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Question Number 57023 by otchereabdullai@gmail.com last updated on 28/Mar/19

$$\mathrm{There}\mathrm{are}28\mathrm{players}\mathrm{in}\mathrm{a}\mathrm{national}$$$$\mathrm{football}\mathrm{team}.14\mathrm{play}\mathrm{midfield}\mathrm{and}$$$$\mathrm{defence},15\mathrm{play}\mathrm{defence}\mathrm{and}\mathrm{attack}$$$$\mathrm{and}3\mathrm{play}\mathrm{midfield}\mathrm{only}.\mathrm{the}\mathrm{number}$$$$\mathrm{of}\mathrm{players}\mathrm{who}\mathrm{play}\mathrm{attack}\mathrm{only}\mathrm{is}$$$$\mathrm{twice}\mathrm{those}\mathrm{who}\mathrm{play}\mathrm{defence}\mathrm{only},\mathrm{and}$$$$\mathrm{the}\mathrm{number}\mathrm{who}\mathrm{play}\mathrm{defence}$$$$\mathrm{is}\mathrm{equal}\mathrm{to}\mathrm{those}\mathrm{who}\mathrm{play}\mathrm{attack}.\mathrm{If}$$$$18\mathrm{play}\mathrm{midfield}\mathrm{represent}\mathrm{the}$$$$\mathrm{information}\mathrm{on}\mathrm{a}\mathrm{venn}\mathrm{diagram}.$$$$\mathrm{Find}$$$$\mathrm{i}.\mathrm{how}\mathrm{many}\mathrm{play}\mathrm{at}\mathrm{most}\mathrm{two}\mathrm{games}$$$$\mathrm{ii}.\mathrm{how}\mathrm{many}\mathrm{play}\mathrm{neither}\mathrm{attack}\mathrm{nor}$$$$\mathrm{defence}$$$$\mathrm{iii}.\mathrm{how}\mathrm{many}\mathrm{play}\mathrm{either}\mathrm{midfield}\mathrm{or}$$$$\mathrm{attack}.$$

Answered by MJS last updated on 29/Mar/19

$$\mathrm{the}\mathrm{venn}\mathrm{diagram}\mathrm{has}\mathrm{got}7\mathrm{areas}$$$$$$$$\left(1\right)m+a+d+ma+md+ad+mad=28$$$$\left(2\right)md+mad=14\Rightarrow md=14-mad$$$$\left(3\right)ad+mad=15\Rightarrow ad=15-mad$$$$\left(4\right)m=3$$$$\left(5\right)a=2d$$$$\left(6\right)d+md+ad+mad=a+ma+ad+mad\Rightarrow$$$$\Rightarrow d+md=a+ma$$$$\left(7\right)m+ma+md+mad=18$$$$$$$$$$$$\left(1\right)3+2d+d+ma+14-mad+15-mad+mad=28\Rightarrow$$$$\Rightarrow 3d+ma-mad=-4$$$$\left(6\right)d+14-mad=2d+ma\Rightarrow$$$$\Rightarrow d+ma+mad=14$$$$\left(7\right)3+ma+14-mad+mad=18\Rightarrow ma=1$$$$$$$$$$$$\left(1\right)3d-mad=-5$$$$\left(6\right)d+mad=13$$$$\Rightarrow d=2;mad=11$$$$$$$$m=3$$$$a=4$$$$d=2$$$$ma=1$$$$md=3$$$$ad=4$$$$mad=11$$$$$$$$\mathrm{i}.28-mad=17$$$$\mathrm{ii}.m=3$$$$\mathrm{iii}.m+a+md+ad=14$$

Commented by otchereabdullai@gmail.com last updated on 29/Mar/19

$$\mathrm{thanks}\mathrm{prof}$$

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