Difference between revisions of "1988 AJHSME Problems/Problem 7"
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We estimate the first thing to be <math>2.5</math>, the second thing to be <math>8</math>, and the third thing to be <math>10</math>. We now have <math>2.5\cdot 8\cdot 10=25\cdot 8=200\Rightarrow \mathrm{(B)}</math>. | We estimate the first thing to be <math>2.5</math>, the second thing to be <math>8</math>, and the third thing to be <math>10</math>. We now have <math>2.5\cdot 8\cdot 10=25\cdot 8=200\Rightarrow \mathrm{(B)}</math>. | ||
| − | Note: The real answer is 200. | + | Note: The real answer is 200.78971902 ~ xxsc |
==See Also== | ==See Also== | ||
Revision as of 13:01, 12 November 2019
Problem
is closest to
Solution
We estimate the first thing to be 
, the second thing to be 
, and the third thing to be 
. We now have 
.
Note: The real answer is 200.78971902 ~ xxsc
See Also
| 1988 AJHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 6 |
Followed by Problem 8 | |
| 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
