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Question Number 114236 by Aina Samuel Temidayo last updated on 18/Sep/20

70% of the employees in a multinational corporation have VCD players, 75% have microwave ovens, 80% have ACs and 85% have washing machines. At least what percentage of employees has all four gadgets?

$$\mathrm{70\%}\:\mathrm{of}\:\mathrm{the}\:\mathrm{employees}\:\mathrm{in}\:\mathrm{a} \\ $$$$\mathrm{multinational}\:\mathrm{corporation}\:\mathrm{have}\:\mathrm{VCD} \\ $$$$\mathrm{players},\:\mathrm{75\%}\:\mathrm{have}\:\mathrm{microwave}\:\mathrm{ovens}, \\ $$$$\mathrm{80\%}\:\mathrm{have}\:\mathrm{ACs}\:\mathrm{and}\:\mathrm{85\%}\:\mathrm{have}\:\mathrm{washing} \\ $$$$\mathrm{machines}.\:\mathrm{At}\:\mathrm{least}\:\mathrm{what}\:\mathrm{percentage}\:\mathrm{of} \\ $$$$\mathrm{employees}\:\mathrm{has}\:\mathrm{all}\:\mathrm{four}\:\mathrm{gadgets}? \\ $$

Answered by Dat_Das last updated on 18/Sep/20

10%

$$\mathrm{10\%} \\ $$

Answered by 1549442205PVT last updated on 18/Sep/20

Denote VCD players unit byV M− the microwave ovens A− the ACs W− the washing machines m−number of the employees have only V n−number of the employees have only M p−number of the employees have only A w−number of the employees have only W a−number of the employees have only two gadgets VM b−number of the employees have only two gadgets VA c−number of the employees have only two gadgets VW d−number of the employees have only two gadgets MA e−number of the employees have only two gadgets MW f−number of the employees have only two gadgets AW x−number of the employees have only three gadgets MAV y−number of the employees have only three gadgets VAW z−number of the employees have only three gadgets VMW t−number of the employees have only three gadgets VMA T−number of the employees have all four gadgets. Then from the hypothesis we have m+n+p+q+a+b+c+d+e+f +x+y+z+t+T=100(1) m+a+b+c+z+y+t+T=70(2) n+a+d+e+x+z+t+T=75(3) p+b+d+f+x+y+t+T=80(4) q+c+e+f+x+y+z+T=85(5) Adding up four above equalities we get m+n+p+q+2(a+b+c+d+e+f) +3(x+y+z+t)+4T=310(6) Multiplying two sides of (1) by 3 then substracting from (6)we get −2(m+n+p+q)−(a+b+c+d+e+f)+T=10 ⇒T=10+2(m+n+p+q)+(a+b+c+d+e+f) ⇒T≥10(%) The equality ocurrs if and only if m=n=p=q=a=b=c=d=e=f=0 which means nobody has only one gadgets and also nobody has only two gadgets all have at least three gadgets up Answer :at least 10% of the employees have all four gadgets

$$\mathrm{Denote}\:\:\mathrm{VCD}\:\mathrm{players}\:\mathrm{unit}\:\mathrm{byV} \\ $$$$\mathrm{M}−\:\:\mathrm{the}\:\mathrm{microwave}\:\mathrm{ovens} \\ $$$$\mathrm{A}−\:\mathrm{the}\:\:\mathrm{ACs} \\ $$$$\mathrm{W}−\:\mathrm{the}\:\mathrm{washing}\:\mathrm{machines} \\ $$$$\mathrm{m}−\mathrm{number}\:\mathrm{of}\:\mathrm{the}\:\mathrm{employees}\:\mathrm{have} \\ $$$$\mathrm{only}\:\mathrm{V} \\ $$$$\mathrm{n}−\mathrm{number}\:\mathrm{of}\:\mathrm{the}\:\mathrm{employees}\:\mathrm{have} \\ $$$$\mathrm{only}\:\mathrm{M} \\ $$$$\mathrm{p}−\mathrm{number}\:\mathrm{of}\:\mathrm{the}\:\mathrm{employees}\:\mathrm{have} \\ $$$$\mathrm{only}\:\mathrm{A} \\ $$$$\mathrm{w}−\mathrm{number}\:\mathrm{of}\:\mathrm{the}\:\mathrm{employees}\:\mathrm{have} \\ $$$$\mathrm{only}\:\mathrm{W} \\ $$$$\mathrm{a}−\mathrm{number}\:\mathrm{of}\:\mathrm{the}\:\mathrm{employees}\:\mathrm{have} \\ $$$$\mathrm{only}\:\mathrm{two}\:\mathrm{gadgets}\:\mathrm{VM} \\ $$$$\mathrm{b}−\mathrm{number}\:\mathrm{of}\:\mathrm{the}\:\mathrm{employees}\:\mathrm{have} \\ $$$$\mathrm{only}\:\mathrm{two}\:\mathrm{gadgets}\:\mathrm{VA} \\ $$$$\mathrm{c}−\mathrm{number}\:\mathrm{of}\:\mathrm{the}\:\mathrm{employees}\:\mathrm{have}\: \\ $$$$\mathrm{only}\:\mathrm{two}\:\mathrm{gadgets}\:\mathrm{VW} \\ $$$$\mathrm{d}−\mathrm{number}\:\mathrm{of}\:\mathrm{the}\:\mathrm{employees}\:\mathrm{have}\:\mathrm{only}\:\mathrm{two}\:\mathrm{gadgets} \\ $$$$\mathrm{MA} \\ $$$$\mathrm{e}−\mathrm{number}\:\mathrm{of}\:\mathrm{the}\:\mathrm{employees}\:\mathrm{have}\:\mathrm{only}\:\mathrm{two}\:\mathrm{gadgets} \\ $$$$\mathrm{MW} \\ $$$$\mathrm{f}−\mathrm{number}\:\mathrm{of}\:\mathrm{the}\:\mathrm{employees}\:\mathrm{have}\:\mathrm{only}\:\mathrm{two}\:\mathrm{gadgets} \\ $$$$\mathrm{AW} \\ $$$$\mathrm{x}−\mathrm{number}\:\mathrm{of}\:\mathrm{the}\:\mathrm{employees}\:\mathrm{have}\:\mathrm{only} \\ $$$$\mathrm{three}\:\mathrm{gadgets}\:\mathrm{MAV} \\ $$$$\mathrm{y}−\mathrm{number}\:\mathrm{of}\:\mathrm{the}\:\mathrm{employees}\:\mathrm{have}\:\mathrm{only} \\ $$$$\mathrm{three}\:\mathrm{gadgets}\:\mathrm{VAW} \\ $$$$\mathrm{z}−\mathrm{number}\:\mathrm{of}\:\mathrm{the}\:\mathrm{employees}\:\mathrm{have}\:\mathrm{only} \\ $$$$\mathrm{three}\:\mathrm{gadgets}\:\mathrm{VMW} \\ $$$$\mathrm{t}−\mathrm{number}\:\mathrm{of}\:\mathrm{the}\:\mathrm{employees}\:\mathrm{have}\:\mathrm{only} \\ $$$$\mathrm{three}\:\mathrm{gadgets}\:\mathrm{VMA} \\ $$$$\mathrm{T}−\mathrm{number}\:\mathrm{of}\:\mathrm{the}\:\mathrm{employees}\:\mathrm{have}\: \\ $$$$\mathrm{all}\:\mathrm{four}\:\mathrm{gadgets}. \\ $$$$\mathrm{Then}\:\mathrm{from}\:\mathrm{the}\:\mathrm{hypothesis}\:\mathrm{we}\:\mathrm{have} \\ $$$$\mathrm{m}+\mathrm{n}+\mathrm{p}+\mathrm{q}+\mathrm{a}+\mathrm{b}+\mathrm{c}+\mathrm{d}+\mathrm{e}+\mathrm{f} \\ $$$$+\mathrm{x}+\mathrm{y}+\mathrm{z}+\mathrm{t}+\mathrm{T}=\mathrm{100}\left(\mathrm{1}\right) \\ $$$$\mathrm{m}+\mathrm{a}+\mathrm{b}+\mathrm{c}+\mathrm{z}+\mathrm{y}+\mathrm{t}+\mathrm{T}=\mathrm{70}\left(\mathrm{2}\right) \\ $$$$\mathrm{n}+\mathrm{a}+\mathrm{d}+\mathrm{e}+\mathrm{x}+\mathrm{z}+\mathrm{t}+\mathrm{T}=\mathrm{75}\left(\mathrm{3}\right) \\ $$$$\mathrm{p}+\mathrm{b}+\mathrm{d}+\mathrm{f}+\mathrm{x}+\mathrm{y}+\mathrm{t}+\mathrm{T}=\mathrm{80}\left(\mathrm{4}\right) \\ $$$$\mathrm{q}+\mathrm{c}+\mathrm{e}+\mathrm{f}+\mathrm{x}+\mathrm{y}+\mathrm{z}+\mathrm{T}=\mathrm{85}\left(\mathrm{5}\right) \\ $$$$\mathrm{Adding}\:\mathrm{up}\:\mathrm{four}\:\mathrm{above}\:\mathrm{equalities}\:\mathrm{we}\:\mathrm{get} \\ $$$$\mathrm{m}+\mathrm{n}+\mathrm{p}+\mathrm{q}+\mathrm{2}\left(\mathrm{a}+\mathrm{b}+\mathrm{c}+\mathrm{d}+\mathrm{e}+\mathrm{f}\right) \\ $$$$+\mathrm{3}\left(\mathrm{x}+\mathrm{y}+\mathrm{z}+\mathrm{t}\right)+\mathrm{4T}=\mathrm{310}\left(\mathrm{6}\right) \\ $$$$\mathrm{Multiplying}\:\mathrm{two}\:\mathrm{sides}\:\mathrm{of}\:\left(\mathrm{1}\right)\:\mathrm{by}\:\mathrm{3}\:\mathrm{then} \\ $$$$\mathrm{substracting}\:\mathrm{from}\:\left(\mathrm{6}\right)\mathrm{we}\:\mathrm{get} \\ $$$$−\mathrm{2}\left(\mathrm{m}+\mathrm{n}+\mathrm{p}+\mathrm{q}\right)−\left(\mathrm{a}+\mathrm{b}+\mathrm{c}+\mathrm{d}+\mathrm{e}+\mathrm{f}\right)+\mathrm{T}=\mathrm{10} \\ $$$$\Rightarrow\mathrm{T}=\mathrm{10}+\mathrm{2}\left(\mathrm{m}+\mathrm{n}+\mathrm{p}+\mathrm{q}\right)+\left(\mathrm{a}+\mathrm{b}+\mathrm{c}+\mathrm{d}+\mathrm{e}+\mathrm{f}\right) \\ $$$$\Rightarrow\mathrm{T}\geqslant\mathrm{10}\left(\%\right) \\ $$$$\mathrm{The}\:\mathrm{equality}\:\mathrm{ocurrs}\:\mathrm{if}\:\mathrm{and}\:\mathrm{only}\:\mathrm{if} \\ $$$$\mathrm{m}=\mathrm{n}=\mathrm{p}=\mathrm{q}=\mathrm{a}=\mathrm{b}=\mathrm{c}=\mathrm{d}=\mathrm{e}=\mathrm{f}=\mathrm{0} \\ $$$$\mathrm{which}\:\mathrm{means}\:\mathrm{nobody}\:\mathrm{has}\:\mathrm{only}\:\mathrm{one} \\ $$$$\mathrm{gadgets}\:\mathrm{and}\:\mathrm{also}\:\mathrm{nobody}\:\mathrm{has}\:\mathrm{only} \\ $$$$\mathrm{two}\:\mathrm{gadgets}\:\mathrm{all}\:\mathrm{have}\:\mathrm{at}\:\mathrm{least}\:\mathrm{three} \\ $$$$\mathrm{gadgets}\:\mathrm{up} \\ $$$$\mathrm{Answer}\::\mathrm{at}\:\mathrm{least}\:\mathrm{10\%}\:\mathrm{of}\:\mathrm{the}\:\mathrm{employees} \\ $$$$\mathrm{have}\:\mathrm{all}\:\mathrm{four}\:\mathrm{gadgets} \\ $$

Commented by Aina Samuel Temidayo last updated on 19/Sep/20

Thanks.

$$\mathrm{Thanks}. \\ $$

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