Scale problem

eddy2017

eddy2017

Elite Member
Joined
Oct 27, 2017
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Pat wanted to make a scale drawing with 1cm= 4ft. How many centimeters should she use to represent the dimension x in the drawing.
Let me see how far along I can go without having to tap your amazing brains. I am attaching a drawing that comes with the exercise. :)
I have to find the value of x in feet.
Given
Scale is 1 cm =4ft
44 ft total length
to the left I have 12 ft
to the right I have 13.5
how much is x in ft=?

So, I will add `12 + 13.5 =25.5`
so, the rest is x,
`x=18.5ft`
I know the dimension of x now, but she wants to know how many centimeters should she use to represent 18.5 ft in the drawing.
well,
1foot=30.48 cm

I am at a loss what to do next. I will need a hint.
Thanks,
eddy
 

Attachments

  • Drawing (1).png
    Drawing (1).png
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[math]44\;ft\cdot\dfrac{1\;scale\;cm}{4\;ft} = 11\;scale\;cm[/math]
[math]13.5\;ft\cdot\dfrac{1\;scale\;cm}{4\;ft} = 3.375\;scale\;cm[/math]
[math]12\;ft\cdot\dfrac{1\;scale\;cm}{4\;ft} = 3\;scale\;cm[/math]
[math]18.5\;ft\cdot\dfrac{1\;scale\;cm}{4\;ft} = 4.625\;scale\;cm[/math]
Checking: 3.375 + 3 + 4.625 = 11 -- Yup.

Why do we care about "1 foot=30.48 cm"? That's real life. We're scaling.
 
eddy2017 said:
Pat wanted to make a scale drawing with 1cm= 4ft. How many centimeters should she use to represent the dimension x in the drawing.
Let me see how far along I can go without having to tap your amazing brains. I am attaching a drawing that comes with the exercise. :)
I have to find the value of x in feet.
Given
Scale is 1 cm =4ft
44 ft total length
to the left I have 12 ft
to the right I have 13.5
how much is x in ft=?
Click to expand...
So, I will add `12 + 13.5 =25.5`
so, the rest is x,
`x=18.5ft`
I know the dimension of x now, but she wants to know how many centimeters should she use to represent 18.5 ft in the drawing.
well,
1foot=30.48 cm .......................................Not for this problem!!

You were given

Scale is 1 cm = 4ft

4ft → 1 cm

then

1 ft → 1/4 cm = 0.25 cm

18.5 ft → 0.25 * 18.5 → = ? ................................Don't forget to check your answer (how?!)
 
Last edited by a moderator:
1 cm =4ft
Multiply both sides by whatever you need to multiply 4 by to get 18.5. Let this unknown value be y. Can you write and solve the necessary eq to find y?
 
"Jomo, post: 523645, member: 46610"
1 cm =4ft .........................Incorrect .....................?:p:D:ROFLMAO::devilish:
 
tkhunny said:
[math]44\;ft\cdot\dfrac{1\;scale\;cm}{4\;ft} = 11\;scale\;cm[/math]
[math]13.5\;ft\cdot\dfrac{1\;scale\;cm}{4\;ft} = 3.375\;scale\;cm[/math]
[math]12\;ft\cdot\dfrac{1\;scale\;cm}{4\;ft} = 3\;scale\;cm[/math]
[math]18.5\;ft\cdot\dfrac{1\;scale\;cm}{4\;ft} = 4.625\;scale\;cm[/math]
Checking: 3.375 + 3 + 4.625 = 11 -- Yup.

Why do we care about "1 foot=30.48 cm"? That's real life. We're scaling.
Click to expand...
 
tkhunny said:
[math]44\;ft\cdot\dfrac{1\;scale\;cm}{4\;ft} = 11\;scale\;cm[/math]
[math]13.5\;ft\cdot\dfrac{1\;scale\;cm}{4\;ft} = 3.375\;scale\;cm[/math]
[math]12\;ft\cdot\dfrac{1\;scale\;cm}{4\;ft} = 3\;scale\;cm[/math]
[math]18.5\;ft\cdot\dfrac{1\;scale\;cm}{4\;ft} = 4.625\;scale\;cm[/math]
Checking: 3.375 + 3 + 4.625 = 11 -- Yup.
Click to expand...
 
Jomo said:
1 cm =4ft
Multiply both sides by whatever you need to multiply 4 by to get 18.5. Let this unknown value be y. Can you write and solve the necessary eq to find y?
Click to expand...
4FT=121.92cm but I was told that is irrelevant information here.
 
tkhunny said:
[math]44\;ft\cdot\dfrac{1\;scale\;cm}{4\;ft} = 11\;scale\;cm[/math]
[math]13.5\;ft\cdot\dfrac{1\;scale\;cm}{4\;ft} = 3.375\;scale\;cm[/math]
[math]12\;ft\cdot\dfrac{1\;scale\;cm}{4\;ft} = 3\;scale\;cm[/math]
[math]18.5\;ft\cdot\dfrac{1\;scale\;cm}{4\;ft} = 4.625\;scale\;cm[/math]
Checking: 3.375 + 3 + 4.625 = 11 -- Yup.

Why do we care about "1 foot=30.48 cm"? That's real life. We're scaling.
Click to expand...
Thanks, I get everything except the 11, where is the 11 coming from?.
 
Subhotosh Khan said:
1 cm =4ft .........................Incorrect .....................?:p:D:ROFLMAO::devilish:
Click to expand...
OK, if that is how you want to play:
18.5 ft → 0.25 * 18.5 → = ? .........Incorrect......????????

It should be 18.5 ft → 0.25 cm* 18.5 → = ?
I was going to let that go. See what I get for being nice.!!!
 
Subhotosh Khan said:
So, I will add `12 + 13.5 =25.5`
so, the rest is x,
`x=18.5ft`
I know the dimension of x now, but she wants to know how many centimeters should she use to represent 18.5 ft in the drawing.
well,
1foot=30.48 cm .......................................Not for this problem!!

You were given

Scale is 1 cm = 4ft

4ft → 1 cm

then

1 ft → 1/4 cm = 0.25 cm

18.5 ft → 0.25 * 18.5 → = ? ................................Don't forget to check your answer (how?!)
Click to expand...
=4.625 cm
 
eddy2017 said:
Thanks, I get everything except the 11, where is the 11 coming from?.
Click to expand...
Look at the first line of tkhunny's response (#2) with pencil and paper and write it down.

then think 44/4 = ?
 
eddy2017 said:
I am sure Mr Jomo meant the information given in the problem where1cm taken to scale is 4 ft.
Click to expand...
Thank you! Finally someone is nice to me.

Would you believe that every year all the other helpers on this forum can not dispute that they get a 10% raise in their salary. I have never gotten a raise even once!
 
Jomo said:
Thank you! Finally someone is nice to me.

Would you believe that every year all the other helpers on this forum can not dispute that they get a 10% raise in their salary. I have never gotten a raise even once!
Click to expand...
I got a note from Ted stating that my 100% raise was passed on to you.....
 
tkhunny said:
[math]44\;ft\cdot\dfrac{1\;scale\;cm}{4\;ft} = 11\;scale\;cm[/math]
[math]13.5\;ft\cdot\dfrac{1\;scale\;cm}{4\;ft} = 3.375\;scale\;cm[/math]
[math]12\;ft\cdot\dfrac{1\;scale\;cm}{4\;ft} = 3\;scale\;cm[/math]
[math]18.5\;ft\cdot\dfrac{1\;scale\;cm}{4\;ft} = 4.625\;scale\;cm[/math]
Checking: 3.375 + 3 + 4.625 = 11 -- Yup.

Why do we care about "1 foot=30.48 cm"? That's real life. We're scaling.
Click to expand...
 
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