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A painting contractor tells the owner of a large custom-built cupboard that the painting cost will be proportional to the square of a cupboard's surface area as measured in square-m. The owner dismantles the cupboard into 4 modules in the ratio 3:5:6:8 of its surface area, such that no new surfaces are required to be painted. If the owner's savings turn out to be Rs 1400 due to the dismantling, what was original painting cost quoted by the contractor?

Solution

✅ Correct Option: 2

Let the total surface area =A= A sq. m and the original painting cost =kA2= kA^2


The 4 modules have surface areas in ratio 3:5:6:83 : 5 : 6 : 8

Sum of ratio parts =3+5+6+8=22= 3 + 5 + 6 + 8 = 22

So the surface areas of the modules are 3A22\frac{3A}{22}, 5A22\frac{5A}{22}, 6A22\frac{6A}{22}, 8A22\frac{8A}{22}


Cost after dismantling =k(3A22)2+k(5A22)2+k(6A22)2+k(8A22)2= k\left(\frac{3A}{22}\right)^2 + k\left(\frac{5A}{22}\right)^2 + k\left(\frac{6A}{22}\right)^2 + k\left(\frac{8A}{22}\right)^2

=kA2484(9+25+36+64)= \frac{kA^2}{484}(9 + 25 + 36 + 64)

=kA2×134484= kA^2 \times \frac{134}{484}


Savings == Original Cost −- New Cost

1400=kA2−kA2×1344841400 = kA^2 - kA^2 \times \frac{134}{484}

1400=kA2(484−134484)1400 = kA^2\left(\frac{484 - 134}{484}\right)

1400=kA2×3504841400 = kA^2 \times \frac{350}{484}

kA2=1400×484350kA^2 = 1400 \times \frac{484}{350}

kA2=4×484kA^2 = 4 \times 484

kA2=1936kA^2 = 1936


The original painting cost == Rs 19361936

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