Answer :
[tex]In\ the\ first\ jug\ is\ \frac{3}{4}l\ olive\ oil.\\\\In\ the\ second\ jug\ is\ \frac{4}{5}l\ olive\ oil\\\\\frac{3}{4}l+\frac{4}{5}l\\\\\frac{3}{4}\to the\ denominator\ is\ 4\\\\list\ the\ multiples\ of\ 4:\ 0;\ 4;\ 8;\ 12;\ 16;\ \fbox{20};\ 24;\ 28;...\\\\\frac{4}{5}\to the\ denominator\ is\ 5\\\\list\ the\ multiples\ of\ 5:\ 0;\ 5;\ 10;\ 15;\ \fbox{20};\ 25;\ 30;...\\\\Least\ Cammon\ Denominator\ of\ \frac{3}{4}\ and\ \frac{4}{5}\ is\ 20[/tex]
[tex]\frac{3}{4}=\frac{3\cdot5}{4\cdot5}=\frac{15}{20}\\\\\frac{4}{5}=\frac{4\cdot4}{5\cdot4}=\frac{16}{20}\\-----------------\\\\\frac{3}{4}l+\frac{4}{5}l=\frac{15}{20}l+\frac{16}{20}l=\frac{15+16}{20}l=\frac{31}{20}l=1\frac{11}{20}l=1.55l\leftarrow solution[/tex]
[tex]\frac{3}{4}=\frac{3\cdot5}{4\cdot5}=\frac{15}{20}\\\\\frac{4}{5}=\frac{4\cdot4}{5\cdot4}=\frac{16}{20}\\-----------------\\\\\frac{3}{4}l+\frac{4}{5}l=\frac{15}{20}l+\frac{16}{20}l=\frac{15+16}{20}l=\frac{31}{20}l=1\frac{11}{20}l=1.55l\leftarrow solution[/tex]
[tex]The\ first\ jug:\\\\\frac{oil}{vinegar} = \frac{3}{1} \ \ \ \Rightarrow\ \ \ the\ volume\ of\ the\ oil= \frac{3}{4} \ of\ the\ jug\\\\.\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \Rightarrow\ \ \ the\ volume\ of\ oil\ =\frac{3}{4} \ litre\\\\The\ second\ jug:\\\\\frac{oil}{vinegar} = \frac{4}{1} \ \ \ \Rightarrow\ \ \ the\ volume\ of\ the\ oil= \frac{4}{5} \ of\ the\ jug\\\\.\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \Rightarrow\ \ \ the\ volume\ of\ oil\ =\frac{4}{5} \ litre\\\\[/tex]
[tex]The\ larger\ jug:\\\\ \frac{3}{4} \ litre+ \frac{4}{5}\ litre=0.75+0.8\ [litre]=1.55\ [litre]\\\\Ans.\ the\ volume\ of\ the\ oil\ is\ 1.55\ litres.[/tex]
[tex]The\ larger\ jug:\\\\ \frac{3}{4} \ litre+ \frac{4}{5}\ litre=0.75+0.8\ [litre]=1.55\ [litre]\\\\Ans.\ the\ volume\ of\ the\ oil\ is\ 1.55\ litres.[/tex]
