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Difference between revisions of "1985 AJHSME Problems/Problem 21"

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==Solution==
 
==Solution==
  
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If he gets a <math>10\% </math> raise, then his new salary is <math>1.1</math> of what it was before. He gets <math>4</math> raises, so his new salary is <math>1.1^4</math> of the previous salary. We want to know the percent of increase from <math>1</math>, so we just need <math>1.1^4-1</math> as a percent.
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Assume his salary is originally <math>100</math> dollars. Then, in the next year, he would have <math>110</math> dollars, and in the next, he would have <math>121</math> dollars. The next year he would have <math>133.1</math> dollars and in the final year, he would have <math>146.41</math>. As the total increase is greater than <math>45\%</math>, the answer is <math>\boxed{\text{E}}</math>.  
  
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If you know your powers of <math>11</math> well, then you can just do <math>1.4641-1=.4641>45\% </math>, so the answer is <math>\boxed{\text{E}}</math>
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~sakshamsethi
  
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If you don't know this, then first we note that <math>1.1^4</math> clearly does not have a whole number percent representation, so the only possible choices are <math>\text{A}</math> and <math>\text{E}</math>. We also have <cmath>1.1^4=1.21^2>1.2^2=1.44</cmath>
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== Solution 2 ==
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Let Mr.Green's salary be <math>x</math> dollars.
  
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so the percent is greater than <math>40\% </math> and the only choice left is <math>\boxed{\text{E}}</math>.
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Notice Mr.Green's salary is a geometric sequence with common ratio <math>\frac{11}{10}</math>
  
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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

Problem

Mr. Green receives a $10\%$ raise every year. His salary after four such raises has gone up by what percent?

$\text{(A)}\ \text{less than }40\% \qquad \text{(B)}\ 40\% \qquad \text{(C)}\ 44\% \qquad \text{(D)}\ 45\% \qquad \text{(E)}\ \text{more than }45\%$

Solution

Assume his salary is originally $100$ dollars. Then, in the next year, he would have $110$ dollars, and in the next, he would have $121$ dollars. The next year he would have $133.1$ dollars and in the final year, he would have $146.41$. As the total increase is greater than $45\%$, the answer is $\boxed{\text{E}}$.

~sakshamsethi

Solution 2

Let Mr.Green's salary be $x$ dollars.

Notice Mr.Green's salary is a geometric sequence with common ratio $\frac{11}{10}$

Apply the general term formula for geometric sequences, after four years, Mr.Green's salary becomes $x \cdot (\frac{11}{10})^4$ which is $\frac{14641}{10000}x$.

$\frac{4641}{10000}$ is more than $45\%$, thus the answer is $\boxed{\textbf{(E)}\text{more than } 45\%}$

~ 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


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