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Difference between revisions of "1955 AHSME Problems/Problem 15"
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If we plug in <math>r</math> for the smaller radius, we get <math>r:\sqrt{3}r</math> The difference between the two sides would be (approximately) <math>1.732r - r = \textbf{(D)} 0.73r</math>
 
If we plug in <math>r</math> for the smaller radius, we get <math>r:\sqrt{3}r</math> The difference between the two sides would be (approximately) <math>1.732r - r = \textbf{(D)} 0.73r</math>
 
== See Also ==
 
== See Also ==
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{{AHSME box|year=1955|num-b=14|num-a=16}}
  
 
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Latest revision as of 09:44, 15 February 2021

Problem 15

The ratio of the areas of two concentric circles is $1: 3$. If the radius of the smaller is $r$, then the difference between the radii is best approximated by:

$\textbf{(A)}\ 0.41r \qquad \textbf{(B)}\ 0.73 \qquad \textbf{(C)}\ 0.75 \qquad \textbf{(D)}\ 0.73r \qquad \textbf{(E)}\ 0.75r$

Solution

Observe that there are 3 feet in 1 yard, 9 square feet in 1 square yard, and 27 cubic feet in 1 cubic yard.

This means that if the ratio is $x$ in 1D, it is $x^2$ in 2D, and it is $x^3$ in 3D.

We can apply this thinking into our current question, which would have $1:3$ in the 2D-plane. Since the radius is 1D, we would have to square root both sides to get $1:\sqrt{3}$

If we plug in $r$ for the smaller radius, we get $r:\sqrt{3}r$ The difference between the two sides would be (approximately) $1.732r - r = \textbf{(D)} 0.73r$

See Also

1955 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 14 		Followed by
Problem 16
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30
All AHSME Problems and Solutions


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