Difference between revisions of "1985 AJHSME Problems/Problem 21"
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== Solution 2 == | == Solution 2 == | ||
| + | Let Mr.Green's salary be <math>x</math> dollars. | ||
| + | Notice Mr.Green's salary is a geometric sequence with common ratio <math>\frac{11}{10}</math> | ||
| + | |||
| + | Apply the general term formula for geometric sequences, after four years, Mr.Green's salary becomes <math>x \cdot (\frac{11}{10})^4</math> which is <math>\frac{14641}{10000}x</math>. | ||
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| + | <math>\frac{4641}{10000}</math> is more than <math>45\%</math>, thus the answer is <math>\boxed{\textbf{(E)}\text{more than } 45\%}</math> | ||
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| + | ~ lovelearning999 | ||
==See Also== | ==See Also== | ||
Latest revision as of 15:19, 6 October 2024
Contents
Problem
Mr. Green receives a 
raise every year. His salary after four such raises has gone up by what percent?
Solution
Assume his salary is originally 
dollars. Then, in the next year, he would have 
dollars, and in the next, he would have 
dollars. The next year he would have 
dollars and in the final year, he would have 
. As the total increase is greater than 
, the answer is 
.
~sakshamsethi
Solution 2
Let Mr.Green's salary be 
dollars.
Notice Mr.Green's salary is a geometric sequence with common ratio 
Apply the general term formula for geometric sequences, after four years, Mr.Green's salary becomes 
which is 
.
is more than , thus the answer is 
~ lovelearning999
See Also
| 1985 AJHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 20 |
Followed by Problem 22 | |
| 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. 
