Length of largest tile = H.C.F. of 1517 cm and 902 cm = 41 cm.
Area of each tile = (41 x 41) sq.cm
Required number of tiles = 1517 x (902/41) x 41 = 814
Let original length = x and original breadth = y.
Decrease in area = xy -[(80/100x) * (90/100y)]
=[(xy-(18/25)xy)]
=(7/25)xy
Decrease % =[(7/25)xy * (1/xy) * 100] =28%
Let 'B' be the nuber of bricks.
=> 10 x 4/100 x 5 x 90/100 = 25/100 x 15/100 x 8/100 x B
=> 10 x 20 x 90 = 15 x 2 x B
=> B = 6000
We have: l = 20 ft and lb = 680 sq. ft.
So, b = 34 ft.
Length of fencing = (l + 2b) = (20 + 68) ft = 88 ft.
Area of the square = = 1/2 * 3.8 * 3.8 = 7.22 sq.m
The maximum area of the triangle will come when the given sides are placed at right angles. If we take 8 as the base and 5 as the height the area = ½ x 8 x 5=20
d1 = and d2 = are diameters
surface area of a cube= 6 x (8 x 8) = 384 sq.ft
quantity of paint required=(384/16)=24kg
cost of painting= 36.5 x 24 = Rs.876
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