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...) |
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==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> | ||
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==See Also== | ==See Also== | ||
− | [[ | + | {{AJHSME box|year=1987|num-b=10|num-a=12}} |
+ | [[Category:Introductory Algebra Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 23:52, 4 July 2013
Problem
The sum is between
Solution
Since and
,
Clearly,
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 |
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