Difference between revisions of "1987 AJHSME Problems/Problem 11"
5849206328x (talk | contribs) (New page: ==Problem== The sum <math>2\frac17+3\frac12+5\frac{1}{19}</math> is between <math>\text{(A)}\ 10\text{ and }10\frac12 \qquad \text{(B)}\ 10\frac12 \text{ and } 11 \qquad \text{(C)}\ 11\t...) |
5849206328x (talk | contribs) m |
||
| Line 1: | Line 1: | ||
==Problem== | ==Problem== | ||
| − | The sum <math>2\frac17+3\frac12+5\frac{1}{19}</math> is between | + | The [[sum]] <math>2\frac17+3\frac12+5\frac{1}{19}</math> is between |
<math>\text{(A)}\ 10\text{ and }10\frac12 \qquad \text{(B)}\ 10\frac12 \text{ and } 11 \qquad \text{(C)}\ 11\text{ and }11\frac12 \qquad \text{(D)}\ 11\frac12 \text{ and }12 \qquad \text{(E)}\ 12\text{ and }12\frac12</math> | <math>\text{(A)}\ 10\text{ and }10\frac12 \qquad \text{(B)}\ 10\frac12 \text{ and } 11 \qquad \text{(C)}\ 11\text{ and }11\frac12 \qquad \text{(D)}\ 11\frac12 \text{ and }12 \qquad \text{(E)}\ 12\text{ and }12\frac12</math> | ||
| Line 17: | Line 17: | ||
==See Also== | ==See Also== | ||
| − | [[ | + | {{AJHSME box|year=1987|num-b=10|num-a=12}} |
| + | [[Category:Introductory Algebra Problems]] | ||
Revision as of 14:46, 29 May 2009
Problem
The sum 
is between
Solution
Since 
and 
, ![\[2\frac17+3\frac12+5\frac{1}{19}<2\frac14+3\frac12+5\frac14 =11\]](http://latex.artofproblemsolving.com/5/0/b/50b86140cb45f6c65e2ea10298f1bca02d77069c.png)
Clearly, ![\[2\frac17+3\frac12+5\frac{1}{19}>2+3\frac12+5=10\frac12\]](http://latex.artofproblemsolving.com/7/b/1/7b1235f5f8ab622c6856a48846e5252207e1b322.png)
Thus, the sum is between 
and 
.
See Also
| 1987 AJHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 10 |
Followed by Problem 12 | |
| 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 | ||
