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
Other side= = = = 6ft
Area of closet =(6 x 4.5)sq.ft = 27 sq.ft
We know that,
One square meter is equal to 10,00,000 sq mm.
Given, each mesh is 8mm length and 5mm wide
Therefore for each mesh : 8 x 5 = 40 sq mm.
So, number of meshes =
= 25000.
h = 14 cm, r = 7 cm.
So, l =
Total surface area = =
=
= (3.236 x 154) sq.cm
= 498.35 sq.cm
Let the length of the wire be h.
Radius r =1/2 mm = 1/20 cm
=> h= (66 x 20 x 20) x (7/22) = 8400cm = 84m
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%
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
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