The diagonal of the floor of a rectangular closet is 7½ feet. The shorter side of the closet is 4½ feet. What is the area of the closet in square feet?

Let breadth = x metres.

Then, length = (x + 20) metres.

Perimeter =	❨	5300	❩	m = 200 m.
		26.50

∴ 2[(x + 20) + x] = 200

⟹ 2x + 20 = 100

⟹ 2x = 80

⟹ x = 40.

Hence, length = x + 20 = 60 m.

Let original length = x and original breadth = y.

Decrease in area	= xy - ❨ 80 x x 90 y ❩ 100 100
	= ❨ xy - 18 xy ❩ 25
	= 7 xy. 25

∴ Decrease % =	❨	7	xy x	1	x 100	❩%	= 28%.
		25		xy

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%.
		2		xy

The hands of a clock point in opposite directions (in the same straight line) 11 times in every 12 hours. (Because between 5 and 7 they point in opposite directions at 6 o'clcok only).

So, in a day, the hands point in the opposite directions 22 times.

55 min. spaces are covered in 60 min.

60 min. spaces are covered in	❨	60	x 60	❩min.	= 65	5	min.
		55				11

Loss in 64 min. =	❨	65	5	- 64	❩	=	16	min.
			11				11

Loss in 24 hrs =	❨	16	x	1	x 24 x 60	❩min.	=	32	8	min.
		11		64					11

We have: l = 20 ft and lb = 680 sq. ft.

So, b = 34 ft.

∴ Length of fencing = (l + 2b) = (20 + 68) ft = 88 ft.

100 cm is read as 102 cm.

∴ A₁ = (100 x 100) cm² and A₂ (102 x 102) cm².

(A₂ - A₁) = [(102)² - (100)²]

∴ Percentage error =	❨	404	x 100	❩%	= 4.04%
		100 x 100

Let the side of the square(ABCD) be x metres.

Then, AB + BC = 2x metres.

AC = √2x = (1.41x) m.

Saving on 2x metres = (0.59x) m.

Saving % =	❨	0.59x	x 100	❩%	= 30% (approx.)
		2x

Perimeter = Distance covered in 8 min. =	❨	12000	x 8	❩m = 1600 m.
		60

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) m² = 153600 m².

√l² + b² = √41.

Also, lb = 20.

(l + b)² = (l² + b²) + 2lb = 41 + 40 = 81

⟹ (l + b) = 9.

∴ Perimeter = 2(l + b) = 18 cm.