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Difference between revisions of "1991 AHSME Problems/Problem 9"

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== Solution ==
 
== Solution ==
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<math>\fbox{D}</math> The scale factors for the increases are <math>1+\frac{i}{100}</math> and <math>1+\frac{j}{100}</math>, so the overall scale factor is <math>(1+\frac{i}{100})(1+\frac{j}{100}) = 1 + \frac{i}{100} + \frac{j}{100} + \frac{ij}{100^2}</math>. To convert this to a percentage, we subtract 1 and then multiply by 100, giving <math>i + j + \frac{ij}{100}.</math>
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The scale factors for the increases are <math>1+\frac{i}{100}</math> and <math>1+\frac{j}{100}</math>, so the overall scale factor is <math>(1+\frac{i}{100})(1+\frac{j}{100}) = 1 + \frac{i}{100} + \frac{j}{100} + \frac{ij}{100^2}</math>. To convert this to a percentage, we subtract 1 and then multiply by 100, giving <math>i + j + \frac{ij}{100}</math>, or <math>\boxed{\text{D}}</math>.
  
 
== See also ==
 
== See also ==

Latest revision as of 12:29, 14 March 2023

Problem

From time $t=0$ to time $t=1$ a population increased by $i\%$, and from time $t=1$ to time $t=2$ the population increased by $j\%$. Therefore, from time $t=0$ to time $t=2$ the population increased by

$\text{(A) (i+j)\%} \quad \text{(B) } ij\%\quad \text{(C) } (i+ij)\%\quad \text{(D) } \left(i+j+\frac{ij}{100}\right)\%\quad \text{(E) } \left(i+j+\frac{i+j}{100}\right)\%$

Solution

The scale factors for the increases are $1+\frac{i}{100}$ and $1+\frac{j}{100}$, so the overall scale factor is $(1+\frac{i}{100})(1+\frac{j}{100}) = 1 + \frac{i}{100} + \frac{j}{100} + \frac{ij}{100^2}$. To convert this to a percentage, we subtract 1 and then multiply by 100, giving $i + j + \frac{ij}{100}$, or $\boxed{\text{D}}$.

See also

1991 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 8 		Followed by
Problem 10
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 • 26 • 27 • 28 • 29 • 30
All AHSME Problems and Solutions

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