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# x > 1

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x > 1 [#permalink]  06 Nov 2018, 14:25
Expert's post
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Question Stats:

77% (00:25) correct 22% (00:17) wrong based on 22 sessions

$$x > 1$$

 Quantity A Quantity B $$x(x^2)^4$$ $$(x^3)^3$$

A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.
[Reveal] Spoiler: OA

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Joined: 20 Apr 2016
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Kudos [?]: 559 [1] , given: 101

Re: x > 1 [#permalink]  08 Nov 2018, 03:36
1
KUDOS
Carcass wrote:
$$x > 1$$

 Quantity A Quantity B $$x(x^2)^4$$ $$(x^3)^3$$

A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.

Explanation::

QTY A : $$x(x^2)^4 = x^9$$

QTY B : $$(x^3)^3 = x^9$$

Hence QTY A = QTY B

Option C
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Intern
Joined: 04 Nov 2018
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Kudos [?]: 6 [0], given: 8

Re: x > 1 [#permalink]  10 Nov 2018, 09:36
Carcass wrote:
$$x > 1$$

 Quantity A Quantity B $$x(x^2)^4$$ $$(x^3)^3$$

A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.

Why is it wrong that I did first did $$(x^1 *x^2)$$= $$(x^3)$$ then $$(x^3)^4$$=x^12?
Intern
Joined: 17 Oct 2018
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Kudos [?]: 3 [1] , given: 23

Re: x > 1 [#permalink]  15 Nov 2018, 04:56
1
KUDOS
Mercychee wrote:

Why is it wrong that I did first did $$(x^1 *x^2)$$= $$(x^3)$$ then $$(x^3)^4$$=x^12?

BODMAS/PEMDAS: Solve brackets or parentheses first. The 4 in $$(x^2)^4$$ is a part of the bracket that requires first priority while solving.
Intern
Joined: 07 Aug 2016
Posts: 42
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Kudos [?]: 13 [0], given: 0

Re: x > 1 [#permalink]  19 Nov 2018, 16:02
x (x^2)^4 = x * x^ 8 = x^9

(x^3)^3 = x^9

Both quantities are equal.

Re: x > 1   [#permalink] 19 Nov 2018, 16:02
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