Seema is painting the walls and ceiling of a hall whose dimensions are

1.	Seema is painting the walls and ceiling of a hall whose dimensions are 10 m x 15 m x 5 m, from each can pf paint 120m2 of area is painted. How many cans of paint she need to pain the hall?
Answer» $\large\underline{\sf{Solution-}}$ Dimensions of Room :- Length of Room = 10 m Breadth of Room = 15 m Height of Room = 5 m We know, that $\boxed{ \sf \: Area_{(walls)} \: = \: 2(length + breadth) \times height}$ So, On PLUGGING the VALUES, we have $\rm :\longmapsto\:Area_{(walls)} = 2(15 + 10) \times 5$ $\rm :\longmapsto\:Area_{(walls)} = 10 \times 25$ $\bf\implies \:Area_{(walls)} = 250 \: {m}^{2}$ We know that, $\boxed{ \sf \: Area_{(ceiling)} = length \times breadth}$ On plugging the values, we have $\rm :\longmapsto\:Area_{(ceiling)} = 10 \times 15$ $\bf\implies \:Area_{(ceiling)} = 150 \: {m}^{2}$ Total area to be painted, $\rm :\longmapsto\:Area_{(to \: be \: painted)} = Area_{(walls)} + Area_{(ceiling)}$ $\rm :\longmapsto\:Area_{(to \: be \: painted)} = 250 + 150$ $\bf\implies \:Area_{(to \: be \: painted)} = 400 \: {m}^{2}$ Now, In ONE can, area to be painted = 120 square meter. To PAINT area of 400 square meter, number of cans required, $\rm :\longmapsto\:Number_{(cans)} \: = \dfrac{400}{120} = 3.33 \:$ $\bf\implies \:Number_{(cans)} \: = \: 4 \: cans \: required.$ Additional Information :- $\boxed{ \sf{ \: CSA{(cylinder)} = 2\pi \: rh}}$ $\boxed{ \sf{ \: Volume_{(cuboid)} = lbh}}$ $\boxed{ \sf{ \: Volume_{(cone)} = \dfrac{1}{3} \pi \: {r}^{2} h}}$ $\boxed{ \sf{ \: Volume_{(cube)} = {(edge)}^{3} }}$ $\boxed{ \sf{ \: TSA{(cone)} = {6(edge)}^{2} }}$ $\boxed{ \sf{ \: TSA{(cuboid)} = 2(lb + bh + hl)}}$ $\boxed{ \sf{ \: TSA{(cone)} = \pi \: r(l + r)}}$ $\boxed{ \sf{ \: TSA{(cylinder)} = 2\pi \: r(h + r)}}$ $\boxed{ \sf{ \: CSA{(cone)} = \pi \: rl}}$