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Difference between revisions of "1987 AJHSME Problems/Problem 11"

(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
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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==
  
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[[1987 AJHSME Problems]]
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{{AJHSME box|year=1987|num-b=10|num-a=12}}
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[[Category:Introductory Algebra Problems]]
 +
{{MAA Notice}}

Latest revision as of 22:52, 4 July 2013

Problem

The sum $2\frac17+3\frac12+5\frac{1}{19}$ is between

$\text{(A)}\ 10\text{ and }10\frac12 \qquad \text{(B)}\ 10\frac12 \text{ and } 11 \qquad \text{(C)}\ 11\text{ and }11\frac12 \qquad \text{(D)}\ 11\frac12 \text{ and }12 \qquad \text{(E)}\ 12\text{ and }12\frac12$

Solution

Since $\frac{1}{7}<\frac14$ and $\frac{1}{19}<\frac14$, \[2\frac17+3\frac12+5\frac{1}{19}<2\frac14+3\frac12+5\frac14 =11\]

Clearly, \[2\frac17+3\frac12+5\frac{1}{19}>2+3\frac12+5=10\frac12\]

Thus, the sum is between $10\frac12$ and $11$.

$\boxed{\text{B}}$

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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