Solving Proportion Problems: A Brain-Racking Challenge

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The discussion revolves around solving proportion problems, highlighting three specific examples. The first problem, calculating earnings over 3.5 weeks, was correctly solved by multiplying 850 euros by 3.5, confirming no additional steps were needed. The second problem involved resizing a picture, where the user mistakenly set the height instead of the width to 10 inches, leading to confusion in the ratio setup. The third problem, concerning two triangles, required clarification on the relationship between their dimensions to find the missing piece, emphasizing the need to maintain the correct proportions. Overall, the thread illustrates common pitfalls in solving proportion problems and the importance of accurately setting up ratios.
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This has been racking my brain for awhile, and I know I have it right, but something is missing.

1. A person earns 850 euros in one week. How much will she earn in 3.5 weeks?

This seemed simple enough, so I just multiplied 850 by 3.5 and got $2,975. I know there's one more step, and I can't find it.

2. A picture is 73/4 inches wide and 81/2 inches high. If you wish to enlarge the width to 10 inches, how high will the picture be? Express in fractions instead of decimals.

I converted the fractions into decimals so I can get the fraction answer. I came up with 7.75/8.5 = x/10 = I then got 77.5 over 8.5 and came up with 911/100. Obviously this is wrong, but I can't figure out the step I missed here.

3. There's 5/4 of a triangle and 2/x of another one, I'm supposed to find the missing piece.I come up with 8/5. I'm stuck on where to go from here.

Any help would be appericated.
 
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simplywrite said:
1. A person earns 850 euros in one week. How much will she earn in 3.5 weeks?

This seemed simple enough, so I just multiplied 850 by 3.5 and got $2,975. I know there's one more step, and I can't find it.
This one's just fine---no missing steps.

2. A picture is 73/4 inches wide and 81/2 inches high. If you wish to enlarge the width to 10 inches, how high will the picture be? Express in fractions instead of decimals.

I converted the fractions into decimals so I can get the fraction answer. I came up with 7.75/8.5 = x/10 = I then got 77.5 over 8.5 and came up with 911/100. Obviously this is wrong, but I can't figure out the step I missed here.
You messed up the statement of ratios. It looks like you tried to use:
\frac{W_1}{H_1} = \frac{W_2}{H_2}
But you put H_2 = 10 instead of W_2 = 10.
3. There's 5/4 of a triangle and 2/x of another one, I'm supposed to find the missing piece.I come up with 8/5. I'm stuck on where to go from here.
I don't understand this problem.
 
simplywrite said:
This has been racking my brain for awhile, and I know I have it right, but something is missing.

1. A person earns 850 euros in one week. How much will she earn in 3.5 weeks?

This seemed simple enough, so I just multiplied 850 by 3.5 and got $2,975. I know there's one more step, and I can't find it.
Right. You forgot to account for taxes ;). Just kidding, that looks correct to me.

2. A picture is 73/4 inches wide and 81/2 inches high. If you wish to enlarge the width to 10 inches, how high will the picture be? Express in fractions instead of decimals.

I converted the fractions into decimals so I can get the fraction answer. I came up with 7.75/8.5 = x/10 = I then got 77.5 over 8.5 and came up with 911/100. Obviously this is wrong, but I can't figure out the step I missed here.

I'd suggest staying with fractions if you're comforatable:

So you've got an inital aspect ratio:
\frac{width}{height}=\frac{\frac{73}{4}}{\frac{81}{2}}=\frac{146}{324}=\frac{73}{162}
and then you want so solve for the same aspect ratio with a different width:
\frac{10}{h}=\frac{73}{162}
10\times162=73 h
\frac{1620}{73}=h

Of course, since \frac{73}{4} > 10 and the problem mentions increase to 10 inches, I'm guessing you copied it incorrectly.

3. There's 5/4 of a triangle and 2/x of another one, I'm supposed to find the missing piece.I come up with 8/5. I'm stuck on where to go from here.

Any help would be appericated.
I can't quite follow what you're doing, but I'd guess that you'should be getting \frac{8}{5x} instead of \frac{8}{5}.
 
The exact problem is

I have two triangles, one is 5 inches on the left and 4 inches at the bottom. Another one is 2 inches on the left side and x on the bottom. The idea is to find the missing piece of that one, that's what I was trying to say.
 
simplywrite said:
The exact problem is

I have two triangles, one is 5 inches on the left and 4 inches at the bottom. Another one is 2 inches on the left side and x on the bottom. The idea is to find the missing piece of that one, that's what I was trying to say.

Can you fill in the spaces below?
5 is to 4 as _ is to _
 
5 is to 4 as 2 is to x.

One triange is 5 inches on the left, and 4 at the bottom. Another is 2 inches to the left. The idea is to find the x at the bottom of the other triange.
 
Let me redo this one:

A picture is 7 3/4 inches wide and is 8 1/2 inches high. If you enlarge the picture by 10 inches, how high will the picture be expressed as a fraction over a decimal?
 
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