Question

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Q.24. A school ordered $ 800 worth of burgers. Some of the burgers cost $ 2 each and the others cost $ 3 each. If the number of burgers costing $ 2 were $\frac{7}{6}$ of the number of the burgers costing $ 3, then how many burgers were ordered altogether?

Answer 1

Dear student,
Here is the solution of your asked query:
$$\begin{gathered} \mathrm{Let}\mathrm{number}\mathrm{of}\mathrm{burgers}\mathrm{costing}\&3\mathrm{are}\mathrm{x}. \\ \mathrm{Then},\mathrm{number}\mathrm{of}\mathrm{burgers}\mathrm{dollar}\mathrm{costing}\&2=\frac{7}{6}\times \mathrm{number}\mathrm{of}\mathrm{burgers}\mathrm{costing}\&3=\frac{7\mathrm{x}}{6} \\ \mathrm{So}\mathrm{value}\mathrm{of}\mathrm{x}\mathrm{burgers}\mathrm{costing}\&3\mathrm{each}=\mathrm{x}\times 3=3\mathrm{x} \\ \mathrm{And}\mathrm{value}\mathrm{of}\frac{7\mathrm{x}}{6}\mathrm{burgers}\mathrm{costing}\&2\mathrm{each}=\frac{7\mathrm{x}}{6}\times 2=\frac{7\mathrm{x}}{3} \\ \mathrm{so},\mathrm{total}\mathrm{value}\mathrm{of}\mathrm{burgers}\mathrm{ordered}=3\mathrm{x}+\frac{7\mathrm{x}}{3}=\&800 \\ \Rightarrow \frac{16\mathrm{x}}{3}=800 \\ \Rightarrow \mathrm{x}=\frac{800\times 3}{16}=150 \\ \mathrm{so},\mathrm{number}\mathrm{of}\mathrm{burgers}\mathrm{costing}\&3=150 \\ \mathrm{And}\mathrm{number}\mathrm{of}\mathrm{burgers}\mathrm{ordered}\mathrm{costing}\&3=\frac{7}{6}\times 150=175 \\ \mathrm{Therefore},\mathrm{total}\mathrm{number}\mathrm{of}\mathrm{burgers}\mathrm{ordered}\mathrm{altogether}=150+175=325 \end{gathered}$$
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