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# Maths problem! Watch

1. A metal bar can be melted down to form 320 solid ornaments of height 2.8cm, how many similar ornaments of height 11.2cm could have been formed by the same metal bar?
2. (Original post by CresentHan)
A metal bar can be melted down to form 320 solid ornaments of height 2.8cm, how many similar ornaments of height 11.2cm could have been formed by the same metal bar?
So the total height of all the ornaments will be 2.8cm x 320 = 896 cm

With that 896cm, you want to make 11.2cm tall ornaments.
896 / 11.2 = 80 ornaments(11.2cm)
3. (Original post by lyamlim97)
So the total height of all the ornaments will be 2.8cm x 320 = 896 cm

With that 896cm, you want to make 11.2cm tall ornaments.
896 / 11.2 = 80 ornaments(11.2cm)
An alternative is to work out the scale factor between the height of the ornaments(because they are similar).
To work out the scale factor -> 2.8/11.2. This gives 0.25. So we know the proportion between the height of the ornaments is 0.25

Therefore to find how many ornaments 11.2cm will make, we do 320 * 0.25(because it is similar so you can work out the how many ornaments there are).

However I quite like your method, but is algebraic?
4. (Original post by theBranicAc)
An alternative is to work out the scale factor between the height of the ornaments(because they are similar).
To work out the scale factor -> 2.8/11.2. This gives 0.25. So we know the proportion between the height of the ornaments is 0.25

Therefore to find how many ornaments 11.2cm will make, we do 320 * 0.25(because it is similar so you can work out the how many ornaments there are).

However I quite like your method, but is algebraic?
Yeah, didnt think of that. Yep its algebraic, I like to do things that way.
5. (Original post by lyamlim97)
Yeah, didnt think of that. Yep its algebraic, I like to do things that way.
Oh okay, the only reason I asked is because depending on how many marks the question is, you may not of got full marks as it was such a simple, yet effective, solution.
6. When considering mathematical similarity of 3D objects you have to consider that it will be stretched in 3 dimensions. So if the larger object is 4 times taller it would be 4x4x4 greater volume.

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7. (Original post by CresentHan)
A metal bar can be melted down to form 320 solid ornaments of height 2.8cm, how many similar ornaments of height 11.2cm could have been formed by the same metal bar?
The length scale factor is 2.8/11.2 = 1/4 (= 0.25)

Bars are similar and three-dimensional, hence we need to find the volume scale factor. This is the same as the length scale factor cubed.

Volume scale factor = (1/4)^3 = 1/64

320 * 1/64 = 320/64 = 5 ornaments
8. Which one is correct?
9. (Original post by CresentHan)
Which one is correct?

(Original post by OscarTG)
The length scale factor is 2.8/11.2 = 1/4 (= 0.25)

Bars are similar and three-dimensional, hence we need to find the volume scale factor. This is the same as the length scale factor cubed.

Volume scale factor = (1/4)^3 = 1/64

320 * 1/64 = 320/64 = 5 ornaments
I think he is correct^^
Isn't there a markscheme for this question?
10. (Original post by theBranicAc)
I think he is correct^^
Isn't there a markscheme for this question?
Unfortunately no, I have a mock exam tomorrow and it was a school compiled past paper. My teacher didn't give me the answers to some of the questions
11. Thinking about it now, it does say similar so all dimensions should change.
12. (Original post by lyamlim97)
So the total height of all the ornaments will be 2.8cm x 320 = 896 cm

With that 896cm, you want to make 11.2cm tall ornaments.
896 / 11.2 = 80 ornaments(11.2cm)
(Original post by lyamlim97)
Yeah, didnt think of that. Yep its algebraic, I like to do things that way.
(Original post by OscarTG)
The length scale factor is 2.8/11.2 = 1/4 (= 0.25)

Bars are similar and three-dimensional, hence we need to find the volume scale factor. This is the same as the length scale factor cubed.

Volume scale factor = (1/4)^3 = 1/64

320 * 1/64 = 320/64 = 5 ornaments
See this guy is correct^^, thats why I asked if your method was algerbraic, because of how simple it was.

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