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Ratios problem

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Ratios problem

by gmattesttaker2 » Sat Jul 21, 2012 11:49 pm
Hello,

This is taken from MGMAT Strategy Guide 1 (5th edition). P. 69:

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 cu. meter?

Ans: 8 cu. cm

I am not clear with the solution given in the book. Can you please assist with this? Thanks for your valuable time and help.

Best Regards,
Sri
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by eagleeye » Sat Jul 21, 2012 11:58 pm
gmattesttaker2 wrote:Hello,

This is taken from MGMAT Strategy Guide 1 (5th edition). P. 69:

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 cu. meter?

Ans: 8 cu. cm

I am not clear with the solution given in the book. Can you please assist with this? Thanks for your valuable time and help.

Best Regards,
Sri
Let side of model = m and side of sculpture = s.
Volume of model = m^3. Volume of sculpture = s^3.

We are given that m/s = 0.5cm/1 meter and we need to find m^3.

Now Volume of sculpture = s^3 = 64 cu. meter => s= 64^(1/3) meter = 4 meter.

m/s = 0.5 cm / (1meter)= 0.5*4/(4 meter) = 2cm/(4 meter).

Hence m = 2 cm.

Then volume of sculpture = m^3 = (2cm)^3 = 8 cubic centimeter. :)
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by Stuart@KaplanGMAT » Sun Jul 22, 2012 1:13 am
Hi!

A bit of theory before getting into the calculations.

First, when converting between linear, second order and third order equations (i.e. lengths, areas and volumes), you simply apply the proper exponent to find the new ratio.

For example, if comparing the areas of two squares with sides of ratio of x:y, the areas will be in the ratio of x^2:y^2. If you were to compare the volumes of two cubes, they'd be in the ratio of x^3:y^3.

Second, always remember to carefully convert units!

Applying those rules to this question, we see that the ratio of the sides is .5cm:1m. First, let's convert units.

1m = 100cm, so the ratio is:

.5:100 = 1:200

Next, we see that we're given the volume of the real sculpture, so we need to turn our linear ratio into a 3rd order ratio (since volume is 3-dimensional). So, the ratio of the volumes is:

1^3:200^3 = 1:8,000,000

Finally, we can apply the actual measurement, remembering to convert cubic metres to cubic centimetres.

64 m^3 = 64 * 100cm * 100cm * 100cm = 64,000,000 cm^3

Now, our ratio:

x/1 = 64000000cm^3/8000000
x = 8cm^3

done!
gmattesttaker2 wrote:Hello,

This is taken from MGMAT Strategy Guide 1 (5th edition). P. 69:

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 cu. meter?

Ans: 8 cu. cm

I am not clear with the solution given in the book. Can you please assist with this? Thanks for your valuable time and help.

Best Regards,
Sri
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by gmattesttaker2 » Sun Jul 22, 2012 10:39 am
eagleeye wrote:
gmattesttaker2 wrote:Hello,

This is taken from MGMAT Strategy Guide 1 (5th edition). P. 69:

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 cu. meter?

Ans: 8 cu. cm

I am not clear with the solution given in the book. Can you please assist with this? Thanks for your valuable time and help.

Best Regards,
Sri
Let side of model = m and side of sculpture = s.
Volume of model = m^3. Volume of sculpture = s^3.

We are given that m/s = 0.5cm/1 meter and we need to find m^3.

Now Volume of sculpture = s^3 = 64 cu. meter => s= 64^(1/3) meter = 4 meter.

m/s = 0.5 cm / (1meter)= 0.5*4/(4 meter) = 2cm/(4 meter).

Hence m = 2 cm.

Then volume of sculpture = m^3 = (2cm)^3 = 8 cubic centimeter. :)
Hello Eagleeye,

Thank you very much for your prompt reply. The solution is clear to me now. Thanks again.

Best Regards,
Sri
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by gmattesttaker2 » Sun Jul 22, 2012 10:41 am
Stuart Kovinsky wrote:Hi!

A bit of theory before getting into the calculations.

First, when converting between linear, second order and third order equations (i.e. lengths, areas and volumes), you simply apply the proper exponent to find the new ratio.

For example, if comparing the areas of two squares with sides of ratio of x:y, the areas will be in the ratio of x^2:y^2. If you were to compare the volumes of two cubes, they'd be in the ratio of x^3:y^3.

Second, always remember to carefully convert units!

Applying those rules to this question, we see that the ratio of the sides is .5cm:1m. First, let's convert units.

1m = 100cm, so the ratio is:

.5:100 = 1:200

Next, we see that we're given the volume of the real sculpture, so we need to turn our linear ratio into a 3rd order ratio (since volume is 3-dimensional). So, the ratio of the volumes is:

1^3:200^3 = 1:8,000,000

Finally, we can apply the actual measurement, remembering to convert cubic metres to cubic centimetres.

64 m^3 = 64 * 100cm * 100cm * 100cm = 64,000,000 cm^3

Now, our ratio:

x/1 = 64000000cm^3/8000000
x = 8cm^3

done!
gmattesttaker2 wrote:Hello,

This is taken from MGMAT Strategy Guide 1 (5th edition). P. 69:

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 cu. meter?

Ans: 8 cu. cm

I am not clear with the solution given in the book. Can you please assist with this? Thanks for your valuable time and help.

Best Regards,
Sri
Hello Stuart,

Thank you very much for the thorough explanation. It is clear to me now. Thanks again for your help.

Best Regards,
Sri
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