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Question Number 34739 by chakraborty ankit last updated on 10/May/18

Commented by abdo mathsup 649 cc last updated on 10/May/18

$$5)wehavetheformula$$$$\sum _{k=0}^{n-1}[x+\frac{k}{n}]=\left[nx\right]\Rightarrow$$$$\sum _{k=0}^{98}[\frac{2}{3}+\frac{k}{99}]=[99.\frac{2}{3}]=\left[66\right]=66answerc.$$

Commented by math khazana by abdo last updated on 11/May/18

$$7)2f\left(x\right)+f(1-x)=x^{2}fisapolynom$$$$letputf\left(x\right)={ax}^2+bx+c$$$$\iff 2{ax}^2+2bx+2c+a{(1-x)}^2+b(1-x)+c=x^2$$$$\iff 2{ax}^2{x}^2+2bx+2c+a(x^2-2x+1)+b-bx+c=x^2$$$$\iff 3{ax}^2+(2b-2a-b)x+2c+a+b+c=x^2$$$$\iff 3a=1,b-2a=0,3c+a+b=0$$$$\iff a=\frac{1}{3},b=\frac{2}{3},c=-\frac{a+b}{3}=-\frac{1}{3}\Rightarrow$$$$f\left(x\right)=\frac{1}{3}{x}^2+\frac{2}{3}x-\frac{1}{3}=\frac{x^2+2x-1}{3}answerB.$$

Commented by chakraborty ankit last updated on 13/May/18

$$thankyousir$$

Answered by Rasheed.Sindhi last updated on 13/May/18

$$1.$$$$\mathrm{A}=\{1,2,3,4\},\mathrm{B}=\{2,4,6\}$$$$\mathrm{A}\cap \mathrm{B}\supseteq \mathrm{C}\subseteq \mathrm{A}\cup \mathrm{B}$$$$\mathrm{A}\cap \mathrm{B}=\{2,4\},\mathrm{A}\cup \mathrm{B}=\{1,2,3,4,6\}$$$$\mathrm{C}=\{2,4\}\cup \mathrm{X},\mathrm{\forall }\mathrm{X}\in \mathbb{P}((\mathrm{A}\cup \mathrm{B})-(\mathrm{A}\cap \mathrm{B}))$$$$\mathbb{P}(\{1,2,3,4,6\}-\{2,4\})$$$$=\mathbb{P}(\{1,3,6\}$$$$\mathrm{Number}\mathrm{of}{\mathrm{X}}^{\prime }\mathrm{s}=2^3=8$$$$\mathrm{Number}\mathrm{of}{\mathrm{C}}^{\prime }\mathrm{s}=8[\because \mathrm{C}=\{2,4\}\cup \mathrm{X}]$$$$\left(\mathrm{C}\right)8$$

Answered by rahul 19 last updated on 12/May/18

$$3)\left(A\right)$$$$4)\left(B\right)(4,\mathrm{\infty }).$$$$6)\left(B\right)$$

Answered by Rasheed.Sindhi last updated on 13/May/18

$$2.$$$$\mathrm{Let}\mathrm{the}\mathrm{number}\mathrm{of}\mathrm{families}\mathrm{in}\mathrm{the}\mathrm{town}\mathrm{is}\mathrm{y}.$$$$25\mathrm{\%}\mathrm{of}\mathrm{y}=\mathrm{y}/4\mathrm{having}\mathrm{cell}\mathrm{phone}.$$$$15\mathrm{\%}\mathrm{of}\mathrm{y}=3\mathrm{y}/20\mathrm{having}\mathrm{scooter}.$$$$1500\mathrm{families}\mathrm{having}\mathrm{both}.$$$$----$$$$\left(\mathrm{y}/4\right)-1500\mathrm{having}\mathrm{cell}\mathrm{phones}\mathrm{only}.$$$$\left(3\mathrm{y}/20\right)-1500\mathrm{having}\mathrm{scooters}\mathrm{only}.$$$$\mathrm{Families}\mathrm{having}\mathrm{a}\mathrm{cell}\mathrm{phone}\mathrm{or}\mathrm{a}\mathrm{scooter}:$$$$[\left(\mathrm{y}/4\right)-1500]+1500+[\left(3\mathrm{y}/20\right)-1500]$$$$\frac{\mathrm{y}}{4}+\frac{3\mathrm{y}}{20}-1500=\frac{5\mathrm{y}+3\mathrm{y}-30000}{20}$$$$\mathrm{Having}\mathrm{neither}\mathrm{cell}\mathrm{phone}\mathrm{nor}\mathrm{scooter}$$$$\mathrm{y}-\frac{8\mathrm{y}-30000}{20}=65\mathrm{\%}\mathrm{of}\mathrm{y}$$$$\frac{20\mathrm{y}-8\mathrm{y}+30000}{20}=\frac{13\mathrm{y}}{20}$$$$13\mathrm{y}-12\mathrm{y}=30000$$$$\mathrm{y}=30000$$$$\mathcal{T}otalnumberoffamiliesinthetown$$$$\left(\mathrm{C}\right)30,000$$

Commented by chakraborty ankit last updated on 13/May/18

$$thankyousir$$

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