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 ==
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== 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 \frac{1}{420}</math>
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<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>.
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== Solution 2 ==
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<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 Also ==
 
== See Also ==

Latest revision as of 10:47, 22 May 2014

Problem

Evaluate $29 \frac{27}{28} \times 27\frac{14}{15}$.

Solution 1

$\frac{29 \cdot 28 + 27}{28} \cdot \frac{27 \cdot 15 + 14}{15} = \frac{839 \cdot 419}{420} = \frac{839(420 - 1)}{420} = 839 - \frac{839}{420} = 837 \tfrac{1}{420}$.

Solution 2

$(30-x)(28-y)=840-28x-30y+xy$, so for $x=\tfrac1{28},\ y=\tfrac1{15}$ the result is $840-1-2+\tfrac1{420}=837\tfrac1{420}$.

See Also

1997 PMWC (Problems)
Preceded by
First question
Followed by
Problem I2
I: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
T: 1 2 3 4 5 6 7 8 9 10