A copying machine reduces a 12" line segment to 10.5". At this setting, what would a 16" line segment become?

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- May 29th 2010, 03:20 PMcherryperryScale problem
A copying machine reduces a 12" line segment to 10.5". At this setting, what would a 16" line segment become?

- May 29th 2010, 03:24 PMcherryperry
- May 29th 2010, 06:56 PMDeanSchlarbaum
- May 29th 2010, 07:02 PMProve It
$\displaystyle 12k = 10.5$

$\displaystyle k = \frac{10.5}{12}$

$\displaystyle = \frac{105}{120}$

$\displaystyle = \frac{7}{8}$.

So at the same setting...

$\displaystyle 16k = 16\left(\frac{7}{8}\right)$

$\displaystyle = 14$.

Therefore, this setting will reduce a $\displaystyle 16''$ segment to $\displaystyle 14''$. - May 30th 2010, 08:44 PMDeanSchlarbaum
__cherryperry__: One more time. . .

Depending on how you daugther thinks, there is another way to see this problem. What the question (or ones like it) is asking is for a comparision between two ratios that are equal, one of them known and the other one unknown. So, the problem could be thought of as

$\displaystyle \left(\frac{x}{16}\right)=\left(\frac{10.5}{12}\ri ght)$

Now, make the equality an equation equal to $\displaystyle 0$ --

$\displaystyle \left(\frac{x}{16}-\frac{10.5}{12}\right)=0$

Then "cross multiply" so that --

$\displaystyle {12}{x}={168}$

$\displaystyle {x}={168}$ divided by $\displaystyle {12}={14}$

Here is another examlple --

A 90 gal. container is filled with 22.5 gal. of fluid. How much fluid must be put into a 150 gal. container so that the percentage of fluid in both is equal?

$\displaystyle \left(\frac{x}{150}\right)=\left(\frac{22.5}{90}\r ight)$

$\displaystyle {90}{x}={3375}$

$\displaystyle {x}={3375}$ divided by $\displaystyle {90}={37.5}$

Also, a tip (which you may or may not already know): If you want to print a thread so that your daughter can review, if you click on the "Thread Tools" button it will show, among other things, "Printable Version." If you click on that it will print just the "meat" of the posts, without the advertising in between them.