1995 AJHSME Problems/Problem 5

Problem

Find the smallest whole number that is larger than the sum

$2\dfrac{1}{2}+3\dfrac{1}{3}+4\dfrac{1}{4}+5\dfrac{1}{5}.$

$\text{(A)}\ 14 \qquad \text{(B)}\ 15 \qquad \text{(C)}\ 16 \qquad \text{(D)}\ 17 \qquad \text{(E)}\ 18$

Solution

Solution 1

Adding the whole numbers gives $2 + 3 + 4 + 5 = 14$ .

Adding the fractions gives $\frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} = \frac{30}{60} + \frac{20}{60} + \frac{15}{60} + \frac{12}{60} = \frac{77}{60}$ . This will create one more whole, and a fraction that is less than $1$ . Thus, the smallest whole number that is less than $15$ plus some fractional part is $16$ , and the answer is $\boxed{C}$ .

Solution 2

Convert the fractional parts to decimals, and approximate the answer. $2.5 + 3.33 + 4.25 + 5.2 = 15.28$ , and the answer is $16$ , which is $\boxed{C}$ .

Solution 3 (Speedy)

Clearly, break the mixed numbers into $2+3+4+5+\frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5}$ .

This is the same as $14+(\frac{1}{2} + \frac{1}{3}) + (\frac{1}{4} + \frac{1}{5})$ . Notice that clearly, $\frac{1}{2}+\frac{1}{3}=\frac{5}{6}$ , and $1>\frac{1}{4}+\frac{1}{5}>\frac{1}{6}$ (clearly, $\frac{1}{4}>\frac{1}{6}$ ). This means that the fractional parts is between 1 and 2, so when that is added to 14, the original expression is between 15, and 16. So the least integer that is bigger than this sum is clearly 16. So select $\boxed{C}$ /

~hastapasta

1995 AJHSME (Problems • Answer Key • Resources)
Preceded by Problem 4	Followed by Problem 6
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All AJHSME/AMC 8 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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