If-a-fraction-is-added-to-its-denominator-it-reduces-to-1-2-and-when-the-same-fraction-added-to-numerator-it-also-reduces-to-2-3-a-what-is-the-fraction-b-find-the-square-root-such-that-th

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Question Number 41879 by Tawa1 last updated on 14/Aug/18

$$\mathrm{If}\mathrm{a}\mathrm{fraction}\mathrm{is}\mathrm{added}\mathrm{to}\mathrm{its}\mathrm{denominator},\mathrm{it}\mathrm{reduces}\mathrm{to}\frac{1}{2}\mathrm{and}$$$$\mathrm{when}\mathrm{the}\mathrm{same}\mathrm{fraction}\mathrm{added}\mathrm{to}\mathrm{numerator},\mathrm{it}\mathrm{also}\mathrm{reduces}\mathrm{to}\frac{2}{3}$$$$\left(\mathrm{a}\right)\mathrm{what}\mathrm{is}\mathrm{the}\mathrm{fraction}$$$$\left(\mathrm{b}\right)\mathrm{find}\mathrm{the}\mathrm{square}\mathrm{root}\mathrm{such}\mathrm{that}\mathrm{the}\mathrm{result}\mathrm{of}\mathrm{the}\mathrm{fraction}\mathrm{is}\mathrm{less}$$$$\mathrm{than}1$$$$$$

Answered by tanmay.chaudhury50@gmail.com last updated on 14/Aug/18

$$\frac{p}{q+\frac{p}{q}}=\frac{1}{2}\frac{pq}{p+q^2}=\frac{1}{2}\dots ..eqn1$$$$$$$$\frac{p+\frac{p}{q}}{q}=\frac{2}{3}\frac{pq+p}{q^2}=\frac{2}{3}\dots .eqn2$$$$3p(q+1)=2q^2$$$$p=\frac{2q^2}{3(q+1)}puttingthevalueof?pineqn1$$$$2pq=p+q^2$$$$\frac{4q^2\times q}{3(q+1)}=\frac{2q^2}{3(q+1)}+q^2$$$$\frac{4q^3}{3(q+1)}=\frac{2q^2+3q^3+3q^2}{3(q+1)}$$$$q^3-5q^2=0$$$$q^2(q-5)=0soq=5$$$$\frac{p\times 5}{p+25}=\frac{1}{2}$$$$$$$$10p=p+25p=\frac{25}{9}$$$$fraction=\frac{25}{9\times 5}=\frac{5}{9}$$$$\frac{5}{9+\frac{5}{9}}=\frac{45}{86}$$$$\frac{5+\frac{5}{9}}{9}=\frac{50}{81}$$$$somethingwronginqueztio\dots .$$

Commented by Tawa1 last updated on 14/Aug/18

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}$$

Commented by MJS last updated on 15/Aug/18

$$\mathrm{the}\mathrm{point}\mathrm{is},\mathrm{the}\mathrm{equations}\mathrm{are}\mathrm{solved}\mathrm{for}$$$$p=\frac{25}{9}\wedge q=5\Rightarrow \mathrm{you}\mathrm{cannot}\mathrm{set}\frac{p}{q}=\frac{5}{9}\mathrm{and}\mathrm{insert}$$$$p=5\wedge q=9\mathrm{in}\mathrm{the}\mathrm{original}\mathrm{equation}$$$$$$$$\mathrm{we}\mathrm{must}\mathrm{first}\mathrm{put}p,q\in \mathbb{Z}\mathrm{or}\mathrm{otherwise}{\mathrm{it}}^{\prime }\mathrm{s}\mathrm{no}$$$$\mathrm{regular}\mathrm{fraction}.\mathrm{then}\mathrm{at}\mathrm{the}\mathrm{point}\mathrm{we}\mathrm{get}$$$$p=\frac{25}{9}{\mathrm{it}}^{\prime }\mathrm{s}\mathrm{not}\mathrm{in}\mathrm{our}\mathrm{domain}\mathrm{of}\mathrm{definition}\Rightarrow$$$$\Rightarrow \mathrm{the}\mathrm{problem}\mathrm{has}\mathrm{no}\mathrm{valid}\mathrm{solution}$$

Commented by tanmay.chaudhury50@gmail.com last updated on 15/Aug/18

$$thankyousir\dots$$

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