Difference between revisions of "1994 AHSME Problems/Problem 5"
(Created page with "==Problem== Pat intended to multiply a number by <math>6</math> but instead divided by <math>6</math>. Pat then meant to add <math>14</math> but instead subtracted <math>14</math...") |
(→Solution) |
||
(2 intermediate revisions by one other user not shown) | |||
Line 5: | Line 5: | ||
\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== | ||
+ | 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>. | ||
+ | |||
+ | 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>. | ||
+ | |||
+ | --Solution by [http://www.artofproblemsolving.com/Forum/memberlist.php?mode=viewprofile&u=200685 TheMaskedMagician] | ||
+ | |||
+ | ==See Also== | ||
+ | |||
+ | {{AHSME box|year=1994|num-b=4|num-a=6}} | ||
+ | {{MAA Notice}} |
Latest revision as of 16:26, 9 January 2021
Problem
Pat intended to multiply a number by but instead divided by . Pat then meant to add but instead subtracted . After these mistakes, the result was . If the correct operations had been used, the value produced would have been
Solution
We reverse the operations that he did and then use the correct operations. His end result is . Before that, he subtracted which means that his number after the first operation was . He divided by so his number was .
Now, we multiply by to get . Finally, . Since , our answer is .
--Solution by TheMaskedMagician
See Also
1994 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 4 |
Followed by Problem 6 | |
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.