Difference between revisions of "1951 AHSME Problems/Problem 7"

Revision as of 20:11, 10 April 2013

Problem

An error of $.02''$ is made in the measurement of a line $10''$ long, while an error of only $.2''$ is made in a measurement of a line $100''$ long. In comparison with the relative error of the first measurement, the relative error of the second measurement is:

$\textbf{(A)}\ \text{greater by }.18 \qquad\textbf{(B)}\ \text{the same} \qquad\textbf{(C)}\ \text{less}$ $\textbf{(D)}\ 10 \text{ times as great} \qquad\textbf{(E)}\ \text{correctly described by both (A) and (D)}$

Solution

There error percentage of the first measure is $0.2\%$ . The error percentage of the second measure is also $0.2\%$ . Therefore, the answer is $\textbf{(B)}\ \text{the same}$

1951 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 6	Followed by Problem 8
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All AHSME Problems and Solutions

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 == Solution ==
-{{solution}}
+There error percentage of the first measure is <math>0.2\%</math>. The error percentage of the second measure is also <math>0.2\%</math>. Therefore, the answer is <math>\textbf{(B)}\ \text{the same} </math>
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Difference between revisions of "1951 AHSME Problems/Problem 7"

Revision as of 20:11, 10 April 2013

Problem

Solution

See Also