Question Number 7881 by Rasheed Soomro last updated on 23/Sep/16
$${A}\:{gets}\:\mathrm{1}/\mathrm{5}\:\:{of}\:\:{some}\:\:{amount}, \\ $$$${B}\:\:{gets}\:\mathrm{1}/\mathrm{3}\:\:{of}\:\:{remaining},\:{C} \\ $$$${gets}\:\mathrm{1}/\mathrm{6}\:{of}\:{remaining},\:{D}\:{gets} \\ $$$$\mathrm{1}/\mathrm{7}\:\:{of}\:\:{remaining}\:\:{and}\:{E}\:{gets} \\ $$$${rest}\:{of}\:{amount}.\:{What}\:{fraction} \\ $$$${gets}\:{E}\:? \\ $$
Commented by sandy_suhendra last updated on 23/Sep/16
![A=(1/5) B=(1/3) [1−(1/5)]= (4/(15)) C=(1/6)[1−(1/5)−(4/(15))]=(4/(45)) D=(1/7)[1−(1/5)−(4/(15))−(4/(45))]=((20)/(315)) E=1−(1/5)−(4/(15))−(4/(45))−((20)/(315))=((120)/(315))=(8/(21)) or A=(1/5) so the remaining = 1−(1/5)=(4/5) B=(1/3)×(4/5)=(4/(15)) C=(1/6)×(4/5)=(2/(15)) D=(1/7)×(4/5)=(4/(35)) E=(4/5)−(4/(15))−(2/(15))−(4/(35))=((30)/(105))=(2/7) but I don′t know which one is true](https://www.tinkutara.com/question/Q7886.png)
$${A}=\frac{\mathrm{1}}{\mathrm{5}} \\ $$$${B}=\frac{\mathrm{1}}{\mathrm{3}}\:\left[\mathrm{1}−\frac{\mathrm{1}}{\mathrm{5}}\right]=\:\frac{\mathrm{4}}{\mathrm{15}} \\ $$$${C}=\frac{\mathrm{1}}{\mathrm{6}}\left[\mathrm{1}−\frac{\mathrm{1}}{\mathrm{5}}−\frac{\mathrm{4}}{\mathrm{15}}\right]=\frac{\mathrm{4}}{\mathrm{45}} \\ $$$${D}=\frac{\mathrm{1}}{\mathrm{7}}\left[\mathrm{1}−\frac{\mathrm{1}}{\mathrm{5}}−\frac{\mathrm{4}}{\mathrm{15}}−\frac{\mathrm{4}}{\mathrm{45}}\right]=\frac{\mathrm{20}}{\mathrm{315}} \\ $$$${E}=\mathrm{1}−\frac{\mathrm{1}}{\mathrm{5}}−\frac{\mathrm{4}}{\mathrm{15}}−\frac{\mathrm{4}}{\mathrm{45}}−\frac{\mathrm{20}}{\mathrm{315}}=\frac{\mathrm{120}}{\mathrm{315}}=\frac{\mathrm{8}}{\mathrm{21}} \\ $$$${or} \\ $$$${A}=\frac{\mathrm{1}}{\mathrm{5}}\:{so}\:{the}\:{remaining}\:=\:\mathrm{1}−\frac{\mathrm{1}}{\mathrm{5}}=\frac{\mathrm{4}}{\mathrm{5}} \\ $$$${B}=\frac{\mathrm{1}}{\mathrm{3}}×\frac{\mathrm{4}}{\mathrm{5}}=\frac{\mathrm{4}}{\mathrm{15}} \\ $$$${C}=\frac{\mathrm{1}}{\mathrm{6}}×\frac{\mathrm{4}}{\mathrm{5}}=\frac{\mathrm{2}}{\mathrm{15}} \\ $$$${D}=\frac{\mathrm{1}}{\mathrm{7}}×\frac{\mathrm{4}}{\mathrm{5}}=\frac{\mathrm{4}}{\mathrm{35}} \\ $$$${E}=\frac{\mathrm{4}}{\mathrm{5}}−\frac{\mathrm{4}}{\mathrm{15}}−\frac{\mathrm{2}}{\mathrm{15}}−\frac{\mathrm{4}}{\mathrm{35}}=\frac{\mathrm{30}}{\mathrm{105}}=\frac{\mathrm{2}}{\mathrm{7}} \\ $$$${but}\:{I}\:{don}'{t}\:{know}\:{which}\:{one}\:{is}\:{true} \\ $$
Commented by Rasheed Soomro last updated on 27/Sep/16
![I think your first answer is correct. I have done it slightly in defferent way. A gets (1/5) Remaining after A=1−(1/5)=(4/5) B gets (4/5)×(1/3)=(4/(15)) Remaining after B=(4/5)−(4/(15))=((12−4)/(15))=(8/(15)) C gets (8/(15))×(1/6)=(4/(45)) Remaining after C=(8/(15))−(4/(45))=((24−4)/(45))=((20)/(45))=(4/9) D gets (4/9)×(1/7)=(4/(63)) Remaining after D=(4/9)−(4/(63))=((28−4)/(63))=((24)/(63))=(8/(21)) E gets (8/(21))×1=(8/(21)) [ D gets whole of the remaining] E gets fraction (8/(21))](https://www.tinkutara.com/question/Q7893.png)
$$ \\ $$$$ \\ $$$${I}\:{think}\:{your}\:{first}\:{answer}\:{is}\:{correct}. \\ $$$${I}\:{have}\:{done}\:{it}\:{slightly}\:{in}\:{defferent} \\ $$$${way}. \\ $$$$ \\ $$$${A}\:\:{gets}\:\frac{\mathrm{1}}{\mathrm{5}} \\ $$$${Remaining}\:{after}\:{A}=\mathrm{1}−\frac{\mathrm{1}}{\mathrm{5}}=\frac{\mathrm{4}}{\mathrm{5}} \\ $$$${B}\:{gets}\:\:\:\:\frac{\mathrm{4}}{\mathrm{5}}×\frac{\mathrm{1}}{\mathrm{3}}=\frac{\mathrm{4}}{\mathrm{15}} \\ $$$${Remaining}\:\:{after}\:{B}=\frac{\mathrm{4}}{\mathrm{5}}−\frac{\mathrm{4}}{\mathrm{15}}=\frac{\mathrm{12}−\mathrm{4}}{\mathrm{15}}=\frac{\mathrm{8}}{\mathrm{15}} \\ $$$${C}\:\:{gets}\:\frac{\mathrm{8}}{\mathrm{15}}×\frac{\mathrm{1}}{\mathrm{6}}=\frac{\mathrm{4}}{\mathrm{45}} \\ $$$${Remaining}\:\:{after}\:{C}=\frac{\mathrm{8}}{\mathrm{15}}−\frac{\mathrm{4}}{\mathrm{45}}=\frac{\mathrm{24}−\mathrm{4}}{\mathrm{45}}=\frac{\mathrm{20}}{\mathrm{45}}=\frac{\mathrm{4}}{\mathrm{9}} \\ $$$${D}\:{gets}\:\frac{\mathrm{4}}{\mathrm{9}}×\frac{\mathrm{1}}{\mathrm{7}}=\frac{\mathrm{4}}{\mathrm{63}} \\ $$$${Remaining}\:{after}\:{D}=\frac{\mathrm{4}}{\mathrm{9}}−\frac{\mathrm{4}}{\mathrm{63}}=\frac{\mathrm{28}−\mathrm{4}}{\mathrm{63}}=\frac{\mathrm{24}}{\mathrm{63}}=\frac{\mathrm{8}}{\mathrm{21}} \\ $$$${E}\:{gets}\:\:\frac{\mathrm{8}}{\mathrm{21}}×\mathrm{1}=\frac{\mathrm{8}}{\mathrm{21}}\:\:\:\left[\:{D}\:{gets}\:{whole}\:{of}\:{the}\:{remaining}\right] \\ $$$${E}\:{gets}\:{fraction}\:\frac{\mathrm{8}}{\mathrm{21}} \\ $$
Answered by FilupSmith last updated on 23/Sep/16
$${Total}=\mathrm{1} \\ $$$$\mathrm{1}=\frac{\mathrm{1}}{\mathrm{5}}+\left(\frac{\mathrm{1}}{\mathrm{5}}\right)\frac{\mathrm{1}}{\mathrm{3}}+\left(\frac{\mathrm{1}}{\mathrm{5}}×\frac{\mathrm{1}}{\mathrm{3}}\right)\frac{\mathrm{1}}{\mathrm{6}}+\left(\frac{\mathrm{1}}{\mathrm{5}}×\frac{\mathrm{1}}{\mathrm{3}}×\frac{\mathrm{1}}{\mathrm{6}}×\frac{\mathrm{1}}{\mathrm{7}}\right)+{E} \\ $$$$\mathrm{1}=\frac{\mathrm{1}}{\mathrm{5}}+\frac{\mathrm{1}}{\mathrm{15}}+\frac{\mathrm{1}}{\mathrm{90}}+\frac{\mathrm{1}}{\mathrm{540}}+{E} \\ $$$$\mathrm{1}=\frac{\mathrm{88}}{\mathrm{315}}+{E} \\ $$$${E}=\frac{\mathrm{227}}{\mathrm{315}} \\ $$$$\: \\ $$$${I}\:{am}\:{unsure}\:{if}\:{i}\:{answered}\:{correct} \\ $$