Difference between revisions of "1955 AHSME Problems/Problem 15"

(Created page with "== Problem 15== The ratio of the areas of two concentric circles is <math>1: 3</math>. If the radius of the smaller is <math>r</math>, then the difference between the radii...")
 
Line 13: Line 13:
 
We can apply this thinking into our current question, which would have <math>1:3</math> in the 2D-plane. Since the radius is 1D, we would have to square root both sides to get <math>1:\sqrt{3}</math>
 
We can apply this thinking into our current question, which would have <math>1:3</math> in the 2D-plane. Since the radius is 1D, we would have to square root both sides to get <math>1:\sqrt{3}</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) $1.732r - r \textbf{(D)} 0.73r
+
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 ==
 +
{{AHSME box|year=1955|num-b=3|num-a=5}}
 +
 
 +
{{MAA Notice}}

Revision as of 22:33, 5 August 2020

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 (ProblemsAnswer KeyResources)
Preceded by
Problem 3
Followed by
Problem 5
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


The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png