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> | + | 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 | ||
Line 10: | Line 10: | ||
== Solution == | == Solution == | ||
− | <math>\fbox{D}</math> | + | <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> |
== See also == | == See also == |
Revision as of 17:24, 23 February 2018
Problem
From time to time a population increased by , and from time to time the population increased by . Therefore, from time to time the population increased by
Solution
The scale factors for the increases are and , so the overall scale factor is . To convert this to a percentage, we subtract 1 and then multiply by 100, giving
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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