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Difference between revisions of "1995 AJHSME Problems/Problem 5"
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Convert the fractional parts to decimals, and approximate the answer.  <math>2.5 + 3.33 + 4.25 + 5.2 = 15.28</math>, and the answer is <math>16</math>, which is <math>\boxed{C}</math>.
 
Convert the fractional parts to decimals, and approximate the answer.  <math>2.5 + 3.33 + 4.25 + 5.2 = 15.28</math>, and the answer is <math>16</math>, which is <math>\boxed{C}</math>.
  
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==Solution 3 (Speedy)==
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===Solution 3 (Speedy)===
 
Clearly, break the mixed numbers into <math>2+3+4+5+\frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5}</math>.  
 
Clearly, break the mixed numbers into <math>2+3+4+5+\frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5}</math>.  
  

Latest revision as of 12:06, 1 April 2022

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

See Also

1995 AJHSME (Problems • Answer Key • Resources)
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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