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 $\boxed{C}$.

See Also

1995 AJHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 4
Followed by
Problem 6
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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