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Director  D
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If the scale model of a cube sculpture is 0.5 cm per every 1 m  [#permalink]

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Question Stats: 84% (01:33) correct 16% (01:05) wrong based on 60 sessions

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If the scale model of a cube sculpture is 0.5 cm per every 1 m of the real sculpture, what is the volume of the model, if the volume of the real sculpture is 64 m3?

A. 64 cm3
B. 8 cm3
C. 2 cm3
D. 4 cm3
E. 16 cm3

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If the scale model of a cube sculpture is 0.5 cm per every 1 m  [#permalink]

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shashankism wrote:
If the scale model of a cube sculpture is 0.5 cm per every 1 m of the real sculpture, what is the volume of the model, if the volume of the real sculpture is 64 m3?

A. 64 cm3
B. 8 cm3
C. 2 cm3
D. 4 cm3
E. 16 cm3

Volume of cube $$= x^3$$ units ---------- (x is the side of cube)

Volume of the real sculpture is $$64 m^3$$$$= (4^3)m^3$$

Therefore side of real sculpture is $$4 m$$.

Scale model of a cube sculpture is 0.5 cm per every 1 m of the real sculpture. Let the side of model be x.

$$0.5cm -- 1 m$$
$$x cm -- 4m$$

Cross multiplying the above we get;

$$x = 2 cm$$

Therefore volume of model $$= 2^3 = 8 cm^3$$ . Answer (B)...
Senior Manager  G
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Re: If the scale model of a cube sculpture is 0.5 cm per every 1 m  [#permalink]

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As 0.5 cm represents 1 m, then 1$$m^3$$ is represented by $$0.5^3 cm^3$$ or 0.125 $$cm^3$$

Therefore 64$$m^3$$ is represented by 64 x 0.125 $$cm^3$$ = 8 $$cm^3$$

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Re: If the scale model of a cube sculpture is 0.5 cm per every 1 m  [#permalink]

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shashankism wrote:
If the scale model of a cube sculpture is 0.5 cm per every 1 m of the real sculpture, what is the volume of the model, if the volume of the real sculpture is 64 m3?

A. 64 cm3
B. 8 cm3
C. 2 cm3
D. 4 cm3
E. 16 cm3

Since we have 0.5 cm per every 1 m, we can say:

(0.5 cm)^3 = (1 m)^3

0.125 cm^3 = 1 m^3

Thus, the volume of a model that is 64 m^3 is:

64 m^3 x 0.125 cm^3/1 m^3 = 64 x 0.125 = 8 cm^3

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Re: If the scale model of a cube sculpture is 0.5 cm per every 1 m  [#permalink]

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_________________ Re: If the scale model of a cube sculpture is 0.5 cm per every 1 m   [#permalink] 17 Oct 2018, 20:01
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