1987 AJHSME Problems/Problem 11

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) 12:49, 18 June 2025 (EDT)

1987 AJHSME (Problems • Answer Key • Resources)
Preceded by Problem 10	Followed by Problem 12
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1987 AJHSME Problems/Problem 11

Contents

Problem

Solution 1

Solution 2

See Also