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 · 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

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

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