Difference between revisions of "1950 AHSME Problems/Problem 34"
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==Solution 2== | ==Solution 2== | ||
The radii of the circles are <math>\frac{20}{2\pi}</math> and <math>\frac{25}{2\pi}</math>, respectively. The positive difference is therefore <math>\boxed{\textbf{(D)}\ \dfrac{5}{2\pi}\text{ in}}</math>. | The radii of the circles are <math>\frac{20}{2\pi}</math> and <math>\frac{25}{2\pi}</math>, respectively. The positive difference is therefore <math>\boxed{\textbf{(D)}\ \dfrac{5}{2\pi}\text{ in}}</math>. | ||
+ | |||
+ | ==Solution 3== | ||
+ | let the radius of the circle with the larger circumference be <math>r_2</math> and the circle with the smaller circumference be <math>r_1</math>; calculating the ratio of the two | ||
+ | <cmath>\frac{r_2}{r_1}=\frac{25}{20}=\frac{5}{4}</cmath> | ||
+ | <cmath>4r_2=5r_1</cmath> | ||
+ | <cmath>4(r_2-r_1)=r_1</cmath> | ||
+ | <cmath>r_2-r_1=\frac{r_1}{4}=\frac{\frac{20}{2\pi}}{4}=\frac{10}{4\pi}=\frac{5}{2\pi}</cmath> | ||
==See Also== | ==See Also== |
Revision as of 03:39, 10 June 2020
Contents
[hide]Problem
When the circumference of a toy balloon is increased from inches to inches, the radius is increased by:
Solution
When the circumference of a circle is increased by a percentage, the radius is also increased by the same percentage (or else the ratio of the circumference to the diameter wouldn't be anymore) We see that the circumference was increased by . This means the radius was also increased by . The radius of the original balloon is . With the increase, it becomes . The increase is
Solution 2
The radii of the circles are and , respectively. The positive difference is therefore .
Solution 3
let the radius of the circle with the larger circumference be and the circle with the smaller circumference be ; calculating the ratio of the two
See Also
1950 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 33 |
Followed by Problem 35 | |
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All AHSME Problems and Solutions |
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