Difference between revisions of "1995 AJHSME Problems/Problem 5"
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Adding the whole numbers gives <math>2 + 3 + 4 + 5 = 14</math>. | Adding the whole numbers gives <math>2 + 3 + 4 + 5 = 14</math>. | ||
− | Adding the fractions gives <math>\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}</math>. This will create one more whole, and a fraction that is less than <math>1</math>. Thus, the | + | Adding the fractions gives <math>\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}</math>. This will create one more whole, and a fraction that is less than <math>1</math>. Thus, the smallest whole number that is less than <math>15</math> plus some fractional part is <math>16</math>, and the answer is <math>\boxed{C}</math>. |
===Solution 2=== | ===Solution 2=== | ||
− | 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>. |
==See Also== | ==See Also== |
Revision as of 21:40, 26 August 2017
Problem
Find the smallest whole number that is larger than the sum
Solution
Solution 1
Adding the whole numbers gives .
Adding the fractions gives . This will create one more whole, and a fraction that is less than . Thus, the smallest whole number that is less than plus some fractional part is , and the answer is .
Solution 2
Convert the fractional parts to decimals, and approximate the answer. , and the answer is , which is .
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.