Algebra Puzzle – Challenge 46

Challenge:

If \(\frac{1}{6}\) of a number is \(2 \frac{7}{15}\) less than \(6 \frac{1}{3}\) of that number, \(1 \frac{4}{5} \) of that number is how much less than 4?

A- \(2\frac{4}{15}\)

B- \(\frac{15}{6}\)

C- \(\frac{82}{25}\)

D- \(\frac{33}{15}\)

E- \(\frac{23}{6}\)

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Let x be the number. Then:
\(\frac{1}{6 } x = 6 \frac{1}{3 }x – 2 \frac{7}{15} ⇒ 6 \frac{1}{3 }x – \frac{1}{6 } x = 2 \frac{7}{15 } ⇒ \frac{37}{6 }x = \frac{37}{15 }\)
\(X = \frac{2}{5 }\)
Then:
\(1 \frac{4}{5 } x = \frac{9}{5 } × \frac{2}{5 } = \frac{18}{25 }\)
\(4 – (\frac{18}{25 }) = \frac{37}{25 } =\frac{82}{25 }\)

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