Circumference = No.of revolutions * Distance covered
Distance to be covered in 1 min. = (66 X1000)/60 m = 1100 m.
Circumference of the wheel = 2 x (22/7) x 0.70 m = 4.4 m.
Number of revolutions per min. =(1100/4.4) = 250.
Let the breadth of the given rectangle be x then length is 2x.
thus area of the given rect is
after dec 5cm from length and inc 5cm breadth , new lenght becomes 2x-5 and breadth is x+5.thus new area =(2x-5)(x+5)=
since new area is 75 units greater than original area thus
5x=75+25
5x=100
therefore x=20
hence length of the rectangle is 40 cm.
Area of the room=(544 * 374)
size of largest square tile= H.C.F of 544 & 374 = 34 cm
Area of 1 tile = (34 x 34)
Number of tiles required== [(544 x 374) / (34 x 34)] = 176
ratio =
Area to be plastered =
=
Cost of plastering =
area of the room = 544 x 374 sq.cm
size of largest square tile = H.C.F of 544cm and 374 cm= 34 cm
area of 1 tile = 34x34 sq cm
no. of tiles required = (544 x 374) / (34 x 34) = 176
Let a = 13, b = 14 and c = 15. Then, =21
(s- a) = 8, (s - b) = 7 and (s - c) = 6.
Area =
=
= 84 sq.cm
Let the breadth of floor be 'b' m.
Then, length of the floor is 'l = (b + 25)'
Area of the rectangular floor = l x b = (b + 25) × b
According to the question,
(b + 15) (b + 8) = (b + 25) × b
2b = 120
b = 60 m.
l = b + 25 = 60 + 25 = 85 m.
Area of the floor = 85 × 60 = 5100 sq.m.
Let original length = x and original breadth = y.
Original area = xy.
New length = x/2 and New breadth=3y
New area =
Therefore, Increase in area = New area-original area =
Therefore, Increase % = %
100 cm is read as 102 cm.
A1 = (100*100)Sq.cm
A2 = (102*102)Sq.cm
(A2 - A1) = = (102 + 100) x (102 - 100) = 404 sq.cm.
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