parks is wearing several rubber bracelets. 1/3 of the braclets are tie-dye,1/6 are blue, and 1/3 of the remainder are camouflage. if parks wears 2 camouflage bracelets, how many bracelets does he have on?

[tex]\frac{1}{3} = \frac{2}{6} [/tex] of the bracelets are tie-dye, that is [tex]\boxed{\cdot} \boxed{\cdot} [/tex] or 2 of 6 unit.
[tex]\frac{1}{6}[/tex] are blue, that is [tex]\boxed{\cdot} [/tex] or 1 of 6 unit.

The rest is 6 - 2 - 1 = 3 units.

[tex]\frac{1}{3}[/tex] of the remainder are camouflage, that is [tex]\boxed{ \ \frac{1}{3} \times 3 \ units = 1 \ unit \ }[/tex]

Parks wears 2 camouflage bracelets.

[tex]\boxed{ \ 1 \ unit = 2 \ camouflage \ bracelets \ }[/tex]

Let us count how many bracelets does he have on.

[tex]\boxed{ \ 6 \ unit = \ ? \ camouflage \ bracelets \ }[/tex]

[tex]\boxed{ \ 6 \times 2 = 12 \ }[/tex]

Thus, Parks has 12 bracelets on.

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Quick Steps

Let the sum of all bracelets be N.

Parks wears 2 camouflage bracelets, that is,

[tex]\boxed{ \ \frac{1}{3} \times \bigg( 1 - \frac{1}{3} - \frac{1}{6} \bigg) \times N = 2 \ }[/tex]

[tex]\boxed{ \ \frac{1}{3} \times \bigg( \frac{6}{6} - \frac{2}{6} - \frac{1}{6} \bigg) \times N = 2 \ }[/tex]

[tex]\boxed{ \ \frac{1}{3} \times \frac{3}{6} \times N = 2 \ }[/tex]

[tex]\boxed{ \ \frac{1}{6} \times N = 2 \ }[/tex]

[tex]\boxed{ \ N = 2 \div \frac{1}{6} \ }[/tex]

[tex]\boxed{ \ N = 2 \times \frac{6}{1} \ }[/tex]

[tex]\boxed{\boxed{ \ N = 12 \ }}[/tex]

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