Question Number 119463 by TANMAY PANACEA last updated on 24/Oct/20
$${one}\:{cube}\:{of}\:{side}\:{length}\:\mathrm{3}{cm}\:{is}\:{painted}\:{by}\:{three}\:{colour} \\ $$$${say}\:\left({red}\:{blue}\:{and}\:{black}\right)\:{opposte}\:{face}\:{same}\:{colour} \\ $$$${Now}\:{cube}\:{is}\:{cut}\:{into}\:\mathrm{27}\:{small}\:{cube} \\ $$$$\left.{i}\right){How}\:{many}\:{cube}\:{has}\:{three}\:{face}\:{coloured} \\ $$$$\left.{ii}\right){How}\:{many}\:{cube}\:{two}\:{face}\:{coloured} \\ $$$$\left.{iii}\right){How}\:{many}\:{cube}\:\:{has}\:{no}\:{colour}\:{in}\:{its}\:{face} \\ $$
Commented by TANMAY PANACEA last updated on 25/Oct/20
Commented by TANMAY PANACEA last updated on 25/Oct/20
$$\left.\mathrm{1}\right){three}\:{face}\:{coloured}=\mathrm{8}\:\left({cube}\:{at}\:{corner}\right) \\ $$$$\left.\mathrm{2}\right){see}\:{top}\:{most}\:{layer}\:{and}\:{bottom}\:{most}\:{layer} \\ $$$$\mathrm{4}×\mathrm{2}=\mathrm{8}\:{cube}\:{two}\:{face}\:{coloured}. \\ $$$${middle}\:{layer}\:\mathrm{4}\:{cube}\:{of}\:{corner} \\ $$$${so}\:{total}\:=\mathrm{8}+\mathrm{4}=\mathrm{12} \\ $$$$\left.\mathrm{3}\right){see}\:{topmost}\:{and}\:{bottom}\:{most}\:{face}\:{centred}\:{cube} \\ $$$${one}\:{face}\:{coloured}\:{and}\:{middle}\:{layer}\:\mathrm{4}\:{cube} \\ $$$$=\mathrm{2}+\mathrm{4}=\mathrm{6}\:{cube}\:{one}\:{face}\:{coloured} \\ $$$$\left.\mathrm{4}\right){noface}\:{coloured}=\mathrm{1}\:{cube} \\ $$$$\boldsymbol{{so}} \\ $$$$\boldsymbol{{three}}\left(\boldsymbol{{o}}\mathrm{3}\right)\boldsymbol{{face}}\:\boldsymbol{{coloured}}=\mathrm{8} \\ $$$$\boldsymbol{{two}}\:\boldsymbol{{face}}\:\boldsymbol{{coloured}}=\mathrm{12} \\ $$$$\boldsymbol{{one}}\:\boldsymbol{{face}}\:\boldsymbol{{coloured}}=\mathrm{6} \\ $$$$\boldsymbol{{no}}\:\boldsymbol{{face}}\:\boldsymbol{{coloured}}=\mathrm{1} \\ $$$$\mathrm{8}+\mathrm{12}+\mathrm{6}+\mathrm{1}=\mathrm{27}\: \\ $$
Answered by som(math1967) last updated on 25/Oct/20
$$\left.\mathrm{i}\right)\mathrm{8}\:\mathrm{cubes} \\ $$$$\left.\mathrm{ii}\right)\mathrm{8}\:\mathrm{cubes} \\ $$$$\left.\mathrm{iii}\right)\:\mathrm{1cube} \\ $$$$ \\ $$
Commented by som(math1967) last updated on 25/Oct/20
$$\mathrm{Thik}\:\mathrm{bollam}? \\ $$