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Question Number 35088 by Raj Singh last updated on 15/May/18

a son and father do a work in  24 day . if both work together  and father work last 6 day only  then they how much do

$$\boldsymbol{{a}}\:\boldsymbol{{son}}\:\boldsymbol{{and}}\:\boldsymbol{{father}}\:\boldsymbol{{do}}\:\boldsymbol{{a}}\:\boldsymbol{{work}}\:\boldsymbol{{in}} \\ $$$$\mathrm{24}\:\boldsymbol{{day}}\:.\:\boldsymbol{{if}}\:\boldsymbol{{both}}\:\boldsymbol{{work}}\:\boldsymbol{{together}} \\ $$$$\boldsymbol{{and}}\:\boldsymbol{{father}}\:\boldsymbol{{work}}\:\boldsymbol{{last}}\:\mathrm{6}\:\boldsymbol{{day}}\:\boldsymbol{{only}} \\ $$$$\boldsymbol{{then}}\:\boldsymbol{{they}}\:\boldsymbol{{how}}\:\boldsymbol{{much}}\:\boldsymbol{{do}} \\ $$

Commented by Rasheed.Sindhi last updated on 15/May/18

Data seems incomplete.

$$\mathrm{Data}\:\mathrm{seems}\:\mathrm{incomplete}. \\ $$

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

let father do (1/f) portion in a day  son do (1/(s ))portion in a day  so 24((1/f)+(1/s))=1  (1/f)+(1/s)=(1/(24))  in last 6 day father do (6/f) portion  remaining work (1−(6/f))  (6/f)+(6/s)=(6/(24))  (6/f)=(1/4)−(6/s)  remaining work =(1−(6/f))  =(1−(1/4)+(6/s))  =(3/4)+(6/s)  so remaining work>(3/4)

$${let}\:{father}\:{do}\:\frac{\mathrm{1}}{{f}}\:{portion}\:{in}\:{a}\:{day} \\ $$$${son}\:{do}\:\frac{\mathrm{1}}{{s}\:}{portion}\:{in}\:{a}\:{day} \\ $$$${so}\:\mathrm{24}\left(\frac{\mathrm{1}}{{f}}+\frac{\mathrm{1}}{{s}}\right)=\mathrm{1} \\ $$$$\frac{\mathrm{1}}{{f}}+\frac{\mathrm{1}}{{s}}=\frac{\mathrm{1}}{\mathrm{24}} \\ $$$${in}\:{last}\:\mathrm{6}\:{day}\:{father}\:{do}\:\frac{\mathrm{6}}{{f}}\:{portion} \\ $$$${remaining}\:{work}\:\left(\mathrm{1}−\frac{\mathrm{6}}{{f}}\right) \\ $$$$\frac{\mathrm{6}}{{f}}+\frac{\mathrm{6}}{{s}}=\frac{\mathrm{6}}{\mathrm{24}} \\ $$$$\frac{\mathrm{6}}{{f}}=\frac{\mathrm{1}}{\mathrm{4}}−\frac{\mathrm{6}}{{s}} \\ $$$${remaining}\:{work}\:=\left(\mathrm{1}−\frac{\mathrm{6}}{{f}}\right) \\ $$$$=\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{6}}{{s}}\right) \\ $$$$=\frac{\mathrm{3}}{\mathrm{4}}+\frac{\mathrm{6}}{{s}} \\ $$$${so}\:{remaining}\:{work}>\frac{\mathrm{3}}{\mathrm{4}} \\ $$

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