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1989 AJHSME Problems/Problem 4: Difference between revisions

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<math>\text{(A)}\ .2 \qquad \text{(B)}\ 2 \qquad \text{(C)}\ 20 \qquad \text{(D)}\ 200 \qquad \text{(E)}\ 2000</math>
<math>\text{(A)}\ .2 \qquad \text{(B)}\ 2 \qquad \text{(C)}\ 20 \qquad \text{(D)}\ 200 \qquad \text{(E)}\ 2000</math>


==Solution==
==Solution 1==


<math>401</math> is around <math>400</math> and <math>.205</math> is around <math>.2 </math> so the [[fraction]] is approximately <cmath>\frac{400}{.2}=2000\rightarrow \boxed{\text{E}}</cmath>
<math>401</math> is around <math>400</math> and <math>.205</math> is around <math>.2 </math> so the [[fraction]] is approximately <cmath>\frac{400}{.2}=2000\rightarrow \boxed{\text{E}}</cmath>
==Solution 2==
Using scientific notation estimated to the first significant digit, we see the numerator rounds to <math>\frac{4*10^2}{2*10^-1}</math>.
Subtracting the exponents gives <math>2-(-1)=3</math>. The answer choice that has <math>3</math> zeros to the right of the significant digit is <math>\boxed{\text{E}}</math>


==See Also==
==See Also==

Revision as of 00:19, 28 October 2024

Problem

Estimate to determine which of the following numbers is closest to $\frac{401}{.205}$.

$\text{(A)}\ .2 \qquad \text{(B)}\ 2 \qquad \text{(C)}\ 20 \qquad \text{(D)}\ 200 \qquad \text{(E)}\ 2000$

Solution 1

$401$ is around $400$ and $.205$ is around $.2$ so the fraction is approximately \[\frac{400}{.2}=2000\rightarrow \boxed{\text{E}}\]

Solution 2

Using scientific notation estimated to the first significant digit, we see the numerator rounds to $\frac{4*10^2}{2*10^-1}$. Subtracting the exponents gives $2-(-1)=3$. The answer choice that has $3$ zeros to the right of the significant digit is $\boxed{\text{E}}$


See Also

1989 AJHSME (Problems • Answer Key • Resources)
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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