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Difference between revisions of "1988 AHSME Problems/Problem 10"

(→‎Solution)
m (Fixed some typos)
 
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If added together, we have:
 
If added together, we have:
 
<cmath>2.43865+0.00312=2.44177</cmath>
 
<cmath>2.43865+0.00312=2.44177</cmath>
−
This rounds to <math>2.4</math>. If they subtracted, we have:
+
This rounds to <math>2.44</math>. If they subtracted, we have:
 
<cmath>2.43865-0.00312=2.43553.</cmath>
 
<cmath>2.43865-0.00312=2.43553.</cmath>
−
This rounds to <math>2.4</math>. Therefore, we have the answer to be <math>\fbox{\textbf{(D)} 2.4}</math>.
+
This rounds to <math>2.44</math>. Therefore, we have the answer to be <math>\fbox{\textbf{(D)} 2.44}</math>.
  
 
== See also ==
 
== See also ==

Latest revision as of 17:11, 26 February 2018

Problem

In an experiment, a scientific constant $C$ is determined to be $2.43865$ with an error of at most $\pm 0.00312$. The experimenter wishes to announce a value for $C$ in which every digit is significant. That is, whatever $C$ is, the announced value must be the correct result when $C$ is rounded to that number of digits. The most accurate value the experimenter can announce for $C$ is

$\textbf{(A)}\ 2\qquad \textbf{(B)}\ 2.4\qquad \textbf{(C)}\ 2.43\qquad \textbf{(D)}\ 2.44\qquad \textbf{(E)}\ 2.439$

Solution

If added together, we have: \[2.43865+0.00312=2.44177\] This rounds to $2.44$. If they subtracted, we have: \[2.43865-0.00312=2.43553.\] This rounds to $2.44$. Therefore, we have the answer to be $\fbox{\textbf{(D)} 2.44}$.

See also

1988 AHSME (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
All AHSME Problems and Solutions

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