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[Hot Lava Mama]

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Edited August 11, 2017 by Hot Lava Mama

[Arcadia]

I can't figure out where the solution my daughter came up with for this problem is wrong. Can you help me?

 

Question:

You have an 8-by-10 inch photo of a soccer player that must be reduced to a size of 1.6 by 2 inches for the school newsletter. What percent does the photo need to be reduced to in order for it to fit in the allotted space?

 

I knew immediately "how" to solve it (using ratios of percent) to get the solution of: the photo needs to be 20% of the original photo.

...

My first thought was that maybe we could do it that way since the relative reduction of area size (maybe) should be the same as the reduction of the side lengths. Maybe that is where we went wrong.

 

The reduction of the side lengths is such that the reduced length would be 20% of original side lengths keeping the aspect ratio if you are using a computer photo editor.

 

The reduction in area is by 96% as your daughter has calculated so the reduced area is 4% of the original area. 20% (length) x 20% (length) = 4% (area)

[wendyroo]

The problem she is working in square inches and you are working in inches.  I don't know which one the problem is asking for.

 

If you think about it, her answer is absolutely correct for how much the area of the picture needs to be reduced.  8" x 10" is 80 sq. inches.  A 20% reduction would be 64 sq. inches, but 1.6" x 2" is only 3.2 sq. inches.  Much more than a 20% reduction.  (80-3.2)/80 = 96% reduction in area.

 

Wendy

[hornblower]

I think the way she set it up what she figured out is what percentage the new photo is of the old photo, not the percent of the reduction which it seems the question is asking. 

So instead: 

Original Area - Original Area times x = New Area    ...... where x is the percentage reduction. 

80-80x=3.2

 

subtract 80 from both sides

-80x=-76.8

divide by 80


x=.96 

96 percent. The original photo must be reduced by 96%. 

[EKS]

I'm going to be honest.  I think this is a terrible problem as written because it assumes that the student knows how a photocopier thinks or has enough experience with a photocopier to know what you have to input to reduce a copy by a certain amount.

 

The problem should state that a photocopier uses the length of the sides to determine percent reduction.

 

I would explain the thing about the length of the sides and then have her try again.

[wendyroo]

I'm going to be honest.  I think this is a terrible problem as written because it assumes that the student knows how a photocopier thinks or has enough experience with a photocopier to know what you have to input to reduce a copy by a certain amount.

 

The problem should state that a photocopier uses the length of the sides to determine percent reduction.

 

I would explain the thing about the length of the sides and then have her try again.

 

I agree.  

 

Though it would be a very interesting problem if it explained copier reduction and then said the picture had to fit into a 2.75" by 4" space.  Then the student would have to figure out that each dimension required a different percent reduction and that in order to maintain the aspect ratio both dimensions would have to be reduced by the larger percentage even though the picture would not end up neatly filling the space.

 

That is a real world application of percent reduction.

 

Wendy

[regentrude]

I can't figure out where the solution my daughter came up with for this problem is wrong.  Can you help me?

 

Question:

You have an 8-by-10 inch photo of a soccer player that must be reduced to a size of 1.6 by 2 inches for the school newsletter.  What percent does the photo need to be reduced to in order for it to fit in the allotted space?

 

I knew immediately "how" to solve it (using ratios of percent) to get the solution of: the photo needs to be 20% of the original photo.

 

Here is how my daughter set up the problem:

 

[(8 inches x 10 inches) - ( 1.6 inches x 2 inches)] = N% x (8 inches by 10 inches)

 

So, it seems like she is comparing the area of the full size to the area of the reduced size, and then equating it with the percent of the full size. 

 

My first thought was that maybe we could do it that way since the relative reduction of area size (maybe) should be the same as the reduction of the side lengths.  Maybe that is where we went wrong.

 

Or, is the right side of her equation wrong?  Maybe that is the problem.

 

When she worked out her problem, she got 96%. 

 

Any ideas of what went wrong?  I can't explain to her why this doesn't work.

 

Hot Lava Mama

 

Her calculation is correct, but she is calculating the area BY which the photo has to be reduced (original area minus new area as percentage of original area). It is correct that the photo must be reduced by 96%. Which means it must be reduced TO 4%.

[Hot Lava Mama]

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Edited August 11, 2017 by Hot Lava Mama

[regentrude]

Hmm.  I was thinking that, too, but the answer book has the correct answer as "20%".  Thus, they say that the new size is 20% of the original size (or, it has to be reduced by 80%).  It still seems like I should be able to get there with her set up of the problem. 

 

The length of each SIDE has to be reduced to 20%. Which reduced the area to 4%.

 

It sounds as if the problem is with an imprecise phrasing of the question. It is important to state exactly what the student is supposed to calculate. "What percent does the photo have to be reduced to?" is unclear. It has to be reduced to 4% of its area, which requires a reduction of each side to 20%. 

 

ETA: If she wants to calculate side length percentage, she needs to take the square root of the 4% she obtains for the reduction factor of the area, because area is side length squared. That gives 20%.

Edited July 9, 2016 by regentrude