Difference between revisions of "1994 AHSME Problems/Problem 5"

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\textbf{(D)}\ \text{between 800 and 1000} \qquad\textbf{(E)}\ \text{greater than 1000}</math>
 
\textbf{(D)}\ \text{between 800 and 1000} \qquad\textbf{(E)}\ \text{greater than 1000}</math>
 
==Solution==
 
==Solution==
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We reverse the operations that he did and then use the correct operations. His end result is <math>16</math>. Before that, he subtracted <math>14</math> which means that his number after the first operation was <math>30</math>. He divided by <math>6</math> so his number was <math>180</math>.
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Now, we multiply <math>180</math> by <math>6</math> to get <math>1080</math>. Finally, <math>1080+14=1094</math>. Since <math>1094>1000</math>, our answer is <math>\boxed{\textbf{(E)}\ \text{greater than 1000}}</math>.
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--Solution by [http://www.artofproblemsolving.com/Forum/memberlist.php?mode=viewprofile&u=200685 TheMaskedMagician]
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==See Also==
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{{AHSME box|year=1994|num-b=4|num-a=6}}
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{{MAA Notice}}

Latest revision as of 17:26, 9 January 2021

Problem

Pat intended to multiply a number by $6$ but instead divided by $6$. Pat then meant to add $14$ but instead subtracted $14$. After these mistakes, the result was $16$. If the correct operations had been used, the value produced would have been

$\textbf{(A)}\ \text{less than 400} \qquad\textbf{(B)}\ \text{between 400 and 600} \qquad\textbf{(C)}\ \text{between 600 and 800} \\ \textbf{(D)}\ \text{between 800 and 1000} \qquad\textbf{(E)}\ \text{greater than 1000}$

Solution

We reverse the operations that he did and then use the correct operations. His end result is $16$. Before that, he subtracted $14$ which means that his number after the first operation was $30$. He divided by $6$ so his number was $180$.

Now, we multiply $180$ by $6$ to get $1080$. Finally, $1080+14=1094$. Since $1094>1000$, our answer is $\boxed{\textbf{(E)}\ \text{greater than 1000}}$.

--Solution by TheMaskedMagician

See Also

1994 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 4
Followed by
Problem 6
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