Let original length = x and original breadth = y. Original area = xy. New length = x . 2 New breadth = 3y. New area = ❨ x x 3y ❩ = 3 xy. 2 2 ∴ Increase % = ❨ 1 xy x 1 x 100 ❩% = 50%
Length = (40 x 10 ) dm = 400 dm.Breadth = (15 x 10 ) dm = 150 dm.Area of veranda = (400 x 150 ) dm2Area of one stone = (6 x 5 ) dm2? Required number of stones = (400 x 150) /(6 x 5) = 2000
Let the parallel sides be 3a and 5a.So Area of trapezium = 1/2 x sum of parallel side x perpendicular distance between them.? 1/2 (3a +5a) x 12 = 384? 8a = 64? a =8? Smaller side = 3x = 3 x 8 = 24 cm.
Let the radius of the park be r, then?r + 2r = 288(? + 2)r = 288? [(22/7) + 2]r = 288? r = (288 x 7)/36 = 56 ? Area of the park = (1/2)?r2= (1/2) x (22/7) x 56 x 56= 4928
Area of the room =(544 x 374) cm2size of largest square tile = H.C.F. of 544 & 374 = 34 cm Area of 1 tile = (34 x 34) cm2? Least number of tiles required = (544 x 374) / (34 x 34) = 176
Distance covered in one revolution = total distance travelled / total number of revolution. = ( 88 x 1000) / 1000 m = 88 mWe know that the distance covered in one revolution = circumference of the wheel.? ?d = 88? 22d / 7 = 88? d = 28 m
EXAMIANS