hi Bunuel - can you explain why ab^(-1)=6^(-1) yields b/a = 6?

Derkus · Answer

It would make it much clearer if you edited the original question to indicate for statement two "a * b^(-1)". I got confused and thought the whole term ab was taken to the negative first power.

Bunuel · Answer

In this case it would be (ab)^(-1), not ab^(-1). Still edited.

Bunuel · Answer

a*b^{(-1)}=6^{(-1)} ;

a*\frac{1}{b}=\frac{1}{6} ;

\frac{a}{b}=\frac{1}{6} ;

\frac{b}{a}=6 .

Hope it's clear.

bumpbot · Answer

Automated notice from GMAT Club BumpBot: 

A member just gave Kudos to this thread, showing it’s still useful. I’ve bumped it to the top so more people can benefit. Feel free to add your own questions or solutions.

 This post was generated automatically.

What is the value of a^(-2)*b^(-3)?

Prep Toolkit

Top 5 GMAT Debrief Videos

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards

GMAT Data Sufficiency (DS) Questions

What is the value of a^(-2)*b^(-3)?

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards