Movie Earnings and Relative Ticket Sizes

Question

Given that movie A made 308 million and a 1 by 3 cm ticket represented A and movie B made 609 million, assuming that prices have been adjusted for inflation and ticket B is the same shape as ticket A, what would the size of ticket B be?

I said if the area of A was 3 cm which represented 308 million than ticket B would just need to be about 1.97 $cm^2$ larger i.e. have an area of about 5.91. I got this because since its a graphical representation then ticket B should be proportional to ticket A. Not really sure if it's right so if anyone would check it out and let me know, thanks!

Answer 1

Making the areas proportional, as you've suggested, is the most reasonable way to represent the data; however I've also seen in some places scaling just one axis proportionally. If you did this, you would make the long side $3\times \frac{609}{308}=5.93$cm, and the other side $1\times \frac{609}{309}=1.98$cm.

Answer 2

Movie A: $308$ Units, $1cm$ by $3$$cm$: Resultant Area: $3cm^2$
Movie A: $609$ Units
$\frac{609}{308} = 1.977$
$\therefore$ $Area\;B = 1.977 \cdot Area\;A$
$\therefore$ $Area\;B = 5.931$

You can then work out the dimensions of $Area\;B$. You know that the rectangles are proportional ("same shape"), so one side is three times as long as the other. $x$ will be the length of one side, so $3x$ will be the length of another.

$x \cdot 3x = 5.931$
$3x^2 = 5.931$
$x = \sqrt{1.977}$
$x = \sqrt{1.977}$
$x = 1.4061$
$3x = 4.218$

$\therefore$ the dimensions of Ticket B will be $1.4061cm$ by $4.218cm$.

Graphical Representation in this way is a weird idea, but that should be right.

Answer 3

As you say, the ratio of areas should be $\frac {609}{308}\approx 1.977$. Note that this ratio has no units: both ticket areas have the same unit: cm$^2$ so the ratio has none. To get the dimensions for ticket B, you need to multiply each dimension of ticket A by the square root of this, $\sqrt {1.977}\approx 1.406$ so the ticket size is $1.406 \times 4.218$ cm.