Difference between revisions of "1997 PMWC Problems/Problem I1"
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Evaluate <math>29 \frac{27}{28} \times 27\frac{14}{15}</math>. | Evaluate <math>29 \frac{27}{28} \times 27\frac{14}{15}</math>. | ||
| − | == Solution == | + | == Solution 1 == |
| − | <math>\frac{29 \cdot 28 + 27}{28} \cdot \frac{27 \cdot 15 + 14}{15} = \frac{839 \cdot 419}{420} = \frac{839(420 - 1)}{420} = 839 - \frac{839}{420} = 837 \ | + | <math>\frac{29 \cdot 28 + 27}{28} \cdot \frac{27 \cdot 15 + 14}{15} = \frac{839 \cdot 419}{420} = \frac{839(420 - 1)}{420} = 839 - \frac{839}{420} = 837 \tfrac{1}{420}</math>. |
| + | == Solution 2 == | ||
| + | <math>(30-x)(28-y)=840-28x-30y+xy</math>, so for <math>x=\tfrac1{28},\ y=\tfrac1{15}</math> the result is <math>840-1-2+\tfrac1{420}=837\tfrac1{420}</math>. | ||
| − | == See | + | == See Also == |
{{PMWC box|year=1997|before=First question|num-a=I2}} | {{PMWC box|year=1997|before=First question|num-a=I2}} | ||
[[Category:Introductory Algebra Problems]] | [[Category:Introductory Algebra Problems]] | ||
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