Perimeter = Distance covered in 8 min. = (12000/60) * 8 = 1600m
Let length = 3x metres and breadth = 2x metres.
Then, 2(3x + 2x) = 1600 or x = 160.
Length = 480 m and Breadth = 320 m.
Area = (480 x 320) = 153600
100 cm is read as 102 cm.
A1 = (100 x 100) and A2 =(102 x 102)
(A2 - A1) = [(102 x 102)-(100 x 100) ]= 404
Percentage error =[404/(100 x 100 )] x 100%= 4.04%
2(l+b)/b = 5/1
=> 2l + 2b = 5b
=> 3b = 2l
=> b =(2/3) x l
Then, Area = 216 sq.cm
l x b = 216
=> l x [(2/3)x l] =216
=> l x l = 324
=> l = 18 cm.
We have: (l - b) = 23 and 2(l + b) = 206 or (l + b) = 103.
Solving the two equations, we get: l = 63 and b = 40.
Area = (l x b) = (63 x 40) sq.m
= 2520 sq.m
Let original length = x and original breadth = y.
Original area = xy.
New length = x/2 and New breadth = 3y.
New area = (x/2 * 3y) = (3/2 )xy
Therefore, Increase % = [(1/2)xy * (1/xy) * 100] % = 50%
Area to be plastered= [2(l + b) x h] + (l x b)
= [2(25 + 12) x 6] + (25 x 12)
= (444 + 300)
= 744
Cost of plastering = Rs. 744 x (75/100)= Rs. 558
Volume of water displaced = (3 x 2 x 0.01) = 0.06 cu.m.
Mass of man = Volume of water displaced x Density of water
= (0.06 x 1000) kg
= 60 kg.
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