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Percentage problem. Need Help.

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by aman88 » Tue Dec 18, 2012 1:41 am
Alex is trying to figure out which television to buy. Both televisions have rectangular screen that measure 30 inches on the diagonal. The widescreen TV has an aspect ratio (width/length) of 2:1. The standard TV has an aspect ratio of 4:3. The screen area of widescreen TV is approximately what percent of the screen area of the standard TV?

A: 72
B: 75
C: 80
D: 83
E: 87

Please help me out on this problem. I tried to find their lengths and widths by using the diagonal formula (because the diagonal is same for both) but couldn't find the answer.

Is there some other approach for this Q?

OA: D

Thanks.
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by Sam_hellboy » Tue Dec 18, 2012 3:24 am
Wide screen TV- let length n breadth are 2x & x.
sqroot of ( (2x)^2 + x^2) = 30
by solving this we get x = 6sqrt5
hence length = 12*sqrt5, breadth = 6*sqrt5
Area = 72*5

The standard TV -
let length n breadth are 4y & 3y.
sqroot of ( (4y)^2 + 93y)^2) = 30
by solving this we get y = 6
hence length = 4y = 24, breadth = 3y = 18
Area = 24*18
Now if you calculate % it will come around 83.33%
So ans is option D
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by aman88 » Tue Dec 18, 2012 3:50 am
Right right! Damn! I was doing such a silly mistake.

Thanks man! :)
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