Question Number 4443 by 123456 last updated on 27/Jan/16
![find all x,y∈Z such that x∈[0,50] y∈[0,50] x+y=k,k∈[0,50] (x/(x+y))=((99)/(100))](https://www.tinkutara.com/question/Q4443.png)
$$\mathrm{find}\:\mathrm{all}\:{x},{y}\in\mathbb{Z}\:\mathrm{such}\:\mathrm{that} \\ $$$${x}\in\left[\mathrm{0},\mathrm{50}\right] \\ $$$${y}\in\left[\mathrm{0},\mathrm{50}\right] \\ $$$${x}+{y}={k},{k}\in\left[\mathrm{0},\mathrm{50}\right] \\ $$$$\frac{{x}}{{x}+{y}}=\frac{\mathrm{99}}{\mathrm{100}} \\ $$
Answered by RasheedSindhi last updated on 27/Jan/16
![^(Rasheed Soomro) (x/(x+y))=((99)/(100))...........Given 100x=99x+99y x=99y...................(i) x+y=k,k∈[0,50]......Given x=k−y.................(ii) From (i) & (ii): y=0.01k & x=0.99k k=0⇒x+y=0 k=50⇒.01×50+0.99×50=50 (x,y ∈{0}∪Z^+ ⇒ k∈{0}∪Z^+ ) ∀ k∈[0,50], both conditions satisfied. ∴ (x,y)=(0.99k,0.01k) ∀k∈{0,1,2,...,50} (Total 51 pairs)](https://www.tinkutara.com/question/Q4444.png)
$$\:\:^{{Rasheed}\:{Soomro}} \\ $$$$\frac{{x}}{{x}+{y}}=\frac{\mathrm{99}}{\mathrm{100}}………..{Given} \\ $$$$\mathrm{100}{x}=\mathrm{99}{x}+\mathrm{99}{y} \\ $$$${x}=\mathrm{99}{y}……………….\left({i}\right) \\ $$$${x}+{y}={k},{k}\in\left[\mathrm{0},\mathrm{50}\right]……{Given} \\ $$$${x}={k}−{y}……………..\left({ii}\right) \\ $$$${From}\:\left({i}\right)\:\&\:\left({ii}\right): \\ $$$${y}=\mathrm{0}.\mathrm{01}{k}\:\:\&\:{x}=\mathrm{0}.\mathrm{99}{k} \\ $$$${k}=\mathrm{0}\Rightarrow{x}+{y}=\mathrm{0} \\ $$$${k}=\mathrm{50}\Rightarrow.\mathrm{01}×\mathrm{50}+\mathrm{0}.\mathrm{99}×\mathrm{50}=\mathrm{50} \\ $$$$\:\:\:\:\:\left({x},{y}\:\in\left\{\mathrm{0}\right\}\cup{Z}^{+} \Rightarrow\:{k}\in\left\{\mathrm{0}\right\}\cup{Z}^{+} \right) \\ $$$$\forall\:{k}\in\left[\mathrm{0},\mathrm{50}\right],\:{both}\:{conditions}\:{satisfied}. \\ $$$$\therefore\:\left({x},{y}\right)=\left(\mathrm{0}.\mathrm{99}{k},\mathrm{0}.\mathrm{01}{k}\right)\:\forall{k}\in\left\{\mathrm{0},\mathrm{1},\mathrm{2},…,\mathrm{50}\right\} \\ $$$$\left({Total}\:\mathrm{51}\:{pairs}\right) \\ $$