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(1/2)^(-3)*(1/4)^(-2)*(1/16)^(-1)

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(1/2)^(-3)*(1/4)^(-2)*(1/16)^(-1)  [#permalink]

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New post 29 Dec 2012, 04:58
5
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A
B
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Question Stats:

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\((\frac{1}{2})^{-3}(\frac{1}{4})^{-2}(\frac{1}{16})^{-1}=\)


(A) \((\frac{1}{2})^{(-48)}\)

(B) \((\frac{1}{2})^{(-11)}\)

(C) \((\frac{1}{2})^{(-6)}\)

(D) \((\frac{1}{8})^{(-11)}\)

(E) \((\frac{1}{8})^{(-6)}\)


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Re: (1/2)^(-3)*(1/4)^(-2)*(1/16)^(-1)  [#permalink]

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New post 29 Dec 2012, 05:01
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Walkabout wrote:
\((\frac{1}{2})^{-3}(\frac{1}{4})^{-2}(\frac{1}{16})^{-1}=\)

(A) (1/2)^(-48)
(B) (1/2)^(-11)
(C) (1/2)^(-6)
(D) (1/8)^(-11)
(E) (1/8)^(-6)


\((\frac{1}{2})^{-3}(\frac{1}{4})^{-2}(\frac{1}{16})^{-1}=2^3*4^2*16=2^3*2^4*2^4=2^{11}=(\frac{1}{2})^{-11}\).

Answer: B.
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Re: (1/2)^(-3)*(1/4)^(-2)*(1/16)^(-1)  [#permalink]

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New post 20 Mar 2014, 23:21
7
Answer = B


Keep the base as 2 & modify the powers accordingly

\((\frac{1}{2})^{-3}(\frac{1}{4})^{-2}(\frac{1}{16})^{-1}\)

= \((\frac{1}{2})^{-3}(\frac{1}{2})^{-4}(\frac{1}{2})^{-4}\)

With the base the same, add the powers

-3-4-4 = -11

\((\frac{1}{2})^{-11}\)
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Re: (1/2)^(-3)*(1/4)^(-2)*(1/16)^(-1)  [#permalink]

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New post 29 Dec 2012, 05:14
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Walkabout wrote:
\((\frac{1}{2})^{-3}(\frac{1}{4})^{-2}(\frac{1}{16})^{-1}=\)

(A) (1/2)^(-48)
(B) (1/2)^(-11)
(C) (1/2)^(-6)
(D) (1/8)^(-11)
(E) (1/8)^(-6)


whenevr you see a fraction raised to a negative power think reciprocal

so we have : 2^3 * 4^2 * 16^1

Consolidate: 2^3 * 2^4 * 2^4

Clearly as you can see we have 2^11 only B fits

otherwise take the reciprocal of 2^11 ----> 1/2^-11

B is the best
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Re: (1/2)^(-3)*(1/4)^(-2)*(1/16)^(-1)  [#permalink]

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New post 03 Dec 2014, 06:20
Resolved it as 16 x 2^3 x 4^2
After 2:20sec :(
Simplified it as 1/2
B
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Re: (1/2)^(-3)*(1/4)^(-2)*(1/16)^(-1)  [#permalink]

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New post 15 Jul 2016, 03:41
1
Walkabout wrote:
\((\frac{1}{2})^{-3}(\frac{1}{4})^{-2}(\frac{1}{16})^{-1}=\)

(A) (1/2)^(-48)
(B) (1/2)^(-11)
(C) (1/2)^(-6)
(D) (1/8)^(-11)
(E) (1/8)^(-6)


We start by using the negative exponent rule. When a fractional base is raised to a negative exponent, we can rewrite the expression (without the negative exponent) by flipping the fraction and making the exponent positive. For example, (1/2)^-3 = 2^3

We are using the negative exponent rule because it’s not only easier to deal with positive exponents, but also when we flip the fractional base, the fraction becomes an integer.

(1/2)^-3 = 2^3

(1/4)^-2 = 4^2 = (2 x 2)^2 = (2^2)^2 = 2^4

(1/16)^-1 = 16^1 = (2 x 2 x 2 x 2)^1 = (2^4)^1 = 2^4

We multiply each term in the expression, obtaining:

2^3 x 2^4 x 2^4

Remember, since the bases are the same, we keep the base and add the exponents. We are left with:

2^(3+4+4) = 2^11

Finally, since our answer choices are expressed in fractional form, we once again have to use the negative exponent rule. To convert a base with a positive exponent, take the reciprocal of the base and change the positive exponent to its negative counterpart. Using the rule we get:

2^11 = (1/2)^-11

The answer is B.
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Re: (1/2)^(-3)*(1/4)^(-2)*(1/16)^(-1)  [#permalink]

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New post 30 Nov 2016, 16:48
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Walkabout wrote:
\((\frac{1}{2})^{-3}(\frac{1}{4})^{-2}(\frac{1}{16})^{-1}=\)

(A) (1/2)^(-48)
(B) (1/2)^(-11)
(C) (1/2)^(-6)
(D) (1/8)^(-11)
(E) (1/8)^(-6)


Nice rule: \((\frac{a}{b})^{-n} = (\frac{b}{a})^{n}\)

\((\frac{1}{2})^{-3}(\frac{1}{4})^{-2}(\frac{1}{16})^{-1} = (\frac{2}{1})^{3}(\frac{4}{1})^{2}(\frac{16}{1})^{1}\)

\(= (2^{3})(4^{2})(16^{1})\)

\(= (2^{3})(2^{4})(2^{4})\)

\(= 2^{11}\)

\(= (2/1)^{11}\)

\(= (1/2)^{-11}\)

Answer: B
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Re: (1/2)^(-3)*(1/4)^(-2)*(1/16)^(-1)  [#permalink]

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New post 07 Dec 2016, 16:28
(2)^3 (4)^2 (16)^1
Simplify to base 2
(2)^3 (2)^4 (2)^4
2^11
(1/2)^-11
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Re: (1/2)^(-3)*(1/4)^(-2)*(1/16)^(-1)  [#permalink]

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New post 22 Aug 2017, 03:16
I understood the method but I am just wondering if we could do this way - i tried and got the wrong answer

(1/2)^(-3)* (1/4)^(-2)*(1/16)^(-1)

= 1/(1/2)^3 * 1/(1/4)^2* 1/(1/16)^1

=1/(1/8)*1/(1/16)*1/(1/16)
=8/1*16/1*16/1


Please let me know why cant we do this way?
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Re: (1/2)^(-3)*(1/4)^(-2)*(1/16)^(-1)  [#permalink]

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New post 22 Aug 2017, 03:21
santro789 wrote:
I understood the method but I am just wondering if we could do this way - i tried and got the wrong answer

(1/2)^(-3)* (1/4)^(-2)*(1/16)^(-1)

= 1/(1/2)^3 * 1/(1/4)^2* 1/(1/16)^1

=1/(1/8)*1/(1/16)*1/(1/16)
=8/1*16/1*16/1


Please let me know why cant we do this way?


You can and you'll get the same answer: 8/1*16/1*16/1 = 2^11 = (1/2)^(-11)
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Re: (1/2)^(-3)*(1/4)^(-2)*(1/16)^(-1)  [#permalink]

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New post 24 Aug 2017, 20:39
1
Hello Moderators,

Do you think it would be good to make the answer choices math friendly - just like the main stem? This is OG question and it would be good to see as it would appear in the real exam.

Thanks in advance

Walkabout wrote:
\((\frac{1}{2})^{-3}(\frac{1}{4})^{-2}(\frac{1}{16})^{-1}=\)

(A) (1/2)^(-48)
(B) (1/2)^(-11)
(C) (1/2)^(-6)
(D) (1/8)^(-11)
(E) (1/8)^(-6)

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Re: (1/2)^(-3)*(1/4)^(-2)*(1/16)^(-1)  [#permalink]

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New post 24 Aug 2017, 20:44
susheelh wrote:
Hello Moderators,

Do you think it would be good to make the answer choices math friendly - just like the main stem? This is OG question and it would be good to see as it would appear in the real exam.

Thanks in advance

Walkabout wrote:
\((\frac{1}{2})^{-3}(\frac{1}{4})^{-2}(\frac{1}{16})^{-1}=\)

(A) (1/2)^(-48)
(B) (1/2)^(-11)
(C) (1/2)^(-6)
(D) (1/8)^(-11)
(E) (1/8)^(-6)

_______________
Edited. Thank you.
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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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Re: (1/2)^(-3)*(1/4)^(-2)*(1/16)^(-1)  [#permalink]

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New post 27 Aug 2018, 02:21
where can I find similar questions?
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Re: (1/2)^(-3)*(1/4)^(-2)*(1/16)^(-1)  [#permalink]

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New post 27 Aug 2018, 06:27
onyx12102 wrote:
where can I find similar questions?

Just click on the Tags " Exponents/Powers" you will be redirected here

https://gmatclub.com/forum/search.php?s ... &tag_id=60

Hope this helps.
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Re: (1/2)^(-3)*(1/4)^(-2)*(1/16)^(-1) &nbs [#permalink] 27 Aug 2018, 06:27
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