Question

# The dimensions of a rectangular box are in the ratio of 2 : 3 : 4 and the difference between the cost of covering it with sheet of paper at the rates of Rs 8 and Rs 9.50 per m2 is Rs. 1248. Find the dimensions of the box.

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Solution

## $\mathrm{Suppose}\mathrm{that}\mathrm{the}\mathrm{dimensions}\mathrm{be}\mathrm{x}\mathrm{multiple}\mathrm{of}\mathrm{each}\mathrm{other}.\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{The}\mathrm{dimensions}\mathrm{are}\mathrm{in}\mathrm{the}\mathrm{ratio}2:3:4.\phantom{\rule{0ex}{0ex}}\mathrm{Hence},\mathrm{length}=2\mathrm{x}\mathrm{m}\phantom{\rule{0ex}{0ex}}\mathrm{Breadth}=3\mathrm{x}\mathrm{m}\phantom{\rule{0ex}{0ex}}\mathrm{Height}=4\mathrm{x}\mathrm{m}\phantom{\rule{0ex}{0ex}}\mathrm{So},\mathrm{total}\mathrm{surface}\mathrm{area}\mathrm{of}\mathrm{the}\mathrm{rectangular}\mathrm{box}=2×\left(\mathrm{length}×\mathrm{breadth}+\mathrm{breadth}×\mathrm{height}+\mathrm{length}×\mathrm{height}\right)\phantom{\rule{0ex}{0ex}}=2×\left(2\mathrm{x}×3\mathrm{x}+3\mathrm{x}×4\mathrm{x}+2\mathrm{x}×4\mathrm{x}\right)\phantom{\rule{0ex}{0ex}}=2×\left(6{\mathrm{x}}^{2}+12{\mathrm{x}}^{2}+8{\mathrm{x}}^{2}\mathrm{\right)}\phantom{\rule{0ex}{0ex}}=2×\left(26{\mathrm{x}}^{2}\mathrm{\right)}\phantom{\rule{0ex}{0ex}}=52{\mathrm{x}}^{2}{\mathrm{m}}^{2}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Also},\mathrm{the}\mathrm{cost}\mathrm{of}\mathrm{covering}\mathrm{the}\mathrm{box}\mathrm{with}\mathrm{paper}\mathrm{at}\mathrm{the}\mathrm{rate}\mathrm{Rs}8/{\mathrm{m}}^{2}\mathrm{and}\mathrm{Rs}9.50/{\mathrm{m}}^{2}\mathrm{is}\mathrm{Rs}1248.\phantom{\rule{0ex}{0ex}}\mathrm{Here},\mathrm{the}\mathrm{total}\mathrm{cost}\mathrm{of}\mathrm{covering}\mathrm{the}\mathrm{box}\mathrm{at}\mathrm{a}\mathrm{rate}\mathrm{of}\mathrm{Rs}8/{\mathrm{m}}^{2}=8×52{\mathrm{x}}^{2}=\mathrm{Rs}416{\mathrm{x}}^{2}\phantom{\rule{0ex}{0ex}}\mathrm{And}\mathrm{the}\mathrm{total}\mathrm{cost}\mathrm{of}\mathrm{covering}\mathrm{the}\mathrm{box}\mathrm{at}\mathrm{a}\mathrm{rate}\mathrm{of}\mathrm{Rs}9.50/{\mathrm{m}}^{2}=9.50×52{\mathrm{x}}^{2}=\mathrm{Rs}494{\mathrm{x}}^{2}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Now},\mathrm{total}\mathrm{cost}\mathrm{of}\mathrm{covering}\mathrm{the}\mathrm{box}\mathrm{at}\mathrm{the}\mathrm{rate}\mathrm{Rs}9.50/{\mathrm{m}}^{2}-\mathrm{total}\mathrm{cost}\mathrm{of}\mathrm{covering}\mathrm{the}\mathrm{box}\mathrm{at}\mathrm{the}\mathrm{rate}\mathrm{Rs}8/{\mathrm{m}}^{2}=1248\phantom{\rule{0ex}{0ex}}⇒494{\mathrm{x}}^{2}-416{\mathrm{x}}^{2}=1248\phantom{\rule{0ex}{0ex}}⇒78{\mathrm{x}}^{2}=1248\phantom{\rule{0ex}{0ex}}⇒{\mathrm{x}}^{2}\mathrm{=}\frac{1248}{78}=16\phantom{\rule{0ex}{0ex}}⇒\mathrm{x}=\sqrt{16}=4\phantom{\rule{0ex}{0ex}}\mathrm{Hence},\mathrm{length}\mathrm{of}\mathrm{the}\mathrm{rectangular}\mathrm{box}=2×\mathrm{x}=2×4=8\mathrm{m}\phantom{\rule{0ex}{0ex}}\mathrm{Breadth}=3×\mathrm{x}=3×4=12\mathrm{m}\phantom{\rule{0ex}{0ex}}\mathrm{Height}=4×\mathrm{x}=4×4=16\mathrm{m}$

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