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

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== Problem ==
 
== Problem ==
  
From time <math>t=0</math> to time <math>t=1</math> a population increased by <math>I\%</math>, and from time <math>t=1</math> to time <math>t=2</math> the population increased by <math>j\%</math>. Therefore, from time <math>t=0</math> to time <math>t=2</math> the population increased by
+
From time <math>t=0</math> to time <math>t=1</math> a population increased by <math>i\%</math>, and from time <math>t=1</math> to time <math>t=2</math> the population increased by <math>j\%</math>. Therefore, from time <math>t=0</math> to time <math>t=2</math> the population increased by
  
 
<math>\text{(A) (i+j)\%} \quad
 
<math>\text{(A) (i+j)\%} \quad

Revision as of 16:26, 28 September 2014

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

$\fbox{D}$

See also

1991 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 8
Followed by
Problem 10
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All AHSME Problems and Solutions

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