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Difference between revisions of "1987 AJHSME Problems/Problem 11"
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<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>
  
−
==Solution==
+
==Solution 1==
  
 
Since <math>\frac{1}{7}<\frac14</math> and <math>\frac{1}{19}<\frac14</math>, <cmath>2\frac17+3\frac12+5\frac{1}{19}<2\frac14+3\frac12+5\frac14 =11</cmath>
 
Since <math>\frac{1}{7}<\frac14</math> and <math>\frac{1}{19}<\frac14</math>, <cmath>2\frac17+3\frac12+5\frac{1}{19}<2\frac14+3\frac12+5\frac14 =11</cmath>
Line 14: Line 14:
  
 
<math>\boxed{\text{B}}</math>
 
<math>\boxed{\text{B}}</math>
 +
==Solution 2==
 +
Using a common denominator, we get:
  
 +
<math>2\frac{1 \times 2 \times 19}{2 \times 7 \times 19} + 3\frac{1 \times 19 \times 7}{2 \times 7 \times 19} + 5\frac{1 \times 2 \times 7}{2 \times 17 \times 19}</math>
 +
 +
<math>= 10\frac {2 \times 19 + 1 \times 19 \times 7 + 2 \times 7}{2 \times 7 \times 19}</math>
 +
 +
Since <math>2 \times 19 + 2 \times 7 = 52</math>, which is clearly less than <math>19 \times 7</math>, but <math>\frac{19 \times 7}{2 \times 7 \ 19} = \frac{1}{2}</math>, the fractional part of the mixed number must be between <math>\frac{1}{2}</math> and <math>1</math>. Thus:
 +
 +
<math>10\frac{1}{2} < 10\frac{2 \times 19 + 1 \times 19 \ times 7 + 2 \times 7}{2 \times 7 \times 19} < 10 + 1 = 11</math>
 +
 +
This gives us <math>\boxed {\textbf {(B) } 10\frac12 \text{ and } 11}</math> as the answer.
 +
 +
[[User:Anabel.disher|Anabel.disher]] ([[User talk:Anabel.disher|talk]]) 12:48, 18 June 2025 (EDT)
 
==See Also==
 
==See Also==
  

Revision as of 12:48, 18 June 2025

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 1

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 $11$.

$\boxed{\text{B}}$

Solution 2

Using a common denominator, we get:

$2\frac{1 \times 2 \times 19}{2 \times 7 \times 19} + 3\frac{1 \times 19 \times 7}{2 \times 7 \times 19} + 5\frac{1 \times 2 \times 7}{2 \times 17 \times 19}$

$= 10\frac {2 \times 19 + 1 \times 19 \times 7 + 2 \times 7}{2 \times 7 \times 19}$

Since $2 \times 19 + 2 \times 7 = 52$, which is clearly less than $19 \times 7$, but $\frac{19 \times 7}{2 \times 7 \ 19} = \frac{1}{2}$, the fractional part of the mixed number must be between $\frac{1}{2}$ and $1$. Thus:

$10\frac{1}{2} < 10\frac{2 \times 19 + 1 \times 19 \ times 7 + 2 \times 7}{2 \times 7 \times 19} < 10 + 1 = 11$

This gives us $\boxed {\textbf {(B) } 10\frac12 \text{ and } 11}$ as the answer.

Anabel.disher (talk) 12:48, 18 June 2025 (EDT)

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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. AMC Logo.png

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