Question-107122

Question Number 107122 by aurpeyz last updated on 08/Aug/20

Answered by 1549442205PVT last updated on 09/Aug/20

Denote a,b,c be number of persons get medal in only one category dance,dramatics,music respectively m−be number of persons got exactly medal in two categories dance and dramatics n−in two categories dance and music p−in two categories dramatics and music Then from the hypothesis we have: { ((a+b+c+m+n+p+4=45(1))),((a+m+n+4=36 (2))),((b+m+p+4=12 (3))),((c+n+p+4=18 (4))) :} Adding up three equations (2)(3)(4) and simplification we get a+b+c+2(m+n+p)=54(5) from (1)we have a+b+c+m+n+p=41(6) Substracting (6) from (5) we get m+n+p=13 which is realy number of persons got exactly medals in two categories.Thus,we choose answer D:13

$$\mathrm{Denote}\:\mathrm{a},\mathrm{b},\mathrm{c}\:\mathrm{be}\:\mathrm{number}\:\mathrm{of}\:\mathrm{persons}\:\mathrm{get} \\ $$$$\mathrm{medal}\:\mathrm{in}\:\mathrm{only}\:\mathrm{one}\:\mathrm{category} \\ $$$$\mathrm{dance},\mathrm{dramatics},\mathrm{music}\:\mathrm{respectively} \\ $$$$\mathrm{m}−\mathrm{be}\:\mathrm{number}\:\mathrm{of}\:\mathrm{persons}\:\mathrm{got} \\ $$$$\mathrm{exactly}\:\mathrm{medal}\:\mathrm{in}\:\mathrm{two}\:\mathrm{categories}\:\mathrm{dance} \\ $$$$\mathrm{and}\:\mathrm{dramatics} \\ $$$$\mathrm{n}−\mathrm{in}\:\mathrm{two}\:\mathrm{categories}\:\mathrm{dance}\:\mathrm{and}\:\mathrm{music} \\ $$$$\mathrm{p}−\mathrm{in}\:\mathrm{two}\:\mathrm{categories}\:\mathrm{dramatics}\:\:\mathrm{and}\:\mathrm{music} \\ $$$$\mathrm{Then}\:\mathrm{from}\:\mathrm{the}\:\mathrm{hypothesis}\:\mathrm{we}\:\mathrm{have}: \\ $$$$\begin{cases}{\mathrm{a}+\mathrm{b}+\mathrm{c}+\mathrm{m}+\mathrm{n}+\mathrm{p}+\mathrm{4}=\mathrm{45}\left(\mathrm{1}\right)}\\{\mathrm{a}+\mathrm{m}+\mathrm{n}+\mathrm{4}=\mathrm{36}\:\left(\mathrm{2}\right)}\\{\mathrm{b}+\mathrm{m}+\mathrm{p}+\mathrm{4}=\mathrm{12}\:\left(\mathrm{3}\right)}\\{\mathrm{c}+\mathrm{n}+\mathrm{p}+\mathrm{4}=\mathrm{18}\:\left(\mathrm{4}\right)}\end{cases} \\ $$$$\mathrm{Adding}\:\mathrm{up}\:\mathrm{three}\:\mathrm{equations}\:\left(\mathrm{2}\right)\left(\mathrm{3}\right)\left(\mathrm{4}\right) \\ $$$$\mathrm{and}\:\mathrm{simplification}\:\mathrm{we}\:\mathrm{get} \\ $$$$\mathrm{a}+\mathrm{b}+\mathrm{c}+\mathrm{2}\left(\mathrm{m}+\mathrm{n}+\mathrm{p}\right)=\mathrm{54}\left(\mathrm{5}\right) \\ $$$$\mathrm{from}\:\left(\mathrm{1}\right)\mathrm{we}\:\mathrm{have} \\ $$$$\mathrm{a}+\mathrm{b}+\mathrm{c}+\mathrm{m}+\mathrm{n}+\mathrm{p}=\mathrm{41}\left(\mathrm{6}\right) \\ $$$$\mathrm{Substracting}\:\left(\mathrm{6}\right)\:\mathrm{from}\:\left(\mathrm{5}\right)\:\mathrm{we}\:\mathrm{get} \\ $$$$\mathrm{m}+\mathrm{n}+\mathrm{p}=\mathrm{13}\:\mathrm{which}\:\mathrm{is}\:\mathrm{realy}\:\mathrm{number}\:\mathrm{of} \\ $$$$\mathrm{persons}\:\mathrm{got}\:\mathrm{exactly}\:\mathrm{medals}\:\mathrm{in}\:\mathrm{two}\: \\ $$$$\mathrm{categories}.\mathrm{Thus},\mathrm{we}\:\mathrm{choose}\:\mathrm{answer} \\ $$$$\mathrm{D}:\mathrm{13} \\ $$

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