A) Rs.102, Rs.118, Rs.113
B) Rs.105, Rs.120, Rs.108
C) Rs.85, Rs.125, Rs.123
D) None of these
Correct Answer: B
Solution :
Let salaries of A, B, C be Rs. x, y, z respectively Together their amount = x + y+ z = 333. Now, Expenditure of A = 80% \[SI=\frac{2500\times 4\times 5}{100}=Rs.500\]Savings = 20% Expenditure of B = 85% \[2P=P{{\left( 1+\frac{R}{100} \right)}^{3}}\]Savings = 15% Expenditure of C = 75% \[1+\frac{R}{100}={{2}^{\frac{1}{3}}}\]Savings = 25% \[16P=P{{\left( 1+\frac{R1}{100} \right)}^{n}}\]Savings of\[A=\frac{20x}{100},B=\frac{15y}{100},C=\frac{25z}{100}\] Also given savings are \[7:6:9\] \[\therefore \] \[\frac{20x}{700}=\frac{15y}{600}=\frac{25z}{900}\] \[\Rightarrow \] \[\frac{20x}{7}=\frac{15y}{6}=\frac{25z}{9}\] Now, \[\frac{20x}{7}=\frac{15y}{6}\Rightarrow 8x=7y=x=\frac{7y}{8}\]and\[\frac{15y}{6}=\frac{25z}{9}\] \[{{2}^{\frac{1}{3}}}={{2}^{\frac{4}{n}}}\] \[\frac{9y}{10}=z\] \[1\text{ }year=614.55-578.40=Rs.36.15.\] \[R=\frac{100\times I}{P\times T}=\frac{100\times 3.615}{578.40\times 1}=6\frac{1}{4}%\] \[85\times \frac{12}{20}=51\] \[\frac{7}{8}y+y+\frac{9}{10}y=333\] \[\Rightarrow \]\[\therefore \] \[\Rightarrow \] \[=5\times 24=Rs.120.\] \[=6\times 25=Rs.150\] \[x=\frac{7}{8}(120)=105,\,\,z=\frac{9}{10}(120)=108\] Hence, salary of A = Rs. 105, B = Rs.120, C = Rs.108.
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