The difference between the length and breadth of a rectangle is 23 m. If its perimeter is 206 m, then its area is:

2(l + b)	=	5
b		1

⟹ 2l + 2b = 5b

⟹ 3b = 2l

b =	2	l
	3

Then, Area = 216 cm²

⟹ l x b = 216

⟹l x	2	l	= 216
	3

⟹ l² = 324

⟹ l = 18 cm.

Length of largest tile = H.C.F. of 1517 cm and 902 cm = 41 cm.

Area of each tile = (41 x 41) cm².

∴ Required number of tiles =	❨	1517 x 902	❩	= 814.
		41 x 41

Area of the park = (60 x 40) m² = 2400 m².

Area of the lawn = 2109 m².

∴ Area of the crossroads = (2400 - 2109) m² = 291 m².

Let the width of the road be x metres. Then,

60x + 40x - x² = 291

⟹ x² - 100x + 291 = 0

⟹ (x - 97)(x - 3) = 0

⟹ x = 3.

√l² + b² = √41.

Also, lb = 20.

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

⟹ (l + b) = 9.

∴ Perimeter = 2(l + b) = 18 cm.

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.

Area to be plastered	= [2(l + b) x h] + (l x b)
	= {[2(25 + 12) x 6] + (25 x 12)} m2
	= (444 + 300) m2
	= 744 m2.

∴ Cost of plastering = Rs.	❨	744 x	75	❩	= Rs. 558.
			100

Let original length = x metres and original breadth = y metres.

Original area = (xy) m².

New length =	❨	120	x	❩m	=	❨	6	x	❩m.
		100					5

New breadth =	❨	120	y	❩m	=	❨	6	y	❩m.
		100					5

New Area =	❨	6	x x	6	y	❩m2	=	❨	36	xy	❩m2.
		5		5					25

The difference between the original area = xy and new-area 36/25 xy is

= (36/25)xy - xy

= xy(36/25 - 1)

= xy(11/25) or (11/25)xy

∴ Increase % =	❨	11	xy x	1	x 100	❩%	= 44%.
		25		xy

The word 'OPTICAL' contains 7 different letters.

When the vowels OIA are always together, they can be supposed to form one letter.

Then, we have to arrange the letters PTCL (OIA).

Now, 5 letters can be arranged in 5! = 120 ways.

The vowels (OIA) can be arranged among themselves in 3! = 6 ways.

∴ Required number of ways = (120 x 6) = 720.

We may have(1 black and 2 non-black) or (2 black and 1 non-black) or (3 black).

∴ Required number of ways	= (3C1 x 6C2) + (3C2 x 6C1) + (3C3)
	= ❨ 3 x 6 x 5 ❩ + ❨ 3 x 2 x 6 ❩ + 1 2 x 1 2 x 1
	= (45 + 18 + 1)
	= 64.

Number of ways of selecting (3 consonants out of 7) and (2 vowels out of 4)

	= (7C3 x 4C2)
	= ❨ 7 x 6 x 5 x 4 x 3 ❩ 3 x 2 x 1 2 x 1
	= 210.

Number of groups, each having 3 consonants and 2 vowels = 210.

Each group contains 5 letters.

Number of ways of arranging 5 letters among themselves	= 5!
	= 5 x 4 x 3 x 2 x 1
	= 120.

∴ Required number of ways = (210 x 120) = 25200.