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Preeti Dabral 3 years ago
cost of coffee=250 per kg. cost of chicory =75 per kg coffee: chicory 5:2 if the weight of mixture is 1 kg.
{tex}\begin{aligned} & \frac{5 \mathrm{x}+2 \mathrm{x}}{7}=1 \\ & \Rightarrow 7 \mathrm{x}=7 \\ & \Rightarrow \mathrm{x}=1 \\ & \therefore \text { weight of coffee }=\frac{5 \mathrm{x}}{7}=\frac{5}{7} \mathrm{~kg} \\ & \text { weight of chicory }=\frac{2 \mathrm{x}}{7}=\frac{2}{7} \mathrm{~kg} \\ & \text { cost of } \frac{5}{7} \mathrm{~kg} \text { coffee }=\frac{5}{7} \times 250=\frac{1250}{7} \\ & \text { cost of } \frac{2}{7} \mathrm{~kg} \text { chicory }=\frac{2}{7} \times 75=\frac{150}{7} \\ & \text { total cost of } 1 \mathrm{~kg} \text { mixture }=\frac{1250}{7}+\frac{150}{7} \\ & \text { cost price of } 1 \mathrm{~kg} \text { mixture }=\frac{1400}{7}=200 \text { Rs } \\ & \text { selling of } 1 \mathrm{~kg} \text { minute }=230 \mathrm{Rs} \\ & \text { gain or loss }=\frac{\text { SP }-\mathrm{CP}}{\text { CP }} \times 100=\frac{230-200}{200} \times 100 \\ & =\frac{30}{2}=15 \% \end{aligned}{/tex}
since its positive, the mixture was sold at a gain of 15%.
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Preeti Dabral 3 years ago
{tex}{\left\{ {{{\left( {\frac{1}{3}} \right)}^{ - 1}} - {{\left( {\frac{1}{4}} \right)}^{ - 1}}} \right\}^{ - 1}}{/tex}
{tex} = {\left( {\frac{{{1^{ - 1}}}}{{{3^{ - 1}}}} - \frac{{{1^{ - 1}}}}{{{4^{ - 1}}}}} \right)^{ - 1}} = {\left( {\frac{{{3^1}}}{{{1^1}}} - \frac{{{4^1}}}{{{1^1}}}} \right)^{ - 1}}{/tex}
{tex} = {\left( {\frac{3}{1} - \frac{4}{1}} \right)^{ - 1}}{/tex} = (3 – 4)–1
{tex} = {( - 1)^{ - 1}} = \frac{1}{{{{( - 1)}^1}}}{/tex}
{tex} = \frac{1}{{( - 1)}} = - 1{/tex}
Posted by Hashwanth Ramesh 3 years ago
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Preeti Dabral 3 years ago
cube 1 :
side = 4 cm
volume = a³
volume = 4³ = 64 cm³
cube 2 :
side = 1.5 m
volume = 1.5³ = 3.375 m³
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Simran Singh 2 years, 11 months ago
0Thank You