Difference between revisions of "1989 AJHSME Problems/Problem 4"

m (Solution 2)
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==Solution 2==
 
==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>.
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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>
 
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 00:20, 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 (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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