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Question Number 210972 by ujah last updated on 25/Aug/24

Answered by TonyCWX08 last updated on 25/Aug/24

a.   n(C only) = 18−6−3=9    b.  n(One subject only)  =n(C only) + n(P  only) + n(M only)  =9+5+(22−6−2)  =14+14  =28    c.  n(Two subject only)  =3+2  =5

$${a}.\: \\ $$$${n}\left({C}\:{only}\right)\:=\:\mathrm{18}−\mathrm{6}−\mathrm{3}=\mathrm{9} \\ $$$$ \\ $$$${b}. \\ $$$${n}\left({One}\:{subject}\:{only}\right) \\ $$$$={n}\left({C}\:{only}\right)\:+\:{n}\left({P}\:\:{only}\right)\:+\:{n}\left({M}\:{only}\right) \\ $$$$=\mathrm{9}+\mathrm{5}+\left(\mathrm{22}−\mathrm{6}−\mathrm{2}\right) \\ $$$$=\mathrm{14}+\mathrm{14} \\ $$$$=\mathrm{28} \\ $$$$ \\ $$$${c}. \\ $$$${n}\left({Two}\:{subject}\:{only}\right) \\ $$$$=\mathrm{3}+\mathrm{2} \\ $$$$=\mathrm{5} \\ $$

Commented by TonyCWX08 last updated on 25/Aug/24

$$ \\ $$

Answered by mm1342 last updated on 25/Aug/24

32−16=16  ⇒x+y+z=16   &  z+y=9  &  y+x=14  ⇒x=7  &  y=7  &  z=2  ⇒ans  (a): 2      (b) : 14    (c): 12

$$\mathrm{32}−\mathrm{16}=\mathrm{16} \\ $$$$\Rightarrow{x}+{y}+{z}=\mathrm{16}\:\:\:\&\:\:{z}+{y}=\mathrm{9}\:\:\&\:\:{y}+{x}=\mathrm{14} \\ $$$$\Rightarrow{x}=\mathrm{7}\:\:\&\:\:{y}=\mathrm{7}\:\:\&\:\:{z}=\mathrm{2} \\ $$$$\Rightarrow{ans} \\ $$$$\left({a}\right):\:\mathrm{2}\:\:\:\:\:\:\left({b}\right)\::\:\mathrm{14}\:\:\:\:\left({c}\right):\:\mathrm{12} \\ $$$$ \\ $$

Commented by mm1342 last updated on 25/Aug/24

Commented by TonyCWX08 last updated on 25/Aug/24

okay

$${okay} \\ $$

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