Difference between revisions of "1997 AJHSME Problems/Problem 4"

 
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We need a number of words between <math>30\times150=4500</math> and <math>45\times150=6750</math>.  The only such choice is <math>\boxed{\textbf{(E)}}</math>.
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==Problem==
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Julie is preparing a speech for her class.  Her speech must last between one-half hour and three-quarters of an hour.  The ideal rate of speech is 150 words per minute.  If Julie speaks at the ideal rate, which of the following number of words would be an appropriate length for her speech?
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<math>\text{(A)}\ 2250 \qquad \text{(B)}\ 3000 \qquad \text{(C)}\ 4200 \qquad \text{(D)}\ 4350 \qquad \text{(E)}\ 5650</math>
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==Solution==
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Since there are <math>60</math> minutes in an hour, we need a number of words between <math>(\frac{1}{2}\times 60)\times150=4500</math> and <math>(\frac{3}{4}\times60)\times150=6750</math>.  The only such choice is <math>\boxed{\textbf{(E)}}</math>.
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== See also ==
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{{AJHSME box|year=1997|num-b=3|num-a=5}}
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* [[AJHSME]]
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* [[AJHSME Problems and Solutions]]
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* [[Mathematics competition resources]]
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{{MAA Notice}}

Latest revision as of 00:26, 5 July 2013

Problem

Julie is preparing a speech for her class. Her speech must last between one-half hour and three-quarters of an hour. The ideal rate of speech is 150 words per minute. If Julie speaks at the ideal rate, which of the following number of words would be an appropriate length for her speech?


$\text{(A)}\ 2250 \qquad \text{(B)}\ 3000 \qquad \text{(C)}\ 4200 \qquad \text{(D)}\ 4350 \qquad \text{(E)}\ 5650$


Solution

Since there are $60$ minutes in an hour, we need a number of words between $(\frac{1}{2}\times 60)\times150=4500$ and $(\frac{3}{4}\times60)\times150=6750$. The only such choice is $\boxed{\textbf{(E)}}$.

See also

1997 AJHSME (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
All AJHSME/AMC 8 Problems and Solutions

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