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

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==Solution==
 
==Solution==
  
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Assume his salary is 100 dollars then the next year he would have 110 dollars then the next year he would have 121 dollars then the next year he would have 133.1 dollars so therefore it is A
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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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so the percent is smaller than <math>40\% </math> and the only choice left is <math>\boxed{\text{A}}</math>.
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~sakshamsethi
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It is 4 raises, not 3.
 
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Thus, the correct answer is a 46.41 percent increase, or <math>\boxed{\text{E}}</math>.
 
  
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== Solution 2 ==
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Let Mr.Green's salary be <math>x</math> dollars.
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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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