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Difference between revisions of "1959 AHSME Problems/Problem 1"

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==Solution==
 
==Solution==
 
Note that increasing the length of each edge by <math>50</math>% with result in a cube that is similar to the original cube with scale factor <math>1.5</math>. Therefore, the surface area will increase by a factor of <math>1.5^2</math>, or <math>2.25</math>. Converting this back into a percent, the percent increase will be <math>125</math>%. Therefore, the answer is <math>\boxed{\textbf{B}}</math>.
 
Note that increasing the length of each edge by <math>50</math>% with result in a cube that is similar to the original cube with scale factor <math>1.5</math>. Therefore, the surface area will increase by a factor of <math>1.5^2</math>, or <math>2.25</math>. Converting this back into a percent, the percent increase will be <math>125</math>%. Therefore, the answer is <math>\boxed{\textbf{B}}</math>.
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== See also ==
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{{AHSME 50p box|year=1959|before=First Question|num-a=2}}
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{{MAA Notice}}

Latest revision as of 21:56, 25 February 2018

Problem 1

Each edge of a cube is increased by $50$%. The percent of increase of the surface area of the cube is: $\textbf{(A)}\ 50 \qquad\textbf{(B)}\ 125\qquad\textbf{(C)}\ 150\qquad\textbf{(D)}\ 300\qquad\textbf{(E)}\ 750$

Solution

Note that increasing the length of each edge by $50$% with result in a cube that is similar to the original cube with scale factor $1.5$. Therefore, the surface area will increase by a factor of $1.5^2$, or $2.25$. Converting this back into a percent, the percent increase will be $125$%. Therefore, the answer is $\boxed{\textbf{B}}$.

See also

1959 AHSC (Problems • Answer Key • Resources)
Preceded by
First Question 		Followed by
Problem 2
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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