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

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What is the value of a^(-2)*b^(-3)?  [#permalink]

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Updated on: 30 Jan 2015, 06:04
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What is the value of $$a^{(-2)}*b^{(-3)}$$?

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

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

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Originally posted by monirjewel on 11 Nov 2010, 10:02.
Last edited by Bunuel on 30 Jan 2015, 06:04, edited 3 times in total.
Edited the question.
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Re: What is the value of a^(-2)*b^(-3)?  [#permalink]

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11 Nov 2010, 11:07
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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.

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Re: What is the value of a^(-2)*b^(-3)?  [#permalink]

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11 Nov 2010, 12:51
1
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,

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What is the value of a^(-2)*b^(-3)?  [#permalink]

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22 Apr 2011, 08:55
dc123 wrote:
anyone??

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

1) (a^-3)(b^-2) = (36^-1)

2) ab^-1 = 6^-1

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)
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Re: What is the value of a^(-2)*b^(-3)?  [#permalink]

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22 Apr 2011, 10:14
2
dc123 wrote:
What is the value of a^-2b^-3

1) a^-3b^-2 = 36^-1

2) ab^-1 = 6^-1

the Ans says C but doesnt only B work?

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"
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Re: What is the value of a^(-2)*b^(-3)?  [#permalink]

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03 May 2011, 01:01
a. gives values (1,6) and (1,-6)
b gives values (1,6) and (-1,-6)

a+b gives (1,6) hence C
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Re: What is the value of a^(-2)*b^(-3)?  [#permalink]

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03 May 2011, 17:48
dc123 wrote:
What is the value of a^-2b^-3

1) a^-3b^-2 = 36^-1

2) ab^-1 = 6^-1

the Ans says C but doesnt only B work?

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)
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Re: What is the value of a^(-2)*b^(-3)?  [#permalink]

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22 Feb 2012, 02:39
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?
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Re: What is the value of a^(-2)*b^(-3)?  [#permalink]

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22 Feb 2012, 02:49
1
wizard wrote:
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?

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 $$b=\sqrt[3]{9}$$.

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.
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Re: What is the value of a^(-2)*b^(-3)?  [#permalink]

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16 Aug 2012, 00:32
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

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Re: What is the value of a^(-2)*b^(-3)?  [#permalink]

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13 Sep 2012, 07:53
Bunuel wrote:
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) $$ab^{-1}=6$$ --> $$\frac{b}{a}=\frac{1}{6}$$. Not sufficient.

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

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.
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Re: What is the value of a^(-2)*b^(-3)?  [#permalink]

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13 Sep 2012, 07:59
fameatop wrote:
Bunuel wrote:
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) $$ab^{-1}=6$$ --> $$\frac{b}{a}=\frac{1}{6}$$. Not sufficient.

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

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.

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
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Re: What is the value of a^(-2)*b^(-3)?  [#permalink]

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13 Jul 2013, 16:11
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
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Re: What is the value of a^(-2)*b^(-3)?  [#permalink]

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13 Jul 2013, 23:41
alphabeta1234 wrote:
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

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.
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Re: What is the value of a^(-2)*b^(-3)?  [#permalink]

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29 Jan 2015, 18:12
Hi Bunuel - can you explain why ab^(-1)=6^(-1) yields b/a = 6?
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Re: What is the value of a^(-2)*b^(-3)?  [#permalink]

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29 Jan 2015, 20:47
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.
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Re: What is the value of a^(-2)*b^(-3)?  [#permalink]

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30 Jan 2015, 06:04
Derkus wrote:
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.

In this case it would be (ab)^(-1), not ab^(-1). Still edited.
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Re: What is the value of a^(-2)*b^(-3)?  [#permalink]

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30 Jan 2015, 06:06
cg0588 wrote:
Hi Bunuel - can you explain why ab^(-1)=6^(-1) yields b/a = 6?

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

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

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

$$\frac{b}{a}=6$$.

Hope it's clear.
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