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A square has an area of (0.027)^(-2/3) square meters. The pe

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A square has an area of (0.027)^(-2/3) square meters. The pe

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GMATH practice exercise (Quant Class 12)

A square has an area of (0.027)^(-2/3) square meters. The perimeter of this square, in meters, belongs to which of the following intervals?

(A) [0,10.5]
(B) [10.5, 11.5]
(C) [11.5, 12.5]
(D) [12.5, 13.5]
(E) [13.5, 20]

Answer: ___(D)__

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Fabio Skilnik :: GMATH method creator ( Math for the GMAT)
English-speakers :: https://www.gmath.net
Portuguese-speakers :: https://www.gmath.com.br

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fskilnik@GMATH wrote:
GMATH practice exercise (Quant Class 12)

A square has an area of (0.027)^(-2/3) square meters. The perimeter of this square, in meters, belongs to which of the following intervals?

(A) [0,10.5]
(B) [10.5, 11.5]
(C) [11.5, 12.5]
(D) [12.5, 13.5]
(E) [13.5, 20]
$$? = 4L$$
$${L^2} = {\left( {{3^3} \cdot {{10}^{ - 3}}} \right)^{ - {2 \over 3}}} = {\left[ {{{\left( {{3^3} \cdot {{10}^{ - 3}}} \right)}^{ - {1 \over 3}}}} \right]^2} = {\left( {{3^{ - 1}} \cdot {{10}^1}} \right)^2}$$
$$L > 0\,\,\, \Rightarrow \,\,\,\,\,L = {3^{ - 1}} \cdot {10^1} = {{10} \over 3}$$
$$? = 4L = {{39 + 1} \over 3} = 13{1 \over 3}\,\,\,\,\, \Rightarrow \,\,\,\,\,\left( {\rm{D}} \right)$$


We follow the notations and rationale taught in the GMATH method.

Regards,
Fabio.

_________________
Fabio Skilnik :: GMATH method creator ( Math for the GMAT)
English-speakers :: https://www.gmath.net
Portuguese-speakers :: https://www.gmath.com.br

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fskilnik@GMATH wrote:
GMATH practice exercise (Quant Class 12)

A square has an area of (0.027)^(-2/3) square meters. The perimeter of this square, in meters, belongs to which of the following intervals?

(A) [0,10.5]
(B) [10.5, 11.5]
(C) [11.5, 12.5]
(D) [12.5, 13.5]
(E) [13.5, 20]
$$? = 4L$$
$${L^2} = {\left( {{3^3} \cdot {{10}^{ - 3}}} \right)^{ - {2 \over 3}}} = {\left[ {{{\left( {{3^3} \cdot {{10}^{ - 3}}} \right)}^{ - {1 \over 3}}}} \right]^2} = {\left( {{3^{ - 1}} \cdot {{10}^1}} \right)^2}$$
$$L > 0\,\,\, \Rightarrow \,\,\,\,\,L = {3^{ - 1}} \cdot {10^1} = {{10} \over 3}$$
$$? = 4L = {{39 + 1} \over 3} = 13{1 \over 3}\,\,\,\,\, \Rightarrow \,\,\,\,\,\left( {\rm{D}} \right)$$


We follow the notations and rationale taught in the GMATH method.

Regards,
Fabio.

_________________
Fabio Skilnik :: GMATH method creator ( Math for the GMAT)
English-speakers :: https://www.gmath.net
Portuguese-speakers :: https://www.gmath.com.br

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