(C) The area of the floor is $4\, m \times 3\, m = 12\, m^2$.Since $1\, m = 100\, cm$,then $1\, m^2 = 10,000\, cm^2$.Therefore,the area of the floor i

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(C) The area of the floor is $4\, m \times 3\, m = 12\, m^2$.Since $1\, m = 100\, cm$,then $1\, m^2 = 10,000\, cm^2$.Therefore,the area of the floor in square centimeters is $12 \times 10,000 = 120,000\, cm^2$.The area of one rectangular tile is $8\, cm \times 6\, cm = 48\, cm^2$.The number of tiles required is the total area of the floor divided by the area of one tile.Number of tiles $= \frac{120,000}{48} = 2,500$.

Competitive Quantitative Aptitude Mensuration Measurement of Area

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The dimensions of the floor of a rectangular hall are $4\, m \times 3\, m$. The floor of the hall is to be tiled fully with $8\, cm \times 6\, cm$ rectangular tiles without breaking the tiles into smaller sizes. The number of tiles required is:

A
$4800$
B
$2600$
C
$2500$
D
$2400$

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Measurement of Area

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In $\Delta ABC$ and $\Delta PQR$,$\angle B = \angle Q$ and $\angle C = \angle R$. $M$ is the mid-point of side $QR$. If $AB : PQ = 7 : 4$,then $\frac{\text{area}(\Delta ABC)}{\text{area}(\Delta PMR)}$ is:

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The dimensions of the floor of a rectangular hall are $4\, m \times 3\, m$. The floor of the hall is to be tiled fully with $8\, cm \times 6\, cm$ rectangular tiles without breaking the tiles into smaller sizes. The number of tiles required is:

Similar Questions

The area of a trapezium is $2500 \, m^2$. One of its parallel sides is $75 \, m$. If the distance between the two parallel sides is $40 \, m$,then find the length of the other parallel side in $m$.

The length of a rectangle is $2 \, cm$ more than its breadth. The perimeter is $48 \, cm$. The area of the rectangle (in $cm^2$) is?

The surface area of a spherical part of a bowl with a flat circular detachable cover,excluding the cover,is $616 \, cm^2$. The area of the cover is $38.5 \, cm^2$. What is the volume of the bowl? (in $cm^3$)

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In $\Delta ABC$ and $\Delta PQR$,$\angle B = \angle Q$ and $\angle C = \angle R$. $M$ is the mid-point of side $QR$. If $AB : PQ = 7 : 4$,then $\frac{\text{area}(\Delta ABC)}{\text{area}(\Delta PMR)}$ is:

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