At what time between 9 and 10 o'clock will the hands of a watch be together?

Angle traced by hour hand in 12 hrs. = 360°.

Angle traced by it in	11	hrs =	❨	360	x	11	❩	°	= 110°.
	3			12		3

Angle traced by minute hand in 60 min. = 360°.

Angle traced by it in 40 min. =	❨	360	x 40	❩	°	= 240°.
		60

∴ Required angle (240 - 110)° = 130°.

Angle traced by hour hand in 12 hrs = 360°.

Angle traced by hour hand in 5 hrs 10 min. i.e.,	31	hrs =	❨	360	x	31	❩	°	= 155°.
	6			12		6

Angle traced by hour hand in	125	hrs =	❨	360	x	125	❩	°	= 312	1	°	.
	12			12		12				2

Angle traced by minute hand in 25 min =	❨	360	x 25	❩	°	= 150°.
		60

∴ Reflex angle = 360° -	❨	312	1	- 150	❩	°	= 360° - 162	1	°	= 197	1°	.
			2					2			2

At 5 o'clock, the hands are 25 min. spaces apart.

To be at right angles and that too between 5.30 and 6, the minute hand has to gain (25 + 15) = 40 min. spaces.

55 min. spaces are gained in 60 min.

40 min. spaces are gained in	❨	60	x 40	❩min	=	43	7	min.
		55					11

∴ Required time = 43	7	min. past 5.
	11

Angle traced by the hour hand in 6 hours =	❨	360	x 6	❩	°	= 180°.
		12

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

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.

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

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 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.