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Difference between revisions of "1951 AHSME Problems/Problem 10"

(Created page with "== Problem == Of the following statements, the one that is incorrect is: <math> \textbf{(A)}\ \text{Doubling the base of a given rectangle doubles the area.}</math> <math> \text...")
 
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== Solution ==  
 
== Solution ==  
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The well-known area formula for a circle is <math>A = \pi r^2</math>, so doubling the radius will result it quadrupling the area (since <math>A' = \pi (2r)^2 = 4 \pi r^2 = 4A</math>). Statement <math>\boxed{\textbf{(C)}}</math> is therefore incorrect, and is the correct answer choice.
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Statements <math>\textbf{(A)}</math> and <math>\textbf{(B)}</math> are evidently correct, since in triangles the area is directly proportional to both the base and the height (<math>A = \frac{1}{2} bh</math>). Statement <math>\textbf{(D)}</math> is also correct: let <math>q = \frac{a}{b}</math>. Then <math>q' = \frac{a \div 2}{2b} = \frac{a}{4b} = \frac{q}{4}</math>, which is a change from <math>q</math>. Finally, statement <math>\textbf{(E)}</math> is true: <math>2x < x</math> if <math>x < 0</math>, as doubling a negative number makes it even more negative (and therefore less than it originally was).
  
 
== See Also ==
 
== See Also ==

Revision as of 23:07, 23 February 2013

Problem

Of the following statements, the one that is incorrect is:

$\textbf{(A)}\ \text{Doubling the base of a given rectangle doubles the area.}$ $\textbf{(B)}\ \text{Doubling the altitude of a triangle doubles the area.}$ $\textbf{(C)}\ \text{Doubling the radius of a given circle doubles the area.}$ $\textbf{(D)}\ \text{Doubling the divisor of a fraction and dividing its numerator by 2 changes the quotient.}$ $\textbf{(E)}\ \text{Doubling a given quantity may make it less than it originally was.}$

Solution

The well-known area formula for a circle is $A = \pi r^2$, so doubling the radius will result it quadrupling the area (since $A' = \pi (2r)^2 = 4 \pi r^2 = 4A$). Statement $\boxed{\textbf{(C)}}$ is therefore incorrect, and is the correct answer choice.


Statements $\textbf{(A)}$ and $\textbf{(B)}$ are evidently correct, since in triangles the area is directly proportional to both the base and the height ($A = \frac{1}{2} bh$). Statement $\textbf{(D)}$ is also correct: let $q = \frac{a}{b}$. Then $q' = \frac{a \div 2}{2b} = \frac{a}{4b} = \frac{q}{4}$, which is a change from $q$. Finally, statement $\textbf{(E)}$ is true: $2x < x$ if $x < 0$, as doubling a negative number makes it even more negative (and therefore less than it originally was).

See Also

1951 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 9 		Followed by
Problem 11
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 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50
All AHSME Problems and Solutions
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